Proof. Suppose $T,\Gamma\vdash\theta$, meaning a formal proof $(\theta_1,\ldots,\theta_n)$ of $\theta$ from $\Gamma$ exists. Let $k$ in $1,\ldots,n$, and suppose for inductive hypothesis that $T,\Gamma\vDash\theta_j$ for all $j$ in $1,\ldots,k-1$, then:

• If $\theta_k$ is an axiom or a member of $\Gamma$, then trivially, $T,\Gamma\vDash\theta_k$.
• If $\theta_k$ is obtained from MP, then for some $i,j$ in $1,\ldots,k-1$, $\theta_j$ is $(\theta_i\to\theta_k)$. Let $\mathcal M$ be a model of $T$ that satisfies every member of $\Gamma$, then $\mathcal M$ satisfies $\theta_i$ and $\theta_j$. Let $\sigma_k$ be an assignment function of $\theta_k$, let $\sigma_j$ be an extension of $\sigma_k$ on $\text{free}(\theta_j)$, and let $\sigma_i$ be the restriction of $\sigma_j$ on $\text{free}(\theta_i)$. Then $\mathcal M$ satisfies $\theta_i$ with respect to $\sigma_i$ and $\theta_j$ with respect to $\sigma_j$. Suppose $(a_1,\ldots,a_m)$ are occurrences of free variables in $\theta_i$ and $(b_1,\ldots,b_l)$ are occurrences of free variables in $\theta_k$ in order of occurrence, then $(a_1,\ldots,a_m,b_1,\ldots,b_l)$ are occurrences of free variables in $\theta_j$ in order of occurrence. And both $\mathcal F_{\theta_i}(\sigma_i(a_1),\ldots,\sigma_i(a_m))$ and $\mathcal F_{\theta_j}(\sigma_j(a_1),\ldots,\sigma_j(b_l))$ are $\top$. Note that $\mathcal F_{\theta_i}(\sigma_j(a_1),\ldots,\sigma_j(a_m))$ is $\top$. If $\mathcal F_{\theta_k}(\sigma_k(b_1),\ldots,\sigma_k(b_l))$ is $\bot$, then $\mathcal F_{\theta_k}(\sigma_j(b_1),\ldots,\sigma_j(b_l))$ is $\bot$, implying $\mathcal F_{\theta_j}(\sigma_j(a_1),\ldots,\sigma_j(b_l))$ is $\bot$, a contradiction. Thus we have that $\mathcal F_{\theta_k}(\sigma_k(b_1),\ldots,\sigma_k(b_l))$ is $\top$. Hence $\mathcal M$ satisfies $\theta_k$. Therefore $T,\Gamma\vDash\theta_k$.
• If $\psi$ is obtained from UG, then for some variable symbol $x$ and some $j$ in $1,\ldots,k-1$, $\theta_k$ is $\forall x\theta_j$. Let $\mathcal M$ be a model of $T$ that satisfies every member of $\Gamma$, then $\mathcal M$ satisfies $\theta_j$. Let $\sigma_k$ be an assignment function of $\theta_k$. Suppose $(a_1,\ldots,a_m)$ are occurrences of free variables in $\theta_j$ and $(b_1,\ldots,b_l)$ are occurrences of free variables in $\theta_k$ in order of occurrence. Let $u$ be an object in the universe determined by $\mathcal M$, and let $\sigma_j$ be
  • an extension of $\sigma_k$ on $\text{free}(\theta_j)$ that maps $x$ to $u$, if $x$ is in $\text{free}(\theta_j)$;
  • $\sigma_k$, if $x$ is not in $\text{free}(\theta_j)$.
  Then $\mathcal M$ satisfies $\theta_j$ with respect to $\sigma_j$. Thus $\mathcal F_{\theta_j}(\sigma_j(a_1),\ldots,\sigma_j(a_m))$ is $\top$. Note that $(a_1,\ldots,a_m)$ matches $(b_1,\ldots,b_l)$ at the corresponding indexes and is $x$ at other indexes, thus $(\sigma_j(a_1),\ldots,\sigma_j(a_m))$ matches $(\sigma_k(b_1),\ldots,\sigma_k(b_l))$ at the corresponding indexes and is $u$ at other indexes. And we have that $\mathcal F_{\theta_k}(\sigma_k(b_1),\ldots,\sigma_k(b_l))$ is $\top$. Hence $\mathcal M$ satisfies $\theta_k$. Therefore $T,\Gamma\vDash\theta_k$.

In all cases, $T,\Gamma\vDash\theta_j$ for all $j$ in $1,\ldots,k$, which concludes the inductive step. By induction, $T,\Gamma\vDash\theta$. $\blacksquare$

Proof. Fix a formal proof of $\theta$ from $\Gamma\cup\{\psi\}$ such that for every free variable $x$ in $\psi$, there is no quantifier $\forall x$ in the formal proof. For any formula $\varphi$ in the formal proof, assume the inductive hypothesis that $T,\Gamma\vdash(\psi\to\varphi')$ for every formula $\varphi'$ before $\varphi$ in the formal proof, then there are four cases for $\varphi$:

• If $\varphi$ is an axiom or a member of $\Gamma$, then $T,\Gamma\vdash\varphi$, and since $T,\Gamma\vdash(\varphi\to(\psi\to\varphi))$ by LA1, we have $T,\Gamma\vdash(\psi\to\varphi)$ by MP.
• If $\varphi$ is the same formula as $\psi$, since $T\vdash(\varphi\to\varphi)$ by LA1, we have $T\vdash(\psi\to\varphi)$.
• If $\varphi$ is obtained from MP, then there exists some $\varphi^*$ such that both $\varphi^*$ and $(\varphi^*\to\varphi)$ have appeared earlier in the formal proof, and by inductive hypothesis we have $T,\Gamma\vdash(\psi\to\varphi^*)$ and $T,\Gamma\vdash(\psi\to(\varphi^*\to\varphi))$. Since $T,\Gamma\vdash((\psi\to(\varphi^*\to\varphi))\to((\psi\to\varphi^*)\to(\psi\to\varphi)))$ by LA1, applying MP twice, we get $T,\Gamma\vdash(\psi\to\varphi)$.
• If $\varphi$ is obtained from UG, then $\varphi$ takes the form $\forall x\varphi^*$ for some variable symbol $x$ and some formula $\varphi^*$ that appeared earlier in the formal proof. Note that $x$ is not a free variable in $\psi$. By inductive hypothesis, we have $T,\Gamma\vdash(\psi\to\varphi^*)$, and by applying UG, we also have $T,\Gamma\vdash\forall x(\psi\to\varphi^*)$. Since we have $T,\Gamma\vdash(\forall x(\psi\to\varphi^*)\to(\forall x\psi\to\forall x\varphi^*))$ by LA3, applying MP, we get $T,\Gamma\vdash(\forall x\psi\to\forall x\varphi^*)$. Since $x$ does not occur free in $\psi$, by LA4 we have $T,\Gamma\vdash(\psi\to\forall x\psi)$. Since by LA1 we have $T,\Gamma\vdash((\psi\to\forall x\psi)\to((\forall x\psi\to\forall x\varphi^*)\to(\psi\to\forall x\varphi^*)))$, applying MP twice, we get $T,\Gamma\vdash(\psi\to\forall x\varphi^*)$, or $T,\Gamma\vdash(\psi\to\varphi)$.

We have shown that in every possible case, $T,\Gamma\vdash(\psi\to\varphi)$. This concludes the inductive step. By induction, we have $T,\Gamma\vdash(\psi\to\theta)$. $\blacksquare$

Proof. Suppose $\cfrac{\varphi_1\quad\ldots\quad\varphi_n}{\psi}$ is a deduced rule. Take $\Gamma\cup\{\bar\varphi_1,\ldots,\bar\varphi_n\}$ as premises, then there exists a formal proof such that all of $\varphi_1,\ldots,\varphi_n$ appeared in it. Since $\cfrac{\varphi_1\quad\ldots\quad\varphi_n}{\psi}$ is a deduced rule, there exists a formal proof of $\psi$ and hence $\bar\psi$; that is, $T,\Gamma\cup\{\bar\varphi_1,\ldots,\bar\varphi_n\}\vdash\bar\psi$. We will discuss by cases:

• If $n$ is equal to $0$, then $T,\Gamma\vdash\bar\psi$. Hence, given $\Gamma$ as premises, there exists a formal proof of $\bar\psi$ constructed from any formal proof, so $\cfrac{}{\bar\psi}$ is a deduced rule.
• If $n$ is equal to $1$, since $T,\Gamma\cup\{\bar\varphi_1\}\vdash\bar\psi$, by deduction theorem, we have $T,\Gamma\vdash(\bar\varphi_1\to\bar\psi)$. Hence, given $\Gamma$ as premises, there exists a formal proof of $(\bar\varphi_1\to\bar\psi)$ constructed from any formal proof, so $\cfrac{}{(\bar\varphi_1\to\bar\psi)}$ is a deduced rule.
• If $n$ is greater than $1$, since $T,\Gamma\cup\{\bar\varphi_1,\ldots,\bar\varphi_{n-1}\}\cup\{\bar\varphi_n\}\vdash\bar\psi$, by deduction theorem, we have $T,\Gamma\cup\{\bar\varphi_1,\ldots,\bar\varphi_{n-1}\}\vdash(\bar\varphi_n\to\bar\psi)$. For any natural number $i$ from $1$ to $n-1$, take the following as an inductive hypothesis: $T,\Gamma\cup\{\bar\varphi_1,\ldots,\bar\varphi_i\}\vdash(\varphi\to\bar\psi)$, where $\varphi$ is $\bar\varphi_n$ if $i=n-1$, or $(\bar\varphi_{i+1}\land\ldots\land(\bar\varphi_{n-1}\land\bar\varphi_n)\ldots)$ if otherwise. Given $i$ and the inductive hypothesis, by deduction theorem, we have $T,\Gamma\cup\{\bar\varphi_1,\ldots,\bar\varphi_{i-1}\}\vdash(\bar\varphi_i\to(\varphi\to\bar\psi))$. By LA1, we have $((\psi_1\to(\psi_2\to\psi_3))\to((\psi_1\land\psi_2)\to\psi_3))$ for arbitrary formulas $\psi_1,\psi_2,\psi_3$, so by MP we have $T,\Gamma\cup\{\bar\varphi_1,\ldots,\bar\varphi_{i-1}\}\vdash((\bar\varphi_i\land\varphi)\to\bar\psi)$. This concludes the inductive step. Since the inductive hypothesis for $i=n-1$ is satisfied, by induction, we have $T,\Gamma\vdash((\bar\varphi_1\land\ldots\land(\bar\varphi_{n-1}\land\bar\varphi_n)\ldots)\to\bar\psi)$. By LA1, we have $((\psi_1\land\ldots\land\psi_n)\to(\psi_1\land\ldots\land(\psi_{n-1}\land\psi_n)\ldots))$ for arbitrary natural number $n$ and arbitrary formulas $\psi_1,\ldots,\psi_n$. so by HS we have $T,\Gamma\vdash((\bar\varphi_1\land\ldots\land\bar\varphi_n)\to\bar\psi)$. Hence, given $\Gamma$ as premises, there exists a formal proof of $((\bar\varphi_1\land\ldots\land\bar\varphi_n)\to\bar\psi)$ constructed from any formal proof, so $\cfrac{}{((\bar\varphi_1\land\ldots\land\bar\varphi_n)\to\bar\psi)}$ is a deduced rule. $\blacksquare$

Proof. Let $x,y,z$ be distinct variable symbols not occurring in $t_1,t_2,t_3$.

• For E1: $$ \begin{matrix} 1 & (t_1=t_1) & LA5 \end{matrix} $$
• For E2: $$ \begin{matrix} 1 & (x=x) & LA5 \\ 2 & ((x=y)\to((x=x)\to(y=x))) & LA6 \\ 3 & ((x=x)\land((x=y)\to((x=x)\to(y=x)))) & CI(1,2) \\ 4 & (((x=x)\land((x=y)\to((x=x)\to(y=x))))\to((x=y)\to(y=x))) & LA1 \\ 5 & ((x=y)\to(y=x)) & MP(3,4) \\ 6 & ((t_1=y)\to(y=t_1)) & ST(5) \\ 7 & ((t_1=t_2)\to(t_2=t_1)) & ST(6) \end{matrix} $$
• For E3:
  Suppose we have E2 as a deduced rule: $$ \begin{matrix} 1 & (((x=y)\land(y=z))\to(x=y)) & LA1 \\ 2 & ((x=y)\to(y=x)) & E2() \\ 3 & ((y=x)\to(((x=y)\land(y=z))\to((x=x)\land(x=z)))) & LA6 \\ 4 & (((x=y)\land(y=z))\to(y=x)) & HS(1,2) \\ 5 & (((x=y)\land(y=z))\to(((x=y)\land(y=z))\to((x=x)\land(x=z)))) & HS(4,3) \\ 6 & ((((x=y)\land(y=z))\to(((x=y)\land(y=z))\to((x=x)\land(x=z))))\to(((x=y)\land(y=z))\to((x=x)\land(x=z)))) & LA1 \\ 7 & (((x=y)\land(y=z))\to((x=x)\land(x=z))) & MP(5,6) \\ 8 & (((x=y)\land(y=z))\to(((x=x)\land(x=z))\to(x=z))) & LA1 \\ 9 & ((((x=y)\land(y=z))\to((x=x)\land(x=z)))\to(((x=y)\land(y=z))\to(x=z))) & ID(8) \\ 10 & (((x=y)\land(y=z))\to(x=z)) & MP(7,9) \\ 11 & (((t_1=y)\land(y=z))\to(t_1=z)) & ST(10) \\ 12 & (((t_1=t_2)\land(t_2=z))\to(t_1=z)) & ST(11) \\ 13 & (((t_1=t_2)\land(t_2=t_3))\to(t_1=t_3)) & ST(12) \end{matrix} $$

Proof. For these ones, we will only list the key steps. Note that $\varphi[y/x][x/y]$ is the same formula as $\varphi$, and $\varphi[y/x][z/y]$ is the same formula as $\varphi[z/x]$. $$(\forall y(\varphi[y/x]\leftrightarrow(x=y))\to((x=x)\to\varphi))$$ $$(((x=x)\to\varphi)\to\varphi)$$ $$(\forall y(\varphi[y/x]\leftrightarrow(x=y))\to(\varphi[y/x]\to(x=y)))$$ $$(\forall y(\varphi[y/x]\leftrightarrow(x=y))\to\forall y(\varphi[y/x]\to(x=y)))$$ $$(\forall y(\varphi[y/x]\leftrightarrow(x=y))\to(\varphi\land\forall y(\varphi[y/x]\to(x=y))))$$ $$((x=y)\to(\varphi\to\varphi[y/x]))$$ $$(\varphi\to((x=y)\to\varphi[y/x]))$$ $$(\forall y(\varphi[y/x]\to(x=y))\to(\varphi[y/x]\to(x=y)))$$ $$((\varphi\land\forall y(\varphi[y/x]\to(x=y)))\to(\varphi[y/x]\leftrightarrow(x=y)))$$ $$((\varphi\land\forall y(\varphi[y/x]\to(x=y)))\to\forall y(\varphi[y/x]\leftrightarrow(x=y)))$$ $$(\forall y(\varphi[y/x]\leftrightarrow(x=y))\leftrightarrow(\varphi\land\forall y(\varphi[y/x]\to(x=y))))$$ $$(\exists x\forall y(\varphi[y/x]\leftrightarrow(x=y))\leftrightarrow\exists x(\varphi\land\forall y(\varphi[y/x]\to(x=y))))$$ And we have the first formula.
$$(\exists x(\varphi\land\forall y(\varphi[y/x]\to(x=y)))\to\exists x\varphi)$$ $$(\exists x\forall y(\varphi[y/x]\leftrightarrow(x=y))\to\exists x\varphi)$$ $$(\forall y(\varphi[y/x]\leftrightarrow(x=y))\to((\varphi[y/x]\leftrightarrow(x=y))\land(\varphi[z/x]\leftrightarrow(x=z))))$$ $$(((\varphi[y/x]\leftrightarrow(x=y))\land(\varphi[z/x]\leftrightarrow(x=z)))\to((\varphi[y/x]\land\varphi[z/x])\to(y=z)))$$ $$(\forall y(\varphi[y/x]\leftrightarrow(x=y))\to((\varphi[y/x]\land\varphi[z/x])\to(y=z)))$$ $$(\forall y(\varphi[y/x]\leftrightarrow(x=y))\to\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z)))$$ $$(\exists x\forall y(\varphi[y/x]\leftrightarrow(x=y))\to\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z)))$$ $$(\exists x\forall y(\varphi[y/x]\leftrightarrow(x=y))\to(\exists x\varphi\land\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z))))$$ $$(\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z))\to((\varphi[y/x]\land\varphi)\to(x=y)))$$ $$(((\varphi[y/x]\land\varphi)\to(x=y))\to(\varphi\to(\varphi[y/x]\to(x=y))))$$ $$((\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z))\land\varphi)\to(\varphi[y/x]\to(x=y)))$$ $$((\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z))\land\varphi)\to\forall y(\varphi[y/x]\to(x=y)))$$ $$(\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z))\to(\varphi\to\forall y(\varphi[y/x]\to(x=y))))$$ $$((\varphi\to\forall y(\varphi[y/x]\to(x=y)))\to(\varphi\to(\varphi\land\forall y(\varphi[y/x]\to(x=y)))))$$ $$((\varphi\land\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z)))\to(\varphi\land\forall y(\varphi[y/x]\to(x=y))))$$ $$(\exists x(\varphi\land\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z)))\to\exists x(\varphi\land\forall y(\varphi[y/x]\to(x=y))))$$ Let $A$ denote $\varphi$ and $B$ denote $\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z))$. $$(\forall x\neg(A\land B)\to\forall x(B\to\neg A))$$ $$(\forall x(B\to\neg A)\to(\forall xB\to\forall x\neg A))$$ $$(B\to\forall xB)$$ $$((\forall xB\to\forall x\neg A)\to(B\to\forall x\neg A))$$ $$((B\to\forall x\neg A)\to\neg(B\land\neg\forall x\neg A))$$ $$(\forall x\neg(A\land B)\to\neg(B\land\neg\forall x\neg A))$$ $$((B\land\neg\forall x\neg A)\to\neg\forall x\neg(A\land B))$$ $$((\exists x\varphi\land\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z)))\to\exists x(\varphi\land\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z))))$$ $$((\exists x\varphi\land\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z)))\to\exists x\forall y(\varphi[y/x]\leftrightarrow(x=y)))$$ $$(\exists x\forall y(\varphi[y/x]\leftrightarrow(x=y))\leftrightarrow(\exists x\varphi\land\forall y\forall z((\varphi[y/x]\land\varphi[z/x])\to(y=z))))$$ And we have the second formula. $\blacksquare$

Proof. By induction on the inductive steps to obtain $\varphi$ from $\psi$, $\varphi'$ and $\varphi''$ are the same tuple of symbols. Since $\varphi''$ is clearly a formula, $\varphi'$ is also a formula, and they are the same formula. $\blacksquare$

Proof. By the lemma above, $\varphi'$ is obtainable by applying the inductive steps to obtain $\varphi$ from $\psi$ on $\psi'$. We will denote these inductive steps as $\varphi_0,\ldots,\varphi_n$ for $\varphi$ and $\varphi'_0,\ldots,\varphi'_n$ for $\varphi'$, for some natural number $n$. For each $i$ from $1$ to $n$, suppose we have $(\varphi_{i-1}\leftrightarrow\varphi'_{i-1})$, then:

• 1. If $\varphi_i$ is $\neg\varphi_{i-1}$, then $\varphi'_i$ is $\neg\varphi'_{i-1}$. And we have $(\neg\varphi_{i-1}\leftrightarrow\neg\varphi'_{i-1})$ by contraposition.
• 2. If $\varphi_i$ is $(\varphi_{i-1}\to\theta)$ for some formula $\theta$, then $\varphi'_i$ is $(\varphi'_{i-1}\to\theta)$. If $(\varphi_{i-1}\to\theta)$, then by $(\varphi_{i-1}\leftrightarrow\varphi'_{i-1})$ we have $(\varphi'_{i-1}\to\theta)$. Similarly, if $(\varphi'_{i-1}\to\theta)$ then $(\varphi_{i-1}\to\theta)$. Thus we have $((\varphi_{i-1}\to\theta)\leftrightarrow(\varphi'_{i-1}\to\theta))$.
• 3. If $\varphi_i$ is $(\theta\to\varphi_{i-1})$ for some formula $\theta$, then $\varphi'_i$ is $(\theta\to\varphi'_{i-1})$. If $(\theta\to\varphi_{i-1})$, then by $(\varphi_{i-1}\leftrightarrow\varphi'_{i-1})$ we have $(\theta\to\varphi'_{i-1})$. Similarly, if $(\theta\to\varphi'_{i-1})$ then $(\theta\to\varphi_{i-1})$. Thus we have $((\theta\to\varphi_{i-1})\leftrightarrow(\theta\to\varphi'_{i-1}))$.
• 4. If $\varphi_i$ is $\forall x\varphi_{i-1}$ for some variable symbol $x$, then $\varphi'_i$ is $\forall x\varphi'_{i-1}$. By $(\varphi_{i-1}\leftrightarrow\varphi'_{i-1})$, we have $\forall x(\varphi_{i-1}\to\varphi'_{i-1})$ and $\forall x(\varphi'_{i-1}\to\varphi_{i-1})$, thus $(\forall x\varphi_{i-1}\leftrightarrow\forall x\varphi'_{i-1})$.

In all cases, we have $(\varphi_i\leftrightarrow\varphi'_i)$. Thus by induction, we have $((\varphi_0\leftrightarrow\varphi'_0)\to(\varphi_n\leftrightarrow\varphi'_n))$. Since $\varphi_0,\varphi'_0,\varphi_n,\varphi'_n$ are $\psi,\psi',\varphi,\varphi'$ respectively, we have $((\psi\leftrightarrow\psi')\to(\varphi\leftrightarrow\varphi'))$. $\blacksquare$

Proof. Let $\psi$ be the range of the quantifier $\forall x$, and let $\psi'$ be $\psi[y/x]$, then the formula $\varphi''$ obtained by replacing the subformula $\forall x\psi$ of $\varphi$ by $\forall y\psi'$ is the same formula as $\varphi'$. Clearly, we have $(\forall x\psi\to\psi')$ and $(\forall y\psi'\to\psi)$. Thus we have $(\forall y\forall x\psi\to\forall y\psi')$ and $(\forall x\forall y\psi'\to\forall x\psi)$. Note that we also have $(\forall x\psi\to\forall y\forall x\psi)$ and $(\forall y\psi'\to\forall x\forall y\psi')$. Therefore, we have $(\forall x\psi\to\forall y\psi')$. And by the proposition above, we have $(\varphi\leftrightarrow\varphi'')$, and thus $(\varphi\leftrightarrow\varphi')$. Note that by induction, all occurrences of variable symbols other than free occurrences of $x$ in $\psi$ preserve their freeness or boundedness in $\psi'$. Thus $\forall y\psi'$ keeps the freeness or boundedness of every occurrence of variable symbol as in $\forall x\psi$, and their free occurrences match. By the lemma above, $\varphi''$ is obtainable by applying the inductive steps to obtain $\varphi$ from $\forall x\psi$ on $\forall y\psi'$. And by induction, $\varphi''$, and thus $\varphi'$, keeps the freeness or boundedness of every occurrence of variable symbol as in $\varphi$. $\blacksquare$

Proof. Let $\forall v_1,\ldots,\forall v_k$ be quantifiers in $\varphi$ ordered by order of occurrence, let $V_1,\ldots,V_k$ be collections of occurrences in $\varphi$ such that occurrences in $V_i$ are exactly the occurrences in $\varphi$ bounded by $\forall v_i$. Define $\forall v'_1,\ldots,\forall v'_k$ and $V'_1,\ldots,V'_k$ similarly for $\theta$. Then $\forall v_i$ and $\forall v'_i$ are at the same position, and $V_i$ and $V'_i$ occupy the same positions. Let $u_1,\ldots,u_k$ be distinct variable symbols not occurring in $\varphi$ or $\theta$. Let $\varphi_0$ be $\varphi$, and recursively obtain $\varphi_i$ by replacing $\forall v_i$ by $\forall u_i$ and all occurrences in $V_i$ by $u_i$ in $\varphi_{i-1}$. Suppose for inductive hypothesis that $\theta_0\leftrightarrow\theta_{i-1}$ and $\varphi_{i-1}$ keeps the freeness or boundedness of every occurrence of variable symbol as in $\varphi$, while having $\forall v_1,\ldots,\forall v_{i-1}$ and $V_1,\ldots,V_{i-1}$ replaced from $\varphi$, then by the lemma above, we have $\varphi_{i-1}\leftrightarrow\varphi_i$ and thus $\varphi_0\leftrightarrow\varphi_i$, and $\varphi_i$ keeps the freeness or boundedness of every occurrence of variable symbol as in $\varphi_{i-1}$ and thus $\varphi$, while having $\forall v_1,\ldots,\forall v_i$ and $V_1,\ldots,V_i$ replaced from $\varphi$. Since the base case is trivial, by induction we have $\varphi_0\leftrightarrow\varphi_k$. Define $\theta_0,\ldots,\theta_k$ similarly for $\theta$, then $\theta_0\leftrightarrow\theta_k$. By induction, $\varphi_k$ and $\theta_k$ are the same tuple of symbols and thus the same formula, hence $\varphi_0\leftrightarrow\theta_0$, and therefore $\varphi\leftrightarrow\theta$. $\blacksquare$

Proof. The formula we are trying to prove is $\exists!X\forall u\neg(u\in X)$. We will be skipping obvious steps in this proof. $$ \begin{matrix} 1 & \forall w_1\forall X\exists Y\forall u((u\in Y)\leftrightarrow((u\in X)\land\neg(w_1=w_1))) & ZF3 \\ 2 & \exists Y\forall u((u\in Y)\leftrightarrow((u\in X)\land\neg(w_1=w_1))) \\ 3 & \forall u((u\in f(w_1,X))\leftrightarrow((u\in X)\land\neg(w_1=w_1))) & \text{Witness }f \\ 4 & ((u\in f(w_1,X))\leftrightarrow((u\in X)\land\neg(w_1=w_1))) \\ 5 & (w_1=w_1) \\ 6 & \neg((u\in X)\land\neg(w_1=w_1)) \\ 7 & \neg(u\in f(w_1,X)) & \text{From }4,6 \\ 8 & \forall u\neg(u\in f(w_1,X)) \\ 9 & \exists X\forall u\neg(u\in X) \\ 10 & \forall u\neg(u\in\mathcal E) & \text{Witness }\mathcal E \\ 11 & \neg(u\in\mathcal E) \\ 12 & \forall X\forall Y(\forall u((u\in X)\leftrightarrow(u\in Y))\to(X=Y)) & ZF1 \\ 13 & (\forall u((u\in\mathcal E)\leftrightarrow(u\in Y))\to(\mathcal E=Y)) \\ 14 & \forall u((\neg(u\in\mathcal E)\land\neg(u\in Y))\to((u\in\mathcal E)\leftrightarrow(u\in Y))) & \text{Tautology} \\ 15 & (\forall u(\neg(u\in\mathcal E)\land\neg(u\in Y))\to\forall u((u\in\mathcal E)\leftrightarrow(u\in Y))) \\ 16 & (\neg(u\in\mathcal E)\to(\neg(u\in Y)\to(\neg(u\in\mathcal E)\land\neg(u\in Y)))) & \text{Tautology} \\ 17 & (\neg(u\in Y)\to(\neg(u\in\mathcal E)\land\neg(u\in Y))) & \text{From }11,16 \\ 18 & \forall u(\neg(u\in Y)\to(\neg(u\in\mathcal E)\land\neg(u\in Y))) \\ 19 & (\forall u\neg(u\in Y)\to\forall u(\neg(u\in\mathcal E)\land\neg(u\in Y))) \\ 20 & (\forall u\neg(u\in Y)\to(\mathcal E=Y)) & \text{From }13,15,19 \\ 21 & ((\mathcal E=Y)\to(\forall u\neg(u\in\mathcal E)\to\forall u\neg(u\in Y))) \\ 22 & ((\forall u\neg(u\in\mathcal E)\land((\mathcal E=Y)\to(\forall u\neg(u\in\mathcal E)\to\forall u\neg(u\in Y))))\to((\mathcal E=Y)\to\forall u\neg(u\in Y))) & \text{Tautology} \\ 23 & (\forall u\neg(u\in\mathcal E)\land((\mathcal E=Y)\to(\forall u\neg(u\in\mathcal E)\to\forall u\neg(u\in Y)))) & \text{From }10,21 \\ 24 & ((\mathcal E=Y)\to\forall u\neg(u\in Y)) & \text{From }22,23 \\ 25 & (\forall u\neg(u\in Y)\leftrightarrow(\mathcal E=Y)) & \text{From }20,24 \\ 26 & \forall Y(\forall u\neg(u\in Y)\leftrightarrow(\mathcal E=Y)) \\ 27 & \exists X\forall Y(\forall u\neg(u\in Y)\leftrightarrow(X=Y)) \\ 28 & \exists!X\forall u\neg(u\in X) \end{matrix} $$

Proof. Suppose $x=y$. Since we have $(u\in x)\leftrightarrow(u\in x)$, by $LA6$, we have $(u\in x)\leftrightarrow(u\in y)$. Thus $(x=y)\to\forall u((u\in x)\leftrightarrow(u\in y))$ and hence $\forall x\forall y((x=y)\to\forall u((u\in x)\leftrightarrow(u\in y)))$.

Proof. Suppose $a=b$. Note that either $a\in X$ or $a\notin X$, thus either $(a\in X)\land(a=b)$ or $(a\notin X)\land(a=b)$. If $(a\in X)\land(a=b)$, then $b\in X$, thus $(a\in X)\leftrightarrow(b\in X)$. If $(a\notin X)\land(a=b)$, then $b\notin X$, thus $(a\in X)\leftrightarrow(b\in X)$. Hence $(a=b)\to((a\in X)\leftrightarrow(b\in X))$, and we have $\forall a\forall b((a=b)\to\forall X((a\in X)\leftrightarrow(b\in X)))$. $\blacksquare$

Proof. In first-order language, this means

• $\exists!X\forall u((u\in X)\leftrightarrow((u=x_1)\lor\ldots\lor(u=x_n)))$ if $n$ is non-zero, and
• $\exists!X\forall u(u\notin X)$ if $n$ is $0$.

The case $0$ is proven already, which is instantiated by the empty set. Suppose the case $n$ is proven. Given sets $x_1,\ldots,x_{S(n)}$, there exists some set $X_n$ such that a set $u$ is in $X_n$ if and only if $u$ is equal to one of $x_1,\ldots,x_n$. By axiom of pairing, there exists a set $Y_{S(n)}$ such that for all $u$, $u\in Y_{S(n)}$ if and only if $u=x_{S(n)}$ or $u=x_{S(n)}$, if and only if $u=x_{S(n)}$. Again by axiom of pairing, there exists a set $Z_{S(n)}$ such that for all $u$, $u\in Z_{S(n)}$ if and only if $u=X_n$ or $u=Y_{S(n)}$. By axiom of union, there exists a set $X_{S(n)}$ such that for all $u$, $u\in X_{S(n)}$ if and only if for some $w\in Z_{S(n)}$, $u\in w$. If a set $u$ is in $X_{S(n)}$, then for some $w\in Z_{S(n)}$, $u\in w$. Then $w=X_n$ or $w=Y_{S(n)}$, thus $u\in X_n$ or $u\in Y_{S(n)}$.

• If $n$ is $0$, then $u\notin X_n$, thus $u\in Y_{S(n)}$ , implying $u=x_{S(n)}$, which is one of $x_1,\ldots,x_{S(n)}$.
• If $n$ is non-zero:
  • if $u\in X_n$, then $(u=x_1)\lor\ldots\lor(u=x_n)$, thus $u$ is in one of $x_1,\ldots,x_{S(n)}$;
  • if $u\in Y_{S(n)}$, then $u=x_{S(n)}$, which is one of $x_1,\ldots,x_{S(n)}$.

In all cases, $u$ is in one of $x_1,\ldots,x_{S(n)}$. For the other direction, suppose $u$ is in one of $x_1,\ldots,x_{S(n)}$, which means $(u=x_1)\lor\ldots\lor(u=x_{S(n)})$.

• If $n$ is $0$, then we have $u=x_{S(n)}$, thus $u\in Y_{S(n)}$ Note that $Y_{S(n)}\in Z_{S(n)}$, thus for some $w\in Z_{S(n)}$, $u\in w$. And we have $u\in X_{S(n)}$.
• If $n$ is non-zero, then either $(u=x_1)\lor\ldots\lor(u=x_n)$ or $u=x_{S(n)}$. We show above that if $u=x_{S(n)}$, then $u\in X_{S(n)}$. If $(u=x_1)\lor\ldots\lor(u=x_n)$, then $u\in X_n$. Note that $X_n\in Z_{S(n)}$, thus for some $w\in Z_{S(n)}$, $u\in w$. And we have $u\in X_{S(n)}$.

In all cases, $u\in X_{S(n)}$. We have shown that a set $u$ is in $X_{S(n)}$ if and only if $u$ is equal to one of $x_1,\ldots,x_{S(n)}$. By axiom of extensionality, this set is unique. Hence the case $S(n)$ is proven. By meta-logical induction, given any meta-logical natural number $n$, the case $n$ is proven. $\blacksquare$

Proof. Suppose for contradiction that there exists $x$ such that $x\in x$. Then by axiom of foundation, for some $u\in\{x\}$, no set is in both $\{x\}$ and $u$. But then we have $u=x$ and $x\in x$, thus $x\in\{x\}$ and $x\in u$, a contradiction.

Proof. Suppose $u\in x$ and $x\in X$, then $u\in\bigcup X$. Thus $x\in X$ implies $x\subseteq\bigcup X$. $\blacksquare$

Proof. If $x\in a\cup b$, then for some $c\in\{a,b\}$, $x\in c$. Then $c=a$ or $c=b$, thus $x\in a$ or $x\in b$. Suppose $x\in a$ or $x\in b$. If $x\in a$, then $x\in a$ and $a\in\{a,b\}$, thus $x\in a\cup b$. Similarly, if $x\in b$ then $x\in a\cup b$. $\blacksquare$

Proof. Suppose $x\in X$ and $u\in\bigcap X$, then $u\in x$. Thus $x\in X$ implies $\bigcap X\subseteq x$. $\blacksquare$

Proof. If $x\in a\cap b$, then for all $c\in\{a,b\}$, $x\in c$. Since $a\in\{a,b\}$ and $b\in\{a,b\}$, we have $x\in a$ and $x\in b$. Suppose $x\in a$ and $x\in b$. If $c\in\{a,b\}$, then $c=a$ or $c=b$. In both cases, we have $x\in c$. Thus $x\in a\cap b$. $\blacksquare$

Proof. Note that $p=(a,b)=\{\{a\},\{a,b\}\}$, implying $p$ is non-empty. Clearly, $\cup p=\{a,b\}$ and $\cap p=\{a\}$. So $\cup\cap p=\cup\{a\}=a$.

• If $a=b$, then $\cup p=\{a\}$, and $\cup p=\cap p$, so we have $(\cup p\neq\cap p)\to(u\notin\cap p)$, implying $\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}=\cup p=\{a\}$, so $\cup\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}=a=b$.
• If $a\neq b$, then $\cup p\neq\cap p$.
  • If $u=a$, then $u\in\cap p$, so $u\notin\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}$.
  • If $u=b$, then $u\in\cup p$ and $u\notin\cap p$, so $u\in\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}$.
  • If $u\neq a$ and $u\neq b$, then $u\notin\cup p$, so $u\notin\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}$.
  Hence, $u\in\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}$ if and only if $u=b$, implying $\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}=\{b\}$ and thus $\cup\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}=b$.

Therefore, we conclude that $\cup\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}=b$. $\blacksquare$

Proof. Suppose $(a,b)=(c,d)$. Let $p$ denote $(a,b)$ and $q$ denote $(c,d)$. Then $a=\cup\cap p=\cup\cap q=c$ and $b=\cup\{u\in\cup p|((\cup p\neq\cap p)\to(u\notin\cap p))\}=\cup\{u\in\cup q|((\cup q\neq\cap q)\to(u\notin\cap q))\}=d$. $\blacksquare$

Proof. Suppose $p\in A\times B$, then there exist $a\in A$ and $b\in B$ such that $p=(a,b)$, meaning that $p$ is an ordered pair of $A$ and $B$. If $p$ is an ordered pair of $A$ and $B$, then there exist $a\in A$ and $b\in B$ such that $p=(a,b)$. Thus $\{a\},\{a,b\}\in\mathcal P(A\cup B)$, and hence $p=(a,b)=\{\{a\},\{a,b\}\}\in\mathcal P(\mathcal P(A\cup B))$. Therefore, $p\in A\times B$. $\blacksquare$

Proof. Let $u\in f$, then $u\in X\times Y$, so for some $x\in X$ and $y\in Y$, $u=(x,y)$. Then we have $(x,y)\in f$, implying $g(x)=f(x)=y$. Note that $(x,g(x))\in g$, hence $(x,y)\in g$ and thus $u\in g$. By a symmetric argument, if $u\in g$, then $u\in f$. Therefore $f=g$. $\blacksquare$

Proof. Let $y\in\{y\in Y|\exists x\in X,f(x)=y\}$. Then for some $x\in X$, $f(x)=y$. Since $f\subseteq X\times Z$ and $(x,f(x))\in f$, we have $y=f(x)\in Z$. Thus $y\in\{y\in Z|\exists x\in X,f(x)=y\}$. By a symmetric argument, we have the other direction. $\blacksquare$

Proof. Suppose $f$ is a function from $X$ to $Z$. Let $y\in V$, then for some $x\in X$, $f(x)=y$. Since $f\subseteq X\times Z$ and $(x,f(x))\in f$, we have $y=f(x)\in Z$. Thus $V\subseteq Z$.

Suppose $V\subseteq Z$. We only need to show that $f\in\mathcal P(X\times Z)$. Let $u\in f$. Since $f\subseteq X\times Y$, there exists $x\in X$ and $y\in Y$ such that $u=(x,y)$, thus $(x,y)\in f$. Since $x\in X$, we have $f(x)=y$. Then $y\in\{y\in Y|\exists x\in X,f(x)=y\}=V\subseteq Z$. Thus $u=(x,y)\in X\times Z$. And we have shown that $f\in\mathcal P(X\times Z)$. $\blacksquare$

Proof. Clearly $f\in\mathcal P(X\times Y)$. Let $x'\in X$, then there exists a unique $y^*\in Y$ that satisfies $\varphi(x',y^*)$. Thus $(x',y^*)\in X\times Y$, and $(x,y)=(x',y^*)$ implies $x'=x$ and $y^*=y$, which implies $\varphi(x',y)\to\varphi(x,y)$ and $\varphi(x',y^*)\to\varphi(x',y)$, and since we have $\varphi(x',y^*)$, we have $\varphi(x,y)$. We have shown that $(x',y^*)\in f$. Now suppose $(x',y'')\in f$, then $y''\in Y$. Since $\forall x\forall y((x,y)=(x',y'')\to\varphi(x,y))$, we have $\varphi(x',y'')$. And by uniqueness of $y\in Y$ that satisfies $\varphi(x',y)$, we have $y^*=y''$. We have shown that there exists a unique $y'$ such that $(x',y')\in f$. Thus $f$ is a function from $X$ to $Y$. $\blacksquare$

Proof. Note that for all $x'\in X$, $y(x')\in Y$ and $y(x')=y(x')$. Trivially, if $z\in Y$ and $z=y(x')$, then $z=y(x')$. Thus given $x'\in X$, there exists a unique $y'\in Y$ such that $y'=y(x')$. Then we can define a function $f:X\to Y$ by $y'=y(x')$ with $x'$ as input and $y'$ as output. Let $x\in X$, then $(x,y(x))\in X\times Y$, and if $(x',y')=(x,y(x))$, then $y'=y(x)=y(x')$. Hence $(x,y(x))\in f$, and therefore $f(x)=y(x)$. $\blacksquare$

Proof. Clearly, for all $x\in X$, there exists a unique $y'$ such that $y'=y(x)$. Then by axiom schema of replacement, there exists $Y$ such that for all $x\in X$, there exists $y'\in Y$ such that $y'=y(x)$, implying $y(x)\in Y$. Then by the above proposition, we can define a function $f:X\to Y$ such that for all $x\in X$, $f(x)=y(x)$. $\blacksquare$

Proof. Suppose $u\in\{y(x):\varphi(x)\}$. Then there exists $x\in X$ such that $f(x)=u$. Since $x\in X$, we have $\varphi(x)$ and $f(x)=y(x)$, thus $y(x)=u$.

Suppose there exists $x$ such that $\varphi(x)$ and $y(x)=u$. Since $\varphi(x)$, we have $x\in X$, thus $f(x)=y(x)=u$ and $u=f(x)\in Y$. Hence $u\in\{y(x):\varphi(x)\}$. $\blacksquare$

Proof. Let $u\in\{y_1(x):\varphi(x)\}$, then there exists $x$ such that $\varphi(x)$ and $y_1(x)=u$, thus $y_2(x)=y_1(x)=u$, hence $u\in\{y_2(x):\varphi(x)\}$. By a symmetric argument, if $u\in\{y_2(x):\varphi(x)\}$ then $u\in\{y_1(x):\varphi(x)\}$. Therefore $\{y_1(x):\varphi(x)\}=\{y_2(x):\varphi(x)\}$. $\blacksquare$

Proof. Since $x\in U$ and $f(x)=f(x)$, there exists $u\in U$ such that $f(u)=f(x)$, thus $f(x)\in f(U)$. $\blacksquare$

Proof. If $y\in f(V)$, then there exists $x\in V$ such that $f(x)=y$, thus there exists $x\in U$ such that $f(x)=y$, hence $y\in f(U)$. $\blacksquare$

Proof. If $x\in f^{-1}(T)$, then $x\in X$ and $f(x)\in T$, thus $x\in X$ and $f(x)\in S$, hence $x\in f^{-1}(S)$. $\blacksquare$

Proof. Let $x\in U$, then $x\in X$ and $f(x)\in f(U)$, thus $x\in f^{-1}(f(U))$.

Let $y\in f(f^{-1}(V))$, then there exists $x\in f^{-1}(V)$ such that $f(x)=y$, thus $y=f(x)\in V$. $\blacksquare$

Proof. Suppose $f$ is bijective, then it is both injective and surjective. By surjectivity, there exists $x\in X$ such that $f(x)=y$. Now suppose we have $x'\in X$ such that $f(x')=y$, then by injectivity, we have $x=x'$. Thus there exists a unique $x\in X$ such that $f(x)=y$.

Suppose for all $y\in Y$, there exists a unique $x\in X$ such that $f(x)=y$. Let $y\in Y$, then there trivially exists $x\in X$ such that $f(x)=y$, thus $f$ is surjective. Now suppose $x_1,x_2\in X$ and $f(x_1)=f(x_2)$. Since $f(x_1)\in Y$, there exists a unique $x\in X$ such that $f(x)=f(x_1)$. Since $x_1\in X$ and $f(x_1)=f(x_1)$, by uniqueness we have $x_1=x$. Since $x_2\in X$ and $f(x_2)=f(x_1)$, by uniqueness we have $x_2=x$. Thus $x_1=x_2$. And we have shown that $f$ is injective. Since $f$ is both injective and surjective, it is bijective. $\blacksquare$

Proof. Let $g=\{u\in(Y\times X)|\forall y\forall x((y,x)=u\to f(x)=y)\}$ Clearly, $g$ is a subset of $Y\times X$. Let $y'\in Y$, by surjectivity of $f$, there exists $x^*\in X$ such that $f(x^*)=y'$. If $(y,x)=(y',x^*)$, then $y=y'$ and $x=x^*$, thus $f(x)=y$. We have shown that $(y',x^*)\in g$. Suppose $(y',x'')\in g$, then $f(x'')=y'$, and by injectivity of $f$, $x''=x^*$. We have shown that there exists a unique $x'$ such that $(y',x')\in g$. Therefore, $g$ is a function from $Y$ to $X$. $\blacksquare$

Proof. Let $x\in X$, then $f(x)\in Y$, so $(f(x),x)\in Y\times X$. Since if $(y',x')=(f(x),x)$, then $f(x')=f(x)=y'$, we have $(f(x),x)\in f^{-1}$, thus $f^{-1}(f(x))=x$.

Let $y\in Y$, then there exists $x\in X$ such that $f(x)=y$. Hence $f(f^{-1}(y))=f(f^{-1}(f(x)))=f(x)=y$. $\blacksquare$

Proof. Let $f:X\to Y$ be bijective.

Suppose $y_1,y_2\in Y$ and $f^{-1}(y_1)=f^{-1}(y_2)$. Then $y_1=f(f^{-1}(y_1))=f(f^{-1}(y_2))=y_2$. Thus $f^{-1}$ is injective.

Suppose $x\in X$, then $f(x)\in Y$ and $f^{-1}(f(x))=x$. Thus $f^{-1}$ is surjective. $\blacksquare$

Proof. Since $f^{-1}:Y\to X$ is bijective, $(f^{-1})^{-1}$ is a function from $X$ to $Y$. Let $x\in X$, then there exists $y\in Y$ such that $f^{-1}(y)=x$, thus $$(f^{-1})^{-1}(x)=(f^{-1})^{-1}(f^{-1}(y))=y=f(f^{-1}(y))=f(x)$$ Therefore $(f^{-1})^{-1}=f$. $\blacksquare$

Proof. Suppose $x_1,x_2\in X$ and $f(x_1)=f(x_2)$, then $x_1=g(f(x_1))=g(f(x_2))=x_2$. Thus $f$ is injective. Let $y\in Y$, then $g(y)\in X$ and $f(g(y))=y$. Thus $f$ is surjective. We have shown that $f$ is bijective.

Let $y\in Y$, then there exists $x\in X$ such that $f(x)=y$, thus $g(y)=g(f(x))=x=f^{-1}(f(x))=f^{-1}(y)$. Therefore $g=f^{-1}$. $\blacksquare$

Proof. Clearly, $f|_U$ is a function from $U$ to $V$, which is the range of $f|_U$. Let $x_1,x_2\in U$ and $f|_U(x_1)=f|_U(x_2)$, then $f(x_1)=f|_U(x_1)=f|_U(x_2)=f(x_2)$, then by injectivity of $f$, $x_1=x_2$. Thus $f|_U:U\to V$ is injective. Let $y\in V$, then $y\in\{v\in Y|\exists u\in U,f|_U(u)=v\}$, thus there exists $x\in U$ such that $f|_U(x)=y$, implying $f|_U:U\to V$ is surjective.

Note that both $(f|_U)^{-1}$ and $(f^{-1})|_V$ are functions from $V$. Let $y\in V$, then there exists $x\in U$ such that $f|_U(x)=y$, and thus $f(x)=f|_U(x)=y$. Hence $$(f|_U)^{-1}(y)=(f|_U)^{-1}(f|_U(x))=x=f^{-1}(f(x))=f^{-1}(y)=(f^{-1})|_V(y)$$ Therefore $(f|_U)^{-1}=(f^{-1})|_V$. $\blacksquare$

Proof. Let $x\in\{u\in X|f(u)\in V\}$. Then $x\in X$ and $f(x)\in V$, and we have $f^{-1}(f(x))=x$. Hence $x\in\{u\in X|\exists v\in V, f^{-1}(v)=u\}=\{u\in X|\exists v\in V, f^{-1}|_V(v)=u\}$.

Let $x\in\{u\in X|\exists v\in V, f^{-1}|_V(v)=u\}$, then $x\in X$ and for some $v\in V$, $f^{-1}(v)=x$. Thus $f(x)=f(f^{-1}(v))=v\in V$. Hence $x\in\{u\in X|f(u)\in V\}$. $\blacksquare$

Proof. Let $x\in U$, then $f(x)\in f(U)$, thus $x=f^{-1}(f(x))\in f^{-1}(f(U))$. Let $x\in f^{-1}(f(U))$, then $x\in X$ and for some $y\in f(U)$, $f^{-1}(y)=x$, thus $y\in Y$ and for some $u\in U$, $f(u)=y$. Hence $x=f^{-1}(f(u))=u\in U$. We have shown that $f^{-1}(f(U))=U$, and by a symmetric argument, we have $f(f^{-1}(V))=(f^{-1})^{-1}(f^{-1}(V))=V$. $\blacksquare$

Proof. Let $h=\{u\in (f^{-1}(C)\times D)|\forall x\forall y(u=(x,y)\to(y=g(f(x))))\}$. Clearly, $h$ is a subset of $f^{-1}(C)\times D$. Suppose $x\in f^{-1}(C)$, then $f(x)\in C$ and $g(f(x))\in D$, thus $(x,g(f(x)))\in f^{-1}(C)\times D$. Since if $(x,g(f(x)))=(x',y')$, then $y'=g(f(x))=g(f(x'))$, we have $(x,g(f(x)))\in h$. Suppose $(x,y)\in h$, then if $(x,y)=(x',y')$, we have $y=y'=g(f(x'))=g(f(x))$. We have shown that there exists a unique $y$ such that $(x,y)\in h$. Therefore $h$ is a function from $f^{-1}(C)$ to $D$. $\blacksquare$

Proof. $h\circ(g\circ f)$ is a function from $(g\circ f)^{-1}(E)$ to $F$ and $(h\circ g)\circ f$ is a function from $f^{-1}(g^{-1}(E))$ to $F$. Let $u\in(g\circ f)^{-1}(E)$, then $u\in f^{-1}(C)$ and $(g\circ f)(u)\in E$, thus $(g\circ f)(u)=g(f(u))$ and $g(f(u))\in E$. Since $f(u)\in C$, $f(u)\in g^{-1}(E)$. Since $u\in A$, $u\in f^{-1}(g^{-1}(E))$. For the other direction, let $u\in f^{-1}(g^{-1}(E))$, then $f(u)\in g^{-1}(E)$ and $g(f(u))\in E$. Since $g^{-1}(E)\subseteq C$, $u\in f^{-1}(C)$, thus $(g\circ f)(u)=g(f(u))$ and $(g\circ f)(u)\in E$. Hence $u\in(g\circ f)^{-1}(E)$. We have shown that $(g\circ f)^{-1}(E)$=$f^{-1}(g^{-1}(E))$, and we will denote these sets as $S$. Let $u\in S$, then $u\in f^{-1}(C)$ and $f(u)\in g^{-1}(E)$ as shown above, thus $$(h\circ(g\circ f))(u)=h((g\circ f)(u))=h(g(f(u)))=(h\circ g)(f(u))=((h\circ g)\circ f)(u)$$ Hence $h\circ(g\circ f)=(h\circ g)\circ f$. $\blacksquare$

Proof. $g\circ f$ is clearly a function from $X$ to $Z$. Suppose $x_1,x_2\in X$ and $(g\circ f)(x_1)=(g\circ f)(x_2)$, then $g(f(x_1))=g(f(x_2))$, thus by injectivity of $g$, $f(x_1)=f(x_2)$, and by injectivity of $f$, $x_1=x_2$. Hence $g\circ f$ is injective. Let $z\in Z$, then by surjectivity of $g$, there exists $y\in Y$ such that $g(y)=z$, and by surjectivity of $f$, there exists $x\in X$ such that $f(x)=y$, thus $(g\circ f)(x)=g(f(x))=g(y)=z$. Hence $g\circ f$ is surjective. We have shown that $g\circ f$ is bijective.

Clearly, both $(g\circ f)^{-1}$ and $f^{-1}\circ g^{-1}$ are functions from $Z$ to $X$. Let $z\in Z$, then there exists $x\in X$ such that $(g\circ f)(x)=z$, thus $$(g\circ f)^{-1}(z)=(g\circ f)^{-1}((g\circ f)(x))=x=f^{-1}(f(x))=f^{-1}(g^{-1}(g(f(x))))=(f^{-1}\circ g^{-1})((g\circ f)(x))=(f^{-1}\circ g^{-1})(z)$$ Therefore $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$. $\blacksquare$

Proof. Denote a witness of the axiom by $S^*$, which is a constant symbol, then we have $$\text{(1) }\forall z(\forall y\neg(y\in z)\to(z\in S^*))$$ and $$\text{(2) }\forall x((x\in S^*)\to\forall X(\forall u((u\in X)\leftrightarrow((u\in x)\lor(u=x)))\to(X\in S^*)))$$ From (1), we have $$(\forall y\neg(y\in\emptyset)\to(\emptyset\in S^*))$$ Since, by definition of $\emptyset$, we have $$\forall y\neg(y\in\emptyset)$$ we conclude that $$\text{(3) }(\emptyset\in S^*)$$ From (2), by LA2, we have $$((x\in S^*)\to(\forall u((u\in(x\cup\{x\}))\leftrightarrow((u\in x)\lor(u=x)))\to((x\cup\{x\})\in S^*)))$$ Clearly, we have $$\forall u((u\in(x\cup\{x\}))\leftrightarrow((u\in x)\lor(u=x)))$$ thus $$((x\in S^*)\to((x\cup\{x\})\in S^*))$$ and hence $$\text{(4) }\forall x((x\in S^*)\to((x\cup\{x\})\in S^*))$$ Therefore, from (3) and (4), $$\exists S((\emptyset\in S)\land\forall x((x\in S)\to((x\cup\{x\})\in S)))$$ $\blacksquare$

Proof. $\emptyset$ is in every set in $I$, so $\emptyset\in N$. Suppose a set $x$ is in $N$, then $x$ is in every set in $I$, so $(x\cup\{x\})$ is in every set in $I$, so $(x\cup\{x\})$ is in $N$. $\blacksquare$

Proof. Given $n\in N$, clearly $n\in S(n)$, thus $S(n)\neq0$. $\blacksquare$

Proof. Suppose $\varphi(0)$ and for every $n\in N$, $\varphi(n)$ implies $\varphi(S(n))$. Consider $\{u\in N|\varphi(u)\}$, denoted $A$. Since $0\in N$ and $\varphi(0)$, we have $0\in A$. If $n\in A$, then $n\in N$ and $\varphi(n)$, so $S(n)\in N$ and $\varphi(S(n))$, which means $S(n)\in A$. Hence we have $\mathcal I(A)$. Using the same notations as above, since $A\subseteq N\subseteq C$, we have $A\in \mathcal P(C)$. Since we also have $\mathcal I(A)$, we know $A\in I$, which then implies $N\subseteq A$, and hence $N=A$. Then we conclude that for every $n\in N$ we have $\varphi(n)$. $\blacksquare$

Proof. We will use induction theorem. For the base case, since $0$ is $\emptyset$, we have $0\subset N$. Now take as inductive hypothesis that $n\in N$ and $n\subset N$. If $x\in S(n)$, then either $x=n$ or $x\in n$. If $x=n$, then $x\in N$. If $x\in n$, since $n\subset N$, we have $x\in N$. Thus $S(n)\subseteq N$. Since $n\in N$, we have $S(n)\in N$, but then since $S(n)\notin S(n)$, we have $S(n)\subset N$. This concludes the inductive step. By induction theorem, for every $n\in N$ we have $n\subset N$. $\blacksquare$

Proof. We will use induction theorem. For the base case, since $0$ is $\emptyset$, we have $\neg(x\in 0)$, so for all $x$, $x\in 0$ implies $x\subset 0$. Now take as inductive hypothesis that $n\in N$ and for all $x$, $x\in n$ implies $x\subset n$. If $x\in S(n)$, then either $x=n$ or $x\in n$:

• If $x=n$, since no set contains itself, we have $n\notin n$, $n\in S(n)$, and $n\subseteq S(n)$, which together imply $n\subset S(n)$, and hence $x\subset S(n)$.
• If $x\in n$, then we have $x\subset n$. Since $n\subseteq S(n)$, we have $x\subset S(n)$.

In both cases we have $x\subset S(n)$. Hence for all $x$, $x\in S(n)$ implies $x\subset S(n)$. This concludes the inductive step. By induction theorem, for every $n\in N$ we have that for all $x$, $x\in n$ implies $x\subset n$. $\blacksquare$

Proof. Let $\varphi(n)$ denote the formula "if $n\neq0$, then there exists $m\in N$ such that $S(m)=n$". Trivially, we have $\varphi(0)$. Suppose $n\in N$ and $\varphi(n)$, we have $S(n)=S(n)$, thus there exists $m\in N$ such that $S(m)=S(n)$, implying we have $\varphi(S(n))$. By induction, for all $n\in N$, we have $\varphi(n)$. $\blacksquare$

Proof. Fix $n\in N$ and suppose $\varphi(0)$ and for all $k\in n$, $\varphi(k)$ implies $\varphi(S(k))$. Let $\psi(k)$ be the formula "if $k\in n$, then $\varphi(k)$". Since $\varphi(0)$, we have $\psi(0)$. Suppose $k\in N$ and $\psi(k)$.

• If $S(k)\in n$, then $S(k)\subset n$, thus $k\in n$, and we have $\varphi(k)$, hence we also have $\varphi(S(k))$, and thus $\psi(S(k))$.
• If $S(k)\notin n$, then we have $\psi(S(k))$.

Then by induction, for all $k\in N$, we have $\psi(k)$.

• If $n=0$, then by $\varphi(0)$, we have $\varphi(n)$.
• If $n\neq0$, then for some $m\in N$ we have $S(m)=n$, implying $m\in n$. Since $m\in N$, we have $\psi(m)$, and since $m\in n$, we have $\varphi(m)$, and $\varphi(m)$ implies $\varphi(S(m))$. Hence we have $\varphi(S(m))$, which is $\varphi(n)$.

$\blacksquare$

Proof. Suppose $n,m\in N$ and $S(n)=S(m)$, then $((n\cup\{n\})=(m\cup\{m\}))$, so either $n=m$ or $n\in m$, and either $m=n$ or $m\in n$. Suppose $n\neq m$, then $n\in m$ and $m\in n$, but by the above lemma, we then have $n\subset m$ and $m\subset n$, and hence $n\subseteq m$ and $m\subseteq n$, which then implies $m=n$, a contradiction. Therefore, $n,m\in N$ and $S(n)=S(m)$ imply $n=m$. This proves injectivity of $S$. $\blacksquare$

Proof. We will first prove that if $m,n\in N$ then we have at least one of $m\in n$, $n\in m$, $m=n$. Denote that we have at least one of $m\in n$, $n\in m$, $m=n$ by $\varphi(m,n)$, denote $\{s\in N|\varphi(x,s)\}$ by $C(x)$, and denote $\{r\in N|C(r)=N\}$ by $I$.

Since $0=0$, we have $\varphi(0,0)$, so $0\in C(0)$. Suppose $a\in N$ and $a\in C(0)$, then $\varphi(0,a)$,

• if $0\in a$, then $0\in S(a)$, so $\varphi(0,S(a))$;
• if $a\in 0$, we have a contradiction, hence $\varphi(0,S(a))$;
• if $a=0$, then $0\in S(a)$, so $\varphi(0,S(a))$.

And we conclude that $a\in N$ and $a\in C(0)$ imply $S(a)\in C(0)$. So by induction, $C(0)=N$, hence $0\in I$.

Suppose $a\in N$ and $a\in I$, then $C(a)=N$, so $\varphi(a,0)$,

• if $a\in 0$, we have a contradiction, hence $\varphi(S(a),0)$;
• if $0\in a$, then $0\in S(a)$, so $\varphi(S(a),0)$;
• if $a=0$, then $0\in S(a)$, so $\varphi(S(a),0)$.

And we conclude that $0\in C(S(a))$. Suppose $b\in N$ and $b\in C(S(a))$, then $\varphi(S(a),b)$,

• if $S(a)\in b$, then $S(a)\in S(b)$, so $\varphi(S(a),S(b))$;
• if $b\in S(a)$, then either $b=a$ or $b\in a$,
  • if $b=a$, then $S(a)=S(b)$, so $\varphi(S(a),S(b))$;
  • if $b\in a$, then $b\subset a$; note that $C(a)=N$, we have $\varphi(a,S(b))$,
    • if $a\in S(b)$, then either $a=b$ or $a\in b$,
      • if $a=b$, then $S(a)=S(b)$, so $\varphi(S(a),S(b))$;
      • if $a\in b$, then $a\subset b$, a contradiction, hence $\varphi(S(a),S(b))$;
    • if $S(b)\in a$, then $S(b)\in S(a)$, so $\varphi(S(a),S(b))$;
    • if $a=S(b)$, then $S(b)\in S(a)$, so $\varphi(S(a),S(b))$;
• if $S(a)=b$, then $S(a)\in S(b)$, so $\varphi(S(a),S(b))$.

And we conclude that $b\in N$ and $b\in C(S(a))$ imply $S(b)\in C(S(a))$. So by induction, $C(S(a))=N$, hence $S(a)\in I$. Note that we are working under the assumption that $a\in N$ and $a\in I$, so we conclude that $a\in N$ and $a\in I$ imply $S(a)\in I$. Again by induction, we have that $I=N$. If $m,n\in N$, then $m\in I$, or $C(m)=N$, so $n\in C(m)$, or $\varphi(m,n)$, we have at least one of $m\in n$, $n\in m$, $m=n$.

We now prove that if $m,n\in N$, then we have at most one of $m\in n$, $n\in m$, $m=n$. Suppose $m,n\in N$, then $m\in n$ implies $m\subset n$, and $n\in m$ implies $n\subset m$. Clearly, we have at most one of $m\subset n$, $n\subset m$, $m=n$. Hence we have at most one of $m\in n$, $n\in m$, $m=n$.

Combining the above results, we conclude that $m,n\in N$ implies that we have at least and at most one of $m\in n$, $n\in m$, $m=n$; that is, we have exactly one of $m\in n$, $n\in m$, $m=n$. $\blacksquare$

Proof. Fix $m,n,k\in N$.

Suppose $m\lt n$ and $n\lt k$, then $m\in n$ and $n\in k$, so $m\subset n$ and $n\subset k$, hence $m\subset k$.

• If $m\in k$, then we have $m\lt k$;
• if $k\in m$, then $k\lt m$, so $k\subset m$, a contradiction;
• if $k=m$, we have a contradiction.

Hence we conclude that $m\lt k$. $\blacksquare$

Proof. Suppose $M\subseteq N$ is non-empty. Suppose for contradiction that it has no least element. Then clearly $0\notin M$. Let $n\in N$. Suppose for inductive hypothesis that for all $k\le n$, $k\notin M$. Then if $S(n)\in M$, it would be the least element of $M$. Thus for all $k\le S(n)$, $k\notin M$. By induction, for all $n\in N$, for all $k\le n$, $k\notin M$. But then $M=\emptyset$, a contradiction. Hence $M$ has a least element. $\blacksquare$

Proof. Fix $m,n\in N$, suppose $m\lt n$ and $\neg(S(m)\le n)$, then neither $S(m)\in n$ nor $S(m)=n$, but we have exactly one of $S(m)\in n$, $n\in S(m)$, $S(m)=n$, so we have $n\in S(m)$, then either $n\in m$ or $n=m$, but then we already have $m\in n$, a contradiction. Hence we have either $\neg(m\lt n)$ or $S(m)\le n$, which is just "$m\lt n$ implies $S(m)\le n$". Hence for all $m,n\in N$, if $m\lt n$, then $S(m)\le n$. $\blacksquare$

Proof. Consider $\{A\in\mathcal P(N\times X)|(((0,a)\in A)\land\forall n\forall x(((n,x)\in A)\to((S(n),f(x))\in A)))\}$, denoted $I$. Clearly $N\times X$ is in $I$, so $I$ is non-empty. So we can define $u$ to be $\cap I$, then $u$ is a subset of $N\times X$, and clearly,

• $(0,a)\in u$, and
• $(n,x)\in u$ implies $(S(n),f(x))\in u$.

Now consider $\{n\in N|\exists!x((n,x)\in u)\}$, denoted $C$.

Suppose $0\notin C$, since $(0,a)\in u$, there exists some $b\neq a$ such that $(0,b)\in u$. Denote $u\setminus\{(0,b)\}$ by $u'$. Since $(0,a)\in u$ and $(0,a)\notin\{(0,b)\}$, $(0,a)\in u'$; if $(n,x)\in u'$, then $(n,x)\in u$, so $(S(n),f(x))\in u$, but then, since $n\in N$, $S(n)\neq0$, so $(S(n),f(x))\notin\{(0,b)\}$, and hence $(S(n),f(x))\in u'$. We have shown that $u'\in I$, so $u\subseteq u'$, a contradiction. Therefore, $0\in C$.

Suppose there exists some $n^*\in C$ such that $S(n^*)\notin C$, then $n^*\in N$ and for some unique $x^*$ we have $(n^*,x^*)\in u$, thus $(S(n^*),f(x^*))\in u$. Since $S(n^*)\in N$, in order for $S(n^*)$ to not be in $C$, there has to be some $y\neq f(x^*)$, such that $(S(n^*),y)\in u$. Denote $u\setminus\{(S(n^*),y)\}$ by $u'$. Since $(0,a)\in u$ and $(0,a)\notin\{(S(n^*),y)\}$, $(0,a)\in u'$. If $(n,x)\in u'$, then $(n,x)\in u$, so $(S(n),f(x))\in u$.

• Suppose $n=n^*$, then $x=x^*$ by uniqueness, so $f(x)\neq y$.
• Suppose $n\neq n^*$, since $S$ is injective, we have $S(n)\neq S(n^*)$.

In both cases we have $(S(n),f(x))\notin\{(S(n^*),y)\}$, and we conclude that $(S(n),f(x))\in u'$. We have shown that $u'\in I$, so $u\subseteq u'$, a contradiction. Therefore, $n\in C$ implies $S(n)\in C$. And we have that for all $n\in N$, $n\in C$ implies $S(n)\in C$.

By induction, for all $n\in N$, $n\in C$. Since $u$ is a subset of $N\times X$, and for all $n\in N$, there exists a unique $x$ such that $(n,x)\in u$, $u$ is a function from $N$ to $X$. Since $(0,a)\in u$, we have $u(0)=a$. let $n\in N$, then $(n,u(n))\in u$, thus $(S(n),f(u(n)))\in u$, implying $u(S(n))=f(u(n))$.

Now suppose $u'$ is a function from $N$ to $X$ such that $u'(0)=a$ and for all $n\in N$, $u'(S(n))=f(u'(n))$. Then $u'(0)=a=u(0)$, and given $n\in N$, if $u'(n)=u(n)$, then $u'(S(n))=f(u'(n))=f(u(n))=u(S(n))$. Thus by induction, for all $n\in N$, $u'(n)=u(n)$. Hence $u'=u$. $\blacksquare$

Proof. Suppose $n\in N$, we immediately have $$n+0=S_n(0)=n$$ So for all $n\in N$, we have $n+0=S_n(0)=n$.

For the other equality, we use induction. For base case, we have $$0+0=0$$ For inductive step, suppose $n\in N$ and $0+n=n$, then $$0+S(n)=S(0+n)=S(n)$$ By induction, for all $n\in N$, we have $0+n=n$. $\blacksquare$

Proof. Fix $m,n\in N$. For base case, we have $$(m+n)+0=m+n=m+(n+0)$$ For inductive step, suppose $k\in N$ and $(m+n)+k=m+(n+k)$, then $$(m+n)+S(k)=S((m+n)+k)=S(m+(n+k))=m+S(n+k)=m+(n+S(k))$$ By induction, for all $k\in N$, we have $(m+n)+k=m+(n+k)$. Hence for all $m,n,k\in N$, we have $(m+n)+k=m+(n+k)$. This proves associativity.

For base case, suppose $n\in N$, we have $$0+n=n=n+0$$ So for all $n\in N$ we have $0+n=n+0$. The inductive step itself requires an induction, as follows. Suppose $m\in N$ and for all $n\in N$ we have $m+n=n+m$, for base case of the inner induction, we have $$S(m)+0=S(m)=0+S(m)$$ For inductive step of the inner induction, suppose $n\in N$ and $S(m)+n=n+S(m)$, then $$S(m)+S(n)=S(S(m)+n)=S(n+S(m))=S(S(n+m))=S(S(m+n))=S(m+S(n))=S(S(n)+m)=S(n)+S(m)$$ By induction, for all $n\in N$, we have $S(m)+n=n+S(m)$. Note that we are working under the assumption that $m\in N$ and for all $n\in N$ we have $m+n=n+m$. We have shown that $m\in N$ and for all $n\in N$ we have $m+n=n+m$ together imply for all $n\in N$ we have $S(m)+n=n+S(m)$. This completes the inductive step. By induction, for all $m\in N$, for all $n\in N$, we have $m+n=n+m$. Hence for all $m,n\in N$, we have $m+n=n+m$. This proves commutativity. $\blacksquare$

Proof. Fix $m,n\in N$. For base case, if $m\lt n$, since $m+0=m=0+m$ and $n+0=n=0+n$, we have that $m+0\lt n+0$ and $0+m\lt 0+n$. So if $m\lt n$, then $m+0\lt n+0$ and $0+m\lt 0+n$.

For inductive step, suppose $k\in N$ and if $m\lt n$, then $m+k\lt n+k$ and $k+m\lt k+n$. If $m\lt n$, we have $m+k\lt n+k$ by inductive hypothesis, thus $S(m+k)\le n+k\lt S(n+k)$. Since $S(m+k)=m+S(k)$ and $S(n+k)=n+S(k)$, we have $m+S(k)\lt n+S(k)$; since $m+S(k)=S(k)+m$ and $n+S(k)=S(k)+n$, we also have $S(k)+m\lt S(k)+n$.

By induction, we have that for all $k\in N$, if $m\lt n$, then $m+k\lt n+k$ and $k+m\lt k+n$. Hence, for all $m,n,k\in N$, if $m\lt n$, then $m+k\lt n+k$ and $k+m\lt k+n$. $\blacksquare$

Proof. Suppose $m+n=0$. If $m\neq0$, then $m\gt0$, thus $m+n\gt0+n=n\ge0$, a contradiction. If $n\neq0$, then $n\gt0$, thus $m+n\gt m+0=m\ge0$, a contradiction. The other direction is trivial. $\blacksquare$

Proof. Fix $m\in N$. For base case, if $m\le0$, then $m=0$, thus $m+0=0+0=0$. If $m+k=0$ for some $k\in N$, then $k=m+k=0$.

For inductive step, suppose $n\in N$ and if $m\le n$, then there exists a unique $k\in N$ such that $m+k=n$. If $m\le S(n)$, then either $m=S(n)$ or $m\le n$.

• If $m=S(n)$, then $m+0=S(n)$. If $m+k=S(n)$ for some $k\in N$, then $m+k=m+0$, implying $k=0$.
• If $m\le n$, then there exists a unique $k\in N$ such that $m+k=n$. Thus $m+S(k)=S(m+k)=S(n)$. If $m+l=S(n)$ for some $l\in N$, then $m+l=m+S(k)$, implying $l=S(k)$.

Thus if $m\le S(n)$, then there exists a unique $k\in N$ such that $m+k=S(n)$. By induction, for all $n\in N$, then there exists a unique $k\in N$ such that $m+k=n$. $\blacksquare$

Proof. Suppose $n\in N$, we immediately have $$n0=P_n(0)=0$$ So for all $n\in N$, we have $n0=0$.

For the other equality, we use induction. For base case, we have $$0\times 0=0$$ For inductive step, suppose $n\in N$ and $0n=0$, then $$0S(n)=0+0n=0+0=0$$ By induction, for all $n\in N$, we have $0n=0$. $\blacksquare$

Proof. Suppose $n\in N$, we have $$n1=nS(0)=n+n0=n+0=n$$ So for all $n\in N$, we have $n1=n$.

For the other equality, we use induction. For base case, we have $$1\times 0=0$$ For inductive step, suppose $n\in N$ and $1n=n$, then $$1S(n)=1+1n=1+n=n+1=n+S(0)=S(n+0)=S(n)$$ By induction, for all $n\in N$, we have $1n=n$. $\blacksquare$

Proof. Fix $m,n\in N$. For base case, we have $$m(n+0)=mn=mn+0=mn+m0$$ For inductive step, suppose $k\in N$ and $m(n+k)=mn+mk$, then $$m(n+S(k))=mS(n+k)=m+m(n+k)=m+(mn+mk)=(m+mn)+mk=(mn+m)+mk=mn+(m+mk)=mn+mS(k)$$ By induction, for all $k\in N$, we have $m(n+k)=mn+mk$.

Fix $m,n\in N$. For base case, we have $$(m+n)0=0=0+0=m0+n0$$ For inductive step, suppose $k\in N$ and $(m+n)k=mk+nk$, then $$(m+n)S(k)=(m+n)+(m+n)k=(m+n)+(mk+nk)=(n+m)+(mk+nk)=((n+m)+mk)+nk=(n+(m+mk))+nk=((m+mk)+n)+nk=(m+mk)+(n+nk)=mS(k)+nS(k)$$ By induction, for all $k\in N$, we have $(m+n)k=mk+nk$. $\blacksquare$

Proof. Fix $m,n\in N$. For base case, we have $$(mn)0=0=m0=m(n0)$$ For inductive step, suppose $k\in N$ and $(mn)k=m(nk)$, then $$(mn)S(k)=mn+(mn)k=mn+m(nk)=m(n+nk)=m(nS(k))$$ By induction, for all $k\in N$, we have $(mn)k=m(nk)$. Hence for all $m,n,k\in N$, we have $(mn)k=m(nk)$. This proves associativity.

For base case, suppose $n\in N$, we have $$0n=0=n0$$ So for all $n\in N$ we have $0n=n0$. The inductive step itself requires an induction, as follows. Suppose $m\in N$ and for all $n\in N$ we have $mn=nm$, for base case of the inner induction, we have $$S(m)0=0=0S(m)$$ For inductive step of the inner induction, suppose $n\in N$ and $S(m)n=nS(m)$, then $$S(m)S(n)=S(m)+S(m)n=S(m)+nS(m)=nS(m)+S(m)=nS(m)+1S(m)=(n+1)S(m)=(n+S(0))S(m)=S(n+0)S(m)=S(n)S(m)$$ By induction, for all $n\in N$, we have $S(m)n=nS(m)$. Note that we are working under the assumption that $m\in N$ and for all $n\in N$ we have $mn=nm$. We have shown that $m\in N$ and for all $n\in N$ we have $mn=nm$ together imply for all $n\in N$ we have $S(m)n=nS(m)$. This completes the inductive step. By induction, for all $m\in N$, for all $n\in N$, we have $mn=nm$. Hence for all $m,n\in N$, we have $mn=nm$. This proves commutativity. $\blacksquare$

Proof. Fix $m,k\in N$ such that $k\neq0$. For base case, if $m\lt 0$, we trivially have $mk\lt 0k$ and $km\lt k0$.

For inductive step, suppose $n\in N$ and if $m\lt n$, then $mk\lt nk$ and $km\lt kn$. If $m\lt S(n)$, then $m\le n$, thus $km=mk\le nk\lt k+nk=k+kn=kS(n)=S(n)k$.

By induction, for all $n\in N$, if $m\lt n$, then $mk\lt nk$ and $km\lt kn$. $\blacksquare$

Proof. If $m\neq0$ and $n\neq0$, then $n\gt0$, thus $mn\gt m0=0$, implying $mn\neq0$. Hence if $mn=0$ then $m=0$ or $n=0$. The other direction is trivial. $\blacksquare$

Proof. Fix $n\in N$ such that $n\neq0$. For base case, we have $0\lt n$ and $0=n0+0$.

For inductive step, suppose $m\in N$ and there exists a pair of natural numbers $q,r$ such that $r\lt n$ and $m=nq+r$. Then $S(r)\le n$.

• If $S(r)\lt n$, then $S(r)\lt n$ and $S(m)=S(nq+r)=nq+S(r)$.
• If $S(r)=n$, then $0\lt n$ and $S(m)=nq+S(r)=nq+n=nS(q)+0$.

By induction, for all $m\in N$, there exists a pair of natural numbers $q,r$ such that $r\lt n$ and $m=nq+r$.

Given $m,n\in N$ such that $n\neq0$, if for some $q,r,q',r'\in N$, we have $r\lt n$, $m=nq+r$, $r'\lt n$, and $m=nq'+r'$, then $nq+r=nq'+r'$. Assume $q\gt q'$, then $q'+(q-q')=q$, thus $q-q'\gt0$ and $nq=nq'+n(q-q')$, implying $n(q-q')\gt0$ and hence $nq\gt nq'$. If $r\ge r'$, then $nq+r\ge nq+r'\gt nq'+r'$, a contradiction. Thus $r\lt r'$. Then $r'=r+(r'-r)$, and we have $(nq'+r)+n(q-q')=nq+r=nq'+r'=(nq'+r)+(r'-r)$, implying $n(q-q')=r'-r$. Since $q-q'\gt0$, we have $q-q'\ge1$, thus $n(q-q')\ge n1=n\gt r'\ge r'-r$, a contradiction. By a symmetric argument, $q\lt q'$ also leads to a contradiction. Hence $q=q'$. Then we have $nq=nq'$, thus $r=r'$. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. Since $m\bmod n\lt n$ and $m\bmod n=n0+m\bmod n$, we have $(m\bmod n)\bmod n=m\bmod n$.

Since $m=n(m\divsymbol n)+(m\bmod n)$ and $k=n(k\divsymbol n)+(k\bmod n)$, we have $$m+k=n((m\divsymbol n)+(k\divsymbol n))+((m\bmod n)+(k\bmod n))$$ Since $$(m\bmod n)+(k\bmod n)=n(((m\bmod n)+(k\bmod n))\divsymbol n)+(((m\bmod n)+(k\bmod n))\bmod n)$$ and $((m\bmod n)+(k\bmod n))\bmod n\lt n$, we have $(m+k)\bmod n=((m\bmod n)+(k\bmod n))\bmod n$.

Since $m=n(m\divsymbol n)+(m\bmod n)$ and $k=n(k\divsymbol n)+(k\bmod n)$, we have $$mk=n(n(m\divsymbol n)(k\divsymbol n)+(m\divsymbol n)(k\bmod n)+(m\bmod n)(k\divsymbol n))+(m\bmod n)(k\bmod n)$$ Since $$(m\bmod n)(k\bmod n)=n(((m\bmod n)(k\bmod n))\divsymbol n)+(((m\bmod n)(k\bmod n))\bmod n)$$ and $((m\bmod n)(k\bmod n))\bmod n\lt n$, we have $(mk)\bmod n=((m\bmod n)(k\bmod n))\bmod n$. $\blacksquare$

Proof. If $k\lt m$, then for some $l\in N$, we have $k+l=m$. Thus $$ (n(k\divsymbol n)+(k\bmod n))+(n(l\divsymbol n)+(l\bmod n)) =k+l =m =n(m\divsymbol n)+((k+l)\bmod n) =n(m\divsymbol n)+((k\bmod n)+(l\bmod n))\bmod n$$ Since $$(k\bmod n)+n(l\divsymbol n)+(l\bmod n)\ge(k\bmod n)+(l\bmod n)\ge((k\bmod n)+(l\bmod n))\bmod n$$ we have $n(k\divsymbol n)\le n(m\divsymbol n)$. Since $n\neq0$, we have $k\divsymbol n\le m\divsymbol n$. $\blacksquare$

Proof. For base case, $1^0=E_1(0)=1$. For inductive step, suppose $k\in N$ and $1^k=1$, then $1^{S(k)}=1\times 1^k=1\times 1=1$. By induction, for all $k\in N$, $1^k=1$. $\blacksquare$

Proof. The first equation is proven as follows. Fix $m,n\in N$. For base case, we have $$(mn)^0=1=1\times 1=m^0n^0$$ For inductive step, suppose $k\in N$ and $(mn)^k=m^kn^k$, then $$(mn)^{S(k)}=(mn)(mn)^k=(mn)(m^kn^k)=(nm)(m^kn^k)=((nm)m^k)n^k=(n(mm^k))n^k=((mm^k)n)n^k=(mm^k)(nn^k)=m^{S(k)}n^{S(k)}$$ By induction, for all $k\in N$, we have $(mn)^k=m^kn^k$. Hence for all $m,n,k\in N$, we have $(mn)^k=m^kn^k$.

The second equation is proven as follows. Fix $m,n\in N$. For base case, we have $$m^{n+0}=m^n=m^n1=m^nm^0$$ For inductive step, suppose $k\in N$ and $m^{n+k}=m^nm^k$, then $$m^{n+S(k)}=m^{S(n+k)}=mm^{n+k}=m(m^nm^k)=(mm^n)m^k=(m^nm)m^k=m^n(mm^k)=m^nm^{S(k)}$$ By induction, for all $k\in N$, we have $m^{n+k}=m^nm^k$. Hence for all $m,n,k\in N$, we have $m^{n+k}=m^nm^k$.

The third equation is proven as follows. Fix $m,n\in N$. For base case, we have $$m^{n0}=m^0=1={(m^n)}^0$$ For inductive step, suppose $k\in N$ and $m^{nk}={(m^n)}^k$, then $$m^{nS(k)}=m^{n+nk}=m^nm^{nk}=(m^n){(m^n)}^k={(m^n)}^{S(k)}$$ By induction, for all $k\in N$, we have $m^{nk}={(m^n)}^k$. Hence for all $m,n,k\in N$, we have $m^{nk}={(m^n)}^k$. $\blacksquare$

Proof. Suppose $m\in N\setminus\{0\}$. Then $m^0=1\neq0$ and suppose $n\in N$ and $m^n\neq0$, then $m^{S(n)}=mm^n\neq0$. By induction, for all $n\in N$, $m^n\neq0$. If $m\in N$ and $n=0$, we have $m^n=1\neq0$. We have shown that given $m,n\in N$, if $m^n=0$ then $m=0$ and $n\neq0$.

Suppose $m=0$. For base case, if $0\neq0$ then $m^0=0$. Suppose $n\in N$ and if $n\neq0$ then $m^n=0$. Then $m^{S(n)}=mm^n=0$. By induction, for all $n\in N$, if $n\neq0$ then $m^n=0$. We have shown that given $m,n\in N$, if $m=0$ and $n\neq0$ then $m^n=0$. $\blacksquare$

Proof. Fix $m,n\in N$ such that $m\lt n$. For base case, if $0\neq0$, then $m^0\lt n^0$. For inductive step, suppose $k\in N$ and if $k\neq0$, then $m^k\lt n^k$. Then $m^k\le n^k$. Since $m\lt n$, $n\neq0$, thus $n^k\neq0$. Then we have $m^{S(k)}=mm^k\le mn^k\lt nn^k=n^{S(k)}$. By induction, for all $k\in N$, if $k\neq0$, then $m^k\lt n^k$. $\blacksquare$

Proof. Fix $m,k\in N$ such that $k\gt 1$. For base case, if $m\lt 0$, then $k^m\lt k^0$. For inductive step, suppose $n\in N$ and if $m\lt n$, then $k^m\lt k^n$. If $m\lt S(n)$, then $m\le n$, thus $k^m\le k^n$. Since $k\neq0$, $k^n\neq0$, thus $k^n\lt kk^n=k^{S(n)}$. By induction, for all $n\in N$, if $m\lt n$, then $k^m\lt k^n$. $\blacksquare$

Proof. We will first show that, if $m,n$ are natural numbers, then there is a bijection between $m$ and $n$ if and only if $m=n$. Clearly, $\emptyset$, as an empty function, is a bijection between $0$ and $0$. Suppose $n\neq0$, then $n$ is not empty, so there is no bijection between $0$ and $n$. We have that, as a base case for induction, there is a bijection between $0$ and $n$ if and only if $n=0$. Now take as inductive hypothesis that for all natural number $k\le m$, there is a bijection between $k$ and $n$ if and only if $n=k$. Then for all $n\le m$, there is no bijection between $m+1$ and $n$. Trivially, there is a bijection between $m+1$ and $m+1$. The only case remaining is between $m+1$ and $n$ such that $n\gt m+1$. Suppose $f$ is a bijection from $m+1$ to $n$ with $n\gt m+1$. Then the restriction $f|_m$ of $f$ on $m$ is a bijection from $m$ to $n\setminus\{f(m)\}$. Define $g:n-1\to n\setminus\{f(m)\}$ such that $g(k)=k$ if $k\lt f(m)$ and $g(k)=k+1$ otherwise, then $g$ is clearly a bijection. Thus, $g^{-1}\circ f|_m$ is a bijection from $m$ to $n-1$. But $m\neq n-1$, so there is no bijection between $m$ and $n-1$, a contradiction. Hence, there is no bijection between $m+1$ and $n$ such that $n\gt m+1$. This concludes the inductive step. And we have shown that, for all natural numbers $m,n$, there is a bijection between $m$ and $n$ if and only if $m=n$.

Suppose $X$ is a finite set. Let $m,n$ be natural numbers such that there exist a bijection between $n$ and $X$ and a bijection between $m$ and $X$. Then there is a bijection between $n$ and $m$, implying $m=n$. We have shown that $X$ has a unique cardinality in $N$. $\blacksquare$

Proof. We will first show that, for all natural number $n$, there is no bijection between $n$ and $N$. Since $N$ is non-empty, there is no bijection between $0$ and $N$. Take as inductive hypothesis that there is no bijection between $m$ and $N$. Suppose there is a bijection $f$ from $m+1$ to $N$. Then the restriction $f|_m$ of $f$ on $m$ is a bijection from $m$ to $N\setminus\{f(m)\}$. Define $g:N\to N\setminus\{f(m)\}$ such that $g(k)=k$ if $k\lt f(m)$ and $g(k)=k+1$ otherwise, then $g$ is clearly a bijection. Thus, $g^{-1}\circ f|_m$ is a bijection from $m$ to $N$, a contradiction. Hence, there is no bijection between $m+1$ and $N$. This concludes the inductive step. We have shown that for all natural number $n$, there is no bijection between $n$ and $N$.

Suppose $X$ is a finite set. Let $n=\abs{X}$. If there exists a bijection between $N$ and $X$, then there is a bijection between $n$ and $N$, a contradiction. Therefore, no finite set has a bijection from $N$ to itself. $\blacksquare$

Proof. The base case for $0$ is trivial. Now suppose for every subset $S$ of $n$, there exists $m\le n$ such that there is a bijection from $m$ to $S$. Then for every subset $S'$ of $n+1$, either $n\in S'$ or $n\notin S'$. If $n\notin S'$, then $S'$ is a subset of $n$, so there exists $m\le n\lt n+1$ such that there is a bijection from $m$ to $S'$. If $n\in S'$, then $S'\setminus\{n\}$ is a subset of $n$, so there exists $m\le n$ such that there is a bijection $f$ from $m$ to $S'\setminus\{n\}$. Define $f':m+1\to S'$ by $f'(k)=f(k)$ if $k\in m$ and $f'(m)=n$, then $f'$ is a bijection from $m+1$ to $S'$. Note that $m+1\le n+1$. By induction, for all $n\in N$, for every subset $S$ of $n$, there exists $m\le n$ such that there is a bijection from $m$ to $S$. $\blacksquare$

Proof. Let $f_A$ be a bijection from $n$ to $A$, Let $S$ be the preimage of $f_A$ on $B$, then $S$ is a subset of $n$, so there exists $m\le n$ such that there is a bijection $f_S$ from $m$ to $S$. Now define $f:m\to B$ by $f(k)=(f_A\circ f_S)(k)$, then $f$ is a bijection from $m$ to $B$.

Note that there also exists $l\le n$ such that there is a bijection $f_{n\setminus S}$ from $l$ to $n\setminus S$. Define $g:l\to A\setminus B$ by $g(k)=(f_A\circ f_{n\setminus S})(k)$, then $g$ is a bijection from $l$ to $A\setminus B$. Define $h:m+l\to n$ by:

• if $k\in m$, then $h(k)=(f_A^{-1}\circ f)(k)$;
• if $k\notin m$, then $h(k)=(f_A^{-1}\circ g)(k-m)$.

Then $h$ is a bijection. Hence $n=m+l$, or $l=n-m$. $\blacksquare$

Proof. Let $f_A$ be a bijection from $n$ to $A$ and $f_B$ be a bijection from $m$ to $B$, then there exists $l\le m$ such that there exists a bijection $f_{A\cap B}$ from $l$ to $A\cap B$, and there exists a bijection $f_{A\setminus B}$ from $n-l$ to $A\setminus B$. Now define $f:m+(n-l)\to A\cup B$ by:

• if $k\in m$, then $f(k)=f_B(k)$;
• if $k\notin m$, then $h(k)=f_{A\setminus B}(k-m)$.

Then $f$ is a bijection. Hence $$\abs{A\cup B}=n+m-l=\abs{A}+\abs{B}-\abs{A\cap B}$$ $\blacksquare$

Proof. Suppose for contradiction that such set $X$ exists. By axiom schema of replacement, there exists a set $Y$ such that for all $x\in N$ there is exactly one $y\in Y$ such that $(x,y)\in X$, which defines a function $f:N\to Y$, such that for all $n\in N$, $(n,f(n))\in X$. Thus given $n\in N$, we have $f(S(n))\in f(n)$. Let $U$ be the range of $f$. Then $U$ is clearly non-empty, thus by axiom of foundation, Some $y\in U$ is disjoint from $U$. Since $y\in U$, some $n\in N$ has $f(n)=y$. But then $f(S(n))\in y\cap U$, a contradiction. $\blacksquare$

Proof. Denote the domain of $T$ by $D$. Let $p\in\bigcup_i F_i$, then for some $i\in D$, $p\in F_i\subseteq T_i\times Y\subseteq X\times Y$. Thus $\bigcup_i F_i\subseteq X\times Y$. Let $x\in X$, then for some $i\in D$, $x\in T_i$, thus $(x,F_i(x))\in F_i\subseteq\bigcup_i F_i$. If $(x,y)\in\bigcup_i F_i$, then for some $j\in D$, $(x,y)\in F_j$, thus $x\in T_j$, implying $T_i$ and $T_j$ are not disjoint, hence $i=j$. Now we have $(x,y)\in F_i$, thus $y=F_i(x)$. Hence there exists a unique $y$ such that $(x,y)\in\bigcup_i F_i$. We have shown that $\bigcup_i F_i$ is a function from $X$ to $Y$. Let $j\in D$ and $x\in T_j$, then $(x,\p{\bigcup_i F_i}|_{T_j}(x))\in\p{\bigcup_i F_i}|_{T_j}\subseteq\bigcup_i F_i$. Thus for some $i\in D$, $(x,\p{\bigcup_i F_i}|_{T_j}(x))\in F_i$, implying $x\in T_i$ and hence $i=j$. Hence $(x,\p{\bigcup_i F_i}|_{T_j}(x))\in F_j$, and we have $\p{\bigcup_i F_i}|_{T_j}(x)=F_j(x)$. We conclude that $\p{\bigcup_i F_i}|_{T_j}=F_j$. $\blacksquare$

Proof. Suppose $U\in\{[x]:x\in S\}$, then for some $x\in S$, $U=[x]=\{u\in S|u\sim x\}$, and thus $U\subseteq S$.

Suppose $A,B\in\{[x]:x\in S\}$ and $A\cap B\neq\emptyset$. Then for some $a\in S$, $A=[a]$, and for some $b\in S$, $B=[b]$. Since $A\cap B$ is non-empty, there exists $c\in A\cap B=[a]\cap[b]$. Then $c\sim a$ and $c\sim b$, and thus $a\sim b$. Hence $A=[a]=[b]=B$.

Suppose $u\in\bigcup\{[x]:x\in S\}$, then for some $y\in\{[x]:x\in S\}$, $u\in y$, and for some $x\in S$, $y=[x]$. Thus $u\in[x]\subseteq S$, implying $u\in S$.

Suppose $u\in S$, then $u\in[u]$ and $[u]\in\{[x]:x\in S\}$, thus $u\in\bigcup\{[x]:x\in S\}$. $\blacksquare$

Proof. Let $F$ be a family of pairwise disjoint non-empty sets. Let $X\in F$, then $\exists x(x\in X)$. Let $\mathcal F$ be a witness function defined by $\exists x(x\in X)$, then $\mathcal F(X)\in X$. Now consider the set $\{u\in\bigcup F|\exists Y((Y\in F)\land(u=\mathcal F(Y)))\}$. Let $X\in F$, then $\mathcal F(X)\in X$, thus $\mathcal F(X)\in\bigcup F$ and $\exists Y((Y\in F)\land(\mathcal F(X)=\mathcal F(Y)))$, and we have $\mathcal F(X)\in\{u\in\bigcup F|\exists Y((Y\in F)\land(u=\mathcal F(Y)))\}$. If some set $z$ has $z\in X$ and $z\in\{u\in\bigcup F|\exists Y((Y\in F)\land(u=\mathcal F(Y)))\}$, Then for some $Y\in F$, $z=\mathcal F(Y)$. Since $\mathcal F(Y)\in Y$, we have $z\in Y$. Since $X,Y\in F$ and $z\in X\cap Y$, we have $X=Y$, thus $\mathcal F(X)=\mathcal F(Y)=z$. Hence there exists a unique $x$ such that $x\in\{u\in\bigcup F|\exists Y((Y\in F)\land(u=\mathcal F(Y)))\}$ and $x\in X$, implying that $x\in\{u\in\bigcup F|\exists Y((Y\in F)\land(u=\mathcal F(Y)))\}$ contains exactly one element in common with each of the sets in $F$. $\blacksquare$

Proof. Suppose $F$ is a set such that we have $\forall X(X\in F\to\exists x(x\in X))$. Let $\mathcal F$ be a witness function defined by $\exists x(x\in X)$, then we have $\exists x(x\in X)\to \mathcal F(X)\in X$. If $X\in F$, then $\exists x(x\in X)$, implying $\mathcal F(X)\in X$, and thus $\mathcal F(X)\in\cup F$. Thus we can define a function $f:F\to\cup F$ such that $f(X)=\mathcal F(X)$. Since for all $X\in F$, $f(X)=\mathcal F(X)\in X$, $f$ is a choice function of $F$. $\blacksquare$

Proof. Let $(S,\preceq)$ be a non-empty partially ordered set. Denote the set of all chains in $S$ by $\mathcal C$. Then $(\mathcal C,\subseteq)$ is a partially ordered set. Our main goal is to show that $S$ has a maximal chain, a chain $C^*\in \mathcal C$ such that if $C^*\cup\{s\}$ is a chain then $s\in C^*$ for all $s\in S$.

By axiom of choice, there exists a choice function $f:\mathcal P(X)\setminus\{\emptyset\}\to X$. For each chain $C\in\mathcal C$, denote $\{u\in X|u\notin C\land C\cup\{u\}\in\mathcal C\}$ by $C'$. Then define a function $g:\mathcal C\to\mathcal C$ by

• $g(C)=C$ if $C'=\emptyset$, and
• $g(C)=C\cup\{f(C')\}$ if $C'\neq\emptyset$.

Now we can define towers $\mathcal T$ in $S$ as subsets of $\mathcal C$ with the following properties:

• (1) $\emptyset\in\mathcal T$;
• (2) for all $C$, $C\in\mathcal T$ implies $g(C)\in\mathcal T$;
• (3) if $\mathcal D\subseteq\mathcal T$ is a chain in $\mathcal C$, then $\cup\mathcal D\in\mathcal T$.

$\mathcal C$ is itself a tower in $S$: (1) and (2) are clearly satisfied. For (3), suppose $\mathcal D\subseteq\mathcal T$ is a chain in $\mathcal C$, if $a,b\in\cup\mathcal D$, then for some $A,B\in\mathcal D$ we have $a\in A$ and $b\in B$. Since $\mathcal D$ is a chain in $\mathcal C$, either $A\subseteq B$ or $B\subseteq A$, so either $a,b\in A$ or $a,b\in B$. Since $A$ and $B$ are both chains in $S$, we have $a\preceq b$ or $b\preceq a$. We have shown that $\cup\mathcal D$ is a chain in $S$, so $\cup\mathcal D\in\mathcal C$.

Since there exists a tower in $S$, we can define $\mathcal T_0$ to be the intersection of all towers in $S$. Then $\mathcal T_0$ is also a tower:

• (1) For each tower $\mathcal T$ in $S$, $\emptyset\in\mathcal T$, so $\emptyset\in\mathcal T_0$.
• (2) For all $C$, if $C\in\mathcal T_0$, then for each tower $\mathcal T$ in $S$, $C\in\mathcal T$, so $g(C)\in\mathcal T$, therefore $g(C)\in\mathcal T_0$.
• (3) Suppose $\mathcal D\subseteq\mathcal T_0$ is a chain in $\mathcal C$, then $\mathcal D\subseteq\mathcal T_0\subseteq\mathcal T$ for every tower $\mathcal T$ in $S$, hence $\cup\mathcal D\in\mathcal T$, therefore $\cup\mathcal D\in\mathcal T_0$.

Our sub-goal is to show that $\mathcal T_0$ is a chain in $\mathcal C$.

Denote $\{A\in\mathcal T_0|\forall B\in\mathcal T_0,(A\subseteq B\lor B\subseteq A)\}$ by $\mathcal E$, then $\mathcal E$ is a tower in $S$:

• (1) Since $\emptyset\in\mathcal T_0$ and for each $B\in\mathcal T_0$, $\emptyset\subseteq B$, we know $\emptyset\in\mathcal E$.
• (2) Let $A\in\mathcal E$, Since $A\in\mathcal E\in\mathcal T_0$, and $T_0$ is a tower, we have $g(A)\in T_0$. Denote $\{B\in\mathcal T_0|g(A)\subseteq B\lor B\subseteq A\}$ by $\mathcal F$, then $\mathcal F$ is a tower in $S$:
  • (1) Since $\emptyset\in\mathcal T_0$ and $\emptyset\subseteq A$, we know $\emptyset\in\mathcal F$.
  • (2) Let $B\in\mathcal F$, then either $g(A)\subseteq B$ or $B\subseteq A$. Since $B\in\mathcal F\in\mathcal T_0$, and $T_0$ is a tower, we have $g(B)\in T_0$. Since $A\in\mathcal E$ and $g(B)\in T_0$, we have either $A\subseteq g(B)$ or $g(B)\subseteq A$.
    • If $g(A)\subseteq B$, then $g(A)\subseteq g(B)$, so $g(B)\in\mathcal F$.
    • If $B\subseteq A$:
      • If $A\subseteq g(B)$, we have $B\subseteq A\subseteq g(B)$, so either $A=B$ or $A=g(B)$,
        • if $A=B$, then $g(A)\subseteq g(B)$, so $g(B)\in\mathcal F$;
        • if $A=g(B)$, then $g(B)\subseteq A$, so $g(B)\in\mathcal F$.
      • If $g(B)\subseteq A$, we have $g(B)\in\mathcal F$.
    In all cases, $g(B)\in\mathcal F$.
  • (3) Suppose $\mathcal D\subseteq\mathcal F$ is a chain in $\mathcal C$, then $\mathcal D\subseteq\mathcal F\subseteq\mathcal T_0$; since $\mathcal T_0$ is a tower, we have $\cup\mathcal D\in\mathcal T_0$. For each $B\in\mathcal D$, since $B\in\mathcal F$, either $g(A)\subseteq B$ or $B\subseteq A$.
    • If for all $B\in\mathcal D$, $B\subseteq A$, then for all $u\in\cup\mathcal D$, there exists $D\in\mathcal D$ such that $u\in D$, then $D\subseteq A$, so $u\in A$; therefore, $\cup\mathcal D\subseteq A$.
    • If there exists $B\in\mathcal D$ such that not $B\subseteq A$, then $g(A)\subseteq B\subseteq\cup\mathcal D$.
    In both cases, either $\cup\mathcal D\subseteq A$ or $g(A)\subseteq\cup\mathcal D$, so $\cup\mathcal D\in\mathcal F$.
  Clearly, we have both $\mathcal F\subseteq\mathcal T_0$ and $\mathcal T_0\subseteq\mathcal F$, so $\mathcal T_0=\mathcal F$. So if $B\in\mathcal T_0$, we also have $B\in\mathcal F$, then we have either $g(A)\subseteq B$ or $B\subseteq A\subseteq g(A)$. Therefore, $g(A)\in\mathcal E$.
• (3) Suppose $\mathcal D\subseteq\mathcal E$ is a chain in $\mathcal C$, then $\mathcal D\subseteq\mathcal E\subseteq\mathcal T_0$; since $\mathcal T_0$ is a tower, we have $\cup\mathcal D\in\mathcal T_0$. Now let $B\in\mathcal T_0$. For each $A\in\mathcal D$, since $A\in\mathcal E$, either $A\subseteq B$ or $B\subseteq A$.
  • If for all $A\in\mathcal D$, $A\subseteq B$, then for all $u\in\cup\mathcal D$, there exists $D\in\mathcal D$ such that $u\in D$, then $D\subseteq B$, so $u\in B$; therefore, $\cup\mathcal D\subseteq B$.
  • If there exists $A\in\mathcal D$ such that not $A\subseteq B$, then $B\subseteq A\subseteq\cup\mathcal D$.
  In both cases, either $\cup\mathcal D\subseteq B$ or $B\subseteq\cup\mathcal D$, so $\cup\mathcal D\in\mathcal E$.

Clearly, we have both $\mathcal E\subseteq\mathcal T_0$ and $\mathcal T_0\subseteq\mathcal E$, so $\mathcal T_0=\mathcal E$. Given $A,B\in\mathcal T_0$, we have $A\in\mathcal E$, so $A\subseteq B$ or $B\subseteq A$. Therefore, we have shown that $\mathcal T_0$ is a chain in $\mathcal C$.

Let $C^*=\cup\mathcal T_0$. Since $\mathcal T_0$ is a chain in $\mathcal C$, by property (3) of towers, $C^*=\cup\mathcal T_0\in\mathcal T_0$. Then by property (2) of towers, $g(C^*)\in\mathcal T_0$. Hence, $g(C^*)\subseteq\cup\mathcal T_0=C^*$. Since $g(C^*)\subseteq C^*\subseteq g(C^*)$, we have $g(C^*)=C^*$, implying ${C^*}'=\emptyset$. Let $s\in S$. If $C^*\cup\{s\}$ is a chain, then $s\in C^*$. We have shown that $S$ has a maximal chain, which is $C^*$.

Since every chain in $S$ has an upper bound, $C^*$ has an upper bound $c^*$. Let $s\in S$, if $c^*\preceq s$, then for all $c\in C^*$, $c\preceq c^*\preceq s$, so $c\preceq s$. Since we also have $s\preceq s$, $C^*\cup\{s\}$ is a chain, then $s\in C^*$, hence $s\preceq c^*$, therefore $s=c^*$. We have shown that $c^*$ is a maximal element in $S$. $\blacksquare$

Proof. Let $X$ be a set. Let $W$ be the collection of pairs $(S,\preceq)$ such that $S\subseteq X$ and $\preceq$ well-orders $S$. Define a relation $\preceq_W$ on $W$ by $(S,\preceq)\preceq_W(S',\preceq')$ if and only if

• $S\subseteq S'$,
• $\preceq$ is the restriction of $\preceq'$ on $S$, and
• for all $s\in S$ and $s'\in S'\setminus\{S\}$, $s\preceq' s'$.

Then $\preceq_W$ is clearly a partial order on $W$. Note that $(\emptyset,\emptyset)\in W$.

Now we will show that every chain in $W$ has an upper bound in $W$. Let $C$ be a chain in $W$. Define $\cup C$ by $\cup_{(S,\preceq)\in C} S$ and $\preceq_{\cup C}$ by $\cup_{(S,\preceq)\in C} \preceq$ We will first show that $(\cup C,\preceq_{\cup C})\in W$. Clearly, we have $\cup C\subseteq X$. Now we need to show that $\preceq_{\cup C}$ well-orders $\cup C$:

• $\preceq_{\cup C}$ is a relation on $\cup C$: if $p\in\preceq_{\cup C}$, then for some $(S,\preceq)\in C$, $p\in\preceq$, so $p\in S\times S$, thus $p\in\cup C\times \cup C$.
• $\preceq_{\cup C}$ is a partial order on $\cup C$:
  • Let $x\in\cup C$, then for some $(S,\preceq)\in C$, $x\in S$, so $x\preceq x$, so $x\preceq_{\cup C}x$.
  • Let $x,y\in\cup C$, if $x\preceq_{\cup C}y$ and $y\preceq_{\cup C}x$, then for some $(S,\preceq),(S',\preceq')\in C$, $x\preceq y$ and $y\preceq' x$. Since $C$ is a chain, either $(S,\preceq)\preceq_W(S',\preceq')$ or $(S',\preceq')\preceq_W(S,\preceq)$. So one of $\preceq$ and $\preceq'$ is a restriction of the other, so either $x\preceq y$ and $y\preceq x$, or $x\preceq' y$ and $y\preceq' x$. We have $x=y$ in either case.
  • Let $x,y,z\in\cup C$, if $x\preceq_{\cup C}y$ and $y\preceq_{\cup C}z$, by the same reasoning as above, there exists $(S,\preceq)\in C$ such that $x\preceq y$ and $y\preceq z$, so we have $x\preceq z$, and hence $x\preceq_{\cup C}z$.
• $\preceq_{\cup C}$ is a total order on $\cup C$: let $x,y\in\cup C$, then for some $(S,\preceq),(S',\preceq')\in C$, $x\in S$ and $y\in S'$. Since $C$ is a chain, either $(S,\preceq)\preceq_W(S',\preceq')$ or $(S',\preceq')\preceq_W(S,\preceq)$. So one of $S$ and $S'$ is a subset of the other. So either $x,y\in S$ or $x,y\in S'$, we either have $x\preceq y$ or $y\preceq x$, or have $x\preceq' y$ and $y\preceq' x$. In either case, either $x\preceq_{\cup C}y$ or $y\preceq_{\cup C}x$.
• $\preceq_{\cup C}$ is a well-order on $\cup C$: Let $T$ be a non-empty subset of $\cup C$. Let $t\in T$, then $t\in\cup C$, so for some $(S,\preceq)\in C$, $t\in S$. Let $s$ be a least element of $T\cap S$ given by $\preceq$. Let $t'\in T$. If $t'\in S$, then $s\preceq t'$, so $s\preceq_{\cup C} t'$. If $t'\notin S$, since for some $(S',\preceq')\in C$, $t'\in S'$, and $C$ being a chain implies $(S,\preceq)\preceq_W(S',\preceq')$, we have $s\preceq't'$, and hence $s\preceq_{\cup C}t'$. So $s$ is a least element of $T$ given by $\preceq_{\cup C}$.

Therefore, $(\cup C,\preceq_{\cup C})\in W$.

The next goal is to show that $(\cup C,\preceq_{\cup C})$ is an upper bound of $C$. For all $(S,\preceq)\in C$:

• $S\subseteq \cup C$.
• Clearly, $\preceq\subseteq\preceq_{\cup C}\cap (S\times S)$. Suppose $p\in\preceq_{\cup C}\cap (S\times S)$, then for some $(S',\preceq')\in C$, $p\in\preceq'$. Since $C$ is a chain, one of $\preceq$ and $\preceq'$ is a restriction of the other. If $\preceq$ is the restriction of $\preceq'$ on $S$, since $p\in\preceq'\cap (S\times S)$, $p\in\preceq$. If $\preceq'$ is a restriction of $\preceq$, then $p\in\preceq$. So $p\in\preceq$ in either case. Hence $\preceq=\preceq_{\cup C}\cap (S\times S)$.
• Let $s\in S$ and $s'\in \cup C\setminus\{S\}$, then for some $(S',\preceq')\in C$, $s'\in S'$. Since $C$ is a chain, we have $(S,\preceq)\preceq_W(S',\preceq')$. Note that $s'\in S'\setminus\{S\}$, so $s\preceq's'$, and hence $s\preceq_{\cup C}s'$.

Therefore, $(S,\preceq)\preceq_W(\cup C,\preceq_{\cup C})$.

Since $(W,\preceq_W)$ is a non-empty partially ordered set such that every chain in $W$ has an upper bound in $W$, by Zorn's lemma, $W$ has a maximal element $(S^*,\preceq)$. Suppose for contradiction that $S^*\neq X$. Let $x^*\in X\setminus\{S^*\}$, define a relation $\preceq'$ on $S^*\cup\{x^*\}$ such that $\preceq'$ contains exactly

• $(x^*,x^*)$,
• $(x,x^*)$, for all $x\in S^*$, and
• $p$, for all $p\in\preceq$.

Then $\preceq'$ clearly well-orders $S^*\cup\{x^*\}$. And we have $(S^*,\preceq)\preceq_W(S^*\cup\{x^*\},\preceq')$ but $(S^*,\preceq)\neq(S^*\cup\{x^*\},\preceq')$, a contradiction to $(S^*,\preceq)$ being a maximal element in $W$. Therefore, $S^*=X$, which implies that $X$ is well-orderable by $\preceq$. $\blacksquare$

Proof. Suppose $a,b,c,d\in N$. If $(a,b)\sim(c,d)$, then for all $a',b',c',d'\in N$ such that $(a,b)=(a',b')$ and $(c,d)=(c',d')$, we have $a'+d'=c'+b'$. Since $a=a',b=b',c=c',d=d'$, we have $a+d=c+b$. If $a+d=c+b$, then for all $a',b',c',d'\in N$ such that $(a,b)=(a',b')$ and $(c,d)=(c',d')$, we have $a=a',b=b',c=c',d=d'$, so $a'+d'=c'+b'$. Hence, we have $(a,b)\sim(c,d)$. Therefore, we have $(a,b)\sim(c,d)$ if and only if $a+d=c+b$. $\blacksquare$

Proof. For all $u,v,w\in N\times N$, there exist $a,b,c,d,e,f\in N$ such that $u=(a,b)$, $v=(c,d)$, and $w=(e,f)$.

• Since $a+b=a+b$, we have $u=(a,b)\sim(a,b)=u$. This proves reflexivity of $\sim$.
• If $u\sim v$, then $a+d=c+b$, so $c+b=a+d$, hence $v=(c,d)\sim(a,b)=u$. This proves symmetry of $\sim$.
• If $u\sim v$ and $v\sim w$, then $a+d=c+b$ and $c+f=e+d$, so $(a+f)+(c+d)=(e+b)+(c+d)$, hence $a+f=e+b$, thus $u=(a,b)\sim(e,f)=w$. This proves transitivity of $\sim$. $\blacksquare$

Proof. Fix $x,y\in Z$, suppose we have $a,b,c,d\in N$ such that $x=[(a,b)]$ and $y=[(c,d)]$. Clearly, if $x=y$, then $[(a,b)]=[(c,d)]$, so $a+d=c+b$; if $a+d=c+b$, then $(a,b)\sim(c,d)$, so $[(a,b)]=[(c,d)]$, and hence $x=y$. we have shown that $x=y$ if and only if $a+d=c+b$.

If $x\le y$, then given any $a',b',c',d'\in N$ such that $x=[(a',b')]$ and $y=[(c',d')]$, we have $a'+d'\le c'+b'$. Hence $a+d\le c+b$. Suppose $a+d\le c+b$. Given any $a',b',c',d'\in N$ such that $x=[(a',b')]$ and $y=[(c',d')]$, we have $[(a,b)]=[(a',b')]$ and $[(c,d)]=[(c',d')]$, so $a+b'=a'+b$ and $c+d'=c'+d$. Then we have $(a+d)+(c'+b')=(c+b)+(a'+d')$. Since $a+d\le c+b$, there exists $k\in N$ such that $(a+d)+k=(c+b)$, so $(a+d)+(c'+b')=(a+d)+k+(a'+d')$, thus $(c'+b')=k+(a'+d')$, hence $a'+d'\le c'+b'$. Therefore, $a+d\le c+b$ implies $x\le y$. We have shown that $x\le y$ if and only if $a+d\le c+b$.

By a symmetric argument, we can show that $y\le x$ if and only if $c+b\le a+d$. Since we also have $x\ge y$ if and only if $y\le x$ and $c+b\le a+d$ if and only if $a+d\ge c+b$, we conclude that $x\ge y$ if and only if $a+d\ge c+b$.

If $x\lt y$, then $x\le y$ and $x\neq y$. By $x\le y$, we have $a+d\le c+b$; by $x\neq y$, we have $a+d\neq c+b$. Hence, we have $a+d\le c+b$ and $a+d\neq c+b$, and therefore, $a+d\lt c+b$. Suppose $a+d\lt c+b$. Then we have $a+d\le c+b$, so $x\le y$. Since $a+d\lt c+b$, we have $a+d\neq c+b$, so $x\neq y$. By $x\le y$ and $x\neq y$, we have $x\lt y$. We have shown that $x\lt y$ if and only if $a+d\lt c+b$.

By a symmetric argument, we can show that $y\lt x$ if and only if $c+b\lt a+d$. Since we also have $x\gt y$ if and only if $y\lt x$ and $c+b\lt a+d$ if and only if $a+d\gt c+b$, we conclude that $x\gt y$ if and only if $a+d\gt c+b$. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. There exist $a,b,c,d\in N$ such that $m=[(a,b)]$, $n=[(c,d)]$. Since we have exactly one of $a+d\lt c+b$, $a+d=c+b$, $a+d\gt c+b$, we have exactly one of $m\lt n$, $m=n$, $m\gt n$. $\blacksquare$

Proof. Fix $m,n,k\in Z$, then there exist $a,b,c,d,e,f\in N$ such that $m=[(a,b)]$, $n=[(c,d)]$, and $k=[(e,f)]$.

Suppose $m\lt n$ and $n\lt k$, then $a+d\lt c+b$ and $c+f\lt e+d$. So $a+d+c+f\lt c+b+e+d$, and $a+f\lt e+b$. Hence we conclude that $m\lt k$. $\blacksquare$

Proof. Fix $w\in Z\times Z$, there exist $x',y'\in Z$ such that $w=(x',y')$, and for all $x,y\in Z$ such that $w=(x,y)$, we have $x=x'$ and $y=y'$. Since $x',y'\in Z$, there exist $a',b',c',d'\in N$ such that $x'=[(a',b')]$ and $y'=[(c',d')]$, and for all $a,b,c,d\in N$ such that $x=[(a,b)]$ and $y=[(c,d)]$, we have $[(a,b)]=x=x'=[(a',b')]$ and $[(c,d)]=y=y'=[(c',d')]$, so $a+b'=a'+b$ and $c+d'=c'+d$, $(a+c)+(b'+d')=(a'+c')+(b+d)$, and hence $[(a'+c',b'+d')]=[(a+c,b+d)]$. We have shown that there exists $z\in Z$ that satisfies the above statement, which is $[(a'+c',b'+d')]$. To show uniqueness, for all $z'\in Z$ that satisfies the above statement, we have $z'=[(a'+c',b'+d')]$. Therefore, for all $w\in Z\times Z$, there exists a unique $z\in Z$ that satisfies the above statement, implying it indeed defines a function. $\blacksquare$

Proof. Suppose $z\in Z$, then there exist $a,b\in N$ such that $z=[(a,b)]$. Since we also have $0=[(0,0)]$, we conclude that $z+0=[(a+0,b+0)]=[(a,b)]=z$ and $0+z=[(0+a,0+b)]=[(a,b)]=z$. $\blacksquare$

Proof. For all $m,n,k\in Z$, there exist $a,b,c,d,e,f\in N$ such that $m=[(a,b)]$, $n=[(c,d)]$, and $k=[(e,f)]$.

We have $$(m+n)+k=([(a,b)]+[(c,d)])+[(e,f)]=[((a+c)+e,(b+d)+f)]=[(a+(c+e),b+(d+f))]=[(a,b)]+([(c,d)]+[(e,f)])=m+(n+k)$$ This proves associativity.

We also have $$m+n=[(a,b)]+[(c,d)]=[(a+c,b+d)]=[(c+a,d+b)]=[(c,d)]+[(a,b)]=n+m$$ This proves commutativity. $\blacksquare$

Proof. Let $a,b,c,d,m,n\in N$ such that $y=[(a,b)]$, $z=[(c,d)]$, and $x=[(m,n)]$. If $y\lt z$, then $a+d\lt b+c$, so $(a+m)+(d+n)=(a+d)+(m+n)\lt (b+c)+(m+n)=(c+m)+(b+n)$, then $x+y=y+x=[(a+m,b+n)]\lt[(c+m,d+n)]=z+x=x+z$. $\blacksquare$

Proof. Given $m,n\in Z$, there exist $a,b,c,d\in N$ such that $m=[(a,b)]$ and $n=[(c,d)]$. If $m\lt n$, then $a+d\lt c+b$, thus $(a+1)+d=(a+d)+1=S(a+d)\le c+b$, implying $m+1=[(a+1,b)]\le[(c,d)]=n$. $\blacksquare$

Proof. Suppose $z\in Z$, then there exist $a,b\in N$ such that $z=[(a,b)]$. Denote $[(b,a)]$ by $w$, then $w\in Z$ and we have $z+w=[(a+b,b+a)]=[(a+b,a+b)]$ and $w+z=[(b+a,a+b)]=[(a+b,a+b)]$. Since $(a+b)+0=0+(a+b)$, we have $(a+b,a+b)\sim(0,0)$, so $[(a+b,a+b)]=[(0,0)]=0$. This shows that there exists $w\in Z$, such that $z+w=0$ and $w+z=0$. To show uniqueness, suppose we have $w'\in Z$ such that $z+w'=0$ and $w'+z=0$, then there exist $a',b'\in N$ such that $w'=[(b',a')]$, and we have $[(a+b',b+a')]=[(0,0)]$, so $a+b'=b+a'$. Hence, we have $(b',a')\sim(b,a)$, and thus $w'=[(b',a')]=[(b,a)]=w$. Therefore, there exists a unique $w\in Z$, such that $z+w=0$ and $w+z=0$. $\blacksquare$

Proof. There exist $a,b,c,d\in N$ such that $m=[(a,b)]$ and $n=[(c,d)]$. If $m\lt n$, then $a+d\lt c+b$, thus $-m=[(b,a)]\gt[(d,c)]=-n$. $\blacksquare$

Proof. There exist $a,b\in N$ such that $z=[(a,b)]$. And $--z=--[(a,b)]=-[(b,a)]=[(a,b)]=z$. $\blacksquare$

Proof. $$m+(n-m)=m+(n+(-m))=n+(m+(-m))=n+0=n$$ $\blacksquare$

Proof. Fix $w\in Z\times Z$, there exist $x',y'\in Z$ such that $w=(x',y')$, and for all $x,y\in Z$ such that $w=(x,y)$, we have $x=x'$ and $y=y'$. Since $x',y'\in Z$, there exist $a',b',c',d'\in N$ such that $x'=[(a',b')]$ and $y'=[(c',d')]$, and for all $a,b,c,d\in N$ such that $x=[(a,b)]$ and $y=[(c,d)]$, we have $[(a,b)]=x=x'=[(a',b')]$ and $[(c,d)]=y=y'=[(c',d')]$, so $a+b'=a'+b$ and $c+d'=c'+d$. Then,

• note that either there exists $m\in N$ such that $a+m=a'$, or there exists $m\in N$ such that $a'+m=a$; suppose there exists $m\in N$ such that $a+m=a'$, then $a+b'=a+m+b$, so $b'=m+b$; suppose there exists $m\in N$ such that $a'+m=a$, then $a'+m+b'=a'+b$, so $m+b'=b$;
• similarly, either there exists $n\in N$ such that $c+n=c'$, or there exists $n\in N$ such that $c'+n=c$; suppose there exists $n\in N$ such that $c+n=c'$, then $c+d'=c+n+d$, so $d'=n+d$; suppose there exists $n\in N$ such that $c'+n=c$, then $c'+n+d'=c'+d$, so $n+d'=d$.

There are four possible cases:

• suppose there exists $m\in N$ such that $a+m=a'$, and there exists $n\in N$ such that $c+n=c'$, then $(ac+bd)+(a'd'+b'c')=ac+bd+(a+m)(d+n)+(b+m)(c+n)=ac+bd+ad+an+md+mn+bc+bn+mc+mn=(a+m)(c+n)+(b+m)(d+n)+ad+bc=(a'c'+b'd')+(ad+bc)$;
• suppose there exists $m\in N$ such that $a+m=a'$, and there exists $n\in N$ such that $c'+n=c$, then $(ac+bd)+(a'd'+b'c')=a(c'+n)+b(d'+n)+(a+m)d'+(b+m)c'=ac'+an+bd'+bn+ad'+md'+bc'+mc'=(a+m)c'+(b+m)d'+a(d'+n)+b(c'+n)=(a'c'+b'd')+(ad+bc)$;
• suppose there exists $m\in N$ such that $a'+m=a$, and there exists $n\in N$ such that $c+n=c'$, then $(ac+bd)+(a'd'+b'c')=(a'+m)c+(b'+m)d+a'(d+n)+b'(c+n)=a'c+mc+b'd+md+a'd+a'n+b'c+b'n=a'(c+n)+b'(d+n)+(a'+m)d+(b'+m)c=(a'c'+b'd')+(ad+bc)$;
• suppose there exists $m\in N$ such that $a'+m=a$, and there exists $n\in N$ such that $c'+n=c$, then $(ac+bd)+(a'd'+b'c')=(a'+m)(c'+n)+(b'+m)(d'+n)+a'd'+b'c'=a'c'+a'n+mc'+mn+b'd'+b'n+md'+mn+a'd'+b'c'=a'c'+b'd'+(a'+m)(d'+n)+(b'+m)(c'+n)=(a'c'+b'd')+(ad+bc)$;

In all cases we have $(ac+bd)+(a'd'+b'c')=(a'c'+b'd')+(ad+bc)$, hence $[(a'c'+b'd',a'd'+b'c')]=[(ac+bd,ad+bc)]$. We have shown that there exists $z\in Z$ that satisfies the above statement, which is $[(a'c'+b'd',a'd'+b'c')]$. To show uniqueness, for all $z'\in Z$ that satisfies the above statement, we have $z'=[(a'c'+b'd',a'd'+b'c')]$. Therefore, for all $w\in Z\times Z$, there exists a unique $z\in Z$ that satisfies the above statement, implying it indeed defines a function. $\blacksquare$

Proof. Suppose $z\in Z$, then there exist $a,b\in N$ such that $z=[(a,b)]$. Then $z0=[(a0+b0,a0+b0)]=[(0,0)]=0$ and $0z=[(0a+0b,0b+0a)]=[(0,0)]=0$. $\blacksquare$

Proof. Suppose $z\in Z$, then there exist $a,b\in N$ such that $z=[(a,b)]$. Since we also have $1=[(1,0)]$, we conclude that $z1=[(a1+b0,a0+b1)]=[(a,b)]=z$ and $1z=[(1a+0b,1b+0a)]=[(a,b)]=z$. $\blacksquare$

Proof. There exist $a,b\in N$ such that $n=[(a,b)]$. Then $(-1)n=[(0,1)][(a,b)]=[(b,a)]=-n$. $\blacksquare$

Proof. For all $m,n,k\in Z$, there exist $a,b,c,d,e,f\in N$ such that $m=[(a,b)]$, $n=[(c,d)]$, and $k=[(e,f)]$.

We have $$(mn)k=([(a,b)][(c,d)])[(e,f)]=[(ac+bd,ad+bc)][(e,f)]=[(ace+bde+adf+bcf,acf+bdf+ade+bce)] =[(a,b)][(ce+df,cf+de)]=[(a,b)]([(c,d)][(e,f)])=m(nk)$$ This proves associativity.

We also have $$mn=[(a,b)][(c,d)]=[(ac+bd,ad+bc)]=[(c,d)][(a,b)]=nm$$ This proves commutativity. $\blacksquare$

Proof. For all $m,n,k\in Z$, there exist $a,b,c,d,e,f\in N$ such that $m=[(a,b)]$, $n=[(c,d)]$, and $k=[(e,f)]$.

We have $$m(n+k)=[(a,b)]([(c,d)]+[(e,f)])=[(a,b)][(c+e,d+f)]=[(a(c+e)+b(d+f),a(d+f)+b(c+e))] =[(ac+bd,ad+bc)]+[(ae+bf,af+be)]=mn+mk$$ and $$(m+n)k=k(m+n)=km+kn=mk+nk$$ $\blacksquare$

Proof. Let $a,b,c,d,m,n\in N$ such that $x=[(m,n)]$, $y=[(a,b)]$ and $z=[(c,d)]$.

If $y\lt z$, then $a+d\lt b+c$, so there exists $k\in N$ such that $a+d+k=b+c$ and $k\neq0$, thus $y-z=[(a+d,b+c)]=[(0,k)]$. Hence $xy-xz=x(y-z)=[(nk,mk)]$. If $x\lt 0$, then $m\lt n$. Since $k\neq0$, $mk\lt nk$, thus $xy-xz=[(nk,mk)]\gt0$, and we have $xy\gt xz$.

The other cases are proven similarly. $\blacksquare$

Proof. If $x=0$ or $y=0$, then $\sgn(xy)=0=\sgn(x)\sgn(y)$. If $x\gt0$ and $y\gt0$, then $xy\gt x0=0$, thus $\sgn(xy)=1=\sgn(x)\sgn(y)$. The other cases are proven similarly. $\blacksquare$

Proof. These are trivial. $\blacksquare$

Proof. Suppose $a,c\in Z$ and $b,d\in Z\setminus\{0\}$. If $(a,b)\sim(c,d)$, then for all $a',c'\in Z$ and $b',d'\in Z\setminus\{0\}$ such that $(a,b)=(a',b')$ and $(c,d)=(c',d')$, we have $a'd'=c'b'$. Since $a=a',b=b',c=c',d=d'$, we have $ad=cb$. If $ad=cb$, then for all $a',c'\in Z$ and $b',d'\in Z\setminus\{0\}$ such that $(a,b)=(a',b')$ and $(c,d)=(c',d')$, we have $a=a',b=b',c=c',d=d'$, so $a'd'=c'b'$. Hence, we have $(a,b)\sim(c,d)$. Therefore, we have $(a,b)\sim(c,d)$ if and only if $ad=cb$. $\blacksquare$

Proof. For all $u,v,w\in Z\times Z\setminus\{0\}$, there exist $a,c,e\in Z$ and $b,d,f\in Z\setminus\{0\}$ such that $u=(a,b)$, $v=(c,d)$, and $w=(e,f)$.

• Since $ab=ab$, we have $u=(a,b)\sim(a,b)=u$. This proves reflexivity of $\sim$.
• If $u\sim v$, then $ad=cb$, so $cb=ad$, hence $v=(c,d)\sim(a,b)=u$. This proves symmetry of $\sim$.
• If $u\sim v$ and $v\sim w$, then $ad=cb$ and $cf=ed$, so $(cd)(af)=(cd)(eb)$.
  • If $c=0$, then $ad=0$ and $ed=0$, but $d\neq0$, so $a=0$ and $e=0$, hence $af=eb$, thus $u=(a,b)\sim(e,f)=w$.
  • If $c\neq0$, since $d\neq0$, we have $cd\neq0$, then by $(cd)(af)=(cd)(eb)$, we have $af=eb$, thus $u=(a,b)\sim(e,f)=w$.
  In both cases we have $u\sim w$. This proves transitivity of $\sim$. $\blacksquare$

Proof. Fix $x,y\in Q$, suppose we have $a,c\in Z$ and $b,d\in Z\setminus\{0\}$ such that $x=[(a,b)]$ and $y=[(c,d)]$. Clearly, if $x=y$, then $[(a,b)]=[(c,d)]$, so $ad=cb$; if $ad=cb$, then $(a,b)\sim(c,d)$, so $[(a,b)]=[(c,d)]$, and hence $x=y$. we have shown that $x=y$ if and only if $ad=cb$. Again, if $ad=cb$, then we have $(ad)\sgn(bd)=(cb)\sgn(bd)$; if $(ad)\sgn(bd)=(cb)\sgn(bd)$, since $b,d\neq0$, $bd\neq0$, so $\sgn(bd)\neq0$, hence $ad=cb$. We have shown that $x=y$ if and only if $(ad)\sgn(bd)=(cb)\sgn(bd)$.

If $x\le y$, then given any $a',c'\in Z$ and $b',d'\in Z\setminus\{0\}$ such that $x=[(a',b')]$ and $y=[(c',d')]$, we have $(a'd')\sgn(b'd')\le(c'b')\sgn(b'd')$. Hence $(ad)\sgn(bd)\le(cb)\sgn(bd)$. Suppose $(ad)\sgn(bd)\le(cb)\sgn(bd)$. Given any $a',c'\in Z$ and $b',d'\in Z\setminus\{0\}$ such that $x=[(a',b')]$ and $y=[(c',d')]$, we have $[(a,b)]=[(a',b')]$ and $[(c,d)]=[(c',d')]$, so $ab'=a'b$ and $cd'=c'd$. Since $b,b',d,d'\neq0$, we have $bb',dd'\neq0$, and thus $[(ab',bb')]=x=[(a'b,b'b)]$ and $[(cd',dd')]=y=[(c'd,d'd)]$. Since $bd\neq0$, either $bd\lt0$ or $bd\gt0$. Same for $b'd'$.

• Suppose $bd\gt0$ and $b'd'\gt0$, then $(ab')(dd')=(b'd')(ad)\sgn(bd)\le(b'd')(cb)\sgn(bd)=(cd')(bb')$， so $(a'd')(bd)=(a'b)(dd')=(ab')(dd')\le(cd')(bb')=(c'd)(bb')=(c'b')(bd)$, hence $a'd'\le c'b'$, thus $(a'd')\sgn(b'd')\le(c'b')\sgn(b'd')$.
• Suppose $bd\gt0$ and $b'd'\lt0$, then $(ab')(dd')=(b'd')(ad)\sgn(bd)\ge(b'd')(cb)\sgn(bd)=(cd')(bb')$， so $(a'd')(bd)=(a'b)(dd')=(ab')(dd')\ge(cd')(bb')=(c'd)(bb')=(c'b')(bd)$, hence $a'd'\ge c'b'$, thus $(a'd')\sgn(b'd')\le(c'b')\sgn(b'd')$.
• Suppose $bd\lt0$ and $b'd'\gt0$, then $-1(ab')(dd')=(b'd')(ad)\sgn(bd)\le(b'd')(cb)\sgn(bd)=-1(cd')(bb')$， so $(a'd')(bd)=(a'b)(dd')=(ab')(dd')\ge(cd')(bb')=(c'd)(bb')=(c'b')(bd)$, hence $a'd'\le c'b'$, thus $(a'd')\sgn(b'd')\le(c'b')\sgn(b'd')$.
• Suppose $bd\gt0$ and $b'd'\lt0$, then $-1(ab')(dd')=(b'd')(ad)\sgn(bd)\ge(b'd')(cb)\sgn(bd)=-1(cd')(bb')$， so $(a'd')(bd)=(a'b)(dd')=(ab')(dd')\le(cd')(bb')=(c'd)(bb')=(c'b')(bd)$, hence $a'd'\ge c'b'$, thus $(a'd')\sgn(b'd')\le(c'b')\sgn(b'd')$.

In all cases, we have $(a'd')\sgn(b'd')\le(c'b')\sgn(b'd')$. Therefore, $(ad)\sgn(bd)\le(cb)\sgn(bd)$ implies $x\le y$. We have shown that $x\le y$ if and only if $(ad)\sgn(bd)\le(cb)\sgn(bd)$.

By a symmetric argument, we can show that $y\le x$ if and only if $(cb)\sgn(db)\le(ad)\sgn(db)$. Since we also have $x\ge y$ if and only if $y\le x$ and $(cb)\sgn(db)\le(ad)\sgn(db)$ if and only if $(ad)\sgn(bd)\ge(cb)\sgn(bd)$, we conclude that $x\ge y$ if and only if $(ad)\sgn(bd)\ge(cb)\sgn(bd)$.

If $x\lt y$, then $x\le y$ and $x\neq y$. By $x\le y$, we have $(ad)\sgn(bd)\le(cb)\sgn(bd)$; by $x\neq y$, we have $(ad)\sgn(bd)\neq(cb)\sgn(bd)$. Hence, we have $(ad)\sgn(bd)\le(cb)\sgn(bd)$ and $(ad)\sgn(bd)\neq(cb)\sgn(bd)$, and therefore, $(ad)\sgn(bd)\lt(cb)\sgn(bd)$. Suppose $(ad)\sgn(bd)\lt(cb)\sgn(bd)$. Then we have $(ad)\sgn(bd)\le(cb)\sgn(bd)$, so $x\le y$. Since $(ad)\sgn(bd)\lt(cb)\sgn(bd)$, we have $(ad)\sgn(bd)\neq(cb)\sgn(bd)$, so $x\neq y$. By $x\le y$ and $x\neq y$, we have $x\lt y$. We have shown that $x\lt y$ if and only if $(ad)\sgn(bd)\lt(cb)\sgn(bd)$.

By a symmetric argument, we can show that $y\lt x$ if and only if $(cb)\sgn(db)\lt(ad)\sgn(db)$. Since we also have $x\gt y$ if and only if $y\lt x$ and $(cb)\sgn(db)\lt(ad)\sgn(db)$ if and only if $(ad)\sgn(db)\gt(cb)\sgn(db)$, we conclude that $x\gt y$ if and only if $(ad)\sgn(db)\gt(cb)\sgn(db)$. $\blacksquare$

Proof. Note that $0=[(0,1)]$, and $(0n)\sgn(n1)=0$.

• If $\sgn(n)=1$, then $(m1)\sgn(n1)=m$, then $\sgn(m)=\sgn([(m,n)])$, thus $\sgn([(m,n)])=\sgn(m)\sgn(n)$.
• If $\sgn(n)=-1$, then $(m1)\sgn(n1)=-m$, then $\sgn(m)+\sgn([(m,n)])=0$, thus $\sgn([(m,n)])=\sgn(m)\sgn(n)$.

$\blacksquare$

Proof. There exist $a,c\in Z$ and $b,d\in Z\setminus\{0\}$ such that $m=[(a,b)]$, $n=[(c,d)]$. Since we have exactly one of $ad\lt cb$, $ad=cb$, $ad\gt cb$, regardless of the value of $\sgn(bd)$, we have exactly one of $m\lt n$, $m=n$, $m\gt n$. $\blacksquare$

Proof. Fix $m,n,k\in Q$, then there exist $a,c,e\in Z$ and $b,d,f\in Z\setminus\{0\}$ such that $m=[(a,b)]$, $n=[(c,d)]$, and $k=[(e,f)]$. Then $m=[(adf,bdf)]$, $n=[(bcf,bdf)]$, and $k=[(bde,bdf)]$.

Suppose $m\lt n$ and $n\lt k$, since $bdf\neq0$, we have $\sgn((bdf)(bdf))=1$, then we have $(adf)(bdf)\lt (bcf)(bdf)$ and $(bcf)(bdf)\lt (bde)(bdf)$, hence $(adf)(bdf)\lt (bde)(bdf)$. Since $\sgn((bdf)(bdf))=1$, we conclude that $m\lt k$. $\blacksquare$

Proof. Fix $w\in Q\times Q$, there exist $x',y'\in Q$ such that $w=(x',y')$, and for all $x,y\in Q$ such that $w=(x,y)$, we have $x=x'$ and $y=y'$. Since $x',y'\in Q$, there exist $a',c'\in Z$ and $b',d'\in Z\setminus\{0\}$ such that $x'=[(a',b')]$ and $y'=[(c',d')]$, and for all $a',c'\in Z$ and $b',d'\in Z\setminus\{0\}$ such that $x=[(a,b)]$ and $y=[(c,d)]$, we have $[(a,b)]=x=x'=[(a',b')]$ and $[(c,d)]=y=y'=[(c',d')]$, so $ab'=a'b$ and $cd'=c'd$, $(ad+cb)(b'd')=ab'dd'+cd'bb'=a'bdd'+c'dbb'=(a'd'+c'b')(bd)$ and hence $[(a'd'+c'b',b'd')]=[(ad+cb,bd)]$. Note that $b'd'\neq0$, so $[(a'd'+c'b',b'd')]\in Q$. We have shown that there exists $z\in Q$ that satisfies the above statement, which is $[(a'd'+c'b',b'd')]$. To show uniqueness, for all $z'\in Q$ that satisfies the above statement, we have $z'=[[(a'd'+c'b',b'd')]]$. Therefore, for all $w\in Q\times Q$, there exists a unique $z\in Q$ that satisfies the above statement, implying it indeed defines a function. $\blacksquare$

Proof. Suppose $q\in Q$, then there exist $a\in Z$ and $b\in Z\setminus\{0\}$ such that $q=[(a,b)]$. Since we also have $0=[(0,1)]$, we conclude that $q+0=[(a1+0b,b1)]=[(a,b)]=q$ and $0+q=[(0b+a1,1b)]=[(a,b)]=q$. $\blacksquare$

Proof. For all $m,n,k\in Q$, there exist $a,c,e\in Z$ and $b,d,f\in Z\setminus\{0\}$ such that $m=[(a,b)]$, $n=[(c,d)]$, and $k=[(e,f)]$.

We have $$(m+n)+k=[(ad+cb,bd)]+[(e,f)]=[(adf+cbf+bde,bdf)]=[(a,b)]+[(cf+ed,df)]=m+(n+k)$$ This proves associativity.

We also have $$m+n=[(ad+cb,bd)]=[(cb+ad,db)]=n+m$$ This proves commutativity. $\blacksquare$

Proof. There exist $a,c,e\in Z$ and $b,d,f\in Z\setminus\{0\}$ such that $x=[(a,b)]$, $y=[(c,d)]$, and $z=[(e,f)]$.

If $y\lt z$, then $(cf)\sgn(df)\lt (ed)\sgn(df)$. Note that $x+y=[(ad+cb,bd)]=[(adf+cfb,bdf)]$ and $x+z=[(af+eb,bf)]=[(adf+edb,bdf)]$.

• If $\sgn(df)=1$ and $b\gt0$, then $cf\lt ed$ and $bdf\gt0$, thus $[(adf+cfb,bdf)]\lt[(adf+edb,bdf)]$.
• If $\sgn(df)=1$ and $b\lt0$, then $cf\lt ed$ and $bdf\lt0$, thus $[(adf+cfb,bdf)]\lt[(adf+edb,bdf)]$.
• If $\sgn(df)=-1$ and $b\gt0$, then $cf\gt ed$ and $bdf\lt0$, thus $[(adf+cfb,bdf)]\lt[(adf+edb,bdf)]$.
• If $\sgn(df)=-1$ and $b\lt0$, then $cf\gt ed$ and $bdf\gt0$, thus $[(adf+cfb,bdf)]\lt[(adf+edb,bdf)]$.

Hence $x+y\lt x+z$. $\blacksquare$

Proof. Suppose $q\in Q$, then there exist $a\in Z$ and $b\in Z\setminus\{0\}$ such that $q=[(a,b)]$. Denote $[(-a,b)]$ by $w$, then $w\in Q$ and we have $q+w=[(ab+(-a)b,bb)]=[(0,bb)]=0$ and $w+q=[((-a)b+ab,bb)]=[(0,bb)]=0$. This shows that there exists $w\in Q$, such that $q+w=0$ and $w+q=0$. To show uniqueness, suppose we have $w'\in Q$ such that $q+w'=0$ and $w'+q=0$, then there exist $a'\in Z$ and $b'\in Z\setminus\{0\}$ such that $w'=[(a',b')]$, and we have $[(ab'+a'b,bb')]=[(0,1)]$, so $ab'+a'b=0$, and thus $a'b=(-a)b'$. Hence, we have $(a',b')\sim(-a,b)$, and thus $w'=[(a',b')]=[(-a,b)]=w$. Therefore, there exists a unique $w\in Q$, such that $q+w=0$ and $w+q=0$. $\blacksquare$

Proof. There exist $a,c\in Z$ and $b,d\in Z\setminus\{0\}$ such that $m=[(a,b)]$ and $n=[(c,d)]$. If $m\lt n$, then $(ad)\sgn(bd)\lt (cb)\sgn(bd)$, thus $(-ad)\sgn(bd)\gt (-cb)\sgn(bd)$, hence $-m=[(-a,b)]\gt[(-c,d)]=-n$. $\blacksquare$

Proof. There exist $a\in Z$ and $b\in Z\setminus\{0\}$ such that $q=[(a,b)]$. And $--q=--[(a,b)]=-[(-a,b)]=[(--a,b)]=[(a,b)]=q$. $\blacksquare$

Proof. $$m+(n-m)=m+(n+(-m))=n+(m+(-m))=n+0=n$$ $\blacksquare$

Proof. Fix $w\in Q\times Q$, there exist $x',y'\in Q$ such that $w=(x',y')$, and for all $x,y\in Z$ such that $w=(x,y)$, we have $x=x'$ and $y=y'$. Since $x',y'\in Q$, there exist $a',c'\in Z$ and $b',d'\in Z\setminus\{0\}$ such that $x'=[(a',b')]$ and $y'=[(c',d')]$, and for all $a,c\in Z$ and $b,d\in Z\setminus\{0\}$ such that $x=[(a,b)]$ and $y=[(c,d)]$, we have $[(a,b)]=x=x'=[(a',b')]$ and $[(c,d)]=y=y'=[(c',d')]$, so $ab'=a'b$ and $cd'=c'd$, $(ac)(b'd')=ab'cd'=a'bc'd=(a'c')(bd)$, and hence $[(a'c',b'd')]=[(ac,bd)]$. Note that $b'd'\neq0$, so $[(a'c',b'd')]\in Q$. We have shown that there exists $z\in Q$ that satisfies the above statement, which is $[(a'c',b'd')]$. To show uniqueness, for all $z'\in Q$ that satisfies the above statement, we have $z'=[(a'c',b'd')]$. Therefore, for all $w\in Q\times Q$, there exists a unique $z\in Q$ that satisfies the above statement, implying it indeed defines a function. $\blacksquare$

Proof. Suppose $q\in Q$, then there exist $a\in Z$ and $b\in Z\setminus\{0\}$ such that $q=[(a,b)]$. Since we also have $0=[(0,1)]$, we conclude that $q0=[(a0,b1)]=[(0,b)]=0$ and $0q=[(0a,1b)]=[(0,b)]=0$. $\blacksquare$

Proof. Suppose $q\in Q$, then there exist $a\in Z$ and $b\in Z\setminus\{0\}$ such that $q=[(a,b)]$. Since we also have $1=[(1,1)]$, we conclude that $q1=[(a1,b1)]=[(a,b)]=q$ and $1q=[(1a,1b)]=[(a,b)]=q$. $\blacksquare$

Proof. Suppose $q\in Q$, then there exist $a\in Z$ and $b\in Z\setminus\{0\}$ such that $q=[(a,b)]$. Then we have $(-1)q=[(-1,1)][(a,b)]=[(-a,b)]=-q$. $\blacksquare$

Proof. For all $m,n,k\in Q$, there exist $a,c,e\in Z$ and $b,d,f\in Z\setminus\{0\}$ such that $m=[(a,b)]$, $n=[(c,d)]$, and $k=[(e,f)]$.

We have $$(mn)k=[(ac,bd)][(e,f)]=[(ace,bdf)]=[(a,b)][(ce,df)]=m(nk)$$ This proves associativity.

We also have $$mn=[(ac,bd)]=[(ca,db)]=nm$$ This proves commutativity. $\blacksquare$

Proof. For all $m,n,k\in Q$, there exist $a,c,e\in Z$ and $b,d,f\in Z\setminus\{0\}$ such that $m=[(a,b)]$, $n=[(c,d)]$, and $k=[(e,f)]$.

We have $$m(n+k)=[(a,b)]([(c,d)]+[(e,f)])=[(a,b)][(cf+ed,df)]=[(a(cf+ed),bdf)] =[((ac)(bf)+(ae)(bd),(bd)(bf))]=[(ac,bd)]+[(ae,bf)]=mn+mk$$ and $$(m+n)k=k(m+n)=km+kn=mk+nk$$ $\blacksquare$

Proof. Let $a,c,m\in Z$ and $b,d,n\in Z\setminus\{0\}$ such that $x=[(m,n)]$, $y=[(a,b)]$ and $z=[(c,d)]$.

If $y\lt z$, then $(ad)\sgn(bd)\lt (cb)\sgn(bd)$. Note that $xy=[(ma,nb)]$, $xz=[(mc,nd)]$, and $\sgn((nb)(nd))=\sgn(nn)\sgn(bd)=\sgn(bd)$. If $x\lt0$, then $mn\lt0$, thus $(ma)(nd)\sgn((nb)(nd))=(mn)(ad)\sgn(bd)\gt(mn)(cb)\sgn(bd)=(mc)(nb)\sgn((nb)(nd))$, implying $xy\gt xz$.

The other cases are proven similarly. $\blacksquare$

Proof. If $x=0$ or $y=0$, then $\sgn(xy)=0=\sgn(x)\sgn(y)$. If $x\gt0$ and $y\gt0$, then $xy\gt x0=0$, thus $\sgn(xy)=1=\sgn(x)\sgn(y)$. The other cases are proven similarly. $\blacksquare$

Proof. Suppose $q\in Q\setminus\{0\}$, then there exist $a\in Z$ and $b\in Z\setminus\{0\}$ such that $q=[(a,b)]$. Since $q\neq0$, we have $a1\neq0b$, so $a\neq0$. Denote $[(b,a)]$ by $w$, then $w\in Q$ and we have $qw=[(ab,ba)]=[(1,1)]=1$ and $wq=[(ba,ab)]=[(1,1)]=1$. This shows that there exists $w\in Q$, such that $qw=1$ and $wq=1$. To show uniqueness, suppose we have $w'\in Q$ such that $qw'=1$ and $w'q=1$, then there exist $a'\in Z$ and $b'\in Z\setminus\{0\}$ such that $w'=[(a',b')]$, and we have $[(aa',bb')]=[(1,1)]$, so $aa'=bb'$. Hence, we have $(a',b')\sim(b,a)$, and thus $w'=[(a',b')]=[(b,a)]=w$. Therefore, there exists a unique $w\in Q$, such that $qw=1$ and $wq=1$. $\blacksquare$

Proof. There exist $a,b\in Z\setminus\{0\}$ such that $q=[(a,b)]$. Then $\frac{1}{-q}=\frac{1}{[(-a,b)]}=[(b,-a)]=[(-b,a)]=-[(b,a)]=-\frac{1}{q}$. $\blacksquare$

Proof. Suppose $m,n\in Q^+$, then there exist $a,b,c,d\in Z^+$ such that $m=[(a,b)]$ and $n=[(c,d)]$. If $m\lt n$, then $ad\lt cb$, thus $\frac{1}{m}=[(b,a)]\gt[(d,c)]=\frac{1}{n}$.

Suppose $m,n\in Q^-$, if $m\lt n$, then $-m\gt -n$ and $-m,-n\in Q^+$, thus $-\frac{1}{m}=\frac{1}{-m}\lt\frac{1}{-n}=-\frac{1}{n}$, thus $\frac{1}{m}=--\frac{1}{m}\gt--\frac{1}{n}=\frac{1}{n}$. $\blacksquare$

Proof. There exist $a,b\in Z\setminus\{0\}$ such that $q=[(a,b)]$. Then $\frac{1}{\frac{1}{q}}=\frac{1}{[(b,a)]}=[(a,b)]=q$. $\blacksquare$

Proof. $$m(\frac{n}{m})=m(n\frac{1}{m})=n(m\frac{1}{m})=n1=n$$ $\blacksquare$

Proof. These are trivial. $\blacksquare$

Proof. Let $r=\{u\in Q|u\lt q\}$. Since $q\notin r$, $r\ne Q$. Since $q-1\in r$, $r\ne\emptyset$. Suppose $x,y\in Q$, if $x\lt y$ and $y\in r$, then $x\lt y\lt q$, so $x\lt q$, hence $x\in r$. Suppose $x\in r$, then $x\lt q$, so $x\lt(x+q)/2\lt q$, hence $(x+q)/2\in r$ and $(x+q)/2\gt x$. $\blacksquare$

Proof. Fix $m,n,k\in R$, if $m\lt n$ and $n\lt k$, then $m\subset n$ and $n\subset k$, so $m\subset k$, hence $m\lt k$. $\blacksquare$

Proof. Fix $r\in R$ and $q\in Q^+$. Suppose for all $s\in r$, $s+q\in r$. Let $s\in r$ and $t\in Q$, then $s+\frac{t-s}{q}q=t$. Note that there exists a natural number $n$ (as a rational number) such that $n\ge\frac{t-s}{q}$. Then $s+nq\ge s+\frac{t-s}{q}q=t$. By induction, $s+nq\in r$, thus $t\in r$. But then we have $r=Q$, a contradiction. $\blacksquare$

Proof. Fix $r\in R$ and $q\in Q$. Suppose $q\notin r$ and there exists $s\in r$ such that $q\le s$, then either $q=s$, directly implying $q\in r$, or $q\lt s$, implying $q,s\in Q$, $q\lt s$, and $s\in r$, which together imply $q\in r$; in both cases we have a contradiction. So if $q\notin r$, we have that for all $s\in r$, $q\gt s$. $\blacksquare$

Proof. Suppose there exists $q\in a$ such that $q\notin b$ and there exists $p\in b$ such that $p\notin a$, then for all $p'\in b$, $q\gt p'$, and for all $q'\in a$, $p\gt q'$, so we have both $q\gt p$ and $p\gt q$, a contradiction. Hence either for all $q\in a$, $q\in b$, or for all $p\in b$, $p\in a$, meaning either $a\subseteq b$ or $b\subseteq a$, which implies we have at least one of $a\subset b$, $a=b$, and $b\subset a$. And by properties of sets, clearly, we have exactly one of $a\subset b$, $a=b$, and $b\subset a$. Hence, we have exactly one of $a\lt b$, $a=b$, and $a\gt b$. $\blacksquare$

Proof. Let $\mathscr C$ be a nonempty subset of $R$ having an upper bound $X$. We will first show that $\bigcup\mathscr C$ is a real number:

• Since $\mathscr C$ is non-empty, there exists $A\in\mathscr C$. Since $A$ is a real number, $A$ is non-empty, so there exists $u\in A$, then $u\in\bigcup\mathscr C$, so $\bigcup\mathscr C\neq\emptyset$. Since $\mathscr C$ has an upper bound $X$, for all $A\in\mathscr C$, $A\subseteq X$. Since $X$ is a real number, $X\subset Q$, so there exists $v\in Q$ such that $v\notin X$, then for all $A\in\mathscr C$, $v\notin A$, hence $v\notin \bigcup\mathscr C$, implying $\bigcup\mathscr C\neq Q$.
• Suppose $x,y\in Q$, $x\lt y$, and $y\in \bigcup\mathscr C$, then there exists $A\in\mathscr C$ such that $y\in A$. Since $A$ is a real number, we know $x\in A$, hence $x\in \bigcup\mathscr C$.
• Suppose $x\in \bigcup\mathscr C$, then there exists $A\in\mathscr C$ such that $x\in A$. Since $A$ is a real number, we know there exists $y\in A$ such that $y\gt x$. And also, $y\in \bigcup\mathscr C$.

Hence $\bigcup\mathscr C$ is a real number.

Then we will show that $\bigcup\mathscr C$ is a least upper bound of $\mathscr C$. Suppose $A\in\mathscr C$, if $u\in A$, then $u\in \bigcup\mathscr C$, so $A\le \bigcup\mathscr C$. Hence $\bigcup\mathscr C$ is an upper bound of $\mathscr C$. Now let $Y$ be any upper bound of $\mathscr C$. If $u\in \bigcup\mathscr C$, then there exists $A\in\mathscr C$ such that $u\in A$. Since $Y$ is an upper bound of $\mathscr C$, $A\subseteq Y$, so $u\in Y$. Hence $\bigcup\mathscr C\le Y$. Therefore $\bigcup\mathscr C$ is a least upper bound of $\mathscr C$.

For uniqueness, let $Z$ be a least upper bounds of $\mathscr C$. Suppose for contradiction that $\bigcup\mathscr C\neq Z$, then either $\bigcup\mathscr C\lt Z$ or $\bigcup\mathscr C\gt Z$, implying $\bigcup\mathscr C$ and $Z$ cannot both be least upper bounds, a contradiction. Therefore, we have $\bigcup\mathscr C=Z$. $\blacksquare$

Proof. It suffices to show that $\{u\in Q|\exists (a\in x,b\in y),u=a+b\}$, denoted $z$, is a real number for all $x,y\in R$.

• Since $x,y\neq\emptyset$, there exist $a\in x$ and $b\in y$, then $a+b\in z$, so $z\neq\emptyset$. Since $x,y\neq Q$, there exist $c,d\in Q$ such that $c\notin x$ and $d\notin y$, so for all $a\in x$, $c\gt a$, and for all $b\in y$, $d\gt b$, implying $c+d\gt a+b$. So there do not exist $a\in x,b\in y$ such that $c+d=a+b$, hence $c+d\notin z$, implying $z\neq Q$.
• Suppose $a,b\in Q$, if $a\lt b$ and $b\in z$, then there exist $c\in x,d\in y$ such that $b=c+d$, but then $c-(b-a)\in x$ and $c-(b-a)+d=c-b+a+b-c=a$, so $a\in z$.
• Suppose $a\in z$, then $c\in x,d\in y$ such that $a=c+d$, but then there exist $c'\in x,d'\in y$ such that $c'\gt c$ and $d'\gt d$, so $c'+d'\in z$ and $c'+d'\gt c+d=a$.

Therefore, $z\in R$. $\blacksquare$

Proof. Suppose $m,n,k\in R$.

Suppose $u\in (m+n)+k$, then there exist $q\in m+n$ and $c\in k$ such that $u=q+c$, since there exist $a\in m$ and $b\in n$ such that $q=a+b$, we have $u=(a+b)+c$. Since $a,b,c\in Q$, we have $(a+b)+c=a+(b+c)$. Clearly, $b+c\in n+k$, and $u=a+(b+c)\in m+(n+k)$. By a symmetric argument, suppose $u\in m+(n+k)$, then $u\in (m+n)+k$. Hence $(m+n)+k=m+(n+k)$.

Suppose $u\in m+n$, then there exist $a\in m$ and $b\in n$ such that $u=a+b$. Since $a,b\in Q$, $a+b=b+a$. Clearly, $u=b+a\in n+m$. By a symmetric argument, suppose $u\in n+m$, then $u\in m+n$. Hence $m+n=n+m$. $\blacksquare$

Proof. $0$ as a real number is essentially the set of negative rationals. Suppose $r\in R$. If $u\in r$, then there exists $u'\in r$ such that $u'\gt u$, then $u-u'\in 0$ and $u'+(u-u')=u$, so $u\in r+0$. If $u\in r+0$, then there exist $a\in r$ and $b\in 0$ such that $u=a+b$. Since $b\in 0$, we have $b\lt 0$. So $u-a\lt0$, and $u\lt a$. Since $u\lt a$ and $a\in r$, we have $u\in r$. Hence $r+0=r$, and $0+r=r+0=r$. $\blacksquare$

Proof. Suppose $x,y,z\in R$ and $y\lt z$. Let $u\in x+y$, then for some $a\in x$ and $b\in y$ we have $u=a+b$. Note that $b\in z$, thus $u\in x+z$. We have shown that $x+y\le x+z$. Since $y\lt z$, there exists $c\in z$ such that $c\notin y$, and there exists $d\in z$ such that $c\lt d$. Then $d-c\gt 0$, thus there exists $a\in x$ such that for all $t\in x$, $a+(d-c)\gt t$. Then for all $t\in x$ and $s\in y$, $a+d\gt t+c\gt t+s$, implying $a+d\notin x+y$. Since $a+d\in x+z$, we have $x+y\neq x+z$. Hence $x+y\lt x+z$. $\blacksquare$

Proof. Let $a\in r$ and $b\in Q\setminus r$. Clearly, there exist integer $m,k$ such that $mq\lt a$ and $kq\gt b$, so $m\lt k$, $mq\in r$ and $kq\notin r$. Suppose there does not exist an integer $n$ such that $nq\in r$ but $(n+1)q\notin r$. Then by induction, for every natural number (as a rational number) $i$, $(m+i)q,(m+i+1)q\in r$. But that implies $kq\in r$, a contradiction. Hence, there exists an integer $n$ such that $nq\in r$ but $(n+1)q\notin r$. $\blacksquare$

Proof. Suppose $r\in R$, denote $\{u\in Q|\exists (v\in Q^+), -u-v\notin r\}$ by $s$. We will first show that $s$ is a real number:

• Since $r\ne\emptyset$ and $r\ne Q$, $Q\setminus r\ne\emptyset$ and $Q\setminus r\ne Q$. Let $x\in Q\setminus r$, then $-(-x-1)-1\in Q\setminus r$, so $-x-1\in s$, implying $s\neq\emptyset$. Let $x\in r$, then for all $v\in Q^+$, $-(-x)-v\in r$, so $-x\in Q$ but $-x\notin s$, implying $s\neq Q$.
• Let $x,y\in Q$, suppose $x\lt y$ and $y\in s$, then there exists $v\in Q^+$ such that $-y-v\notin r$. Since $-x-v\gt -y-v$, $-x-v\notin r$, implying $x\in s$.
• Let $x\in s$, then there exists $v\in Q^+$ such that $-x-v\notin r$. So $-(x+v/2)-v/2\notin r$, implying $x+v/2\in s$, and also, $x+v/2\gt x$.

We have shown that $s$ is a real number.

Then we will show that $s$ is indeed an additive inverse of $r$. Note that $r+s=\{u\in Q|\exists (a\in r,b\in s),u=a+b\}$ and $0=\{u\in Q|u\lt 0\}$ (a $0$ is real and the other is rational). If $u\in r+s$, then there exist $a\in r,b\in s$ such that $u=a+b$ and there exists $v\in Q^+$ such that $-b-v\notin r$, so $-b\notin r$. Now we have $-b\gt a$, so $u=a+b\lt 0$, implying $u\in 0$. If $u\in 0$, let $v=-u/2$, then $v\gt0$, and there exists an integer $n$ such that $vn\in r$ and $v(n+1)\notin r$. Let $p=-v(n+2)$, then $-p-v=v(n+1)\notin r$, so $p\in s$. Note that $vn\in r$ and $u=-2v=-v(n+2)+vn=p+vn$. Hence $u\in r+s$. Therefore, $r+s=0$. And $s+r=r+s=0$.

To show uniqueness, suppose $r+s=r+t=0$, then $s=s+0=s+(r+t)=(s+r)+t=0+t=t$. $\blacksquare$

Proof. $-x=(-x)+y+(-y)\gt(-x)+x+(-y)=-y$. $\blacksquare$

Proof. Since $(--r)+(-r)=0=r+(-r)$, by uniqueness of additive inverse, $--r=r$. $\blacksquare$

Proof. Let $\mathscr C'=\{u\in R|\exists(v\in\mathscr C),u=-v\}$, then $\mathscr C'$ is a non-empty subset of $R$ that has an upper bound, so it has a unique supremum $X$. Then $-X$ is the unique infimum of $\mathscr C$. $\blacksquare$

Proof. $$a+(b-a)=a+(b+(-a))=b+(a+(-a))=b+0=b$$ $\blacksquare$

Proof. Since $r\gt 0$, $0\subset r$, so there exists $q\notin 0$ such that $q\in r$, then $q\in Q\setminus 0=\{u\in Q|u\ge 0\}$, so $q\ge0$ and $q\in r$. Since $r$ is real, there exists $u\in r$ such that $u\gt q\ge 0$. $\blacksquare$

Proof. It suffices to show that $\{u\in Q|\exists (a\in x,b\in y),a\gt0\land b\gt0\land u\le ab\}$, denoted $z$, is a real number for all positive real $x,y$.

• Since $x\gt 0$ and $y\gt 0$, there exists $q\in x$ and $p\in y$ such that $q\gt 0$ and $p\gt0$. So $qp\in z$, implying $z\neq\emptyset$. Let $q\in Q\setminus x$ and $p\in Q\setminus y$, then for all $a\in x,b\in y$, $qp\gt ab$, so $qp\notin z$, implying $z\neq Q$.
• Suppose $q,p\in Q$, $q\lt p$, and $p\in z$, then there exist $a\in x,b\in y$ such that $a\gt0$, $b\gt0$ and $p\le ab$, but then $q\lt p\le ab$, so $q\in z$.
• Suppose $q\in z$, then then there exist $a\in x,b\in y$ such that $a\gt0$, $b\gt0$ and $q\le ab$. Since $x,y$ are real, there exist $a'\in x,b'\in y$ such that $a'\gt a\gt0$ and $b'\gt b\gt 0$. Also $q+(a'b'-ab)=a'b'-(ab-q)\le a'b'$ and $q+(a'b'-ab)\gt q$. Hence, $q+(a'b'-ab)\in z$ and $q+(a'b'-ab)\gt q$.

Therefore, $z\in R$. $\blacksquare$

Proof.

• If $x=0$ or $y=0$, we have $xy=0$ by definition, thus $\sgn(xy)=0=\sgn(x)\sgn(y)$.
• If $x\gt0$ and $y\gt0$, then $xy=\{u\in Q|\exists (a\in x,b\in y),a\gt0\land b\gt0\land u\le ab\}$. Note that there exist $a\in x$ and $b\in y$ such that $a,b\in Q^+$, thus $ab\in xy$. Since $ab\notin 0$, implying $xy\gt0$, hence $\sgn(xy)=1=\sgn(x)\sgn(y)$.
• If $x\gt0$ and $y\lt0$, then $xy=-(x\times^+(-y))\lt0$, thus $\sgn(xy)=-1=\sgn(x)\sgn(y)$.
• If $x\lt0$ and $y\gt0$, then $xy=-((-x)\times^+y)\lt0$, thus $\sgn(xy)=-1=\sgn(x)\sgn(y)$.
• If $x\lt0$ and $y\lt0$, then $xy=(-x)\times^+(-y)\gt0$, thus $\sgn(xy)=1=\sgn(x)\sgn(y)$.

$\blacksquare$

Proof. Suppose $m,n,k\in R$ such that $m,n,k\gt 0$.

Suppose $u\in (mn)k$, then there exist $q\in mn$ and $c\in k$ such that $q\gt0$, $c\gt0$ and $u\le qc$, since there exist $a\in m$ and $b\in n$ such that $a\gt0$, $b\gt0$ and $q\le ab$, we have $u\le qc\le (ab)c$. Since $a,b,c\in Q$, we have $(ab)c=a(bc)$. Clearly, $bc\in nk$, and $a(bc)\in m(nk)$. Since $u\le (ab)c=a(bc)$, $u\in m(nk)$. By a symmetric argument, suppose $u\in m(nk)$, then $u\in (mn)k$. Hence $(mn)k=m(nk)$.

Suppose $u\in mn$, then there exist $a\in m$ and $b\in n$ such that $a\gt0$, $b\gt0$ and $u\le ab$. Since $a,b\in Q$, $ab=ba$. Clearly, $ba\in nm$. Since $u\le ab=ba$, $u\in nm$. By a symmetric argument, suppose $u\in nm$, then $u\in mn$. Hence $mn=nm$.

Now suppose $m,n,k\in R$.

• If any of $m,n,k$ is $0$, then $(mn)k=0=m(nk)$.
• If $m\gt0,n\gt0,k\gt0$, we have already shown that $(mn)k=m(nk)$.
• If $m\gt0,n\gt0,k\lt0$, $(mn)k=-((mn)(-k))=-(m(n(-k)))=-(m(--(n(-k))))=-(m(-(nk)))=m(nk)$.
• If $m\gt0,n\lt0,k\gt0$, $(mn)k=-((-(mn))k)=-((--(m(-n)))k)=-((m(-n))k)=-(m((-n)k))=-(m(--((-n)k)))=-(m(-(nk)))=m(nk)$.
• If $m\gt0,n\lt0,k\lt0$, $(mn)k=(-(mn))(-k)=(--(m(-n)))(-k)=(m(-n))(-k)=m((-n)(-k))=m(nk)$.
• If $m\lt0,n\gt0,k\gt0$, $(mn)k=-((-(mn))k)=-((--((-m)n))k)=-(((-m)n)k)=-((-m)(nk))=m(nk)$.
• If $m\lt0,n\gt0,k\lt0$, $(mn)k=(-(mn))(-k)=(--((-m)n))(-k)=((-m)n)(-k)=(-m)(n(-k))=(-m)(--(n(-k)))=(-m)(-(nk))=m(nk)$.
• If $m\lt0,n\lt0,k\gt0$, $(mn)k=((-m)(-n))k=(-m)((-n)k)=(-m)(--((-n)k))=(-m)(-(nk))=m(nk)$.
• If $m\lt0,n\lt0,k\lt0$, $(mn)k=-((mn)(-k))=-(((-m)(-n))(-k))=-((-m)((-n)(-k)))=m((-n)(-k))=m(nk)$.

• If any of $m,n$ is $0$, then $mn=0=nm$.
• If $m\gt0,n\gt0$, we have already shown that $mn=nm$.
• If $m\gt0,n\lt0$, $mn=-(m(-n))=-((-n)m)=nm$.
• If $m\lt0,n\gt0$, $mn=-((-m)n)=-(n(-m))=nm$.
• If $m\lt0,n\lt0$, $mn=(-m)(-n)=(-n)(-m)=nm$.

$\blacksquare$

Proof. Suppose $r\gt0$. $r1=\{u\in Q|\exists (a\in r,b\in 1),a\gt0\land b\gt0\land u\le ab\}$. Let $u\in r$. If $u\le0$, then it is trivially in $r1$. If $u\gt0$, there exists $v\in r$ such that $v\gt u\gt 0$. Then $\frac{u}{v}\in 1$ and $\frac{u}{v}\gt 0$. So $u=v\frac{u}{v}\in r1$. Now let $u\in r1$, then there exist $a\in r,b\in 1$ such that $a\gt0$, $b\gt0$ and $u\le ab$. Since $b\in 1$, $b\lt 1$ (rational). So $u\le ab\lt a$, hence $u\in r$. We have shown that $r=r1$ for $r\gt 0$.

Now suppose $r\lt 0$, then $r1=-((-r)1)=-(-r)=r$. If $r=0$, then $r1=01=0=r$. We have shown that $r=r1$ for all real $r$. Finally, we have $1r=r1=r$ for all real $r$. $\blacksquare$

Proof.

• If $r\gt0$, then $(-1)r=-((-(-1))r)=-(1r)=-r$.
• If $r=0$, then $(-1)r=(-1)0=0=-0=-r$.
• If $r\lt0$, then $(-1)r=(-(-1))(-r)=1(-r)=-r$.

$\blacksquare$

Proof. Suppose $m,n,k\in R$ such that $m,n,k\gt 0$.

Suppose $u\in m(n+k)$, then there exist $a\in m$ and $q\in n+k$ such that $a\gt0$, $q\gt0$ and $u\le aq$, and there exist $b\in n$ and $c\in k$ such that $q=b+c$. Define $b',c'$ by:

• If $b\gt0$ and $c\gt0$, $b'=b$ and $c'=c$.
• If $b\gt0$ and $c\le0$, since $k\gt0$, there exists $v\in c$ such that $v\gt 0$. Then $b+v\gt b+c=q$. So let $b'=b(q/(b+v))$ and $c'=v(q/(b+v))$, we have $b'+c'=q$, $b'\lt b$ and $c'\lt c$, so $b'\in n$ and $c'\in k$. Also, $b'\gt0$ and $c'\gt0$.
• If $b\le0$ and $c\gt0$, we define $b',c'$ by a symmetric definition to the above case.
• If $b\le0$ and $c\le0$, then $q=b+c\le0$, a contradiction.

Now we have $b'\in n$ and $c'\in k$ such that $b'\gt0$, $c'\gt0$ and $q=b'+c'$.

Then $u\le aq=a(b'+c')=ab'+ac'$. Clearly, $ab'\in mn$ and $ac'\in mk$, so $u\in mn+mk$.

Suppose $u\in mn+mk$, there exist $q\in mn$ and $p\in mk$ such that $u=q+p$, there exist $a\in m$ and $b\in n$ such that $a\gt0$, $b\gt0$ and $q\le ab$, and there exist $c\in m$ and $d\in k$ such that $c\gt0$, $d\gt0$ and $p\le cd$. Let $v=a$ if $a\ge c$ and $v=c$ otherwise, then clearly, $b+d\in n+k$, and $v(b+d)\in m(n+k)$. Since $u=q+p\le ab+cd\le v(b+d)$, we have $u\in m(n+k)$.

We have shown that $m(n+k)=mn+mk$ for positive real $m,n,k$.

Now suppose $m,n,k\in R$.

• If any of $m,n,k$ is $0$, then it is trivial that $m(n+k)=mn+mk$.
• If $m\gt0,n\gt0,k\gt0$, we have already shown that $m(n+k)=mn+mk$.
• If $m\gt0,n\gt0,k\lt0$,
  • if $n+k=0$, then $m(n+k)=0=mn+(-(mn))=mn+m(-n)=mn+mk$;
  • if $n+k\gt0$, then $mn=m((n+k)+(-k))=m(n+k)+m(-k)=m(n+k)+(-(mk))$, so $m(n+k)=mn+mk$;
  • if $n+k\lt0$, then $mk=-(m(-k))=-(m(-((n+k)+(-n))))=-(m(-(n+k)+n))=-(m(-(n+k))+mn)=m(n+k)+(-(mn))$, so $m(n+k)=mn+mk$.
• If $m\gt0,n\lt0,k\gt0$, $m(n+k)=m(k+n)=mk+mn=mn+mk$.
• If $m\gt0,n\lt0,k\lt0$, $m(n+k)=-(m(-(n+k)))=-(m((-n)+(-k)))=-(m(-n)+m(-k))=m(--n)+m(--k)=mn+mk$.
• If $m\lt0,n\gt0,k\gt0$, $m(n+k)=-((-m)(n+k))=-((-m)n+(-m)k)=(--m)n+(--m)k=mn+mk$.
• If $m\lt0,n\gt0,k\lt0$,
  • if $n+k=0$, then $m(n+k)=0=mn+(-(mn))=mn+m(-n)=mn+mk$;
  • if $n+k\gt0$, then $mn=-((-m)n)=-((-m)((n+k)+(-k)))=-((-m)(n+k)+(-m)(-k))=(--m)(n+k)+(--m)(-k)=m(n+k)+(-(mk))$, so $m(n+k)=mn+mk$;
  • if $n+k\lt0$, then $mk=(-m)(-k)=(-m)(-((n+k)+(-n)))=(-m)(-(n+k)+n)=(-m)(-(n+k))+(-m)n=m(n+k)+(-(mn))$, so $m(n+k)=mn+mk$.
• If $m\lt0,n\lt0,k\gt0$, $m(n+k)=m(k+n)=mk+mn=mn+mk$.
• If $m\lt0,n\lt0,k\lt0$, $m(n+k)=(-m)(-(n+k))=(-m)((-n)+(-k))=(-m)(-n)+(-m)(-k)=mn+mk$.

Hence $m(n+k)=mn+mk$. And also, $(m+n)k=k(m+n)=km+kn=mk+nk$. $\blacksquare$

Proof. Suppose $x\gt0$ and $y\gt z$, then $y-z\gt0$, thus $xy-xz=x(y-z)\gt0$, hence $xy\gt xz$. The other cases are proven similarly. $\blacksquare$

Proof. Suppose $r\in R^+$, denote $\{u\in Q|u\le0\}\cup\{u\in Q^+|\exists (v\in Q^+), 1/(u+v)\notin r\}$ by $s$. We will first show that $s$ is a real number:

• Trivially, $0\in s$, so $s\neq\emptyset$. Let $x\in r$ such that $x\gt 0$, then for all $v\in Q^+$, $1/(1/x+v)\lt x$, so $1/(1/x+v)\in r$, so $1/x\in Q$ but $1/x\notin s$, implying $s\neq Q$.
• Let $x,y\in Q$, suppose $x\lt y$ and $y\in s$. If $x\le0$, then $x\in s$. Suppose $x\gt0$, then $y\gt0$, and there exists $v\in Q^+$ such that $1/(y+v)\notin r$. Since $1/(x+v)\gt 1/(y+v)$, $1/(x+v)\notin r$, implying $x\in s$.
• Let $x\in s$. If $x\le0$, it suffices to show that $s$ has a positive element. Let $w\in Q\setminus r$, then $w\gt0$, so $1/(2w)\gt0$. Now we have $1/(1/(2w)+1/(2w))=w\notin r$, so $1/(2w)\in s$, and also, $1/(2w)\gt0\ge x$. If $x\gt0$, then there exists $v\in Q^+$ such that $1/(x+v)\notin r$. So $1/((x+v/2)+v/2)\notin r$, implying $x+v/2\in s$, and also, $x+v/2\gt x$.

We have shown that $s$ is a real number. Also, since $s$ contains a positive element, $s$ is positive.

Then we will show that $s$ is indeed an multiplicative inverse of $r$. Note that $rs=\{u\in Q|\exists (a\in r,b\in s),a\gt0\land b\gt0\land u\le ab\}$ and $1=\{u\in Q|u\lt 1\}$ (a $1$ is real and the other is rational). If $u\in rs$, then there exist $a\in r,b\in s$ such that $a\gt0$, $b\gt0$ and $u\le ab$, and there exists $v\in Q^+$ such that $1/(b+v)\notin r$, so $1/b\notin r$. Now we have $1/b\gt a$, so $u\le ab\lt 1$, implying $u\in 1$. If $u\in 1$, then $u\lt 1$. Suppose $u\le0$, then trivially, $u\in rs$. Suppose $u\gt0$. Since $u\lt 1$, there exists some natural number $n$ such that $u\lt m/(m+1)$ for all $m\ge n$. Let $w\in r$ such that $w\gt0$, let $v=w/(n+1)$, then $v\gt0$, and there exists an integer $m$ such that $vm\in r$ and $v(m+1)\notin r$. Now we have $w(m+1)/(n+1)\notin r$ but $w\in r$, so $m\gt n\ge 0$. Hence, $u/vm\lt 1/(v(m+1))$. Then there exists some $l\in Q^+$ such that $u/vm+l=1/(v(m+1))$, then $1/(u/vm+l)=v(m+1)\notin r$, so $u/vm\in s$. We now have $vm\in r$, $u/vm\in s$, $vm\gt0$, $u/vm\gt0$ and $u=vm(u/vm)$, hence $u\in rs$. Therefore, $rs=1$. And $sr=rs=1$.

Now for $r\in R^-$, since $-r\gt0$, we can define $s$ so that $s\gt0$ and $s(-r)=1$, then $(-s)r=(-(-s))(-r)=s(-r)=1$. And $r(-s)=(-s)r=1$.

To show uniqueness, suppose $rs=rt=1$, then $s=s1=s(rt)=(sr)t=1t=t$. $\blacksquare$

Proof. Since $r\neq0$, $-r\neq0$. Since $-r(-\frac{1}{r})=1$, we have $\frac{1}{-r}=-\frac{1}{r}$. $\blacksquare$

Proof. Suppose $r,s\in R^+$, then $\frac{1}{r},\frac{1}{s}\in R^+$. If $r\lt s$, then $\frac{1}{s}=\frac{1}{s}r\frac{1}{r}\lt\frac{1}{s}s\frac{1}{r}=\frac{1}{r}$.

Suppose $r,s\in R^-$, then $\frac{1}{r},\frac{1}{s}\in R^-$. If $r\lt s$, then $\frac{1}{s}=\frac{1}{s}r\frac{1}{r}\lt\frac{1}{s}s\frac{1}{r}=\frac{1}{r}$. $\blacksquare$

Proof. SInce $r\neq0$, $\frac{1}{r}\neq0$. Since $\frac{1}{r}r=1$, $\frac{1}{\frac{1}{r}}=r$. $\blacksquare$

Proof. $$r(\frac{s}{r})=r(s\frac{1}{r})=s(r\frac{1}{r})=s1=s$$ $\blacksquare$

Proof. The proof follows from exponentiation of natural numbers, using inductions. $\blacksquare$

Proof. Clearly, by induction, if $x\ge y$, then $x^n\ge y^n$. Taking the contrapositive, if $x^n\lt y^n$, then $x\lt y$. $\blacksquare$

Proof. Let $E$ be the set of positive real $t$ such that $t^n\lt x$. Since $x\gt0$, $x^2\gt0$, so $x\lt x+x^2=x(1+x)$, thus $x/(1+x)\lt x$. Since $x/(1+x)\lt 1$, we have $(x/(1+x))^n\le x/(1+x)\lt x$, so $x/(1+x)\in E$, implying $E$ is non-empty. Since $x+1\gt 1$, we have $(x+1)^n\ge x+1\gt x$, so for all $t\in E$, $(x+1)^n\gt t^n$, so $x+1\gt t$, implying $E$ is bounded above by $x+1$. Since $E$ is non-empty and bounded above, it has a supremum $y$. Also, $y\ge x/(1+x)\gt 0$.

We will show that $y^n=x$. Given $b\gt a\gt 0$, the identity $b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+\cdots+a^{n-1})$ yields $b^n-a^n\le (b-a)nb^{n-1}$. Assume for contradiction that $y^n\lt x$. Let $h$ be the infimum of $\{\frac{1}{2},\frac{x-y^n}{2n(y+1)^{n-1}}\}$, then clearly $1\gt h\gt0$ and $h\lt\frac{x-y^n}{n(y+1)^{n-1}}$. So $(y+h)^n-y^n\le hn(y+h)^{n-1}\le hn(y+1)^{n-1}\lt x-y^n$. Thus $(y+h)^n\lt x$, implying $y+h\in E$, which is contradictory to $y$ being an upper bound of $E$. Now Assume for contradiction that $y^n\gt x$. Let $k=\frac{y^n-x}{ny^{n-1}}$, then clearly $k\gt 0$ and $k\lt\frac{y^n}{ny^{n-1}}=\frac{y}{n}\le y$. Let $t=y-k$, then $y^n-t^n=y^n-(y-k)^n\le kny^{n-1}=y^n-x$. Thus $t^n\ge x$, implying $t$ is an upper bound of $E$. But $t=y-k\lt y$, contradictory to $y$ being the least upper bound of $E$. Therefore, we conclude that $y^n=x$.

To show uniqueness, suppose $y^n=z^n$, then $y\lt z$ clearly leads to a contradiction, and so does $y\gt z$. Hence $y=z$. $\blacksquare$

Proof. Suppose $[([(m,0)],[(n,0)])]=[([(m',0)],[(n',0)])]$, then $mn'=m'n$. Let $k=mn'=m'n$, since $r^k\in R^+$ and $nn'\in N^+$, there is a unique positive real $s$ such that $s^{nn'}=r^k$. Since $((r^m)^{1/n})^{nn'}=(((r^m)^{1/n})^n)^{n'}=r^{mn'}=r^k$ and $((r^{m'})^{1/n'})^{nn'}=(((r^{m'})^{1/n'})^{n'})^n=r^{m'n}=r^k$, we conclude that $(r^m)^{1/n}=(r^{m'})^{1/n'}$. $\blacksquare$

Proof. If $x=0$ or $y=0$, then $\sqrt{xy}=0=\sqrt x\sqrt y$. Now suppose $x,y$ are both positive, then $\sqrt{x},\sqrt{y},\sqrt{xy}$ are all positive. Since $(\sqrt{xy})^2=xy=(\sqrt{x})^2(\sqrt{y})^2=(\sqrt{x}\sqrt{y})^2$, and $xy$ is positive, we have $\sqrt{xy}=\sqrt x\sqrt y$. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. By definition of real numbers, $Q\setminus r$ is non-empty. Let $a\in Q\setminus r$ and define $q=\{u\in Q|u\lt a+1\}$, then $q$ is a rational number in $R$. Note that for all $u\in r$, $u\lt a\lt a+1$, so $u\in q$, implying $r\subseteq q$. Also note that $a+1/2\in q$ but $a+1/2\notin r$, hence $r\subset q$, or $r\lt q$.

Therefore, there exists a rational number $p\in R$ such that $-r\lt p$, or $r\gt -p$, where $-p$ is also a rational number in $R$. $\blacksquare$

Proof. Since $r\in R^+$, $\{u\in Q|u\lt0\}\subset r$. Let $q\in Q\setminus r$, then $q\ge0$. If $q=0$, then $\{u\in Q|u\lt1\}$ is a non-zero natural number greater than $r$. If $q\gt0$, then there exist non-zero natural numbers $m,n$ (as rational numbers) such that $q=\frac{m}{n}\le m$, hence $\{u\in Q|u\lt m+1\}$ is a non-zero natural number greater than $r$.

Therefore, there exists a non-zero natural number $m\in R$ such that $m\gt 1/r$, or $r\gt 1/m$. $\blacksquare$

Proof. Since $r\lt s$, there exists $x\in s$ such that $x\notin r$. Then there exists $y\in s$ such that $y\gt x$. Note that for all $z\in r$, $y\gt x\gt z$, and thus $z\in\{u\in Q|u\lt y\}$, implying $r\le\{u\in Q|u\lt y\}$. But then $x\in\{u\in Q|u\lt y\}$ but $x\notin r$, hence $r\lt\{u\in Q|u\lt y\}$. Note that for all $t\in\{u\in Q|u\lt y\}$, $t\lt y$, and thus $t\in s$, implying $\{u\in Q|u\lt y\}\le s$. But then $y\in s$ but $y\notin\{u\in Q|u\lt y\}$, hence $\{u\in Q|u\lt y\}\lt s$. And we have $r\lt\{u\in Q|u\lt y\}\lt s$. $\blacksquare$

Proof. The first three are trivial.

Let $u\in f(m)+f(n)$, then there exist $a\in f(m)$ and $b\in f(n)$ such that $u=a+b$, thus $a\lt m$ and $b\lt n$, hence $a+b\lt m+n$, implying $u=a+b\in f(m+n)$. Let $u\in f(m+n)$, then $u\lt m+n$, thus $\frac{(m+n)-u}{2}\gt0$, hence $m-\frac{(m+n)-u}{2}\in f(m)$ and $n-\frac{(m+n)-u}{2}\in f(n)$. Since $(m-\frac{(m+n)-u}{2})+(n-\frac{(m+n)-u}{2})=u$, we have $u\in f(m)+f(n)$.

Since $f(n)+f(-n)=f(0)=0$, we have $f(-n)=-f(n)$.

• If $m=0$ or $n=0$, then $f(mn)=0=f(m)f(n)$.
• If $m\gt0$ and $n\gt0$, then $f(m),f(n)\gt0$. Let $u\in f(m)f(n)$, then there exist $a\in f(m)$ and $b\in f(n)$ such that $a\gt0$, $b\gt0$, and $u\le ab$. Thus $a\lt m$ and $b\lt n$, and we have $ab\lt mn$. Hence $u=ab\in f(mn)$. If $u\in f(mn)$, then $u\lt mn$. Let $v=\sup(u,0)$, then $v\lt mn$ and $0\le v\lt\frac{v+mn}{2}\lt mn$. Denote $\frac{v+mn}{2mn}$ by $w$, then $0\lt w\lt 1$. Thus $wm\in f(m)$, $wn\in f(n)$, and $wm,wn\gt0$. Note that $(mn-v)(mn-v)\gt0$, thus $(mn+v)(mn+v)\gt 4mnv$, and we have $(wm)(wn)=frac{(v+mn)(v+mn)}{4mn}\gt v\ge u$. Hence $u\in f(m)f(n)$. We have shown that $f(mn)=f(m)f(n)$.
• If $m\gt0$ and $n\lt0$, then $-f(mn)=f(m(-n))=f(m)f(-n)=f(m)(-f(n))=-(f(m)f(n))$, thus $f(mn)=f(m)f(n)$.
• If $m\lt0$ and $n\gt0$, then $-f(mn)=f((-m)n)=f(-m)f(n)=(-f(m))f(n)=-(f(m)f(n))$, thus $f(mn)=f(m)f(n)$.
• If $m\lt0$ and $n\lt0$, then $f(mn)=f((-m)(-n))=f(-m)f(-n)=(-f(m))(-f(n))=f(m)f(n)$.

Suppose $n\neq0$, then $f(n)\neq0$ and $f(n)f(\frac{1}{n})=f(1)=1$, thus $f(\frac{1}{n})=\frac{1}{f(n)}$.

The last one is trivial. $\blacksquare$

Proof. Let $z,w,v\in C$, then there exist $a,b,c,d,e,f\in R$ such that $z=(a,b)$, $w=(c,d)$ and $v=(e,f)$.

• $z+(w+v)=(a+(c+e),b+(d+f))=((a+c)+e,(b+d)+f)=(z+w)+v$ and $z+w=(a+c,b+d)=(c+a,d+b)=w+z$.
• $z+(0,0)=(a+0,b+0)=z$.
• $z+(-a,-b)=(a-a,b-b)=(0,0)$ and if $z+w=(0,0)=z+v$, then $w=w+(z+v)=(z+w)+v=v$.
• $z(wv)=(a,b)(ce-df,cf+de)=(ace-adf-bcf-bde,acf+ade+bce-bdf)=(ac-bd,ad+bc)(e,f)=(zw)v$ and $zw=(ac-bd,ad+bc)=(ca-db,cb+da)=wz$.
• $z(1,0)=(a1-b0,a0+b1)=z$.
• If $a^2+b^2=0$, then $a=0$ and $b=0$. Thus if $z\neq(0,0)$, $a^2+b^2\neq0$, and we can define a complex number $u=(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2})$. Note that $zu=(a\frac{a}{a^2+b^2}-b\frac{-b}{a^2+b^2},a\frac{-b}{a^2+b^2}+b\frac{a}{a^2+b^2})=(1,0)$, and if $zw=(1,0)=zv$, then $w=w(zv)=(zw)v=v$.
• $z(w+v)=(a,b)(c+e,d+f)=(ac+ae-bd-bf,ad+af+bc+be)=(ac-bd,ad+bc)+(ae-bf,af+be)=zw+zv$.

$\blacksquare$

Proof. These are trivial. $\blacksquare$

Proof. The proof follows from exponentiation of natural numbers, using inductions. $\blacksquare$

Proof. $i^2=(0,1)(0,1)=(0\times0-1\times1,0\times1+1\times0)=(-1,0)=-1$. $\blacksquare$

Proof. Since $(-i)i=((-1)i)i=(-1)i^2=(-1)(-1)=1$, $\frac{1}{i}=-i$. $\blacksquare$

Proof. There exist $a,b,c,d\in R$ such that $z=(a,b)$ and $w=(c,d)$. Then we have $$\overline{z+w}=\overline{(a+c,b+d)}=(a+c,-(b+d))=(a,-b)+(c,-d)=\overline z+\overline w$$ $$\overline{-z}=(-a,b)=-\overline z$$ $$\overline{zw}=\overline{(ac-bd,ad+bc)}=(ac-bd,-ad-bc)=(a,-b)(c,-d)=\overline z\text{ }\overline w$$ $$\overline{1/z}=\overline{(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2})}=(\frac{a}{a^2+b^2},\frac{b}{a^2+b^2})=1/\overline z$$
For the last equation, we have $$\overline{z^0}=\overline{(1,0)}=(1,0)=(\overline z)^0$$ as base case, and if $\overline{z^n}=(\overline z)^n$ then $$\overline{z^{n+1}}=\overline{zz^n}=\overline z\text{ }\overline{z^n}=\overline z(\overline z)^n=(\overline z)^{n+1}$$ as inductive step. The equation follows from induction. $\blacksquare$

Proof. Let $z=(a,b)$ and $w=(c,d)$. Then $$\abs{zw} =\sqrt{(ac-bd)^2+(ad+bc)^2} =\sqrt{a^2c^2+b^2d^2-2acbd+a^2d^2+b^2c^2+2adbc} =\sqrt{a^2c^2+b^2d^2+a^2d^2+b^2c^2} =\sqrt{(a^2+b^2)(c^2+d^2)} =\sqrt{(a^2+b^2)}\sqrt{(c^2+d^2)} =\abs{z}\abs{w}$$ $\blacksquare$

Proof. Let $z=(a,b)$ and $w=(c,d)$. Then $$0\le(ad-cb)^2$$ $$2abcd\le a^2d^2+c^2b^2$$ $$(ac+bd)^2=a^2c^2+2abcd+b^2d^2\le a^2d^2+c^2b^2+a^2c^2+b^2d^2=(a^2+b^2)(c^2+d^2)$$ $$ac+bd\le\abs{ac+bd}\le\sqrt{(a^2+b^2)(c^2+d^2)}$$ $$\abs{z+w}^2=(a+c)^2+(b+d)^2=2(ac+bd)+a^2+b^2+c^2+d^2\le2\sqrt{(a^2+b^2)(c^2+d^2)}+a^2+b^2+c^2+d^2=(\abs{z}+\abs{w})^2$$ $$\abs{z+w}\le\abs{z}+\abs{w}$$ $\blacksquare$

Proof.

• $f(0)=(0,0)=0$.
• $f(1)=(1,0)=1$.
• $f(r+s)=(r+s,0)=(r,0)+(s,0)=f(r)+f(s)$.
• Since $f(r)+f(-r)=f(0)=0$, thus $f(-r)=-f(r)$.
• $f(rs)=(rs,0)=(r,0)(s,0)=f(r)f(s)$.
• If $r\neq0$, then $f(r)\neq0$. Since $f(r)f(\frac{1}{r})=f(1)=1$, we have $f(\frac{1}{r})=\frac{1}{f(r)}$.
• By induction.
• If $f(r)=f(s)$, then $(r,0)=(s,0)$, implying $r=s$.

$\blacksquare$

Proof. The set $\mathcal P(N\times S)$ contains every tuple of $S$, hence the collection of all tuples of $S$, denoted $S^*$, is a set. Since $\mathcal P(S^*\times S)$ contains every operation (as a function without extra structure) on $S$, the collection of operations on $S$ is a set. The extra structure is attached in the form of a subset of the nested Cartesian product $N\times\mathcal P(S^*\times S)\times\mathcal P(S^*)\times\{S\}$, representing the arity number, the function itself, the domain and the codomain respectively. $\blacksquare$

Proof. Suppose we have a bijective function $f:A\to B$ with an indexed set of operations $(\varphi_i)_{i\in S}$ on $A$ and an indexed set of operations $(\psi_i)_{i\in S}$ on $B$, such that for all $i\in S$, $\varphi_i$ and $\psi_i$ share the same arity number $k$, and for all $x_1,\ldots,x_k\in A$, we have $$f(\varphi_i(x_1,\ldots,x_k))=\psi_i(f(x_1),\ldots,f(x_k))$$ Let $i\in S$, let $k$ be the arity number of $\psi_i$, and let $y_1,\ldots,y_k\in B$, then $$f^{-1}(\psi_i(y_1,\ldots,y_k)) =f^{-1}(\psi_i(f(f^{-1}(y_1)),\ldots,f(f^{-1}(y_k)))) =f^{-1}(f(\varphi_i(f^{-1}(y_1),\ldots,f^{-1}(y_k)))) =\varphi_i(f^{-1}(y_1),\ldots,f^{-1}(y_k))$$ Thus $f^{-1}$ is also a homomorphism, implying $f$ is an isomorphism. $\blacksquare$

Proof. Suppose the operations in question are $(\varphi_i)_{i\in S}$ on $A$, $(\psi_i)_{i\in S}$ on $B$, and $(\tau_i)_{i\in S}$ on $C$, where for all $i\in S$, $\varphi_i$, $\psi_i$ and $\tau_i$ share the same arity number, and the relations in question are $(R_i)_{i\in T}$ on $A$, $(Q_i)_{i\in T}$ on $B$, and $(P_i)_{i\in T}$ on $C$, where for all $i\in T$, $R_i$, $Q_i$ and $P_i$ share the same arity number.

Let $I:A\to A$ be the identity function on $A$, then $I$ is bijective and $I^{-1}=I$. Let $i\in S$, let $k$ be the arity number of $\varphi_i$, and let $x_1,\ldots,x_k\in A$, then $I(\varphi_i(x_1,\ldots,x_k))=\varphi_i(I(x_1),\ldots,I(x_k))$. The relations are trivially preserved by $I$. Thus $I$ is a homomorphism, and $I^{-1}=I$ is also a homomorphism. Hence $I$ is an isomorphism. We have shown that $A\cong A$.

Suppose $A\cong B$, then there exists isomorphic $f:A\to B$. Then $f^{-1}:B\to A$ is bijective and homomorphic, and $(f^{-1})^{-1}=f$ is also homomorphic. Thus $f^{-1}$ is an isomorphism, implying $B\cong A$.

Suppose $A\cong B$ and $B\cong C$, then there exist isomorphic $f:A\to B$ and $g:B\to C$. Note that $g\circ f:A\to C$ is bijective and $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$. Let $i\in S$, let $k$ be the arity number of $\varphi_i$, and let $x_1,\ldots,x_k\in A$, then $(g\circ f)(\varphi_i(x_1,\ldots,x_k))=g(\psi_i(f(x_1),\ldots,f(x_k)))=\tau_i((g\circ f)(x_1),\ldots,(g\circ f)(x_k))$. The relations are trivially preserved by $g\circ f$. Thus $g\circ f$ is a homomorphism. Let $i\in S$, let $k$ be the arity number of $\tau_i$, and let $y_1,\ldots,y_k\in C$, then $(g\circ f)^{-1}(\tau_i(y_1,\ldots,y_k))=f^{-1}(\psi_i(g^{-1}(y_1),\ldots,g^{-1}(y_k)))=\varphi_i((g\circ f)^{-1}(y_1),\ldots,(g\circ f)^{-1}(y_k))$. The relations are trivially preserved by $(g\circ f)^{-1}$. Thus $(g\circ f)^{-1}$ is a homomorphism. Hence $g\circ f$ is an isomorphism. We have shown that $A\cong C$. $\blacksquare$

Proof. Suppose $N\in\mathcal{N}(x)$, then there exists $M\in\tau$ such that $x\in M$ and $M\subseteq N$, so $x\in N$.

Suppose $M\in\mathcal{P}(X)$ and there exists $N\in\mathcal{N}(x)$ such that $N\subseteq M$, then there exists $A\in\tau$ such that $x\in A$ and $A\subseteq N$, hence $A\subseteq M$, so $M\in\mathcal{N}(x)$.

Suppose $M,N\in\mathcal{N}(x)$, then there exist $A\in\tau$ such that $x\in A$ and $A\subseteq M$ and $B\in\tau$ such that $x\in B$ and $B\subseteq N$, so $A\cap B\in\tau$, $x\in A\cap B$ and $A\cap B\subseteq M\cap N$, hence $M\cap N\in\mathcal{N}(x)$.

Suppose $N\in\mathcal{N}(x)$, then there exists $L\in\tau$ such that $x\in L$ and $L\subseteq N$, so $L\in\mathcal{N}(x)$, $L\subseteq N$, and for all $y\in L$, we have $L\in\tau$, $y\in L$ and $L\subseteq N$, implying $N\in\mathcal{N}(y)$. $\blacksquare$

Proof. $\emptyset\subseteq X$ and $X\setminus\emptyset=X$ is open, thus $\emptyset$ is closed. $X\subseteq X$ and $X\setminus X=\emptyset$ is open, thus $X$ is closed.

Let $\mathcal C$ be a non-empty set of closed sets. Then $\cap\mathcal C=X\setminus\cup\{X\setminus S:S\in\mathcal C\}$, which is closed.

Let $U,V$ be closed sets, then $U\cup V=X\setminus((X\setminus U)\cap(X\setminus V))$, which is closed. $\blacksquare$

Proof. If $S$ is open, then for all $p\in S$, $p\in S\subseteq S$ where $S$ is open, thus $S$ is a neighborhood of $p$.

If $S$ is a neighborhood of every point in $S$, then for every $p\in S$, there exists an open set $S_p$ such that $p\in S_p\subseteq S$. And $S=\cup_{p\in S}S_p$, which is open. $\blacksquare$

Proof. Let $T\in\tau_U$, then for some $S\in\tau_X$, $S\cap U=T$. Thus $T\subseteq U$.

Since $\emptyset,X\in\tau_X$, $\emptyset\cap U=\emptyset$, and $X\cap U=U$, $\emptyset,U\in\tau_U$.

Suppose $\mathcal C\subseteq\tau_U$. For all $T\in\mathcal C$, there exists $S\in\tau_X$ such that $S\cap U=T$, thus $\{S\in\tau_X|S\cap U=T\}$ is non-empty. By axiom of choice, we can define a function $f:\mathcal C\to\mathcal P(X)$ such that $f(T)\in\tau_X$ and $f(T)\cap U=T$ for all $T\in\mathcal C$. Note that $\{f(T):T\in\mathcal C\}\subseteq\tau_X$, thus $\bigcup_{T\in\mathcal C}f(T)\in\tau_X$. Let $x\in\bigcup\mathcal C$, then for some $T\in\mathcal C$, $x\in T$, thus $x\in f(T)\cap U$ and hence $x\in\p{\bigcup_{T\in\mathcal C}f(T)}\cap U$. Let $x\in\p{\bigcup_{T\in\mathcal C}f(T)}\cap U$, then for some $S\in\{f(T):T\in\mathcal C\}$, $x\in S$, and $x\in U$. Since for some $T\in\mathcal C$, $f(T)=S$, we have $x\in f(T)\cap U$, thus $x\in T$ and hence $x\in\bigcup\mathcal C$. We have shown that $\bigcup\mathcal C=\p{\bigcup_{T\in\mathcal C}f(T)}\cap U$. And therefore, $\bigcup\mathcal C\in\{S\cap U:S\in\tau_X\}=\tau_U$.

Suppose $T_1,T_2\in\tau_U$. Then for some $S_1\in\tau_X$, $S_1\cap U=T_1$, and for some $S_2\in\tau_X$, $S_2\cap U=T_2$. Note that $S_1\cap S_2\in\tau_X$, and $(S_1\cap S_2)\cap U=(S_1\cap U)\cap(S_2\cap U)=T_1\cap T_2$. Thus $T_1\cap T_2\in\{S\cap U:S\in\tau_X\}=\tau_U$. $\blacksquare$

Proof. Let $V_a$ denote the former and let $V_b$ denote the latter. $V_a$ and $V_b$ clearly shares the same underlying set. Let $\tau$ denote the topology of $X$. Then the topology of $V_a$, denoted $\tau_a$, is $\{S\cap V:S\in\tau\}$. And the topology of $V_b$, denoted $\tau_b$, is $\{T\cap V:T\in\tau_U\}$, where $\tau_U$ denotes $\{S\cap U:S\in\tau\}$. Suppose $A\in\tau_a$, then for some $S\in\tau$, $A=S\cap V$. Note that $S\cap U\in\tau_U$, thus $A=S\cap V=(S\cap U)\cap V\in\tau_b$. Now suppose $A\in\tau_b$, then for some $T\in\tau_U$, $A=T\cap V$, and for some $S\in\tau$, $T=S\cap U$. Thus $A=T\cap V=(S\cap U)\cap V=S\cap V$, implying $A\in\tau_a$. We have shown that $V_a$ and $V_b$ also share the same topology. $\blacksquare$

Proof. This is obvious by definitions of isolated point and limit point. $\blacksquare$

Proof. $\text{Ext}S$ is the union of a collection of open sets and is hence open. Thus $X\setminus\overline S=X\setminus(X\setminus\text{Ext}S)=\text{Ext}S$ is open, implying $\overline S$ is closed. $\blacksquare$

Proof. If $S$ is closed, then $X\setminus S$ is open, implying $X\setminus S=\text{Ext}S$, hence $S=\overline S$. If $S=\overline S$, then $S$ is closed since $\overline S$ is closed. $\blacksquare$

Proof. If $S$ is closed, then $X\setminus S=\text{Ext}S$. Suppose for contradiction that $p\in X$ is a limit point of $S$ such that $p\notin S$. Then $p\in\text{Ext}S$, implying there exists an open subset $U$ of $X$ such that $p\in U$ and $U\subseteq X\setminus S$. Note that $U$ is also a neighborhood of $p$. Since $p$ is a limit point of $S$, there exists $q\in S$ such that $q\in U\setminus\{p\}$. But then $q\in U$ implies $q\notin S$, a contradiction. Thus if $p\in X$ is a limit point of $S$ then $p\in S$.

Suppose $S$ contains its limit points. Let $p\in X\setminus S$, then $p$ is not a limit point of $S$, hence there exists a neighborhood $V$ of $p$ such that for all $q\in S$, $q\notin V\setminus\{p\}$. Since $p\notin S$, $q\neq p$, thus $q\notin V$. Note that $V$ contains an open set $U$ such that $p\in U$. Then we have $q\notin U$. Therefore, $U\subseteq X\setminus S$, implying $p\in\text{Ext}S$. We have shown that $X\setminus S\subseteq\text{Ext}S$, and hence $X\setminus S=\text{Ext}S$, which is open. Therefore, $S$ is closed. $\blacksquare$

Proof. Clearly, $\overline S$ contains $S$. Since $\overline S$ is closed, it contains its limit points, and hence the limit points of $S$. Therefore, $\overline S$ contains $S$ and the limit points of $S$.

If $p\in\overline S$, such that $p\notin S$, suppose for contradiction that $p$ is not a limit point of $S$. Then there exists a neighborhood $V$ of $p$ such that for all $q\in S$, $q\notin V\setminus\{p\}$. Since $p\notin S$, $q\neq p$, thus $q\notin V$. Note that $V$ contains an open set $U$ such that $p\in U$. Then we have $q\notin U$. Therefore, $U\subseteq X\setminus S$, implying $p\in\text{Ext}S$, thus $p\notin\overline S$, a contradiction. Hence, if $p\in\overline S$, then either $p\in S$ or $p$ is a limit point of $S$. $\blacksquare$

Proof. Let $p\in\text{Ext} V$, then for some open subset $B$ of $X$ contained in $X\setminus V$, $p\in B$. Since $B\subseteq X\setminus V\subseteq X\setminus U$, $p\in\text{Ext} U$. Thus $\text{Ext} V\subseteq \text{Ext} U$. Hence $\overline U=X\setminus\text{Ext} U\subseteq X\setminus\text{Ext} V=\overline V$. $\blacksquare$

Proof. Let $\mathcal C$ be an open cover of $U\cup V$, then it is an open cover of both $U$ and $V$, and thus has a finite subcover $\mathcal C_U$ for $U$ and a finite subcover $\mathcal C_V$ for $V$. Then $\mathcal C_U\cup\mathcal C_V$ is a finite subcover of $\mathcal C$ for $U\cup V$. Thus $U\cap V$ is compact. $\blacksquare$

Proof. Suppose $S$ is compact and $U\subseteq S$ is closed. Let $\mathcal C$ be an open cover of $U$, then $\mathcal C\cup\{X\setminus U\}$ is an open cover of $S$ and thus has a finite subcover $\mathcal C'$. Note that $\mathcal C'\setminus\{X\setminus U\}$ is a finite subset of $\mathcal C$ that covers $U$. We have shown that $U$ is compact. $\blacksquare$

Proof. Suppose $S$ is compact in $X$. Let $\mathcal C$ be an open cover of $S$ with respect to $U$. Then $S\subseteq\cup\mathcal C$, and for each $C\in\mathcal C$, there exists an open subset $B_C$ of $X$ such that $C=B_C\cap U$. Note that $S\subseteq\cup\mathcal C\subseteq\cup_{C\in\mathcal C}B_C$. Then $\{B_C:C\in\mathcal C\}$ is an open cover of $S$ with respect to $X$, and thus has a finite subcover $\mathcal B$ such that $\mathcal B=\{B_C:C\in\mathcal D\}$ for some finite $\mathcal D\subseteq\mathcal C$. Let $p\in S$, then $p\in U$ and for some $C\in\mathcal D$, $p\in B_C$, thus $p\in B_C\cap U=C$. Hence $\mathcal D$ covers $S$. We have shown that $S$ is compact in $U$.

Suppose $S$ is compact in $U$. Let $\mathcal C$ be an open cover of $S$ with respect to $X$. Then $\{C\cap U:C\in\mathcal C\}$ is an open cover of $S$ with respect to $U$, and thus has a finite subcover $\mathcal B$ such that $\mathcal B=\{C\cap U:C\in\mathcal D\}$ for some finite $\mathcal D\subseteq\mathcal C$. Let $p\in S$, then $p\in U$ and for some $C\in\mathcal D$, $p\in C\cap U$, thus $p\in C$. Hence $\mathcal D$ covers $S$. We have shown that $S$ is compact in $X$. $\blacksquare$

Proof. Let $U$ be a subspace of $X$. Let $B$ be an open subset of $Y$, and $A$ be the preimage of $f$ on $B$. Then $A$ is open with respect to $X$ and $U\cap A$ is the preimage of $f|_U$ on $B$. Note that $U\cap A$ is open with respect to $U$. Hence $f|_U$ is continuous. $\blacksquare$

Proof. Let $B$ be an open subset of $Z$ and $A$ be the preimage of $f$ on $B$. Then $B=C\cap Z$ for some open subset $C$ of $Y$. Note that $f^{-1}(C)=A$, thus $A$ is an open subset of $X$. Hence $f:X\to Z$ is continuous. $\blacksquare$

Proof. Let $B$ be an open subset of $Y$, and $A$ be the preimage of $f$ on $B$. For all $p\in A$, there exists an open subspace $U_p$ of $X$ containing $p$ such that $f|_{U_p}$ is continuous. Note that $A\cap U_p$ is the preimage of $f|_{U_p}$ on $B$ and is hence open, with respect to $U_p$. Since $U_p$ is open with respect to $X$, $A\cap U_p$ is also open with respect to $X$. Note that $\cup\{A\cap U_p:p\in A\}=A$, which is open with respect to $X$. We have shown that $f$ is continuous. $\blacksquare$

Proof. Let $C$ be an open subset of $Z$. Then the preimage $B$ of $g$ on $C$ is open. Thus the preimage $A$ of $f$ on $B$ is open. Note that $A$ is the preimage of $g\circ f$ on $C$. Hence $g\circ f$ is continuous. $\blacksquare$

Proof. Let $\mathcal B$ be an open cover of $f(U)$. Then $\{f^{-1}(S):S\in\mathcal B\}$ is an open cover of $U$, and thus has a finite subcover $\mathcal A$, and for some finite subset $\mathcal C$ of $\mathcal B$, $\mathcal A=\{f^{-1}(S):S\in\mathcal C\}$. Note that $\mathcal C$ is a finite subcover of $\mathcal B$ for $f(U)$. Thus $f(U)$ is compact. $\blacksquare$

Proof. Since $f$ is continuous, $f|_U:U\to f(U)$ is continuous. Since $f(U)=f|_U(U)$, we will use $f$ to denote $f|_U$ below. Suppose $f(U)$ is disconnected, then $f(U)=A\cup B$ for some disjoint non-empty open subsets $A,B$ of $f(U)$. Thus $f^{-1}(A),f^{-1}(B)$ are disjoint non-empty open subsets of $U$. Let $x\in U$, then $f(x)\in A$ or $f(x)\in B$, thus $x\in f^{-1}(A)$ or $x\in f^{-1}(B)$. Hence $U=f^{-1}(A)\cup f^{-1}(B)$, implying $U$ is disconnected. Therefore, if $U$ is connected, then $f(U)$ is connected. $\blacksquare$

Proof. Suppose $R$ is not connected, then $R=A\cup B$ for some disjoint non-empty open subsets $A,B$ of $R$. Let $a\in A$ and $b\in B$, then $a\neq b$. Assume $a\lt b$. Let $S=\{u\in[a,b]|[a,u]\subseteq A\}$. Then $S$ is non-empty and bounded above, thus it has a supremum $s$ with $s\le b$. Clearly, $[a,s)\subseteq A$, thus $s\notin B$, implying $s\in A$. Then we have $s\in S$ and $s\lt b$. Let $t\in(s,b)$, then $t\notin S$, implying $[a,t]\not\subseteq A$. Since $[a,s]\subseteq A$, we have $(s,t]\not\subseteq A$. And we have a contradiction. By a symmetric argument, the case where $a\gt b$ also leads to a contradiction. $\blacksquare$

Proof. Suppose $[0,1]$ is not connected, then $[0,1]=A\cup B$ for some disjoint non-empty open subsets $A,B$ of $[0,1]$. Define $A'$ by adding $(-\infty,0)$ to $A$ if $0\in A$ and adding $(1,\infty)$ to $A$ if $1\in A$, and define $B'$ similarly, then $R=A'\cup B'$ where $A',B'$ are disjoint non-empty open subsets of $R$, implying $R$ is not connected, a contradiction. $\blacksquare$

Proof. Suppose for all $u,v\in X$ there exists a path from $u$ to $v$, and $X$ is not connected. Then $X=A\cup B$ for some disjoint non-empty open subsets $A,B$ of $X$. Let $u\in A$ and $v\in B$, then there exists a path $f$ from $u$ to $v$. Note that $[0,1]=f^{-1}(A)\cup f^{-1}(B)$ where $f^{-1}(A),f^{-1}(B)$ are disjoint non-empty open subsets of $[0,1]$, implying $[0,1]$ is not connected, a contradiction. $\blacksquare$

Proof. Suppose $X$ is connected. Let $f:X\to\{1,-1\}$ be continuous, then $X=f^{-1}(\{1\})\cup f^{-1}(\{-1\})$ where $f^{-1}(\{1\}),f^{-1}(\{-1\})$ are disjoint open subsets of $X$, implying they cannot be both non-empty. Thus $f$ is constant.

Suppose every continuous function $f:X\to\{1,-1\}$ is constant. If $X$ is not connected, then $X=A\cup B$ for some disjoint non-empty open subsets $A,B$ of $X$. Then $f:X\to\{1,-1\}$ defined by $f(x)=1$ if $x\in A$ and $f(x)=-1$ if $x\in B$ is continuous and non-constant, a contradiction. Thus $X$ is connected. $\blacksquare$

Proof. Let $x,y\in X$, then $0=d(x,x)\le d(x,y)+d(y,x)=2d(x,y)$, so $d(x,y)\ge0$. $\blacksquare$

Proof. If $x\neq y$, then $d(x,y)\neq0$, since we have $d(x,y)\ge0$, we conclude that $d(x,y)\gt0$. If $d(x,y)\gt0$, then $d(x,y)\neq0$, so $x\neq y$. Therefore, $d(x,y)\gt0\leftrightarrow x\neq y$. $\blacksquare$

Proof. Trivially, $\tau\subseteq\mathcal P(X)$.

Clearly, $\emptyset\in\tau$. For all $u\in X$, let $r=1$, then $B_r(u)\subseteq X$, so $X\in\tau$.

Suppose $\mathcal{A}\in\mathcal{P}(\tau)$. Let $p$ in $\cup\mathcal{A}$, then there exists $U\in\mathcal{A}$ such that $p\in U$. Since $U\in\tau$, there exists $r\gt0$ such that $B_r(p)\subseteq U\subseteq\cup\mathcal{A}$, implying $\cup\mathcal{A}\in\tau$.

Suppose $M,N\in\tau$. Let $p\in M\cap N$, then there exist $r_M,r_N\gt0$ such that $B_{r_M}(p)\subseteq M$ and $B_{r_N}(p)\subseteq N$. Let $r=\inf(r_M,r_N)$, then $r\gt0$ and $B_r(p)\subseteq M\cap N$, implying $M\cap N\in\tau$. $\blacksquare$

Proof. Suppose $\tau_X$ is the topology of $X$. Let $\tau_A$ be the topology of the topological subspace $U$ and $\tau_B$ be the standard topology of the metric subspace $U$.

If $S\in\tau_A$, then $S=T\cap U$ for some $T\in\tau_X$. Let $p\in S$. Then there exists an open ball $\{u\in X|d_X(u,p)\lt r\}$ contained by $T$. Note that $\{u\in U|d_U(u,p)\lt r\}=\{u\in X|d_X(u,p)\lt r\}\cap U$, implying $\{u\in U|d_U(u,p)\lt r\}\subseteq T\cap U=S$. Thus there exists an open ball, with respect to the metric subspace $U$, that is centered at $p$ and contained by $S$. Hence $S\in\tau_B$.

If $S\in\tau_B$, then $S\subseteq U$ and for all $p\in S$, there exists $r\gt0$ such that the open ball $\{u\in U|d_U(u,p)\lt r\}$ is contained by $S$. By axiom of choice, we can define a function from $r:S\to R$ such that for all $p\in S$, $r_p\gt0$ and the open ball $\{u\in U|d_U(u,p)\lt r_p\}$ is contained by $S$. Note that $\{u\in U|d_U(u,p)\lt r\}=\{u\in X|d_X(u,p)\lt r\}\cap U$ for all $p\in S$ and $r\gt0$. Thus $\{u\in U|d_U(u,p)\lt r_p\}=\{u\in X|d_X(u,p)\lt r_p\}\cap U$ for all $p\in S$. Clearly, $\cup_{p\in S}\{u\in U|d_U(u,p)\lt r_p\}=S$ and $\cup_{p\in S}\{u\in X|d_X(u,p)\lt r_p\}\in\tau_X$, and since $$\p{\cup_{p\in S}\{u\in X|d_X(u,p)\lt r_p\}}\cap U =\cup_{p\in S}\p{\{u\in X|d_X(u,p)\lt r_p\}\cap U} =\cup_{p\in S}\{u\in U|d_U(u,p)\lt r_p\} =S$$ we conclude that $S\in\tau_A$. $\blacksquare$

Proof. $$a0=a0+0=a0+(a0+(-(a0)))=(a0+a0)+(-(a0))=a(0+0)+(-(a0))=a0+(-(a0))=0$$
Let $ab=0$. We have $a=0$ or $a\neq0$.

• If $a=0$, then we have $a=0$ or $b=0$.
• If $a\neq0$, then $b=\frac{1}{a}ab=0$, thus $a=0$ or $b=0$.

Hence, if $ab=0$, then $a=0$ or $b=0$. $\blacksquare$

Proof. By definition. $\blacksquare$

Proof. If $a-1=c\le b$ or $a-1\le c=b$, this is trivial. Now we suppose $a-1\lt c\lt b$, then $a\le c\lt b$.

Define

• $f:N\to E$ by $f(n)=z_{a+n}$ if $n\le b-a$ and $f(n)=0$ otherwise,
• $g:N\to E$ by $g(n)=z_{a+n}$ if $n\le c-a$ and $g(n)=0$ otherwise,
• $h:N\to E$ by $h(n)=z_{(c+1)+n}$ if $n\le b-(c+1)$ and $h(n)=0$ otherwise.

With addition or multiplication as operation, denoted by $\cdot$, we can define $F,G,H$ such that $F(n)$ represents $f(0)\cdot\ldots\cdot f(n)$, $G(n)$ represents $g(0)\cdot\ldots\cdot g(n)$, and $H(n)$ represents $h(0)\cdot\ldots\cdot h(n)$. By induction, if $k\le c-a$, then $F(k)=G(k)$, so $F(c-a)=G(c-a)$. Again, by induction, if $k\le b-(c+1)$, then $F((c-a+1)+k)=G(c-a)\cdot H(k)$, so $$F(b-a)=F((c-a+1)+(b-(c+1)))=G(c-a)\cdot H(b-(c+1))$$ representing $$z_a\cdot\ldots\cdot z_b=(z_a\cdot\ldots\cdot z_c)\cdot(z_{c+1}\cdot\ldots\cdot z_b)$$ $\blacksquare$

Proof. Since $a\le b$, $b-a\ge0$ and is thus a natural number.

Suppose $j\in b-a$ and $\sum_{k=a}^{a+j}(cz_k)=c\p{\sum_{k=a}^{a+j}z_k}$. Then $$\sum_{k=a}^{a+S(j)}(cz_k)=\sum_{k=a}^{a+j}(cz_k)+cz_{a+S(j)}=c\p{\p{\sum_{k=a}^{a+j}z_k}+z_{a+S(j)}}=c\p{\sum_{k=a}^{a+S(j)}z_k}$$ With the trivial base case, we have that $\sum_{k=a}^b(cz_k)=c\p{\sum_{k=a}^bz_k}$.

Suppose $j\in b-a$ and $\sum_{k=a}^{a+j}(z_k+w_k)=\sum_{k=a}^{a+j}z_k+\sum_{k=a}^{a+j}w_k$. Then $$\sum_{k=a}^{a+S(j)}(z_k+w_k)=\sum_{k=a}^{a+j}(z_k+w_k)+(z_{a+S(j)}+w_{a+S(j)})=\p{\p{\sum_{k=a}^{a+j}z_k}+z_{a+S(j)}}+\p{\p{\sum_{k=a}^{a+j}w_k}+w_{a+S(j)}}=\sum_{k=a}^{a+S(j)}z_k+\sum_{k=a}^{a+S(j)}w_k$$ With the trivial base case, we have that $\sum_{k=a}^b(z_k+w_k)=\sum_{k=a}^bz_k+\sum_{k=a}^bw_k$.

Suppose $j\in b-a$ and $\prod_{k=a}^{a+j}(cz_k)=c^{j+1}\p{\prod_{k=a}^{a+j}z_k}$. Then $$\prod_{k=a}^{a+S(j)}(cz_k)=\p{\prod_{k=a}^{a+j}(cz_k)}\p{cz_{a+S(j)}}=\p{c^{j+1}c}\p{\p{\prod_{k=a}^{a+j}z_k}z_{a+S(j)}}=c^{S(j)+1}\p{\prod_{k=a}^{a+S(j)}z_k}$$ With the trivial base case, we have that $\prod_{k=a}^b(cz_k)=c^{b-a+1}\p{\prod_{k=a}^bz_k}$.

Suppose $j\in b-a$ and $\prod_{k=a}^{a+j}(z_kw_k)=\prod_{k=a}^{a+j}z_k\prod_{k=a}^{a+j}w_k$. Then $$\prod_{k=a}^{a+S(j)}(z_kw_k)=\p{\prod_{k=a}^{a+j}(z_kw_k)}\p{z_{a+S(j)}w_{a+S(j)}}=\p{\p{\prod_{k=a}^{a+j}z_k}z_{a+S(j)}}\p{\p{\prod_{k=a}^{a+j}w_k}w_{a+S(j)}}=\prod_{k=a}^{a+S(j)}z_k\prod_{k=a}^{a+S(j)}w_k$$ With the trivial base case, we have that $\prod_{k=a}^b(z_kw_k)=\prod_{k=a}^bz_k\prod_{k=a}^bw_k$. $\blacksquare$

Proof. $$\vb0=c\vb0-c\vb0=c(\vb0+\vb0)-c\vb0=c\vb0+c\vb0-c\vb0=c\vb0$$ $$0\vb v=0\vb v+\vb0=0\vb v+(0\vb v+(-(0\vb v)))=(0\vb v+0\vb v)+(-(0\vb v))=(0+0)\vb v+(-(0\vb v))=0\vb v+(-(0\vb v))=\vb0$$ $$-\vb v=-\vb v+\vb0=-\vb v+0\vb v=-\vb v+(1+(-1))\vb v=-\vb v+(1\vb v+(-1)\vb v)=-\vb v+(\vb v+(-1)\vb v)=((-\vb v)+\vb v)+(-1)\vb v=(\vb v+(-\vb v))+(-1)\vb v=\vb0+(-1)\vb v=(-1)\vb v+\vb0=(-1)\vb v$$ $\blacksquare$

Proof. Let $\vb u,\vb v,\vb w\in V$. Then

• $d(\vb u,\vb v)=0$ if and only if $\Vert\vb u-\vb v\Vert=0$ if and only if $\vb u-\vb v=0$ if and only if $\vb u=\vb v$;
• $d(\vb u,\vb v)=\Vert\vb u-\vb v\Vert=\Vert-(\vb u-\vb v)\Vert=\Vert\vb v-\vb u\Vert=d(\vb v,\vb u)$;
• $d(\vb u,\vb w)=\Vert\vb u-\vb w\Vert=\Vert(\vb u-\vb v)+(\vb v-\vb w)\Vert\le\Vert\vb u-\vb v\Vert+\Vert\vb v-\vb w\Vert=d(\vb u,\vb v)+d(\vb v,\vb w)$.

$\blacksquare$

Proof. $$\braket{\vb u}{\vb v+\vb w}=\braket{\vb v+\vb w}{\vb u}=\braket{\vb v}{\vb u}+\braket{\vb w}{\vb u}=\braket{\vb u}{\vb v}+\braket{\vb u}{\vb w}$$ $$\braket{\vb u}{c\vb v}=\braket{c\vb v}{\vb u}=c\braket{\vb v}{\vb u}=c\braket{\vb u}{\vb v}$$ $$\braket{\vb u}{\vb0}=\braket{\vb0}{\vb u}=\braket{0\vb u}{\vb u}=0\braket{\vb u}{\vb u}=0$$ $\blacksquare$

Proof. $$\braket{\vb u}{\vb v+\vb w}=\overline{\braket{\vb v+\vb w}{\vb u}}=\overline{\braket{\vb v}{\vb u}}+\overline{\braket{\vb w}{\vb u}}=\braket{\vb u}{\vb v}+\braket{\vb u}{\vb w}$$ $$\braket{\vb u}{c\vb v}=\overline{\braket{c\vb v}{\vb u}}=\overline{c}\overline{\braket{\vb v}{\vb u}}=\overline{c}\braket{\vb u}{\vb v}$$ $$\braket{\vb0}{\vb u}=\braket{0\vb u}{\vb u}=0\braket{\vb u}{\vb u}=0$$ $$\braket{\vb u}{\vb0}=\overline{\braket{\vb0}{\vb u}}=\overline{0}=0$$ $$\overline{\braket{\vb u}{\vb u}}=\braket{\vb u}{\vb u}$$ Thus $$\Im(\braket{\vb u}{\vb u})=0$$ $\blacksquare$

Proof. Suppose $V$ is real. Let $\vb v\in V$ and $c\in R$. Then

• $\Vert\vb v\Vert=\sqrt{\braket{\vb v}{\vb v}}\ge0$;
• $\Vert\vb v\Vert=0$ if and only if $\sqrt{\braket{\vb v}{\vb v}}=0$ if and only if $\vb v=\vb0$;
• $\Vert c\vb v\Vert=\sqrt{\braket{c\vb v}{c\vb v}}=\sqrt{c^2\braket{\vb v}{\vb v}}=\abs{c}\Vert\vb v\Vert$.

Now suppose $V$ is complex. Let $\vb v\in V$ and $c\in C$. Then

• $\Vert\vb v\Vert=\sqrt{\Re(\braket{\vb v}{\vb v})}\ge0$;
• $\Vert\vb v\Vert=0$ if and only if $\sqrt{\Re(\braket{\vb v}{\vb v})}=0$ if and only if $\vb v=\vb0$;
• $\Vert c\vb v\Vert=\sqrt{\Re(\braket{c\vb v}{c\vb v})}=\sqrt{\Re(c\overline c)\Re(\braket{\vb v}{\vb v})}=\abs{c}\Vert\vb v\Vert$.

$\blacksquare$

Proof. For real $V$: $$\Vert c\vb v\Vert=\sqrt{\braket{c\vb v}{c\vb v}}=\sqrt{c^2\braket{\vb v}{\vb v}}=\sqrt{c^2}\sqrt{\braket{\vb v}{\vb v}} =\abs{c}\Vert\vb v\Vert$$
For complex $V$: $$\Vert c\vb v\Vert=\sqrt{\Re(\braket{c\vb v}{c\vb v})}=\sqrt{\Re(c\overline{c})\Re(\braket{\vb v}{\vb v})}=\sqrt{\abs{c}^2}\sqrt{\Re(\braket{\vb v}{\vb v})} =\abs{c}\Vert\vb v\Vert$$ $\blacksquare$

Proof. For real $V$: Note that $$\Vert\vb u+\vb v\Vert^2=\Vert\vb u\Vert^2+2\braket{\vb u}{\vb v}+\Vert\vb v\Vert^2$$ The theorem follows.

For complex $V$: Note that $$\Vert\vb u+\vb v\Vert^2=\Re(\braket{\vb u}{\vb u}+\braket{\vb u}{\vb v}+\braket{\vb v}{\vb u}+\braket{\vb v}{\vb v}) =\Vert\vb u\Vert^2+2\Re(\braket{\vb u}{\vb v})+\Vert\vb v\Vert^2$$ Thus the implication is one-sided. $\blacksquare$

Proof. If $\vb v=\vb 0$, then $$\Vert\vb u\Vert\Vert\vb v\Vert=0=\abs{\braket{\vb u}{\vb v}}$$ Note that in this case, $\vb v$ is a scalar multiple (the scalar being $0$) of $\vb u$.

If $\vb v\neq\vb 0$, then $\Vert\vb v\Vert^2\neq0$. Let $\vb w=\vb u-\cfrac{\braket{\vb u}{\vb v}}{\Vert\vb v\Vert^2}\vb v$, then $\braket{\vb w}{\vb v}=\braket{\vb u}{\vb v}-\cfrac{\braket{\vb u}{\vb v}}{\Vert\vb v\Vert^2}\braket{\vb v}{\vb v}=0$. Thus $$ \Vert\vb u\Vert^2 =\norm{\vb w+\cfrac{\braket{\vb u}{\vb v}}{\Vert\vb v\Vert^2}\vb v}^2 =\Vert\vb w\Vert^2+\abs{\cfrac{\braket{\vb u}{\vb v}}{\Vert\vb v\Vert^2}}^2\Vert\vb v\Vert^2 =\Vert\vb w\Vert^2+\cfrac{\abs{\braket{\vb u}{\vb v}}^2}{\Vert\vb v\Vert^2} \ge\cfrac{\abs{\braket{\vb u}{\vb v}}^2}{\Vert\vb v\Vert^2} $$ implying $\Vert\vb u\Vert\Vert\vb v\Vert\ge\abs{\braket{\vb u}{\vb v}}$. Since $\vb v\neq\vb 0$, "one of $\vb u$ and $\vb v$ is a scalar multiple of the other" is equivalent to "$\vb u=c\vb v$ for some scalar $c$". If $\vb u=c\vb v$ for some scalar $c$, then $$ \Vert\vb u\Vert\Vert\vb v\Vert =\Vert c\vb v\Vert\Vert\vb v\Vert =\abs{c}\Vert\vb v\Vert^2 =\abs{c}\abs{\braket{\vb v}{\vb v}} =\abs{\braket{c\vb v}{\vb v}} =\abs{\braket{\vb u}{\vb v}} $$ If $\Vert\vb u\Vert\Vert\vb v\Vert=\abs{\braket{\vb u}{\vb v}}$, then $\vb w=0$, implying $\cfrac{\braket{\vb u}{\vb v}}{\Vert\vb v\Vert^2}\vb v=\vb u$. $\blacksquare$

Proof. Let $\vb u,\vb v\in V$, then $$\Vert\vb u+\vb v\Vert^2 \le\Vert\vb u\Vert^2+2\abs{\braket{\vb u}{\vb v}}+\Vert\vb v\Vert^2 \le\Vert\vb u\Vert^2+2\Vert\vb u\Vert\Vert\vb v\Vert+\Vert\vb v\Vert^2 =(\Vert\vb u\Vert+\Vert\vb v\Vert)^2$$ implying $\Vert\vb u+\vb v\Vert\le\Vert\vb u\Vert+\Vert\vb v\Vert$.

For real $V$: If $\Vert\vb u+\vb v\Vert=\Vert\vb u\Vert+\Vert\vb v\Vert$, then $\braket{\vb u}{\vb v}=\Vert\vb u\Vert\Vert\vb v\Vert\ge0$, implying $\abs{\braket{\vb u}{\vb v}}=\Vert\vb u\Vert\Vert\vb v\Vert$. Thus one of $\vb u$ and $\vb v$ is a scalar multiple of the other. Since $\braket{\vb u}{\vb v}$ is non-negative, one of $\vb u$ and $\vb v$ is a non-negative multiple of the other. If one of $\vb u$ and $\vb v$ is a non-negative multiple of the other, then we trivially have $\Vert\vb u+\vb v\Vert=\Vert\vb u\Vert+\Vert\vb v\Vert$.

For complex $V$: If $\Vert\vb u+\vb v\Vert=\Vert\vb u\Vert+\Vert\vb v\Vert$, then $\Re(\braket{\vb u}{\vb v})=\Vert\vb u\Vert\Vert\vb v\Vert\ge\abs{\braket{\vb u}{\vb v}}$. Since $\Re(\braket{\vb u}{\vb v})\le\abs{\braket{\vb u}{\vb v}}$, we have $\abs{\braket{\vb u}{\vb v}}=\Vert\vb u\Vert\Vert\vb v\Vert$. Thus one of $\vb u$ and $\vb v$ is a scalar multiple of the other. Since $\Re(\braket{\vb u}{\vb v})\ge0$ and $\Im(\braket{\vb u}{\vb v})=0$, one of $\vb u$ and $\vb v$ is a non-negative real multiple of the other. If one of $\vb u$ and $\vb v$ is a non-negative real multiple of the other, then we trivially have $\Vert\vb u+\vb v\Vert=\Vert\vb u\Vert+\Vert\vb v\Vert$. $\blacksquare$

Proof. Suppose for contradiction that $E$ has no limit point in $K$. Then for all $p\in K$, there exists a neighborhood $V_p$ of $p$ such that no point other than $p$ is both in $E$ and in $V_p$. Note that $V_p$ contains an open set $U_p$ such that $p\in U_p$. And there is at most one point in $U_p$ contained by $E$, depending on whether $p$ is in $E$. Denote $\{U_p:p\in K\}$, constructed with axiom of choice, by $\mathcal C$, then $\mathcal C$ is an open cover of $K$, and hence $E$, but it does not have a finite subcover of $E$, because $E$ is infinite and every member of $\mathcal C$ contains at most one member of $E$. Since $E\subseteq K$, $\mathcal C$ also does not have a finite subcover of $K$, contradicting the fact that $K$ is compact. Therefore, $E$ has a limit point in $K$. $\blacksquare$

Proof. Suppose for contradiction that there exists an open ball $B_r(p)$ that contains finitely many points of $S$, then $B_r(p)\setminus\{p\}$ also contains finitely many points of $S$. Suppose $B_r(p)\setminus\{p\}$ contains $k\in N$ points $\{s_1,\ldots,s_k\}$ of $S$. If $k=0$ we have a contradiction already. Suppose $k\gt0$, let $r^*=\inf_{s\in\{s_1,\ldots,s_k\}}d(s,p)$, then $B_{r^*}(p)\setminus\{p\}$ contains no point of $S$, a contradiction. Hence every open ball $B_r(p)$ contains infinitely many points of $S$. $\blacksquare$

Proof. Suppose $S$ is closed and bounded. Then there exists $r\gt0$ such that $S$ is a subset of the $n$-cell $I^*$ consisting of all points $(x_1,\ldots,x_n)$ where $-r\le x_i\le r$ for $i$ from $1$ to $n$. Note that there exists $\delta\gt0$ such that for all $\vb a,\vb b\in I^*$, $d(\vb a,\vb b)\lt\delta$. Suppose for contradiction that $\mathcal C$ is an open cover of $I^*$ with no finite subcover. Then for every sub-$n$-cell $I$ of $I^*$ such that $\mathcal C$ has no finite subcover of $I$, if we equally divide $I$ into $2^n$ sub-$n$-cells by dividing $I$ into halves in each dimension, then $\mathcal C$ has no finite subcover of at least one of the sub-$n$-cells, because otherwise there exists a finite subcover of $I$. Therefore, by axiom of choice, there exists a function $f$ that maps every "sub-$n$-cell $I$ of $I^*$ such that $\mathcal C$ has no finite subcover of $I$" to a "sub-$n$-cell $I'$ of $I$ obtained by dividing $I$ into halves in each dimension such that $\mathcal C$ has no finite subcover of $I'$". By recursion theorem, we can define a sequence of $n$-cells $(I_n)$ with $I_0=I^*$, and $I_{n+1}=f(I_n)$. By induction, for any natural number $m$,

• for all $k\in N$ such that $k\lt m$, $I_m\subset I_k$;
• $I_m$ is not covered by any finite subset of $\mathcal C$;
• if $\vb a,\vb b\in I_m$, then $d(\vb a,\vb b)\lt2^{-m}\delta$.

For each $n$-cell $I_m$, in each dimension $i$, $I_m$ has boundaries $a_{m,i}\le x_i\le b_{m,i}$. Note that for all $k\in N$ such that $k\lt m$, $a_{k,i}\le a_{m,i}\lt b_{m,i}\le b_{k,i}$. Let $x^*_i=\sup\{a_{m,i}:m\in N\}$, then clearly, for all $m\in N$, $a_{m,i}\le x^*_i\le b_{m,i}$. Now let $\vb x^*=(x^*_1,\ldots,x^*_n)$, then $\vb x^*\in I_m$ for all $m\in N$. Then there exists a member $C$ of $\mathcal C$ such that $\vb x^*\in C$. Since $C$ is open, there exists $r\gt 0$ such that $B_r(\vb x^*)\subseteq C$. But then there exists some $k\in N$ such that $2^{-k}\delta\lt r$, implying $I_k\subseteq C$, a contradiction. Hence every open cover of $I^*$ has a finite subcover, or $I^*$ is compact. Since $S$ is a closed subset of a compact set $I^*$, $S$ is compact.

Now suppose $S$ is compact. Then every infinite subset of $S$ as a limit point in $S$. Suppose for contradiction that $S$ is not bounded. Then for every $k\in N^+$, there exists $\vb p_k\in S$ such that $\Vert\vb p_k\Vert\gt k$. Let $E=\{\vb p_k\in R^n|k\in N^+\}$, then $E$ is infinite, and for every $\vb p\in R^n$ and every $r\gt0$, $B_r(\vb p)$ contains at most finitely many elements of $E$, hence $\vb p$ is not a limit point of $E$, implying $E$ is an infinite subset of $S$ without a limit point in $R^n$ and hence $S$, a contradiction. Therefore, $S$ is bounded.

Suppose for contradiction that $S$ is not closed. Then there exists a limit point $\vb p$ of $S$ such that $p\notin S$. For each $k\in N^+$, there exists $\vb p_k\in S$ such that $d(\vb p_k,\vb p)\lt1/k$. Let $E=\{\vb p_k\in R^n|k\in N^+\}$, then $E$ is infinite, and $\vb p$ is a limit point of $E$. If $\vb r\in R^n$ such that $\vb r\neq\vb p$, then there exists $m\in N^+$ such that $1/m\lt d(\vb p,\vb r)/2$, so for all $k\in N$ such that $k\gt m$, $d(\vb p_k,\vb r)\ge d(\vb p,\vb r)-d(\vb p,\vb p_k)\gt d(\vb p,\vb r)-1/k\gt d(\vb p,\vb r)-1/m\gt d(\vb p,\vb r)/2$. So let $r=d(\vb p,\vb r)/2$, there are at most finitely many elements of $E$ in $B_r(\vb r)$, implying $\vb r$ is not a limit point of $E$. Hence $E$ is an infinite subset of $S$ without a limit point in $S$, a contradiction. Therefore, $S$ is closed. $\blacksquare$

Proof. Suppose $n\lt k$, then both sides are trivially $0$. Suppose $n=k$, then both sides are trivially $1$. Now suppose $n\gt k$, then $$\binom{n-1}{k}+\binom{n-1}{k-1} =\frac{(n-1)!}{k!(n-k-1)!}+\frac{(n-1)!}{(k-1)!(n-k)!} =(n-1)!\p{\frac{n-k}{k!(n-k)!}+\frac{k}{k!(n-k)!}} =(n-1)!\frac{n}{k!(n-k)!} =\frac{n!}{k!(n-k)!} =\binom{n}{k}$$ $\blacksquare$

Proof. For base case, when $n=0$, both side are clearly $1$. Assume for inductive hypothesis that $(z+w)^n=\sum_{k=0}^n\binom{n}{k}z^kw^{n-k}$, then $$(z+w)^{n+1} =(z+w)(z+w)^n =(z+w)\sum_{k=0}^n\binom{n}{k}z^kw^{n-k} =z\p{\sum_{k=0}^{n-1}\binom{n}{k}z^kw^{n-k}+z^n}+w\p{w^n+\sum_{k=1}^n\binom{n}{k}z^kw^{n-k}} =z^{n+1}+\sum_{k=1}^n\binom{n}{k-1}z^kw^{n-k+1}+\sum_{k=1}^n\binom{n}{k}z^kw^{n-k+1}+w^{n+1} $$ $$ =z^{n+1}+\sum_{k=1}^n\p{\binom{n}{k-1}+\binom{n}{k}}z^kw^{n-k+1}+w^{n+1} =z^{n+1}+\sum_{k=1}^n\binom{n+1}{k}z^kw^{n-k+1}+w^{n+1} =\sum_{k=0}^{n+1}\binom{n+1}{k}z^kw^{(n+1)-k}$$ By induction, for all $n\in N$, we have $(z+w)^n=\sum_{k=0}^n\binom{n}{k}z^kw^{n-k}$. $\blacksquare$

Proof. Clearly, suppose $\tau$ has $k$ inversions, and $\sigma$ is an elementary transposition, then $\sigma\tau$ has $k+1$ or $k-1$ inversions, hence the parity of the permutation switches as a result of the operation. Note that the identity permutation is even, hence any composition of an odd number of elementary transpositions is odd and any composition of an even number of elementary transpositions is even. $\blacksquare$

Proof. Suppose $\tau$ is a composition of $m$ elementary transpositions and $\sigma$ is a composition of $n$ elementary transpositions, then $\tau\sigma$ is a composition of $m+n$ elementary transpositions.

• If $m,n$ are both even or both odd, then $m+n$ is even, so $\sgn(\tau\sigma)=1=\sgn(\tau)\sgn(\sigma)$.
• If one of $m,n$ is even and the other is odd, then $m+n$ is odd, so $\sgn(\tau\sigma)=-1=\sgn(\tau)\sgn(\sigma)$.

Hence $\sgn(\tau\sigma)=\sgn(\tau)\sgn(\sigma)$. $\blacksquare$

Proof. Fix $(x_1,\ldots,x_k)\in F^k$ and let $\sigma\in S_k$ be an elementary transposition such that $\sigma_j=j+1$ and $\sigma_{j+1}=j$. Then $\sum_{i=1}^k x_i=\sum_{i=1}^{j-1} x_i+x_j+x_{j+1}+\sum_{i=j+2}^k x_i=\sum_{i=1}^{j-1} x_{\sigma_i}+x_{\sigma_{j+1}}+x_{\sigma_j}+\sum_{i=j+2}^k x_{\sigma_i} =\sum_{i=1}^k x_{\sigma_i}$. Note that $\tau$ is a composition of a finite number of elementary transpositions, thus by induction $\sum_{i=1}^k x_i=\sum_{i=1}^k x_{\tau_i}$. Similarly, we also have $\prod_{i=1}^k x_i=\prod_{i=1}^k x_{\tau_i}$. $\blacksquare$

Proof. By definition, $\sum_{x\in X} f(x)=\sum_{i=0}^{k-1}f(h^*(i))$ where $h^*:k\to X$ is a bijection. Then ${h^*}^{-1}\circ h\in S_k$, thus $\sum_{i=0}^{k-1}f(h^*(i))=\sum_{i=0}^{k-1}f(h^*(({h^*}^{-1}\circ h)(i)))=\sum_{i=0}^{k-1}f(h(i))$. Similarly, we also have $\prod_{x\in X} f(x)=\prod_{i=0}^{k-1}f(h(i))$. $\blacksquare$

Proof. Let $\abs{X}=k$. Let $h_x$ be a bijection from $k$ to $X$, then $h_y=g^{-1}\circ h_x$ is a bijection from $k$ to $Y$. And $h_x^{-1}\circ g\circ h_y\in S_k$. Thus $\sum_{x\in X} f(x)=\sum_{i=0}^{k-1}f(h_x(i))=\sum_{i=0}^{k-1}f(h_x((h_x^{-1}\circ g\circ h_y)(i)))=\sum_{i=0}^{k-1}f(g(h_y(i)))=\sum_{y\in Y} f(g(y))$. Similarly, we have $\prod_{x\in X} f(x)=\prod_{y\in Y} f(g(y))$. $\blacksquare$

Proof. Let $\abs{X}=k$. Since $Y$ is finite, there exists a bijection from $l$ to $Y$ for some $l\in N$. Let $i\in\{1,\ldots,l\}$, then $Y_i\subseteq X$, so for some $m_i\le k$, there exists a bijection from $m_i$ to $Y_i$. Since $Y$ is pairwise disjoint, $\sum m_i=\sum\abs{Y_i}=\abs{\cup Y}=k$. We can naturally define a bijection from $k$ to $X$ such that $X_n={Y_i}_j$ where $n=\sum_{s=1}^{i-1}m_s+j$. Let $M_n$ denote $\sum_{j=1}^nm_j$, suppose $\sum_{i=1}^{M_n}f(X_i)=\sum_{i=1}^n\sum_{j=1}^{m_i}f({Y_i}_j)$, where $n\lt l$, then $$\sum_{i=1}^{M_{n+1}}f(X_i)=\sum_{i=1}^{M_n}f(X_i)+\sum_{i=M_n+1}^{M_{n+1}}f(X_i) =\sum_{i=1}^n\sum_{j=1}^{m_i}f({Y_i}_j)+\sum_{j=1}^{m_{n+1}}f({Y_{n+1}}_j)=\sum_{i=1}^{n+1}\sum_{j=1}^{m_i}f({Y_i}_j)$$ By induction, $$\sum_{x\in X}f(x)=\sum_{i=1}^kf(X_i)=\sum_{i=1}^{M_l}f(X_i)=\sum_{i=1}^l\sum_{j=1}^{m_i}f({Y_i}_j)=\sum_{i=1}^l\sum_{x\in Y_i}f(x)=\sum_{U\in Y}\sum_{x\in U}f(x)$$ Similarly we also have $\prod_{x\in X}f(x)=\prod_{U\in Y}\prod_{x\in U}f(x)$. $\blacksquare$

Proof. Suppose for all $x_1\in X_1,\ldots,x_k\in X_k$, $$\sum_{(x_{k+1},\ldots,x_n)\in X_{k+1}\times\ldots\times X_n}f(x_1,\ldots,x_n)=\sum_{x_{k+1}\in X_{k+1}}\ldots\sum_{x_n\in X_n}f(x_1,\ldots,x_n)$$ Define $\phi:X_k\to\{\{x\}\times(X_{k+1}\times\ldots\times X_n):x\in X_k\}$ by $\phi(x)=\{x\}\times(X_{k+1}\times\ldots\times X_n)$, and for each $x\in X_k$, define $\psi_x:(X_{k+1}\times\ldots\times X_n)\to\{x\}\times(X_{k+1}\times\ldots\times X_n)$ by $\psi_x(x_{k+1},\ldots,x_n)=x,(x_{k+1},\ldots,x_n)$, then $\phi$ is a bijection and each $\psi_x$ is also a bijection. Also, $\{\{x\}\times(X_{k+1}\times\ldots\times X_n):x\in X_k\}$ is finite, pairwise disjoint and $$X_k\times(X_{k+1}\times\ldots\times X_n)=\bigcup\{\{x\}\times(X_{k+1}\times\ldots\times X_n):x\in X_k\}$$ Thus for all $x_1\in X_1,\ldots,x_{k-1}\in X_{k-1}$, $$\sum_{(x_k,\ldots,x_n)\in X_k\times\ldots\times X_n}f(x_1,\ldots,x_n) =\sum_{x,(x_{k+1},\ldots,x_n)\in X_k\times(X_{k+1}\times\ldots\times X_n)}f(x_1,\ldots,x_{k-1},x,x_{k+1},\ldots,x_n) =\sum_{U\in X_k\times(X_{k+1}\times\ldots\times X_n)}\sum_{x,(x_{k+1},\ldots,x_n)\in U}f(x_1,\ldots,x_{k-1},x,x_{k+1},\ldots,x_n) $$ $$ =\sum_{x_k\in X_k}\sum_{x,(x_{k+1},\ldots,x_n)\in\{x_k\}\times(X_{k+1}\times\ldots\times X_n)}f(x_1,\ldots,x_{k-1},x,x_{k+1},\ldots,x_n) =\sum_{x_k\in X_k}\sum_{(x_{k+1},\ldots,x_n)\in X_{k+1}\times\ldots\times X_n}f(x_1,\ldots,x_n) =\sum_{x_k\in X_k}\ldots\sum_{x_n\in X_n}f(x_1,\ldots,x_n)$$ By induction, $$\sum_{(x_1,\ldots,x_n)\in X_1\times\ldots\times X_n}f(x_1,\ldots,x_n)=\sum_{x_1\in X_1}\ldots\sum_{x_n\in X_n}f(x_1,\ldots,x_n)$$ Similarly, $\prod_{(x_1,\ldots,x_n)\in X_1\times\ldots\times X_n}f(x_1,\ldots,x_n)=\prod_{x_1\in X_1}\ldots\prod_{x_n\in X_n}f(x_1,\ldots,x_n)$. $\blacksquare$

Proof. Note that $(x_1,\ldots,x_n)\mapsto(x_{\sigma_1},\ldots,x_{\sigma_n})$ is a bijection from $X_1\times\ldots\times X_n$ to $X_{\sigma_1}\times\ldots\times X_{\sigma_n}$. Thus the inverse $(y_1,\ldots,y_n)\mapsto(y_{\sigma^{-1}_1},\ldots,y_{\sigma^{-1}_n})$ is a bijection from $X_{\sigma_1}\times\ldots\times X_{\sigma_n}$ to $X_1\times\ldots\times X_n$. Denote the map $(x_1,\ldots,x_n)\mapsto(x_{\sigma_1},\ldots,x_{\sigma_n})$ by $\phi$, then $$\sum_{x_1\in X_1}\ldots\sum_{x_n\in X_n}f(x_1,\ldots,x_n) =\sum_{(x_1,\ldots,x_n)\in X_1\times\ldots\times X_n}f(x_1,\ldots,x_n) =\sum_{(y_1,\ldots,y_n)\in X_{\sigma_1}\times\ldots\times X_{\sigma_n}}(f\circ\phi^{-1})(y_1,\ldots,y_n) $$ $$ =\sum_{y_1\in X_{\sigma_1}}\ldots\sum_{y_n\in X_{\sigma_n}}(f\circ\phi^{-1})(y_1,\ldots,y_n) =\sum_{x_{\sigma_1}\in X_{\sigma_1}}\ldots\sum_{x_{\sigma_n}\in X_{\sigma_n}}(f\circ\phi^{-1})(x_{\sigma_1},\ldots,x_{\sigma_n}) =\sum_{x_{\sigma_1}\in X_{\sigma_1}}\ldots\sum_{x_{\sigma_n}\in X_{\sigma_n}}(f\circ\phi^{-1}\circ\phi)(x_1,\ldots,x_n) =\sum_{x_{\sigma_1}\in X_{\sigma_1}}\ldots\sum_{x_{\sigma_n}\in X_{\sigma_n}}f(x_1,\ldots,x_n)$$ Similarly, $\prod_{x_1\in X_1}\ldots\prod_{x_n\in X_n}f(x_1,\ldots,x_n)=\prod_{x_{\sigma_1}\in X_{\sigma_1}}\ldots\prod_{x_{\sigma_n}\in X_{\sigma_n}}f(x_1,\ldots,x_n)$. $\blacksquare$

Proof. Suppose $f=g$. Let $x\in X$, then $(x,f(x))\in f$, thus $(x,f(x))\in g$, hence $f(x)=g(x)$.

Suppose for all $x\in X$, $f(x)=g(x)$. Let $u\in f$, then there exist $x\in X$ and $y\in Y$ such that $u=(x,y)$, thus $(x,y)\in f$ and we have $y=f(x)=g(x)$. Hence $u=(x,g(x))\in g$. By a symmetric argument, if $u\in g$ then $u\in f$. Therefore $f=g$. $\blacksquare$

Proof. Suppose $u\in f$, then there exist $x$ and $y$ such that $x\in X,y\in Y,u=(x,y)$, thus $f$ is a class relation from $X$ to $Y$. Let $x'\in X$, then there exists a unique $y^*$ such that $y^*\in Y$ and $\varphi(x',y^*)$. Note that $x'\in X,y^*\in Y,(x',y^*)=(x',y^*),\varphi(x',y^*)$, thus $(x',y^*)\in f$. Now suppose $(x',y'')\in f$, then there exist $x$ and $y$ such that $x\in X,y\in Y,(x',y'')=(x,y),\varphi(x,y)$, thus $y''\in Y$ and $\varphi(x',y'')$, and by uniqueness we have $y''=y^*$. Therefore, there exists a unique $y'\in Y$ such that $(x',y')\in f$, implying $f$ is a class function from $X$ to $Y$. $\blacksquare$

Proof. Note that for all $x'\in X$, $y(x')\in Y$ and $y(x')=y(x')$. Trivially, if $z\in Y$ and $z=y(x')$, then $z=y(x')$. Thus given $x'\in X$, there exists a unique $y'\in Y$ such that $y'=y(x')$. Then we can define a class function $f:X\to Y$ by $y'=y(x')$ with $x'$ as input and $y'$ as output. Let $x\in X$, then $x\in X,y(x)\in Y,(x,y(x))=(x,y(x)),y(x)=y(x)$, thus $(x,y(x))\in f$, and therefore $f(x)=y(x)$. $\blacksquare$

Proof. For all $x\in X$, we have $x\in A$, so there exists a unique $y$ such that $(x,y)\in F$. By axiom schema of replacement, there exists a set $Y$ such that for all $x\in X$, there exists, and hence uniquely exists, $y\in Y$ such that $(x,y)\in F$. Therefore, $f=\{u\in X\times Y|u\in F\}$ is a function from $X$ to $Y$, and thus a function from $X$. And for all $x\in X$, $f(x)=F(x)$. Uniqueness is trivial. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. Let $\alpha$ be an ordinal. Let $\beta\in\alpha$. Then clearly, $\beta$ is linearly ordered by $\in$. Let $\gamma\in\beta$, if $\kappa\in\gamma$, then $\beta,\kappa\in\alpha$. Since $\alpha$ is linearly ordered by $\in$, we have exactly one among $\beta=\kappa$, $\beta\in\kappa$ and $\kappa\in\beta$. If $\beta\in\kappa$ or $\beta=\kappa$, then we have an infinitely descending $\in$-sequence, a contradiction; thus $\kappa\in\beta$. We have shown that $\gamma\in\beta$ implies $\gamma\subseteq\beta$, so $\beta$ is transitive. Therefore, every member of $\alpha$ is an ordinal. $\blacksquare$

Proof. Let $\alpha$ be an ordinal. Let $\alpha'\subseteq\alpha$ such that $\alpha'\neq\emptyset$, then by axiom of foundation, there exists $\beta\in\alpha'$ such that $\alpha'\cap\beta=\emptyset$. Suppose for contradiction that $\beta$ is not a strictly least element of $\alpha'$, then there exists $\gamma\in\alpha'$ such that $\gamma\in\beta$, a contradiction. Hence $\beta$ is a strictly least element of $\alpha'$. We have shown that every non-empty subset of $\alpha$ has a strictly least element, hence $\alpha$ is strictly well-ordered by $\in$. $\blacksquare$

Proof. Let $\alpha$ be an ordinal. Let $\beta,\gamma\in\alpha$, then $\beta,\gamma$ are ordinals, so $\gamma\in\beta$ or $\beta=\gamma$ implies $\gamma\subseteq\beta$. Now suppose $\gamma\subseteq\beta$, since $\alpha$ is linearly ordered by $\in$, we have exactly one among $\beta=\gamma$, $\beta\in\gamma$ and $\gamma\in\beta$. If $\beta\in\gamma$, then $\beta\subseteq\gamma$, which implies $\beta=\gamma$. Hence $\gamma\subseteq\beta$ if and only if $\gamma\in\beta$ or $\beta=\gamma$. Since $\in$ strictly well-orders $\alpha$, "$\in$ or $=$" well-orders $\alpha$, hence $\subseteq$ well-orders $\alpha$. $\blacksquare$

Proof. Define $A=\{u\in\alpha|u\notin F(u)^+\}$. Suppose for contradiction that $A\neq\emptyset$. Then $A$ has a strictly least element $x^*$. Thus $x^*\notin F(x^*)^+$, which implies $F(x^*)\in x^*$. So $F(F(x^*))\in F(x^*)$, implying $F(x^*)\in A$, a contradiction. Hence $A=\emptyset$, implying for all $x\in\alpha$, $x\in F(x)^+$. $\blacksquare$

Proof. Suppose for contradiction that there exists an order isomorphism $F$ from $\alpha\to\beta$. Since $\beta\in\alpha$, we have $\beta\subseteq\alpha$. So for all $\gamma\in\alpha$, $\gamma\in F(\gamma)^+$. Hence $\beta\in F(\beta)^+$. But then $F(\beta)\in\beta$, a contradiction. Thus $\alpha\ncong\beta$. $\blacksquare$

Proof. Since $\beta\subset\alpha$, $\alpha\setminus\beta$ is non-empty. Let $a$ be the strictly least element of $\alpha\setminus\beta$, then $a\subseteq\beta$. Let $b\in\beta$. Then we have exactly one of $a\in b$, $a=b$, $b\in a$. If $a\in b$ or $a=b$, then $a\in\beta$, a contradiction. Thus $b\in a$. Hence $\beta\subseteq a$, and therefore, $\beta=a\in\alpha$. $\blacksquare$

Proof. Let $F$ be an order isomorphism from $\alpha$ to $\beta$. Let $\gamma\in\alpha$ and let $\delta=\text{im}F|_\gamma$. Clearly, $F|_\gamma$ is an order isomorphism from $\gamma$ to $\delta$. Since $\in$ strictly well-orders $\gamma$, $\in$ strictly well-orders $\delta$. Let $\epsilon\in\delta$ and suppose for contradiction that there exists $\zeta\in\epsilon$ such that $\zeta\notin\delta$. Then $F^{-1}(\zeta)\in\alpha\setminus\gamma$, thus $F^{-1}(\zeta)=\gamma$ or $\gamma\in F^{-1}(\zeta)$, implying $F(\gamma)\in\zeta^+$, and hence $F(\gamma)\in\delta$. But then there exists $\eta\in\gamma$ such that $F(\eta)=F(\gamma)$, a contradiction. Hence $\epsilon\subseteq\delta$. We have shown that $\delta$ is an ordinal, and hence $\gamma\cong\delta$. Note that $\delta\neq\beta$, because otherwise $\gamma\cong\delta\cong\beta\cong\alpha$, a contradiction. Hence $\delta\subset\beta$, and we have $\delta\in\beta$. Now suppose we have $\delta_1,\delta_2\in\beta$ such that $\gamma\cong\delta_1$ and $\gamma\cong\delta_2$, then $\delta_1\cong\delta_2$. Since both $\delta_1\in\delta_2$ and $\delta_2\in\delta_1$ are contradictory, we have $\delta_1=\delta_2$. $\blacksquare$

Proof. Define $F=\{(a,b)\in\alpha^+\times\beta^+|a\cong b\}$. Suppose for contradiction that $\text{dom}F\neq\alpha^+$ and $\text{im}F\neq\beta^+$. Then there exist a strictly least element $a^*$ of $\alpha^+\setminus\text{dom}F$ and a strictly least element $b^*$ of $\beta^+\setminus\text{im}F$. Note that for all $a\in\alpha^+$ such that $a^*\in a$, $a\notin\text{dom}F$, because otherwise $a^*\in\text{dom}F$, a contradiction. Similarly, for all $b\in\beta^+$ such that $b^*\in b$, $b\notin\text{im}F$. Hence $a^*=\text{dom}F$ and $b^*=\text{im}F$.

Suppose $(a,b)\in F$ and $(a,c)\in F$, then $b\cong c$, hence $b=c$, since both $b\in c$ and $c\in b$ are contradictory. Hence $F$ is a function from $a^*$ to $b^*$. Similarly, if $(a,c)\in F$ and $(b,c)\in F$, then $a=b$, implying $F$ is injective. Since $F$ is clearly surjective from $a^*$ to $b^*$, $F:a^*\to b^*$ is a bijection. Suppose $a,b\in a^*$ with $a\in b$, then clearly $F(a)\neq F(b)$. Suppose for contradiction that $F(b)\in F(a)$, then there exists $c\in a$ such that $b\cong F(b)\cong c$, but we have $c\in b$, a contradiction. Hence $F(a)\in F(b)$. Thus $F$ preserves order, and so does $F^{-1}$, implying $F$ is an order isomorphism between $a^*$ and $b^*$. But then we have $a^*\cong b^*$, a contradiction.

Therefore, we have exactly one among:

• $\text{dom}F=\alpha^+$ and $\text{im}F\neq\beta^+$, in which case $\alpha\cong\gamma$ for some $\gamma\in\beta^+$ but not $\alpha\cong\beta$, hence $\alpha\lt\beta$ and not $\alpha\cong\beta$ or $\beta\lt\alpha$;
• $\text{dom}F=\alpha^+$ and $\text{im}F=\beta^+$, in which case $\alpha\cong\beta$ and not $\alpha\lt\beta$ or $\beta\lt\alpha$;
• $\text{dom}F\neq\alpha^+$ and $\text{im}F=\beta^+$, in which case $\beta\cong\gamma$ for some $\gamma\in\alpha^+$ but not $\beta\cong\alpha$, hence $\beta\lt\alpha$ and not $\alpha\cong\beta$ or $\alpha\lt\beta$.

$\blacksquare$

Proof. Define $A=\{\alpha\in X|\neg\varphi(\alpha)\}$. Assume for contradiction that $A\neq\emptyset$. Then $A$ has a strictly least element $\kappa$. Thus for all $\gamma\in\kappa$, we have $\varphi(\gamma)$. By inductive hypothesis, we have $\varphi(\kappa)$, a contradiction. Therefore, $A=\emptyset$, implying $\forall(\alpha\in X),\varphi(\alpha)$. $\blacksquare$

Proof. Let $\varphi(f,\alpha)$ denote the property "$f$ is a function from $\alpha^+$ such that $f(\beta)=G(f|_\beta)$ for all $\beta\in\alpha^+$". Define $A=\{\alpha\in X|\forall f(\neg\varphi(f,\alpha))\}$. Suppose for contradiction that $A\neq\emptyset$. Then $A$ has a strictly least element $\kappa$. Note that for all $\alpha\in\kappa$, there exists $f$ with $\varphi(f,\alpha)$. Thus we can define the class $H=\cup\{f:\exists(\alpha\in\kappa),\varphi(f,\alpha)\}$. Note that each set in $H$ is an ordered pair of the form $(\beta,f(\beta))$ where we have $\varphi(f,\alpha)$ for some $\alpha\in\kappa$ and $\beta\in\alpha^+$. We use $H_S$ to denote the subclass of $H$ consisting of ordered pairs $(a,b)$ with $a\in S$.

Let $\alpha\in\kappa$ and suppose for inductive hypothesis that for all $\beta\in\alpha$, $H_{\{\beta\}}=\{(\beta,G(f|_\beta))\}$ for some $f$ with $\varphi(f,\beta)$. Clearly, there exists $f$ with $\varphi(f,\alpha)$, so $(\alpha,f(\alpha))\in H_{\{\alpha\}}$. Since $\varphi(f,\alpha)$, we have $f(\alpha)=G(f|_\alpha)$, hence $(\alpha,G(f|_\alpha))\in H_{\{\alpha\}}$. Suppose $a^*,b^*\in H_{\{\alpha\}}$, then $a^*=(\alpha,f_1(\alpha)),b^*=(\alpha,f_2(\alpha))$ where $\varphi(f_1,\alpha_1),\varphi(f_2,\alpha_2)$ for some $\alpha_1,\alpha_2\in\kappa$ and $\alpha$ in both $\alpha_1^+$ and $\alpha_2^+$. Since for all $\beta\in\alpha$, $(\beta,f_1(\beta)),(\beta,f_2(\beta))\in H_{\{\beta\}}$, we have $f_1(\beta)=G(f|_\beta)=f_2(\beta)$ for some $f$ with $\varphi(f,\beta)$. Thus $f_1|_\alpha=f_2|_\alpha$, implying $f_1(\alpha)=G(f_1|_\alpha)=G(f_2|_\alpha)=f_2(\alpha)$, and hence $a^*=b^*$. Therefore, $H_{\{\alpha\}}=\{(\alpha,G(f|_\alpha))\}$ for some $f$ with $\varphi(f,\alpha)$. By transfinite induction, for all $\alpha\in\kappa$, $H_{\{\alpha\}}=\{(\alpha,G(f|_\alpha))\}$ for some $f$ with $\varphi(f,\alpha)$.

Now let $\alpha\in\kappa$, then there exists $f$ with $\varphi(f,\alpha)$, so $(\alpha,f(\alpha))\in H$. If $(\alpha,a^*),(\alpha,b^*)\in H$, then $(\alpha,a^*),(\alpha,b^*)\in H_{\{\alpha\}}$, so $a^*=b^*$. Hence for all $\alpha\in\kappa$, there exists a unique $\beta$ such that $(\alpha,\beta)\in H$. By axiom schema of replacement, there exists a set $Y$ such that for all $\alpha\in\kappa$, there exists a unique $\beta\in Y$ with $(\alpha,\beta)\in H$, which defines a function $h:\kappa\to Y$, and $h$ is clearly a function from $\kappa$. Let $\alpha\in\kappa$, then for all $f$ with $\varphi(f,\alpha)$, for all $\beta\in\alpha^+$, $(\beta,f(\beta))\in H$, so $h(\beta)=f(\beta)$, implying $h|_\alpha=f|_\alpha$. Note that $h(\alpha)=G(f|_\alpha)$ for some $f$ with $\varphi(f,\alpha)$, hence $h(\alpha)=G(f|_\alpha)=G(h|_\alpha)$.

Now let $g=h\cup\{(\kappa,G(h))\}$, then $g$ is a function from $\kappa^+$, such that for all $\alpha\in\kappa^+$, if $\alpha\in\kappa$, then $g(\alpha)=h(\alpha)=G(h|_\alpha)=G(g|_\alpha)$; if $\alpha=\kappa$, then $g(\alpha)=g(\kappa)=G(h)=G(g|_\kappa)=G(g|_\alpha)$. Therefore, we have $\varphi(g,\kappa)$, a contradiction. Hence we have $A=\emptyset$, implying for all $\alpha\in X$, there exists $f$ such that $\varphi(f,\alpha)$. Thus by the same logic as above, we can define a function $F$ from $X$ (resembling the function $h$ from $\kappa$ above), such that for all $\alpha\in X$, $F(\alpha)=G(F|_\alpha)$.

Now suppose we have functions $F_1,F_2$ from $X$ such that for all $\alpha\in X$, $F_1(\alpha)=G(F_1|_\alpha)$ and $F_2(\alpha)=G(F_2|_\alpha)$. Let $\alpha\in X$ and take as inductive hypothesis that for all $\beta\in\alpha$, $F_1(\beta)=F_2(\beta)$, then we have $F_1(\alpha)=G(F_1|_\alpha)=G(F_2|_\alpha)=F_2(\alpha)$, hence by transfinite induction, $\forall(\alpha\in X),F_1(\alpha)=F_2(\alpha)$, implying $F_1=F_2$. We have shown that there exists a unique function $F$ from $X$ such that for all $\alpha\in X$, $F(\alpha)=G(F|_\alpha)$. $\blacksquare$

Proof. Let $\gamma\in\beta^+$ and suppose for inductive hypothesis that for all $\delta\in\gamma$, $F_\alpha(\delta)=F_\beta(\delta)$. Then $$F_\alpha(\gamma)=G_\alpha(F_\alpha|_\gamma) =\{F_\alpha(\delta):\delta\in\gamma\} =\{F_\beta(\delta):\delta\in\gamma\} =G_\beta(F_\beta|_\gamma)=F_\beta(\gamma)$$ By transfinite induction, for all $\gamma\in\beta^+$, $F_\alpha(\gamma)=F_\beta(\gamma)$, hence $F_\alpha|_{\beta^+}=F_\beta$. $\blacksquare$

Proof. Let $\beta\in\alpha^+$ and suppose for inductive hypothesis that for all $\gamma\in\beta$, $O(\gamma)=\{O(\kappa):\kappa\in\gamma\}$. Then $$O(\beta)=\{F_\beta(\gamma):\gamma\in\beta\}=\{F_\gamma(\gamma):\gamma\in\beta\}=\{O(\gamma):\gamma\in\beta\}$$ By transfinite induction, for all $\beta\in\alpha^+$, $O(\beta)=\{O(\gamma):\gamma\in\beta\}$, hence $O(\alpha)=\{O(\beta):\beta\in\alpha\}$. $\blacksquare$

Proof. We will first show that $O|_\alpha$ is an order isomorphism between $\alpha$ and $O(\alpha)$.

For surjectivity, let $\kappa\in O(\alpha)$. Since $O(\alpha)=\{O(\beta):\beta\in\alpha\}$, there exists $\beta\in\alpha$ such that $O|_\alpha(\beta)=O(\beta)=\kappa$.

For injectivity, let $\beta,\gamma\in\alpha$. If $\beta\neq\gamma$, then either $\beta\in\gamma$ or $\gamma\in\beta$. If $\gamma\in\beta$, then $O(\gamma)\in\{O(\gamma):\gamma\in\beta\}=O(\beta)$, implying $O(\beta)\neq O(\gamma)$, thus $\beta\in\gamma$ also implies $O(\beta)\neq O(\gamma)$. Hence $O|_\alpha(\beta)=O|_\alpha(\gamma)$ implies $\beta=\gamma$.

For order preservation, let $\beta,\gamma\in\alpha$. We have shown that if $\gamma\in\beta$ then $O(\gamma)\in O(\beta)$. If $\gamma\notin\beta$, then $\beta=\gamma$ or $\beta\in\gamma$, both imply $O(\gamma)\notin O(\beta)$. Hence $\gamma\in\beta$ if and only if $O|_\alpha(\gamma)\in O|_\alpha(\beta)$. Given $\delta',\kappa'\in O(\alpha)$, there exist $\delta,\kappa\in\alpha$ such that $\delta'=O|_\alpha(\delta)$ and $\kappa'=O|_\alpha(\kappa)$. If $\delta'\in\kappa'$, then both $\kappa\in\delta$ and $\delta=\kappa$ are contradictory, hence $O|_\alpha^{-1}(\delta')=\delta\in\kappa=O|_\alpha^{-1}(\kappa')$.

We have shown that $O|_\alpha$ is an order isomorphism, and the last thing to show is that $O(\alpha)$ is an ordinal. Let $\beta'\in O(\alpha)$, then there exists $\beta\in\alpha$ such that $O(\beta)=\beta'$. For all $\gamma'\in\beta'$, there exists $\gamma\in\beta$ such that $O(\gamma)=\gamma'$. Since $\gamma\in\alpha$, $\gamma'\in O(\alpha)$. Hence $\beta'\subseteq O(\alpha)$. We have shown that $O(\alpha)$ is transitive. Clearly, $O(\alpha)$ is linearly ordered by $\in$, thus $O(\alpha)$ is indeed an ordinal. $\blacksquare$

Proof. Suppose $O(\alpha)=O(\beta)$, then $\alpha\cong O(\alpha)\cong O(\beta)\cong \beta$.

Now suppose $\alpha\cong\beta$ and let $f$ be an order isomorphism from $\alpha$ to $\beta$. Then $f\cup{(\alpha,\beta)}$, denoted $f^+$, is an order isomorphism from $\alpha^+$ to $\beta^+$. Let $\gamma\in\alpha^+$ and suppose for inductive hypothesis that for all $\delta\in\gamma$, $O(\delta)=O(f^+(\delta))$. Then $$O(\gamma)=\{O(\delta):\delta\in\gamma\}=\{O(f^+(\delta)):\delta\in\gamma\}=\{O(\delta'):\delta'\in f^+(\gamma)\}=O(f^+(\gamma))$$ By transfinite induction, for all $\gamma\in\alpha^+$, $O(\gamma)=O(f^+(\gamma))$. Hence $O(\alpha)=O(f^+(\alpha))=O(\beta)$. $\blacksquare$

Proof. Since $\alpha\in\text{Ord}$, for some ordinal $\beta$, $O(\beta)=\alpha$, thus $\beta\cong\alpha$, hence $\alpha=O(\beta)=O(\alpha)$. $\blacksquare$

Proof. $$O(\alpha)=\{O(\beta):\beta\in\alpha\}=\{\beta:\beta\in\alpha\}=\alpha$$ $\blacksquare$

Proof. Let $\alpha$ be an ordinal, then $\alpha^+$ is also an ordinal. Suppose $\beta\in\alpha^+$ and for all $\gamma\in\beta$, $\gamma\in\text{Ord}$, then $\beta\in\text{Ord}$. By transfinite induction on $\alpha^+$, for all $\beta\in\alpha^+$, $\beta\in\text{Ord}$. Thus $\alpha\in\text{Ord}$. $\blacksquare$

Proof. If $\alpha\cong\beta$, then $\alpha=O(\alpha)=O(\beta)=\beta$. If $\alpha=\beta$, since $\alpha\cong\alpha$, we have $\alpha\cong\beta$.

If $\alpha\in\beta$, since $\alpha\cong\alpha$, we have $\alpha\lt\beta$. If $\alpha\lt\beta$, then $\alpha\cong\gamma$ for some $\gamma\in\beta$, thus $\alpha=\gamma\in\beta$. $\blacksquare$

Proof. If $\alpha\le\beta$, then $\alpha\in\beta$ or $\alpha=\beta$, in both cases we have $\alpha\subseteq\beta$. If not $\alpha\le\beta$, then $\beta\lt\alpha$, thus $\beta\in\alpha$. Since $\beta\notin\beta$, not $\alpha\subseteq\beta$. $\blacksquare$

Proof. Obviously, $\text{Ord}$ is linearly ordered by $\in$.

Now suppose there exists $x\in\text{Ord}\cap X$. Let $A=\{u\in x^+|u\in\text{Ord}\cap X\}$, then $A$ is a subset of $x^+$ and $x\in A$, hence $A$ has a strictly least element $x^*$. Suppose for contradiction that there exists $u\in\text{Ord}\cap X$ such that $u\in x^*$, then $u\in x^+$ and hence $u\in A$, a contradiction. Hence for all $u\in\text{Ord}\cap X$, $u\notin x^*$. $\blacksquare$

Proof. Suppose for contradiction we have some set $X=\text{Ord}$, then $X$ is clearly an ordinal. But then $X\in\text{Ord}=X$, a contradiction. $\blacksquare$

Proof. Since $X\subset\text{Ord}$, $\cup X$ is linearly ordered by $\in$. Let $x\in\cup X$, then there exists $\alpha\in X$ such that $x\in\alpha$, and thus $x\subset\alpha$. Let $y\in x$, then $y\in\alpha$, hence $y\in\cup X$. Thus $x\subseteq\cup X$. We have shown that $\cup X$ is also transitive, and is thus an ordinal.

Let $\alpha\in X$, then for all $\beta\in\alpha$, $\beta\in\cup X$, thus $\alpha\subseteq\cup X$. Since $\alpha,\cup X\in\text{Ord}$, we have $\alpha\le\cup X$. Hence $\cup X$ is an upper bound of $X$. Now let $Y\in\cup X$, then there exist $\alpha\in X$ such that $Y\in\alpha$, implying $Y$ is not an upper bound of $X$. Therefore, $\cup X$ is a least upper bound of $X$. Let $A,B$ be least upper bounds of $X$ in $\text{Ord}$. Suppose for contradiction that $A\neq B$, then $A\in B$ or $B\in A$, in both cases they cannot both be least upper bounds, a contradiction. Hence $A=B$. $\blacksquare$

Proof. Let $\gamma,\kappa\in\text{Ord}$ such that $\kappa\in\gamma$. Let $\eta\in\kappa^+$ and suppose for inductive hypothesis that for all $\delta\in\eta$, $F_\gamma(\delta)=F_\kappa(\delta)$. Then $$F_\gamma(\eta)=G_\gamma(F_\gamma|_\eta) =\bigcup\{\mathcal P(F_\gamma(\delta)):\delta\in\eta\} =\bigcup\{\mathcal P(F_\kappa(\delta)):\delta\in\eta\} =G_\kappa(F_\kappa|_\eta)=F_\kappa(\eta)$$ By transfinite induction, for all $\eta\in\kappa^+$, $F_\gamma(\eta)=F_\kappa(\eta)$, hence $F_\gamma(\kappa)=F_\kappa(\kappa)$. Hence let $\alpha\in\text{Ord}$, we have $$V_\alpha=F_\alpha(\alpha)=G_\alpha(F_\alpha|_\alpha)=\bigcup\{\mathcal P(F_\alpha(\beta)):\beta\in\alpha\} =\bigcup\{\mathcal P(F_\beta(\beta)):\beta\in\alpha\}=\bigcup\{\mathcal P(V_\beta):\beta\in\alpha\}$$ $\blacksquare$

Proof. If $x\in\bigcup\{V_\beta:\beta\in\alpha\}$, then there exists $\beta\in\alpha$ such that $x\in V_\beta=\bigcup\{\mathcal P(V_\gamma):\gamma\in\beta\}$, hence there exists $\gamma\in\beta$ such that $x\in\mathcal P(V_\gamma)$. But then $\gamma\in\alpha$, hence $x\in\bigcup\{\mathcal P(V_\gamma):\gamma\in\alpha\}=V_\alpha$. $\blacksquare$

Proof. Suppose $\alpha\in\text{Ord}$ and $X\in V_\alpha$. Then there exists $S\in\{\mathcal P(V_\beta):\beta\in\alpha\}$ such that $X\in S$, and there exists $\beta\in\alpha$ such that $\mathcal P(V_\beta)=S$. Thus $X\subseteq V_\beta$ and $V_\beta\subseteq V_\alpha$, implying $X\subseteq V_\alpha$. $\blacksquare$

Proof. Suppose $\alpha\in\text{Ord}$ and $X\subseteq V_\alpha$. Then $X\in\mathcal P(V_\alpha)$ and $\mathcal P(V_\alpha)\in\{\mathcal P(V_\beta):\beta\in\alpha^+\}$. Thus $X\in\bigcup\{\mathcal P(V_\beta):\beta\in\alpha^+\}=V_{\alpha^+}$. $\blacksquare$

Proof. Let $X$ be a set. If for all $x\in X$, $x\in V$, then for all $x\in X$ there exists $\alpha\in\text{Ord}$ such that $x\in V_\alpha$. Thus the class $\text{Ord}\cap\{\alpha:x\in V_\alpha\}$ has a strictly least element $\kappa_x$. Hence $x\mapsto\kappa_x$ defines a function from $X$, which has a set-sized codomain and hence a set-sized range $Y$. Note that $Y\subset\text{Ord}$. So there exists an upper bound $\gamma$ of $Y$ in $\text{Ord}$. Then we have $Y\subseteq\gamma^+$, where $\gamma^+\in\text{Ord}$. Thus $\bigcup\{V_\beta:\beta\in Y\}\subseteq V_{\gamma^+}$. Since for all $x\in X$, $\kappa_x\in Y$ and $x\in V_{\kappa_x}$, we have $x\in\bigcup\{V_\beta:\beta\in Y\}$, and hence $x\in V_{\gamma^+}$. Therefore, $X\in\mathcal P(V_{\gamma^+})$, implying $X\in\bigcup\{\mathcal P(V_\beta):\beta\in\gamma^{++}\}=V_{\gamma^{++}}$. And we conclude that $X\in V$.

We have shown that if $X\notin V$, then there exists $x\in X$ such that $x\notin V$. Thus $\{x\in X|x\notin V\}\neq\emptyset$. Note that there exists a choice function $\mathcal C_X$ of $\mathcal P(X)\setminus\{\emptyset\}$. Hence $\mathcal C_X(\{x\in X|x\notin V\})\in\{x\in X|x\notin V\}$. Suppose for contradiction that there exists a set $A$ such that $A\notin V$. Then we can define a class function $G:V\to V$ such that:

• If $f$ is a function from some $k\in N$:
  • If $k=0$, $G(f)=A$.
  • If $k\neq0$ and $f(k-1)\notin V$, $G(f)=\mathcal C_{f(k-1)}(\{x\in f(k-1)|x\notin V\})$.
  • If $k\neq0$ and $f(k-1)\in V$, $G(f)=\emptyset$.
• If $f$ is not a function from some $k\in N$, then $G(f)=\emptyset$.

Then we can use transfinite recursion to define a function $F$ from $N$ such that $F(k)=G(F|_k)$. Now we have $F(0)=G(F|_0)=A\notin V$, and suppose $F(k)\notin V$, then $F(k+1)=G(F|_{k+1})=\mathcal C_{F(k)}(\{x\in F(k)|x\notin V\})\in\{x\in F(k)|x\notin V\}$, implying $F(k+1)\in F(k)$ and $F(k+1)\notin V$. Therefore, $F$ is an infinitely descending $\in$-sequence, a contradiction. And we conclude that every set is in $V$. $\blacksquare$

Proof. Let $\kappa\in\text{Ord}$, then $\kappa\subset\text{Ord}$, thus for all $\alpha\in\kappa$, $\alpha\in\text{Ord}$, and we have $(\forall(\beta\in\alpha),\varphi(\beta))\to\varphi(\alpha)$. Then by transfinite induction, we have $\forall(\alpha\in\kappa),\varphi(\alpha)$. Note that we also have $(\forall(\beta\in\kappa),\varphi(\beta))\to\varphi(\kappa)$. Replacing the variable $\alpha$ with $\beta$, we have $\forall(\beta\in\kappa),\varphi(\beta)$, and thus $\varphi(\kappa)$. Again, replacing the variable $\kappa$ with $\alpha$, we have $\forall(\alpha\in\text{Ord}),\varphi(\alpha)$. $\blacksquare$

Proof. Let $\kappa\in\text{Ord}$, then by transfinite recursion, there exists a unique function $F_\kappa$ from $\kappa^+$ such that for all $\alpha\in\kappa^+$, $F_\kappa(\alpha)=G(F_\kappa|_\alpha)$. Thus $F_\kappa(\kappa)=G(F_\kappa|_\kappa)$. And we can define a class function $F:\text{Ord}\to V$ such that for all $\alpha\in\text{Ord}$, $F(\alpha)=F_\alpha(\alpha)$.

Let $\alpha,\beta\in\text{Ord}$ such that $\beta\in\alpha$. Let $\gamma\in\beta^+$ and suppose for inductive hypothesis that for all $\delta\in\gamma$, $F_\alpha(\delta)=F_\beta(\delta)$. Then $F_\alpha|_\gamma=F_\beta|_\gamma$, thus $F_\alpha(\gamma)=G(F_\alpha|_\gamma)=G(F_\beta|_\gamma)=F_\beta(\gamma)$. By transfinite induction, for all $\gamma\in\beta^+$, we have $F_\alpha(\gamma)=F_\beta(\gamma)$.

Let $\alpha\in\text{Ord}$. For all $\beta\in\alpha$, we have $\beta\in\text{Ord}$ and $\beta\in\beta^+$, thus $$F_\alpha|_\alpha(\beta)=F_\alpha(\beta)=F_\beta(\beta)=F(\beta)=F\upharpoonright\alpha(\beta)$$ Hence $F_\alpha|_\alpha=F\upharpoonright\alpha$, and therefore $$F(\alpha)=F_\alpha(\alpha)=G(F_\alpha|_\alpha)=G(F\upharpoonright\alpha)$$
Suppose $E:\text{Ord}\to V$ is a class functions such that for all $\alpha\in\text{Ord}$, $E(\alpha)=G(E\upharpoonright\alpha)$. Let $\alpha\in\text{Ord}$ and suppose for inductive hypothesis that for all $\beta\in\alpha$, $E(\beta)=F(\beta)$. Then for all $\beta\in\alpha$, $E\upharpoonright\alpha(\beta)=E(\beta)=F(\beta)=F\upharpoonright\alpha(\beta)$. Thus $E\upharpoonright\alpha=F\upharpoonright\alpha$, and we have $E(\alpha)=G(E\upharpoonright\alpha)=G(F\upharpoonright\alpha)=F(\alpha)$. By transfinite induction, for all $\alpha\in\text{Ord}$, $E(\alpha)=F(\alpha)$. Therefore $E=F$.

Finally, if $E$ is a class such that $E=F$, then $E$ is a class function from $\text{Ord}$ to $V$. Let $\alpha\in\text{Ord}$. Then for all $\beta\in\alpha$, $E\upharpoonright\alpha(\beta)=E(\beta)=F(\beta)=F\upharpoonright\alpha(\beta)$. Thus $E\upharpoonright\alpha=F\upharpoonright\alpha$. Therefore $E(\alpha)=F(\alpha)=G(F\upharpoonright\alpha)=G(E\upharpoonright\alpha)$. $\blacksquare$

Proof. Suppose some $\beta\in\text{Ord}$ has $\alpha\lt\beta\lt\alpha^+$. Then $\alpha\subset\beta\subset\alpha^+$. Thus for some $\gamma\in\alpha^+$, we have $\gamma\notin\beta$, and hence $\gamma\notin\alpha$. But then we have $\gamma=\alpha$, thus $\alpha\notin\beta$. Let $\delta\in\beta$, then $\delta\in\alpha^+$ and $\delta\neq\alpha$, thus $\delta\in\alpha$. Therefore $\beta\subseteq\alpha$, a contradiction. $\blacksquare$

Proof. By the lemma above, $\alpha$ is a maximal element of $\alpha^+$. $\blacksquare$

Proof. Since $\alpha$ is a successor ordinal, it has a maximal element $\beta$, and $\beta\in\text{Ord}$. Thus $\beta^+\subseteq\alpha$. If some $\gamma\in\alpha$ has $\gamma\notin\beta^+$, then neither $\gamma\lt\beta$ not $\gamma=\beta$, implying that $\beta\lt\gamma$, a contradiction. Thus $\beta^+=\alpha$.

Suppose $\beta,\gamma\in\text{Ord}$ with $\beta^+=\alpha$ and $\gamma^+=\alpha$. Then $\beta^+=\gamma^+$. If $\beta\lt\gamma$, then $\gamma\notin\beta^+$, thus $\beta^+\neq\gamma^+$, a contradiction. Similarly, $\gamma\lt\beta$ also leads to a contradiction. Hence $\gamma=\beta$. $\blacksquare$

Proof. Since $O(0)=\{O(\alpha):\alpha\in\emptyset\}=0$, $0\in\text{Ord}$. Trivially, for all $\alpha$, $\alpha$ is a zero ordinal if and only if $\alpha=0$.

By induction, for all $n\in N$, $n\in\text{Ord}$. Again by induction, for all $n\in N$ such that $n\neq0$, $n$ is a successor ordinal.

By induction, $N$ is transitive. Since $N\subset\text{Ord}$, $N=\bigcup N\in\text{Ord}$. Then trivially, $N$ is a limit ordinal. Since every ordinal less than $N$ is either a zero ordinal or a successor ordinal, $N$ is the least limit ordinal. $\blacksquare$

Proof. By transfinite recursion, the class $+_\alpha$ defined by $G_\alpha$ is a class function from $\text{Ord}$ to $V$ such that for all $\beta\in\text{Ord}$, $+_\alpha(\beta)=G_\alpha(+_\alpha\upharpoonright\beta)$. Let $\beta\in\text{Ord}$ and suppose that for all $\gamma\in\beta$, $+_\alpha(\gamma)\in\text{Ord}$.

• If $\beta$ is a zero ordinal, then $+_\alpha(\beta)=G_\alpha(+_\alpha\upharpoonright\emptyset)=\alpha\in\text{Ord}$.
• If $\beta$ is a successor ordinal, then $+_\alpha(\beta)=G_\alpha(+_\alpha\upharpoonright\beta)=\bigcup\{p_1^+:p\in +_\alpha\upharpoonright\beta\}$. Note that for all $p\in +_\alpha\upharpoonright\beta$, there exist $\gamma,\delta$ such that $\gamma\in\beta$ and $p=(\gamma,\delta)$. Thus $p_1=\delta=+_\alpha\upharpoonright\beta(\gamma)=+_\alpha(\gamma)\in\text{Ord}$, and hence $p_1^+\in\text{Ord}$. Then for all $\delta\in\{p_1^+:p\in +_\alpha\upharpoonright\beta\}$, $\delta\in\text{Ord}$, and thus $\{p_1^+:p\in +_\alpha\upharpoonright\beta\}\subset\text{Ord}$. Hence $\bigcup\{p_1^+:p\in +_\alpha\upharpoonright\beta\}\in\text{Ord}$. Therefore, $+_\alpha(\beta)\in\text{Ord}$.
• If $\beta$ is a limit ordinal, then $+_\alpha(\beta)=G_\alpha(+_\alpha\upharpoonright\beta)=\bigcup\{p_1:p\in +_\alpha\upharpoonright\beta\}$. By the same reasoning as above, for all $p\in +_\alpha\upharpoonright\beta$, $p_1\in\text{Ord}$. Then for all $\delta\in\{p_1:p\in +_\alpha\upharpoonright\beta\}$, $\delta\in\text{Ord}$, and thus $\{p_1:p\in +_\alpha\upharpoonright\beta\}\subset\text{Ord}$. Hence $\bigcup\{p_1:p\in +_\alpha\upharpoonright\beta\}\in\text{Ord}$. Therefore, $+_\alpha(\beta)\in\text{Ord}$.

In all cases, $+_\alpha(\beta)\in\text{Ord}$. Thus by transfinite induction, for all $\beta\in\text{Ord}$, $+_\alpha(\beta)\in\text{Ord}$. Hence $+_\alpha$ is a class function from $\text{Ord}$ to $\text{Ord}$. $\blacksquare$

Proof. Let $\alpha\in\text{Ord}$. Suppose $\beta\in\text{Ord}$ and for all $\delta\in\beta$, for all $\gamma\lt\delta$, $\alpha+\gamma\lt\alpha+\delta$.

• If $\beta$ is a zero ordinal, then for all $\gamma\lt\beta$, $\alpha+\gamma\lt\alpha+\beta$.
• If $\beta$ is a successor ordinal, then for some $\delta\in\beta$ we have $\delta^+=\beta$. Thus for all $\gamma\lt\delta$, $\alpha+\gamma\lt\alpha+\delta$. Note that $\alpha+\beta=G_\alpha(+_\alpha\upharpoonright\beta)=\bigcup\{p_1^+:p\in +_\alpha\upharpoonright\beta\}$. And $\alpha+\delta=+_\alpha(\delta)=+_\alpha\upharpoonright\beta(\delta)$. Thus $(\delta,\alpha+\delta)\in +_\alpha\upharpoonright\beta$. And $(\alpha+\delta)^+\in\{p_1^+:p\in +_\alpha\upharpoonright\beta\}$. Thus $(\alpha+\delta)^+\subseteq\bigcup\{p_1^+:p\in +_\alpha\upharpoonright\beta\}$, implying $(\alpha+\delta)^+\le\alpha+\beta$. If $\gamma\in\beta$, then $\gamma\le\delta$, thus $\alpha+\gamma\le\alpha+\delta\lt(\alpha+\delta)^+\le\alpha+\beta$.
• If $\beta$ is a limit ordinal, then $\alpha+\beta=G_\alpha(+_\alpha\upharpoonright\beta)=\bigcup\{p_1:p\in +_\alpha\upharpoonright\beta\}$. Let $\gamma\in\beta$, then there exists $\delta\in\beta$ such that $\gamma\in\delta$, thus $\alpha+\gamma\lt\alpha+\delta$. Note that $\alpha+\delta=+_\alpha\upharpoonright\beta(\delta)$, implying $(\delta,\alpha+\delta)\in +_\alpha\upharpoonright\beta$. Thus $\alpha+\delta\in\{p_1:p\in +_\alpha\upharpoonright\beta\}$, implying $\alpha+\delta\subseteq\alpha+\beta$ and hence $\alpha+\delta\le\alpha+\beta$.

In all cases, we have that for all $\gamma\lt\beta$, $\alpha+\gamma\lt\alpha+\beta$. By transfinite induction, for all $\beta\in\text{Ord}$, for all $\gamma\lt\beta$, $\alpha+\gamma\lt\alpha+\beta$. $\blacksquare$

Proof. $\alpha+0=G_\alpha(+_\alpha\upharpoonright0)=\alpha$.

Note that $\alpha+\beta^+=G_\alpha(+_\alpha\upharpoonright(\beta^+))=\bigcup\{p_1^+:p\in +_\alpha\upharpoonright(\beta^+)\}$. And $\alpha+\beta=+_\alpha(\beta)=+_\alpha\upharpoonright(\beta^+)(\beta)$. Thus $(\beta,\alpha+\beta)\in +_\alpha\upharpoonright(\beta^+)$. And $(\alpha+\beta)^+\in\{p_1^+:p\in +_\alpha\upharpoonright(\beta^+)\}$. If $u\in\{p_1^+:p\in +_\alpha\upharpoonright(\beta^+)\}$, then for some $p\in +_\alpha\upharpoonright(\beta^+)$, $u=p_1^+$. And we have $p_0\in\beta^+$, thus $p_0\le\beta$ and $p_1=+_\alpha\upharpoonright(\beta^+)(p_0)=+_\alpha(p_0)=\alpha+p_0$. Since $\alpha+p_0\le\alpha+\beta$, we have $u=(\alpha+p_0)^+\subseteq(\alpha+\beta)^+$. Hence $(\alpha+\beta)^+=\bigcup\{p_1^+:p\in +_\alpha\upharpoonright(\beta^+)\}=\alpha+\beta^+$.

Suppose $\beta$ is a limit ordinal. Then $\alpha+\beta=G_\alpha(+_\alpha\upharpoonright\beta)=\bigcup\{p_1:p\in +_\alpha\upharpoonright\beta\}$. Thus if $\delta\in\alpha+\beta$, then for some $u\in\{p_1:p\in +_\alpha\upharpoonright\beta\}$, $\delta\in u$, and for some $p\in +_\alpha\upharpoonright\beta$, $u=p_1$, thus $p_0\in\beta$ and $\delta\in p_1$. Note that $p_1=+_\alpha\upharpoonright\beta(p_0)=+_\alpha(p_0)=\alpha+p_0\in\{\alpha+\gamma:\gamma\in\beta\}$. Thus $\delta\in\bigcup_{\gamma\in\beta}(\alpha+\gamma)$. For the other direction, if $\delta\in\bigcup_{\gamma\in\beta}(\alpha+\gamma)$, then for some $u\in\{\alpha+\gamma:\gamma\in\beta\}$, $\delta\in u$, and for some $\gamma\in\beta$, $u=\alpha+\gamma$. Note that $\alpha+\gamma=+_\alpha(\gamma)=+_\alpha\upharpoonright\beta(\gamma)$, thus $(\gamma,u)\in +_\alpha\upharpoonright\beta$, and hence $u\in\{p_1:p\in +_\alpha\upharpoonright\beta\}$. Then we have $\delta\in\alpha+\beta$. $\blacksquare$

Proof. Let $\alpha\in\text{Ord}$. Suppose for all $\beta\in\alpha$, $0+\beta=\beta$.

• If $\alpha$ is a zero ordinal, then $$0+\alpha=0=\alpha$$
• If $\alpha$ is a successor ordinal, then for some $\beta\in\alpha$, $\beta^+=\alpha$, thus $$0+\alpha=0+\beta^+=0+(\beta+0^+)=(0+\beta)+0^+=\beta+0^+=\beta^+=\alpha$$
• If $\alpha$ is a limit ordinal, then $$0+\alpha=\bigcup_{\beta\in\alpha}(0+\beta)=\bigcup_{\beta\in\alpha}\beta=\alpha$$

By transfinite induction, for all $\alpha\in\text{Ord}$, $0+\alpha=\alpha$. $\blacksquare$

Proof. Let $+$ denote ordinal addition and $+_N$ denote natural number addition. Let $m\in N$. For base case, $m+0=m=m+_N0$. Suppose $n\in N$ and $m+n=m+_Nn$, then $m+S(n)=S(m+n)=S(m+_Nn)=m+_NS(n)$. By induction, for all $n\in N$, $m+n=m+_Nn$. Thus for all $m,n\in N$, $m+n=m+_Nn$. Let $p\in N\times N$, then for some $m,n\in N$, $(m,n)=p$, thus $m,n\in\text{Ord}$ and $p\in\text{Ord}\times\text{Ord}$. Then $+(p)=m+n=m+_Nn=+_N(p)$. $\blacksquare$

Proof. Let $\gamma\in\text{Ord}$ and suppose for all $\delta\in\gamma$, for all $\alpha\le\delta$, there exists a unique $\beta\in\text{Ord}$ such that $\alpha+\beta=\delta$.

• If $\gamma$ is a zero ordinal, then for all $\alpha\le\gamma$, there exists a unique $\beta\in\text{Ord}$ such that $\alpha+\beta=\gamma$.
• If $\gamma$ is a successor ordinal, then some $\delta\in\gamma$ has $\delta^+=\gamma$. Thus given $\alpha\le\gamma$, either $\alpha=\gamma$ or $\alpha\le\delta$.
  • If $\alpha=\gamma$, then $\alpha+0=\gamma$.
  • If $\alpha\le\delta$, then some $\beta\in\text{Ord}$ has $\alpha+\beta=\delta$. Thus $\alpha+\beta^+=(\alpha+\beta)^+=\delta^+=\gamma$.
  Uniqueness is trivial.
• If $\gamma$ is a limit ordinal, then $\gamma$ is non-empty and has no maximal element. Given $\alpha\le\gamma$, either $\alpha=\gamma$ or $\alpha\in\gamma$. The case when $\alpha=\gamma$ is trivial. Suppose $\alpha\in\gamma$. Note that for all $\delta\in\{u\in\gamma|\alpha\le u\}$, there exists a unique $\beta\in\text{Ord}$ such that $\alpha+\beta=\delta$. With axiom of replacement, this defines a function $f$ whose range we will denote as $Y$. For all $u\in Y$, some $\delta\in\{u\in\gamma|\alpha\le u\}$ has $f(\delta)=u$. Since $f(\delta)\in\text{Ord}$, $u\in\text{Ord}$. Thus $Y\subset\text{Ord}$, implying $\bigcup Y\in\text{Ord}$, which is also the least upper bound of $Y$. We denote $\bigcup Y$ as $\epsilon$. Let $\zeta\in Y$, then some $\delta\in\{u\in\gamma|\alpha\le u\}$ has $f(\delta)=\zeta$. Note that $\delta^+\in\{u\in\gamma|\alpha\le u\}$ and $$\alpha+f(\delta^+)=\delta^+=(\alpha+f(\delta))^+=(\alpha+\zeta)^+=\alpha+\zeta^+$$ implying $\zeta^+=f(\delta^+)\in Y$, and thus $\zeta\lt\zeta^+\le\epsilon$. Hence $Y\subseteq\epsilon$. Since $\alpha\in\{u\in\gamma|\alpha\le u\}$ and $\alpha+f(\alpha)=\alpha$, we have $f(\alpha)=0$, implying $0\in Y$, and thus $0\in\epsilon$. Hence $\epsilon$ is non-empty. Given $\delta\in\epsilon$, $\delta$ is not an upper bound of $Y$, thus some $\zeta\in Y$ has $\delta\lt\zeta\lt\epsilon$. Hence $\epsilon$ has no maximal element. Therefore $\epsilon$ is a limit ordinal. Then $\alpha+\epsilon=\bigcup_{\delta\in\epsilon}(\alpha+\delta)$. If $\zeta\in\alpha+\epsilon$, then for some $\delta\in\epsilon$, $\zeta\in\alpha+\delta$. Since $\delta$ is not an upper bound of $Y$, for some $x\in\{u\in\gamma|\alpha\le u\}$, $\delta\lt f(x)$, thus $\zeta\lt\alpha+\delta\lt\alpha+f(x)=x\lt\gamma$. Hence $\alpha+\epsilon\le\gamma$. If $\zeta\in\gamma$, then some $x\in\gamma$ has $\alpha,\zeta\in x$. Thus $\zeta\in x=\alpha+f(x)$. Since $f(x)\in Y$, $f(x)\in\epsilon$. Hence $\zeta\in\bigcup_{\delta\in\epsilon}(\alpha+\delta)=\alpha+\epsilon$. Hence $\gamma\le\alpha+\epsilon$. We have shown that $\alpha+\epsilon=\gamma$. Uniqueness is trivial.

By transfinite induction, for all $\gamma\in\text{Ord}$, for all $\alpha\le\gamma$, there exists a unique $\beta\in\text{Ord}$ such that $\alpha+\beta=\gamma$. $\blacksquare$

Proof. Since $\beta$ is a limit ordinal, it is non-empty and has no maximal element. Thus we have some $\gamma\in\beta$, and $\alpha+\gamma\in\alpha+\beta$. Hence $\alpha+\beta$ is non-empty. Note that $\alpha+\beta=\bigcup_{\gamma\in\beta}(\alpha+\gamma)$. If $\delta\in\alpha+\beta$, then for some $\gamma\in\beta$, $\delta\in\alpha+\gamma\in\alpha+\beta$. Thus $\alpha+\beta$ has no maximal element. $\blacksquare$

Proof. Let $\alpha,\beta\in\text{Ord}$. Suppose $\gamma\in\text{Ord}$ and for all $\delta\in\gamma$, $(\alpha+\beta)+\delta=\alpha+(\beta+\delta)$.

• If $\gamma$ is a zero ordinal, then $$(\alpha+\beta)+\gamma=\alpha+\beta=\alpha+(\beta+\gamma)$$
• If $\gamma$ is a successor ordinal, then some $\delta\in\gamma$ has $\delta^+=\gamma$. Thus $$(\alpha+\beta)+\gamma=(\alpha+\beta)+\delta^+=((\alpha+\beta)+\delta)^+=(\alpha+(\beta+\delta))^+=\alpha+(\beta+\delta)^+=\alpha+(\beta+\delta^+)=\alpha+(\beta+\gamma)$$
• If $\gamma$ is a limit ordinal, then $$(\alpha+\beta)+\gamma =\bigcup_{\delta\in\gamma}((\alpha+\beta)+\delta) =\bigcup_{\delta\in\gamma}(\alpha+(\beta+\delta))$$ If $u\in\bigcup_{\delta\in\gamma}(\alpha+(\beta+\delta))$, then for some $\delta\in\gamma$, $u\in\alpha+(\beta+\delta)$. Since $\beta+\delta\in\beta+\gamma$, for some $\epsilon\in\beta+\gamma$, $u\in\alpha+\epsilon$. Thus $u\in\bigcup_{\epsilon\in\beta+\gamma}(\alpha+\epsilon)$. If $u\in\bigcup_{\epsilon\in\beta+\gamma}(\alpha+\epsilon)$, then for some $\epsilon\in\beta+\gamma$, $u\in\alpha+\epsilon$.
  • If $\epsilon\lt\beta$, then $\alpha+\epsilon\lt\alpha+\beta$, thus $u\in\alpha+(\beta+0)$. Since $0\in\gamma$, we have $u\in\bigcup_{\delta\in\gamma}(\alpha+(\beta+\delta))$.
  • If $\epsilon\ge\beta$, then for some $\delta\in\text{Ord}$, $\beta+\delta=\epsilon$. Thus $\alpha+\epsilon=\alpha+(\beta+\delta)$, implying $u\in\alpha+(\beta+\delta)$. Since $\epsilon\in\beta+\gamma$, we have $\beta+\delta\lt\beta+\gamma$, thus $\delta\in\gamma$. Hence $u\in\bigcup_{\delta\in\gamma}(\alpha+(\beta+\delta))$.
  We have shown that $$\bigcup_{\delta\in\gamma}(\alpha+(\beta+\delta))=\bigcup_{\epsilon\in\beta+\gamma}(\alpha+\epsilon)$$ Since $\gamma$ is a limit ordinal, $\beta+\gamma$ is also a limit ordinal. Thus $$\bigcup_{\epsilon\in\beta+\gamma}(\alpha+\epsilon)=\alpha+(\beta+\gamma)$$

By transfinite induction, for all $\gamma\in\text{Ord}$, $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$. $\blacksquare$

Proof. Let $\alpha,\beta\in\text{Ord}$ and $\beta\lt\alpha$. Suppose $\gamma\in\text{Ord}$ and for all $\delta\in\gamma$, $\beta+\delta\le\alpha+\delta$.

• If $\gamma$ is a zero ordinal, then $$\beta+\gamma=\beta\lt\alpha=\alpha+\gamma$$
• If $\gamma$ is a successor ordinal, then for some $\delta\in\gamma$, $\delta^+=\gamma$, thus $$\beta+\gamma=\beta+\delta^+=(\beta+\delta)^+\le(\alpha+\delta)^+=\alpha+\delta^+=\alpha+\gamma$$
• If $\gamma$ is a limit ordinal, then $$\beta+\gamma=\bigcup_{\delta\in\gamma}(\beta+\delta)\le\bigcup_{\delta\in\gamma}(\alpha+\delta)=\alpha+\gamma$$

By transfinite induction, for all $\gamma\in\text{Ord}$, $\beta+\gamma\le\alpha+\gamma$. $\blacksquare$

Proof. By transfinite recursion, the class $\times_\alpha$ defined by $G_\alpha$ is a class function from $\text{Ord}$ to $V$ such that for all $\beta\in\text{Ord}$, $\times_\alpha(\beta)=G_\alpha(\times_\alpha\upharpoonright\beta)$. Let $\beta\in\text{Ord}$ and suppose that for all $\gamma\in\beta$, $\times_\alpha(\gamma)\in\text{Ord}$.

• If $\beta$ is a zero ordinal, then $\times_\alpha(\beta)=G_\alpha(\times_\alpha\upharpoonright\emptyset)=0\in\text{Ord}$.
• If $\beta$ is a successor ordinal, then $\times_\alpha(\beta)=G_\alpha(\times_\alpha\upharpoonright\beta)=\bigcup\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta\}$. Note that for all $p\in \times_\alpha\upharpoonright\beta$, there exist $\gamma,\delta$ such that $\gamma\in\beta$ and $p=(\gamma,\delta)$. Thus $p_1=\delta=\times_\alpha\upharpoonright\beta(\gamma)=\times_\alpha(\gamma)\in\text{Ord}$, and hence $p_1+\alpha\in\text{Ord}$. Then for all $\delta\in\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta\}$, $\delta\in\text{Ord}$, and thus $\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta\}\subset\text{Ord}$. Hence $\bigcup\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta\}\in\text{Ord}$. Therefore, $\times_\alpha(\beta)\in\text{Ord}$.
• If $\beta$ is a limit ordinal, then $\times_\alpha(\beta)=G_\alpha(\times_\alpha\upharpoonright\beta)=\bigcup\{p_1:p\in \times_\alpha\upharpoonright\beta\}$. By the same reasoning as above, for all $p\in \times_\alpha\upharpoonright\beta$, $p_1\in\text{Ord}$. Then for all $\delta\in\{p_1:p\in \times_\alpha\upharpoonright\beta\}$, $\delta\in\text{Ord}$, and thus $\{p_1:p\in \times_\alpha\upharpoonright\beta\}\subset\text{Ord}$. Hence $\bigcup\{p_1:p\in \times_\alpha\upharpoonright\beta\}\in\text{Ord}$. Therefore, $\times_\alpha(\beta)\in\text{Ord}$.

In all cases, $\times_\alpha(\beta)\in\text{Ord}$. Thus by transfinite induction, for all $\beta\in\text{Ord}$, $\times_\alpha(\beta)\in\text{Ord}$. Hence $\times_\alpha$ is a class function from $\text{Ord}$ to $\text{Ord}$. $\blacksquare$

Proof. Let $\alpha\in\text{Ord}$ and suppose for all $\beta\in\alpha$, $0\beta=0$.

• If $\alpha$ is a zero ordinal, then $0\alpha=G_0(\times_0\upharpoonright0)=0$.
• If $\alpha$ is a successor ordinal, then $0\alpha=G_0(\times_0\upharpoonright\alpha)=\bigcup\{p_1+0:p\in\times_0\upharpoonright\alpha\}=\bigcup\{p_1:p\in\times_0\upharpoonright\alpha\}$. Note that for all $p\in\times_0\upharpoonright\alpha$, $p_0\in\alpha$, thus $p_1=\times_0\upharpoonright\alpha(p_0)=\times_0(p_0)=0$. Hence $\bigcup\{p_1:p\in\times_0\upharpoonright\alpha\}=0$.
• If $\alpha$ is a limit ordinal, then $0\alpha=G_0(\times_0\upharpoonright\alpha)=\bigcup\{p_1:p\in\times_0\upharpoonright\alpha\}=0$, for the same reason as above.

By transfinite induction, for all $\alpha\in\text{Ord}$, $0\alpha=0$. $\blacksquare$

Proof. Let $\alpha\in\text{Ord}$ be non-zero. Suppose $\beta\in\text{Ord}$ and for all $\delta\in\beta$, for all $\gamma\lt\delta$, $\alpha\gamma\lt\alpha\delta$.

• If $\beta$ is a zero ordinal, then for all $\gamma\lt\beta$, $\alpha\gamma\lt\alpha\beta$.
• If $\beta$ is a successor ordinal, then for some $\delta\in\beta$ we have $\delta^+=\beta$. Thus for all $\gamma\lt\delta$, $\alpha\gamma\lt\alpha\delta$. Note that $\alpha\beta=G_\alpha(\times_\alpha\upharpoonright\beta)=\bigcup\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta\}$. And $\alpha\delta=\times_\alpha(\delta)=\times_\alpha\upharpoonright\beta(\delta)$. Thus $(\delta,\alpha\delta)\in \times_\alpha\upharpoonright\beta$. And $(\alpha\delta)+\alpha\in\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta\}$. Thus $(\alpha\delta)+\alpha\subseteq\bigcup\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta\}$, implying $(\alpha\delta)+\alpha\le\alpha\beta$. If $\gamma\in\beta$, then $\gamma\le\delta$, thus $\alpha\gamma\le\alpha\delta\lt(\alpha\delta)+\alpha\le\alpha\beta$.
• If $\beta$ is a limit ordinal, then $\alpha\beta=G_\alpha(\times_\alpha\upharpoonright\beta)=\bigcup\{p_1:p\in \times_\alpha\upharpoonright\beta\}$. Let $\gamma\in\beta$, then there exists $\delta\in\beta$ such that $\gamma\in\delta$, thus $\alpha\gamma\lt\alpha\delta$. Note that $\alpha\delta=\times_\alpha\upharpoonright\beta(\delta)$, implying $(\delta,\alpha\delta)\in \times_\alpha\upharpoonright\beta$. Thus $\alpha\delta\in\{p_1:p\in \times_\alpha\upharpoonright\beta\}$, implying $\alpha\delta\subseteq\alpha\beta$ and hence $\alpha\delta\le\alpha\beta$.

In all cases, we have that for all $\gamma\lt\beta$, $\alpha\gamma\lt\alpha\beta$. By transfinite induction, for all $\beta\in\text{Ord}$, for all $\gamma\lt\beta$, $\alpha\gamma\lt\alpha\beta$. $\blacksquare$

Proof. $\alpha0=G_\alpha(\times_\alpha\upharpoonright0)=0$.

Note that $\alpha\beta^+=G_\alpha(\times_\alpha\upharpoonright\beta^+)=\bigcup\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta^+\}$. And $\alpha\beta=\times_\alpha(\beta)=\times_\alpha\upharpoonright\beta^+(\beta)$. Thus $(\beta,\alpha\beta)\in \times_\alpha\upharpoonright\beta^+$. And $\alpha\beta+\alpha\in\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta^+\}$. If $u\in\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta^+\}$, then for some $p\in \times_\alpha\upharpoonright\beta^+$, $u=p_1+\alpha$. And we have $p_0\in\beta^+$, thus $p_0\le\beta$ and $p_1=\times_\alpha\upharpoonright\beta^+(p_0)=\times_\alpha(p_0)=\alpha p_0$. Since $p_0\le\beta$, we have $\alpha p_0\le\alpha\beta$, thus $u=\alpha p_0+\alpha\subseteq\alpha\beta+\alpha$. Hence $\alpha\beta^+=\bigcup\{p_1+\alpha:p\in \times_\alpha\upharpoonright\beta^+\}=\alpha\beta+\alpha$.

Suppose $\beta$ is a limit ordinal. Then $\alpha\beta=G_\alpha(\times_\alpha\upharpoonright\beta)=\bigcup\{p_1:p\in \times_\alpha\upharpoonright\beta\}$. Thus if $\delta\in\alpha\beta$, then for some $u\in\{p_1:p\in \times_\alpha\upharpoonright\beta\}$, $\delta\in u$, and for some $p\in \times_\alpha\upharpoonright\beta$, $u=p_1$, thus $p_0\in\beta$ and $\delta\in p_1$. Note that $p_1=\times_\alpha\upharpoonright\beta(p_0)=\times_\alpha(p_0)=\alpha p_0\in\{\alpha\gamma:\gamma\in\beta\}$. Thus $\delta\in\bigcup_{\gamma\in\beta}(\alpha\gamma)$. For the other direction, if $\delta\in\bigcup_{\gamma\in\beta}(\alpha\gamma)$, then for some $u\in\{\alpha\gamma:\gamma\in\beta\}$, $\delta\in u$, and for some $\gamma\in\beta$, $u=\alpha\gamma$. Note that $\alpha\gamma=\times_\alpha(\gamma)=\times_\alpha\upharpoonright\beta(\gamma)$, thus $(\gamma,u)\in \times_\alpha\upharpoonright\beta$, and hence $u\in\{p_1:p\in \times_\alpha\upharpoonright\beta\}$. Then we have $\delta\in\alpha\beta$. $\blacksquare$

Proof. If $\alpha\neq0$ and $\beta\neq0$, then $\beta\gt0$, thus $\alpha\beta\gt\alpha0=0$, implying $\alpha\beta\neq0$. The converse is trivial. $\blacksquare$

Proof. $$\alpha1=\alpha(0^+)=\alpha0+\alpha=0+\alpha=\alpha$$ $\blacksquare$

Proof. Let $\alpha\in\text{Ord}$ and suppose for all $\beta\in\alpha$, $1\beta=\beta$.

• If $\alpha$ is a zero ordinal, then $$1\alpha=0=\alpha$$
• If $\alpha$ is a successor ordinal, then for some $\beta\in\alpha$, $\beta^+=\alpha$. Thus $$1\alpha=1(\beta^+)=1\beta+1=\beta+1=\beta^+=\alpha$$
• If $\alpha$ is a limit ordinal, then $$1\alpha=\bigcup_{\beta\in\alpha}(1\beta)=\bigcup_{\beta\in\alpha}(\beta)=\alpha$$

By transfinite induction, for all $\alpha\in\text{Ord}$, $1\alpha=\alpha$. $\blacksquare$

Proof. Let $\times$ denote ordinal multiplication and $\times_N$ denote natural number multiplication. Let $m\in N$. For base case, $m\times0=0=m\times_N0$. Suppose $n\in N$ and $m\times n=m\times_Nn$, then $m\times S(n)=(m\times n)+m=(m\times_Nn)+m=m\times_NS(n)$. By induction, for all $n\in N$, $m\times n=m\times_Nn$. Thus for all $m,n\in N$, $m\times n=m\times_Nn$. Let $p\in N\times N$, then for some $m,n\in N$, $(m,n)=p$, thus $m,n\in\text{Ord}$ and $p\in\text{Ord}\times\text{Ord}$. Then $\times(p)=m\times n=m\times_Nn=\times_N(p)$. $\blacksquare$

Proof. Since $\beta$ is a limit ordinal, it is non-empty and has no maximal element. Thus we have some $\gamma\in\beta$, and since $\alpha\neq0$, we have $\alpha\gamma\in\alpha\beta$. Hence $\alpha\beta$ is non-empty. Note that $\alpha\beta=\bigcup_{\gamma\in\beta}(\alpha\gamma)$. If $\delta\in\alpha\beta$, then for some $\gamma\in\beta$, $\delta\in\alpha\gamma\in\alpha\beta$. Thus $\alpha\beta$ has no maximal element. $\blacksquare$

Proof. Let $\alpha,\beta\in\text{Ord}$. If $\alpha=0$, this is trivial. Now suppose $\alpha\neq0$. Suppose $\gamma\in\text{Ord}$ and for all $\delta\in\gamma$, $\alpha(\beta+\delta)=\alpha\beta+\alpha\delta$.

• If $\gamma$ is a zero ordinal, then $$\alpha(\beta+\gamma)=\alpha\beta=\alpha\beta+\alpha\gamma$$
• If $\gamma$ is a successor ordinal, then some $\delta\in\gamma$ has $\delta^+=\gamma$. Thus $$\alpha(\beta+\gamma)=\alpha(\beta+\delta^+)=\alpha(\beta+\delta)^+=\alpha(\beta+\delta)+\alpha =(\alpha\beta+\alpha\delta)+\alpha=\alpha\beta+(\alpha\delta+\alpha)=\alpha\beta+\alpha\delta^+=\alpha\beta+\alpha\gamma$$
• If $\gamma$ is a limit ordinal, then $\beta+\gamma$ and $\alpha\gamma$ are limit ordinals. Thus $$\alpha(\beta+\gamma)=\bigcup_{\delta\in\beta+\gamma}(\alpha\delta)$$ If $u\in\bigcup_{\delta\in\beta+\gamma}(\alpha\delta)$, then for some $\delta\in\beta+\gamma$, $u\in\alpha\delta$.
  • If $\delta\lt\beta$, since $\alpha\neq0$, we have $\alpha\delta\lt\alpha\beta$. Thus $u\in\alpha\beta$. Since $\gamma$ is a limit ordinal, it is non-zero. Thus $\alpha\gamma\neq0$, implying $0\in\alpha\gamma$. Hence $u\in\bigcup_{\zeta\in\alpha\gamma}(\alpha\beta+\zeta)$.
  • If $\delta\ge\beta$, then for some $\epsilon\in\text{Ord}$, $\beta+\epsilon=\delta\lt\beta+\gamma$, thus $\epsilon\lt\gamma$. Since $\alpha\neq0$, we have $\alpha\epsilon\in\alpha\gamma$. Note that $\alpha\delta=\alpha(\beta+\epsilon)=\alpha\beta+\alpha\epsilon$. Thus $u\in\alpha\beta+\alpha\epsilon$. Hence $u\in\bigcup_{\zeta\in\alpha\gamma}(\alpha\beta+\zeta)$.
  For the other direction, let $u\in\bigcup_{\zeta\in\alpha\gamma}(\alpha\beta+\zeta)$. Then for some $\zeta\in\alpha\gamma$, $u\in\alpha\beta+\zeta$. Since $\alpha\gamma=\bigcup_{\epsilon\in\gamma}(\alpha\epsilon)$, for some $\epsilon\in\gamma$, $\zeta\in\alpha\epsilon$. Thus $\alpha\beta+\zeta\lt\alpha\beta+\alpha\epsilon=\alpha(\beta+\epsilon)$. And we have $u\in\alpha(\beta+\epsilon)$ and $\beta+\epsilon\in\beta+\gamma$. Hence $u\in\bigcup_{\delta\in\beta+\gamma}(\alpha\delta)$. We have shown that $$\bigcup_{\delta\in\beta+\gamma}(\alpha\delta)=\bigcup_{\zeta\in\alpha\gamma}(\alpha\beta+\zeta)$$ And since $\alpha\gamma$ is a limit ordinal, $$\bigcup_{\zeta\in\alpha\gamma}(\alpha\beta+\zeta)=\alpha\beta+\alpha\gamma$$

By transfinite induction, for all $\gamma\in\text{Ord}$, $\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma$. $\blacksquare$

Proof. Let $\alpha,\beta\in\text{Ord}$. If $\alpha=0$ or $\beta=0$, this is trivial. Now suppose $\alpha$ and $\beta$ are non-zero. Suppose $\gamma\in\text{Ord}$ and for all $\delta\in\gamma$, $(\alpha\beta)\delta=\alpha(\beta\delta)$.

• If $\gamma$ is a zero ordinal, then $$(\alpha\beta)\gamma=0=\alpha(\beta\gamma)$$
• If $\gamma$ is a successor ordinal, then some $\delta\in\gamma$ has $\delta^+=\gamma$. Thus $$(\alpha\beta)\gamma=(\alpha\beta)\delta^+=(\alpha\beta)\delta+\alpha\beta=\alpha(\beta\delta)+\alpha\beta =\alpha(\beta\delta+\beta)=\alpha(\beta\delta^+)=\alpha(\beta\gamma)$$
• If $\gamma$ is a limit ordinal, then $\beta\gamma$ is a limit ordinal. We have $$(\alpha\beta)\gamma=\bigcup_{\delta\in\gamma}((\alpha\beta)\delta)$$ If $u\in\bigcup_{\delta\in\gamma}((\alpha\beta)\delta)$, then for some $\delta\in\gamma$, $u\in(\alpha\beta)\delta$. Note that $(\alpha\beta)\delta=\alpha(\beta\delta)$. Thus $u\in\alpha(\beta\delta)$. Since $\beta\neq0$, $\beta\delta\in\beta\gamma$. Thus $u\in\bigcup_{\epsilon\in\beta\gamma}(\alpha\epsilon)$. For the other direction, let $u\in\bigcup_{\epsilon\in\beta\gamma}(\alpha\epsilon)$, then for some $\epsilon\in\beta\gamma$, $u\in\alpha\epsilon$. Note that $\beta\gamma=\bigcup_{\delta\in\gamma}(\beta\delta)$. Thus for some $\delta\in\gamma$, $\epsilon\in\beta\delta$. Since $\alpha\neq0$, we have $\alpha\epsilon\lt\alpha(\beta\delta)=(\alpha\beta)\delta$, implying $u\in(\alpha\beta)\delta$. Hence $u\in\bigcup_{\delta\in\gamma}((\alpha\beta)\delta)$. We have shown that $$\bigcup_{\delta\in\gamma}((\alpha\beta)\delta)=\bigcup_{\epsilon\in\beta\gamma}(\alpha\epsilon)$$ Since $\beta\gamma$ is a limit ordinal, $$\bigcup_{\epsilon\in\beta\gamma}(\alpha\epsilon)=\alpha(\beta\gamma)$$

By transfinite induction, for all $\gamma\in\text{Ord}$, $(\alpha\beta)\gamma=\alpha(\beta\gamma)$. $\blacksquare$

Proof. Let $\alpha,\beta\in\text{Ord}$ and $\beta\lt\alpha$. Suppose $\gamma\in\text{Ord}$ and for all $\delta\in\gamma$, $\beta\delta\le\alpha\delta$.

• If $\gamma$ is a zero ordinal, then $$\beta\gamma=0=\alpha\gamma$$
• If $\gamma$ is a successor ordinal, then for some $\delta\in\gamma$, $\delta^+=\gamma$. Thus $$\beta\gamma=\beta\delta^+=\beta\delta+\beta\lt\beta\delta+\alpha\le\alpha\delta+\alpha=\alpha\delta^+=\alpha\gamma$$
• If $\gamma$ is a limit ordinal, then $$\beta\gamma=\bigcup_{\delta\in\gamma}(\beta\delta)\le\bigcup_{\delta\in\gamma}(\alpha\delta)=\alpha\gamma$$

By transfinite induction, for all $\gamma\in\text{Ord}$, $\beta\gamma\le\alpha\gamma$. $\blacksquare$

Proof. By transfinite recursion, the class $\wedge_\alpha$ defined by $G_\alpha$ is a class function from $\text{Ord}$ to $V$ such that for all $\beta\in\text{Ord}$, $\wedge_\alpha(\beta)=G_\alpha(\wedge_\alpha\upharpoonright\beta)$. Let $\beta\in\text{Ord}$ and suppose that for all $\gamma\in\beta$, $\wedge_\alpha(\gamma)\in\text{Ord}$.

• If $\beta$ is a zero ordinal, then $\wedge_\alpha(\beta)=G_\alpha(\wedge_\alpha\upharpoonright\emptyset)=1\in\text{Ord}$.
• If $\beta$ is a successor ordinal, then $\wedge_\alpha(\beta)=G_\alpha(\wedge_\alpha\upharpoonright\beta)=\bigcup\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta\}$. Note that for all $p\in \wedge_\alpha\upharpoonright\beta$, there exist $\gamma,\delta$ such that $\gamma\in\beta$ and $p=(\gamma,\delta)$. Thus $p_1=\delta=\wedge_\alpha\upharpoonright\beta(\gamma)=\wedge_\alpha(\gamma)\in\text{Ord}$, and hence $p_1\alpha\in\text{Ord}$. Then for all $\delta\in\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta\}$, $\delta\in\text{Ord}$, and thus $\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta\}\subset\text{Ord}$. Hence $\bigcup\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta\}\in\text{Ord}$. Therefore, $\wedge_\alpha(\beta)\in\text{Ord}$.
• If $\beta$ is a limit ordinal, then $\wedge_\alpha(\beta)=G_\alpha(\wedge_\alpha\upharpoonright\beta)=\bigcup\{p_1:p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}\}$. By the same reasoning as above, for all $p\in \wedge_\alpha\upharpoonright\beta$, $p_1\in\text{Ord}$. Thus for all $p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}$, $p_1\in\text{Ord}$. Then for all $\delta\in\{p_1:p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}\}$, $\delta\in\text{Ord}$, and thus $\{p_1:p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}\}\subset\text{Ord}$. Hence $\bigcup\{p_1:p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}\}\in\text{Ord}$. Therefore, $\wedge_\alpha(\beta)\in\text{Ord}$.

In all cases, $\wedge_\alpha(\beta)\in\text{Ord}$. Thus by transfinite induction, for all $\beta\in\text{Ord}$, $\wedge_\alpha(\beta)\in\text{Ord}$. Hence $\wedge_\alpha$ is a class function from $\text{Ord}$ to $\text{Ord}$. $\blacksquare$

Proof. Let $\alpha\in\text{Ord}$ and suppose for all $\beta\in\alpha$, if $\beta\neq0$, then $0^\beta=0$.

• If $\alpha$ is a zero ordinal, then $\alpha$ is not non-zero.
• If $\alpha$ is a successor ordinal, then $0^\alpha=G_0(\wedge_0\upharpoonright\alpha)=\bigcup\{p_10:p\in\wedge_0\upharpoonright\alpha\}=\bigcup\{0:p\in\wedge_0\upharpoonright\alpha\}=0$.
• If $\alpha$ is a limit ordinal, then $0^\alpha=G_0(\wedge_0\upharpoonright\alpha)=\bigcup\{p_1:p\in(\wedge_0\upharpoonright\alpha)\setminus\{(0,1)\}\}$. Let $p\in(\wedge_0\upharpoonright\alpha)\setminus\{(0,1)\}$, then $p_0\in\alpha$. If $p_0=0$, then $p_1=\wedge_0\upharpoonright\alpha(p_0)=\wedge_0(p_0)=1$, implying $(0,1)=p\in(\wedge_0\upharpoonright\alpha)\setminus\{(0,1)\}$, a contradiction. Thus $p_0\neq0$, and we have $p_1=0^{p_0}=0$. Hence $\bigcup\{p_1:p\in(\wedge_0\upharpoonright\alpha)\setminus\{(0,1)\}\}=0$.

In all cases, if $\alpha$ is non-zero, then $0^\alpha=0$. By transfinite induction, for all $\alpha\in\text{Ord}$, if $\alpha$ is non-zero, then $0^\alpha=0$. $\blacksquare$

Proof. Let $\alpha\in\text{Ord}$ and suppose for all $\beta\in\alpha$, $1^\beta=1$.

• If $\alpha$ is a zero ordinal, then $1^\alpha=G_1(\wedge_1\upharpoonright0)=1$.
• If $\alpha$ is a successor ordinal, then $1^\alpha=G_1(\wedge_1\upharpoonright\alpha)=\bigcup\{p_11:p\in\wedge_1\upharpoonright\alpha\}=\bigcup\{p_1:p\in\wedge_1\upharpoonright\alpha\}$. If $p\in\wedge_1\upharpoonright\alpha$, then $p_0\in\alpha$, thus $p_1=\wedge_1\upharpoonright\alpha(p_0)=\wedge_1(p_0)=1$. Hence for all $u\in\{p_1:p\in\wedge_1\upharpoonright\alpha\}$, $u=1$. Since $\alpha$ is non-empty, $\wedge_1\upharpoonright\alpha$ is non-empty, thus $\{p_1:p\in\wedge_1\upharpoonright\alpha\}=\{1\}$. And we have $\bigcup\{p_1:p\in\wedge_1\upharpoonright\alpha\}=1$.
• If $\alpha$ is a limit ordinal, then $1^\alpha=G_1(\wedge_1\upharpoonright\alpha)=\bigcup\{p_1:p\in(\wedge_1\upharpoonright\alpha)\setminus\{(0,1)\}\}$. Let $p\in(\wedge_1\upharpoonright\alpha)\setminus\{(0,1)\}$, then $p_0\in\alpha$. Thus $p_1=\wedge_1\upharpoonright\alpha(p_0)=\wedge_1(p_0)=1$. Hence for all $u\in\{p_1:p\in(\wedge_1\upharpoonright\alpha)\setminus\{(0,1)\}\}$, $u=1$. Since $\alpha$ is a limit ordinal, $(\wedge_1\upharpoonright\alpha)\setminus\{(0,1)\}$ is non-empty, thus $\{p_1:p\in(\wedge_1\upharpoonright\alpha)\setminus\{(0,1)\}\}=\{1\}$. And we have $\bigcup\{p_1:p\in(\wedge_1\upharpoonright\alpha)\setminus\{(0,1)\}\}=1$.

In all cases, $1^\alpha=1$. By transfinite induction, for all $\alpha\in\text{Ord}$, $1^\alpha=1$. $\blacksquare$

Proof. Let $\alpha\in\text{Ord}$ be greater than $1$. Suppose $\beta\in\text{Ord}$ and for all $\delta\in\beta$, for all $\gamma\lt\delta$, $\alpha^\gamma\lt\alpha^\delta$.

• If $\beta$ is a zero ordinal, then for all $\gamma\lt\beta$, $\alpha^\gamma\lt\alpha^\beta$.
• If $\beta$ is a successor ordinal, then for some $\delta\in\beta$ we have $\delta^+=\beta$. Thus for all $\gamma\lt\delta$, $\alpha^\gamma\lt\alpha^\delta$. Note that $\alpha^\beta=G_\alpha(\wedge_\alpha\upharpoonright\beta)=\bigcup\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta\}$. And $\alpha^\delta=\wedge_\alpha(\delta)=\wedge_\alpha\upharpoonright\beta(\delta)$. Thus $(\delta,\alpha^\delta)\in \wedge_\alpha\upharpoonright\beta$. And $\alpha^\delta\alpha\in\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta\}$. Thus $\alpha^\delta\alpha\subseteq\bigcup\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta\}$, implying $\alpha^\delta\alpha\le\alpha^\beta$. Note that $\alpha^\delta\ge\alpha^0=1$, thus $\alpha^\delta$ is non-zero. If $\gamma\in\beta$, then $\gamma\le\delta$, thus $\alpha^\gamma\le\alpha^\delta=\alpha^\delta1\lt\alpha^\delta\alpha\le\alpha^\beta$.
• If $\beta$ is a limit ordinal, then $\alpha^\beta=G_\alpha(\wedge_\alpha\upharpoonright\beta)=\bigcup\{p_1:p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}\}$. Let $\gamma\in\beta$, then there exists $\delta\in\beta$ such that $\gamma\in\delta$, thus $\alpha^\gamma\lt\alpha^\delta$. Note that $\alpha^\delta=\wedge_\alpha\upharpoonright\beta(\delta)$, implying $(\delta,\alpha^\delta)\in \wedge_\alpha\upharpoonright\beta$. Since $\gamma\in\delta$, $\delta\neq0$, thus $(\delta,\alpha^\delta)\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}$. Hence $\alpha^\delta\in\{p_1:p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}\}$, implying $\alpha^\delta\subseteq\alpha^\beta$ and hence $\alpha^\delta\le\alpha^\beta$.

In all cases, we have that for all $\gamma\lt\beta$, $\alpha^\gamma\lt\alpha^\beta$. By transfinite induction, for all $\beta\in\text{Ord}$, for all $\gamma\lt\beta$, $\alpha^\gamma\lt\alpha^\beta$. $\blacksquare$

Proof. $\alpha^0=G_\alpha(\wedge_\alpha\upharpoonright0)=1$.

If $\alpha=0$, then trivially, $\alpha^{\beta^+}=0=\alpha^\beta\alpha$. Now suppose $\alpha\neq0$. Note that $\alpha^{\beta^+}=G_\alpha(\wedge_\alpha\upharpoonright\beta^+)=\bigcup\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta^+\}$. And $\alpha^\beta=\wedge_\alpha(\beta)=\wedge_\alpha\upharpoonright\beta^+(\beta)$. Thus $(\beta,\alpha^\beta)\in \wedge_\alpha\upharpoonright\beta^+$. And $\alpha^\beta\alpha\in\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta^+\}$. If $u\in\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta^+\}$, then for some $p\in \wedge_\alpha\upharpoonright\beta^+$, $u=p_1\alpha$. And we have $p_0\in\beta^+$, thus $p_0\le\beta$ and $p_1=\wedge_\alpha\upharpoonright\beta^+(p_0)=\wedge_\alpha(p_0)=\alpha^{p_0}$. Since $p_0\le\beta$ and $\alpha\neq0$, we have $\alpha^{p_0}\le\alpha^\beta$, thus $u=\alpha^{p_0}\alpha\subseteq\alpha^\beta\alpha$. Hence $\alpha^{\beta^+}=\bigcup\{p_1\alpha:p\in \wedge_\alpha\upharpoonright\beta^+\}=\alpha^\beta\alpha$.

Suppose $\beta$ is a limit ordinal. Then $\alpha^\beta=G_\alpha(\wedge_\alpha\upharpoonright\beta)=\bigcup\{p_1:p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}\}$. Thus if $\delta\in\alpha^\beta$, then for some $p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}$, $\delta\in p_1$. Note that $p_0\in\beta$. If $p_0=0$, then $p_1=\wedge_\alpha\upharpoonright\beta(p_0)=\wedge_\alpha(p_0)=\alpha^p_0=1$, implying $(0,1)=p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}$, a contradiction. Thus $p_0\in\beta\setminus\{0\}$, implying $p_1=\alpha^p_0\in\{\alpha^\gamma:\gamma\in\beta\setminus\{0\}\}$. Thus $\delta\in\bigcup_{\gamma\in\beta\setminus\{0\}}(\alpha^\gamma)$. For the other direction, if $\delta\in\bigcup_{\gamma\in\beta\setminus\{0\}}(\alpha^\gamma)$, then for some $\gamma\in\beta\setminus\{0\}$, $\delta\in\alpha^\gamma$. Note that $\alpha^\gamma=\wedge_\alpha(\gamma)=\wedge_\alpha\upharpoonright\beta(\gamma)$, thus $(\gamma,\alpha^\gamma)\in \wedge_\alpha\upharpoonright\beta$. Since $\gamma\neq0$, $(\gamma,\alpha^\gamma)\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}$. and hence $\alpha^\gamma\in\{p_1:p\in(\wedge_\alpha\upharpoonright\beta)\setminus\{(0,1)\}\}$. Then we have $\delta\in\alpha^\beta$. $\blacksquare$

Proof. $$\alpha^1=\alpha^{0^+}=\alpha^0\alpha=1\alpha=\alpha$$ $\blacksquare$

Proof. Let $\wedge$ denote ordinal exponentiation and $\wedge_N$ denote natural number exponentiation. Let $m\in N$. For base case, $m\wedge0=1=m\wedge_N0$. Suppose $n\in N$ and $m\wedge n=m\wedge_Nn$, then $m\wedge S(n)=(m\wedge n)m=(m\wedge_Nn)m=m\wedge_NS(n)$. By induction, for all $n\in N$, $m\wedge n=m\wedge_Nn$. Thus for all $m,n\in N$, $m\wedge n=m\wedge_Nn$. Let $p\in N\times N$, then for some $m,n\in N$, $(m,n)=p$, thus $m,n\in\text{Ord}$ and $p\in\text{Ord}\times\text{Ord}$. Then $\wedge(p)=m\wedge n=m\wedge_Nn=\wedge_N(p)$. $\blacksquare$

Proof. Since $\beta$ is a limit ordinal, it is non-empty and has no maximal element. Thus we have some $\gamma\in\beta$, and since $\alpha\gt1$, we have $\alpha^\gamma\in\alpha^\beta$. Hence $\alpha^\beta$ is non-empty. Note that $\alpha^\beta=\bigcup_{\gamma\in\beta\setminus\{0\}}(\alpha^\gamma)$. If $\delta\in\alpha^\beta$, then for some $\gamma\in\beta\setminus\{0\}$, $\delta\in\alpha^\gamma\in\alpha^\beta$. Thus $\alpha^\beta$ has no maximal element. $\blacksquare$

Proof. Let $\alpha,\beta,\gamma\in\text{Ord}$. Suppose $\alpha=0$.

• If $\beta=0$ and $\gamma=0$, then both sides are $1$.
• If $\beta\neq0$ or $\gamma\neq0$, then $\beta+\gamma\neq0$, thus both sides are $0$.

If $\alpha=1$, then both sides are $1$. Now suppose $\alpha\gt1$. Suppose for all $\delta\in\gamma$, $\alpha^{\beta+\delta}=\alpha^\beta\alpha^\delta$.

• If $\gamma$ is a zero ordinal, then $$\alpha^{\beta+\gamma}=\alpha^\beta=\alpha^\beta\alpha^\gamma$$
• If $\gamma$ is a successor ordinal, then for some $\delta\in\gamma$, $\delta^+=\gamma$. Thus $$\alpha^{\beta+\gamma}=\alpha^{\beta+\delta^+}=\alpha^{(\beta+\delta)^+} =\alpha^{\beta+\delta}\alpha=\p{\alpha^\beta\alpha^\delta}\alpha=\alpha^\beta\p{\alpha^\delta\alpha} =\alpha^\beta\alpha^{\delta^+}=\alpha^\beta\alpha^\gamma$$
• If $\gamma$ is a limit ordinal, then $\beta+\gamma$ and $\alpha^\gamma$ are limit ordinals. Thus $$\alpha^{\beta+\gamma}=\bigcup_{\delta\in(\beta+\gamma)\setminus\{0\}}(\alpha^\delta)$$ and $$\alpha^\beta\alpha^\gamma=\bigcup_{\zeta\in\alpha^\gamma}(\alpha^\beta\zeta)$$ If $u\in\bigcup_{\delta\in(\beta+\gamma)\setminus\{0\}}(\alpha^\delta)$, then for some $\delta\in(\beta+\gamma)\setminus\{0\}$, $u\in\alpha^\delta$. Thus $\delta\in\bigcup_{\epsilon\in\gamma}(\beta+\epsilon)$, implying that for some $\epsilon\in\gamma$, $\delta\in\beta+\epsilon$. Note that $\alpha^\delta\lt\alpha^{\beta+\epsilon}=\alpha^\beta\alpha^\epsilon$, thus $u\in\alpha^\beta\alpha^\epsilon$. Also note that $\alpha^\epsilon\in\alpha^\gamma$, thus $u\in\bigcup_{\zeta\in\alpha^\gamma}(\alpha^\beta\zeta)$. For the other direction, if $u\in\bigcup_{\zeta\in\alpha^\gamma}(\alpha^\beta\zeta)$, then for some $\zeta\in\alpha^\gamma$, $u\in\alpha^\beta\zeta$. Thus $\zeta\in\bigcup_{\epsilon\in\gamma\setminus\{0\}}(\alpha^\epsilon)$, implying for some $\epsilon\in\gamma\setminus\{0\}$, $\zeta\in\alpha^\epsilon$. Note that $\alpha^\beta\ge\alpha^0=1\neq0$. Thus $\alpha^\beta\zeta\in\alpha^\beta\alpha^\epsilon=\alpha^{\beta+\epsilon}$, implying $u\in\alpha^{\beta+\epsilon}$. Since $\epsilon\in\gamma$, $\beta+\epsilon\in\beta+\gamma$. Since $\epsilon\neq0$, $\beta+\epsilon\neq0$. Hence $u\in\bigcup_{\delta\in(\beta+\gamma)\setminus\{0\}}(\alpha^\delta)$. We have shown that $$\bigcup_{\delta\in(\beta+\gamma)\setminus\{0\}}(\alpha^\delta)=\bigcup_{\zeta\in\alpha^\gamma}(\alpha^\beta\zeta)$$

By transfinite induction, for all $\gamma\in\text{Ord}$, $\alpha^{\beta+\gamma}=\alpha^\beta\alpha^\gamma$. $\blacksquare$

Proof. Let $\alpha,\beta,\gamma\in\text{Ord}$. Suppose $\alpha=0$.

• If $\beta=0$ or $\gamma=0$, then both sides are $1$.
• If $\beta\neq0$ and $\gamma\neq0$, then $\beta\gamma\neq0$, thus both sides are $0$.

If $\alpha=1$, then both sides are $1$. Now suppose $\alpha\gt1$. If $\beta=0$, then both sides are $1$. Now suppose $\beta\neq0$. Suppose for all $\delta\in\gamma$, $\alpha^{\beta\delta}=(\alpha^\beta)^\delta$.

• If $\gamma$ is a zero ordinal, then $$\alpha^{\beta\gamma}=1=(\alpha^\beta)^\gamma$$
• If $\gamma$ is a successor ordinal, then for some $\delta\in\gamma$, $\delta^+=\gamma$. Thus $$\alpha^{\beta\gamma}=\alpha^{\beta\delta^+}=\alpha^{\beta\delta+\beta} =\alpha^{\beta\delta}\alpha^\beta=(\alpha^\beta)^\delta\alpha^\beta=(\alpha^\beta)^{\delta^+}=(\alpha^\beta)^\gamma$$
• If $\gamma$ is a limit ordinal, then $\beta\gamma$ is a limit ordinal. Thus $$\alpha^{\beta\gamma}=\bigcup_{\delta\in(\beta\gamma)\setminus\{0\}}(\alpha^\delta)$$ and $$(\alpha^\beta)^\gamma=\bigcup_{\epsilon\in\gamma\setminus\{0\}}(\alpha^\beta)^\epsilon$$ If $u\in\bigcup_{\delta\in(\beta\gamma)\setminus\{0\}}(\alpha^\delta)$, then for some $\delta\in(\beta\gamma)\setminus\{0\}$, $u\in\alpha^\delta$. Thus $\delta\in\bigcup_{\epsilon\in\gamma}(\beta\epsilon)$, implying that for some $\epsilon\in\gamma$, $\delta\in\beta\epsilon$. Then we have $\alpha^\delta\in\alpha^{\beta\epsilon}=(\alpha^\beta)^\epsilon$. Thus $u\in(\alpha^\beta)^\epsilon$. Since $\beta\epsilon$ is non-empty, $\epsilon\neq0$. Hence $u\in\bigcup_{\epsilon\in\gamma\setminus\{0\}}(\alpha^\beta)^\epsilon$. For the other direction, let $u\in\bigcup_{\epsilon\in\gamma\setminus\{0\}}(\alpha^\beta)^\epsilon$. Then for some $\epsilon\in\gamma\setminus\{0\}$, $u\in(\alpha^\beta)^\epsilon$. Note that $(\alpha^\beta)^\epsilon=\alpha^{\beta\epsilon}$, thus $u\in\alpha^{\beta\epsilon}$. Also note that $\beta\epsilon\in\beta\gamma$, and since both $\beta$ and $\epsilon$ are non-zero, we have $\beta\epsilon\in(\beta\gamma)\setminus\{0\}$. Hence $u\in\bigcup_{\delta\in(\beta\gamma)\setminus\{0\}}(\alpha^\delta)$. We have shown that $$\bigcup_{\delta\in(\beta\gamma)\setminus\{0\}}(\alpha^\delta)=\bigcup_{\epsilon\in\gamma\setminus\{0\}}(\alpha^\beta)^\epsilon$$

By transfinite induction, for all $\gamma\in\text{Ord}$, $\alpha^{\beta\gamma}=(\alpha^\beta)^\gamma$. $\blacksquare$

Proof. Let $\alpha,\beta,\gamma\in\text{Ord}$ and $\beta\lt\alpha$. Suppose $\beta=0$, then $\beta^\gamma\le1$ and $\alpha\ge1$.

• If $\alpha=1$, then $\alpha^\gamma=1\ge\beta^\gamma$.
• If $\alpha\gt1$, then $\alpha^\gamma\ge\alpha^0=1\ge\beta^\gamma$.

Now suppose $\beta\neq0$ and for all $\delta\in\gamma$, $\beta^\delta\le\alpha^\delta$.

• If $\gamma$ is a zero ordinal, then $$\beta^\gamma=1=\alpha^\gamma$$
• If $\gamma$ is a successor ordinal, then for some $\delta\in\gamma$, $\delta^+=\gamma$. Note that $\beta^\delta\neq0$. Thus $$\beta^\gamma=\beta^{\delta^+}=\beta^\delta\beta\lt\beta^\delta\alpha\le\alpha^\delta\alpha=\alpha^{\delta^+}=\alpha^\gamma$$
• If $\gamma$ is a limit ordinal, then $$\beta^\gamma=\bigcup_{\delta\in\gamma\setminus\{0\}}(\beta^\delta)\le\bigcup_{\delta\in\gamma\setminus\{0\}}(\alpha^\delta)=\alpha\gamma$$

By transfinite induction, for all $\gamma\in\text{Ord}$, $\beta^\gamma\le\alpha^\gamma$. $\blacksquare$

Proof. We already showed that every natural number is in $\text{Ord}$. Recall that in the set theory section, we showed that a finite set is bijective to a unique natural number. Thus let $n\in N$, for all $m\in n$, there is no bijection between $n$ and $m$. Hence every natural number is a cardinal. $\blacksquare$

Proof. Recall that in the set theory section, we showed that a finite set cannot be bijective to $\omega$. Thus for all $n\in\omega$, there is no bijection between $\omega$ and $n$. Hence $\omega$ is a cardinal. $\blacksquare$

Proof. Suppose there exists a successor cardinal $\kappa$ such that $\kappa\ge\omega$. Then for some $\alpha\in\kappa$, $\alpha^+=\kappa$. Note that the function

• $\beta\mapsto\beta^+$ for all $\beta\lt\omega$;
• $\beta\mapsto\beta$ for all $\omega\le\beta\lt\alpha$;
• $\beta\mapsto0$ for all $\alpha\le\beta\lt\kappa$

is a bijection from $\kappa$ to $\alpha$, a contradiction. Thus every successor cardinal is in $\omega$, and is hence a non-zero natural number. The other direction is trivial. $\blacksquare$

Proof. Let $X$ be a set, then it is strictly well-orderable by some $\lt$. Let $x\in X$ and suppose for all $t\lt x$, $\{u\in X|u\lt t\}$ is order-isomorphic to a unique $\alpha_t\in\text{Ord}$, with $\alpha_t=\{\alpha_u:u\lt t\}$. Then $t\to\alpha_t$ defines a function $F$ from $\{t\in X|t\lt x\}$, with a set-sized range $\{\alpha_t:t\lt x\}$. Given $a,b\in\{t\in X|t\lt x\}$, if $a\neq b$, then $a\lt b$ or $b\lt a$, so $\alpha_a\in\alpha_b$ or $\alpha_b\in\alpha_a$, implying $F(a)\neq F(b)$. Hence $F$ is a bijection from $\{t\in X|t\lt x\}$ to $\{\alpha_t:t\lt x\}$. Clearly, $F$ is also an order-isomorphism. Since $\{\alpha_t:t\lt x\}\subset\text{Ord}$, $\{\alpha_t:t\lt x\}$ is linearly ordered by $\in$. If $\gamma\in\{\alpha_t:t\lt x\}$, then there exists $t\lt x$ such that $\gamma=\alpha_t=\{\alpha_u:u\lt t\}\subseteq\{\alpha_t:t\lt x\}$. Hence $\{\alpha_t:t\lt x\}$ is an ordinal. Since $\{\alpha_t:t\lt x\}\subset\text{Ord}$, $\{\alpha_t:t\lt x\}\in\text{Ord}$. Suppose $\{t\in X|t\lt x\}$ is order-isomorphic to both $\alpha_1,\alpha_2\in\text{Ord}$, then $\alpha_1\cong\alpha_2$, implying $\alpha_1=\alpha_2$. Thus $\{t\in X|t\lt x\}$ is order-isomorphic to a unique $\alpha_x\in\text{Ord}$, with $\alpha_x=\{\alpha_t:t\lt x\}$. Using the same logic as transfinite induction, for all $x\in X$, $\{t\in X|t\lt x\}$ is order-isomorphic to a unique $\alpha_x\in\text{Ord}$, with $\alpha_x=\{\alpha_t:t\lt x\}$. And by the same reasoning as above, $X$ is order-isomorphic to some $\alpha\in\text{Ord}$. Let $\kappa$ be the strictly least element of $\text{Ord}$ bijective to $\alpha$, then $X$ is bijective to $\kappa$, which is a cardinal. Suppose $X$ is bijective to $\kappa_1,\kappa_2\in\text{Car}$, then $\kappa_1 $ and $\kappa_2$ are bijective and $\kappa_1,\kappa_2\in\text{Ord}$. If $\kappa_1\neq\kappa_2$, then $\kappa_1\in\kappa_2$ or $\kappa_2\in\kappa_1$, implying they cannot both be cardinals, a contradiction. Hence $\kappa_1=\kappa_2$. $\blacksquare$

Proof. Suppose $f:X\to Y$ is injective. Let $g$ be obtained from $f$ by swapping all ordered pairs, then $g$ is a function from $f[X]$ to $X$. Since $X$ is non-empty, let $\lt$ strictly orders $X$, then $X$ has a strictly least element $x^*$. Define $h:Y\to X$ such that $h(y)=g(y)$ if $y\in f[X]$ and $h(y)=x^*$ otherwise, then $h$ is a surjection from $Y$ to $X$.

Suppose $f:X\to Y$ is surjective. Let $y\in Y$, then $\{x\in X|f(x)=y\}$ is non-empty. Let $\lt$ strictly orders $X$, then for each $y\in Y$, there exists a unique $x_y\in X$ such that $x_y$ is the strictly least element of $\{x\in X|f(x)=y\}$. Then $y\to x_y$ defines a function from $Y$ to $X$, which is injective. $\blacksquare$

Proof. We use $X\sim Y$ to denote that there exists a bijection from $X$ to $Y$.

If $X=\emptyset$ or $Y=\emptyset$, then both $X$ and $Y$ are empty and thus $X\sim Y$. Now suppose $X$ and $Y$ are both non-empty, then there exists an injection $f$ from $X$ to $Y$ and an injection $g$ from $Y$ to $X$.

Use recursion to define $X_0,Y_0=X,Y$ and $X_{n+1},Y_{n+1}=g[Y_n],f[X_n]$. Since $f$ and $g$ are injective, $X_{n+1}\sim Y_n$ and $Y_{n+1}\sim X_n$. By induction, we have $X_{n+1}\subseteq X_n$ and $Y_{n+1}\subseteq Y_n$. Let $X^*_n=X_n\setminus X_{n+1}$ and $Y^*_n=Y_n\setminus Y_{n+1}$, then $f[X^*_n]=Y^*_{n+1}$ and $g[Y^*_n]=X^*_{n+1}$, thus $X^*_n\sim Y^*_{n+1}$ and $Y^*_n\sim X^*_{n+1}$. Note that, since both $(X^*_n)$ and $(Y^*_n)$ are pairwise disjoint, $X^*_{2k}\cup X^*_{2k+1}\sim Y^*_{2k+1}\cup Y^*_{2k}$, and hence $\cup X^*_n\sim\cup Y^*_n$. Note that $\cup X^*_n$ and $\cap X_n$ are disjoint and $(\cup X^*_n)\cup(\cap X_n)=X$; $\cup Y^*_n$ and $\cap Y_n$ are disjoint and $(\cup Y^*_n)\cup(\cap Y_n)=Y$. And we have $f[\cap X_n]=\cap Y_n$ and $g[\cap Y_n]=\cap X_n$, and hence $\cap X_n\sim\cap Y_n$. Therefore, we have $(\cup X^*_n)\cup(\cap X_n)\sim(\cup Y^*_n)\cup(\cap Y_n)$, implying $X\sim Y$. $\blacksquare$

Proof. Suppose there is an injection from $X$ to $Y$, then there is an injection from $\abs{X}$ to $\abs{Y}$. Suppose for contradiction that $\abs{Y}\in\abs{X}$, then there exists an injection from $\abs{Y}$ to $\abs{X}$. If $X=\emptyset$, then $\abs{Y}\in\abs{X}=\emptyset$, a contradiction. If $X\neq\emptyset$, then $Y\neq\emptyset$, hence there exists a surjection from $\abs{X}$ to $\abs{Y}$. Then there exists a bijection from $\abs{X}$ to $\abs{Y}$. But then $\abs{X}\notin\text{Car}$, a contradiction. Therefore $\abs{X}\le\abs{Y}$.

Suppose there is a surjection from $X$ to $Y$. If $X=\emptyset$, then $Y=\emptyset$ and we immediately have $\abs{X}\ge\abs{Y}$. If $X\neq\emptyset$, then $Y\neq\emptyset$, hence there is an injection from $Y$ to $X$. Then we have $\abs{Y}\le\abs{X}$, or $\abs{X}\ge\abs{Y}$.

If there exists a bijection from $X$ to $Y$, then $\abs{X}\le\abs{Y}$ and $\abs{Y}\le\abs{X}$, hence $\abs{X}=\abs{Y}$. $\blacksquare$

Proof. Since $X$ is a set of cardinals, $X\subset\text{Ord}$, thus $\bigcup X$ is the least upper bound of $X$ in $\text{Ord}$. Suppose for contradiction that for some $\alpha\in\bigcup X$, $\alpha$ and $\bigcup X$ are bijective. Note that $\alpha$ is not an upper bound of $X$, implying for some $\kappa\in X$, $\alpha\lt\kappa$. Then $\abs{\alpha}=\abs{\bigcup X}\ge\abs{\kappa}\ge\abs{\alpha}$, implying $\abs{\alpha}=\abs{\kappa}$, further implying that $\kappa$ is not a cardinal, a contradiction. Hence $\bigcup X$ is a cardinal. $\blacksquare$

Proof. The map $f(x)=\{x\}$ is an injection from $X$ to $\mathcal P(X)$, hence $\abs{X}\le\abs{\mathcal P(X)}$. Suppose we have a bijection $f:X\to\mathcal P(X)$. Let $S$ denote $\{x\in X|x\notin f(x)\}$ and $a$ denote $f^{-1}(S)$. If $a\in S$, then $a\notin f(a)=S$, a contradiction. If $a\notin S$, then $a\in f(a)=S$, also a contradiction. Hence $\abs{X}\lt\abs{\mathcal P(X)}$. $\blacksquare$

Proof. Let $k\in N$, then there exists a unique $a_k\in N$ such that $\sum_{j=1}^{a_k}j\le k\lt\sum_{j=1}^{a_k+1}j$. Define $b_k$ to be $k-\sum_{j=1}^{a_k}j$, then $0\le b_k\le a_k$. The map $k\mapsto(b_k,a_k-b_k)$ from $N$ to $N\times N$ is a bijection. $\blacksquare$

Proof. Since $\omega$ is a cardinal, $\omega=\abs{\omega}=\abs{N}$.

Counting $a\in Z$ by order of absolute value $\abs{a}$ forms a bijection between $N$ and $Z$, hence $\abs{N}=\abs{Z}$.

Clearly, $\abs{Z\setminus\{0\}}=\abs{Z}$. Note that $f((a,b))=a/b$ is a surjection from $Z\times(Z\setminus\{0\})$ to $Q$. We have $\abs{N}=\abs{N\times N}=\abs{Z\times(Z\setminus\{0\})}\ge\abs{Q}$. Clearly, there exists an injection from $N$ to $Q$, hence $\abs{N}=\abs{Q}$. $\blacksquare$

Proof. We will assume $\abs{\alpha}=\omega$. The case when $\alpha$ is finite trivially follows from this case.

Note that there exists a bijection $f:\omega\to\alpha$. Define $g:\omega\to\alpha^+$ such that $g(0)=\alpha$ and $g(n)=f(n-1)$ if $n\neq0$, then $g$ is a bijection. Thus $\abs{\alpha^+}=\omega$.

Let $\gamma$ be an ordinal and suppose for all $\delta\in\gamma$, if $\delta$ is countable, then $\abs{\alpha+\delta}=\omega$. If $\gamma$ is countable, then every $\delta\in\gamma$ is countable.

• If $\gamma$ is a zero ordinal, then $\abs{\alpha+\gamma}=\abs{\alpha}=\omega$.
• If $\gamma$ is a successor ordinal, then for some $\delta\in\gamma$, $\delta^+=\gamma$, and we have $\abs{\alpha+\gamma}=\abs{(\alpha+\delta)^+}=\omega$.
• If $\gamma$ is a limit ordinal, since $\abs{\alpha+\delta}=\omega$ for all $\delta\in\gamma$, there exists a bijection $f_\delta:\omega\to\alpha+\delta$ for each $\delta\in\gamma$. Since $\gamma$ is countable, there exists a bijection $g:\omega\to\gamma$. Define $h:\omega\times\omega\to\alpha+\gamma$ by $h(m,n)=f_{g(m)}(n)$, then $h$ is a surjection. Thus $\abs{\alpha+\gamma}\le\abs{\omega\times\omega}=\omega$. Also, $\abs{\alpha+\gamma}\ge\abs{\alpha}=\omega$. Thus $\abs{\alpha+\gamma}=\omega$.

Hence if $\gamma$ is countable, then $\abs{\alpha+\gamma}=\omega$. By transfinite induction, for all countable ordinal $\gamma$, $\abs{\alpha+\gamma}=\omega$. Hence $\abs{\alpha+\beta}=\omega$.

Let $\gamma$ be an ordinal and suppose for all $\delta\in\gamma$, if $\delta$ is non-zero countable, then $\abs{\alpha\delta}=\omega$. If $\gamma$ is non-zero countable, then every $\delta\in\gamma$ is either zero or non-zero countable.

• If $\gamma$ is a zero ordinal, we have a contradiction.
• If $\gamma$ is a successor ordinal, then for some $\delta\in\gamma$, $\delta^+=\gamma$, and we have $\abs{\alpha\gamma}=\abs{\alpha\delta+\alpha}=\omega$.
• If $\gamma$ is a limit ordinal, since $\abs{\alpha\delta}=\omega$ for all non-zero $\delta\in\gamma$, there exists a bijection $f_\delta:\omega\to\alpha\delta$ for each non-zero $\delta\in\gamma$. Also define $f_0:\omega\to1$ by $f_0(n)=0$ for all $n\in\omega$. Since $\gamma$ is countable, there exists a bijection $g:\omega\to\gamma$. Define $h:\omega\times\omega\to\alpha\gamma$ by $h(m,n)=f_{g(m)}(n)$, then $h$ is a surjection. Thus $\abs{\alpha\gamma}\le\abs{\omega\times\omega}=\omega$. Also, $\abs{\alpha\gamma}\ge\abs{\alpha}=\omega$. Thus $\abs{\alpha\gamma}=\omega$.

Hence if $\gamma$ is non-zero countable, then $\abs{\alpha\gamma}=\omega$. By transfinite induction, for all non-zero countable ordinal $\gamma$, $\abs{\alpha\gamma}=\omega$. Hence $\abs{\alpha\beta}=\omega$ if $\beta\neq0$.

Let $\gamma$ be an ordinal and suppose for all $\delta\in\gamma$, if $\delta$ is non-zero countable, then $\abs{\alpha^\delta}=\omega$. If $\gamma$ is non-zero countable, then every $\delta\in\gamma$ is either zero or non-zero countable.

• If $\gamma$ is a zero ordinal, we have a contradiction.
• If $\gamma$ is a successor ordinal, then for some $\delta\in\gamma$, $\delta^+=\gamma$, and we have $\abs{\alpha^\gamma}=\abs{\alpha^\delta\alpha}=\omega$.
• If $\gamma$ is a limit ordinal, since $\abs{\alpha^\delta}=\omega$ for all non-zero $\delta\in\gamma$, there exists a bijection $f_\delta:\omega\to\alpha^\delta$ for each non-zero $\delta\in\gamma$. Also define $f_0:\omega\to1$ by $f_0(n)=0$ for all $n\in\omega$. Since $\gamma$ is countable, there exists a bijection $g:\omega\to\gamma$. Define $h:\omega\times\omega\to\alpha^\gamma$ by $h(m,n)=f_{g(m)}(n)$, then $h$ is a surjection. Thus $\abs{\alpha^\gamma}\le\abs{\omega\times\omega}=\omega$. Also, $\abs{\alpha^\gamma}\ge\abs{\alpha}=\omega$. Thus $\abs{\alpha^\gamma}=\omega$.

Hence if $\gamma$ is non-zero countable, then $\abs{\alpha^\gamma}=\omega$. By transfinite induction, for all non-zero countable ordinal $\gamma$, $\abs{\alpha^\gamma}=\omega$. Hence $\abs{\alpha^\beta}=\omega$ if $\beta\neq0$.

$\blacksquare$

Proof. The function

• $\alpha\mapsto(0,\alpha)$ for all $\alpha\lt\kappa$

is a bijection from $\kappa$ to $\kappa\bigsqcup0$. Thus $\kappa+0=\abs{\kappa\bigsqcup0}=\kappa$.

$\kappa0=\abs{\kappa\times0}=\abs{0}=0$.

The function

• $\alpha\mapsto(\alpha,0)$ for all $\alpha\lt\kappa$

is a bijection from $\kappa$ to $\kappa\times1$. Thus $\kappa1=\abs{\kappa\times1}=\kappa$.

$\kappa^0=\abs{\{0\to\kappa\}}=\abs{\{0\}}=1$.

The function

• $\alpha\mapsto\{(0,\alpha)\}$ for all $\alpha\lt\kappa$

is a bijection from $\kappa$ to $\{1\to\kappa\}$. Thus $\kappa^1=\abs{\{1\to\kappa\}}=\kappa$.

If $\kappa\neq0$, then $0^\kappa=\abs{\{\kappa\to0\}}=\abs{0}=0$.

$1^\kappa=\abs{\{\kappa\to1\}}=\abs{1}=1$. $\blacksquare$

Proof. We will use the usual notations to denote natural number arithmetics.

Let $m,n\in N$. Given $p\in m\bigsqcup n$, either $p\in\{0\}\times m$ or $p\in\{1\}\times n$.

• If $p\in\{0\}\times m$, then for some $k\in m$, $p=(0,k)$, thus $p_0m+p_1=k\lt m\le m+n$.
• If $p\in\{1\}\times n$, then for some $k\in n$, $p=(1,k)$, thus $p_0m+p_1=m+k\lt m+n$.

In both cases we have $p_0m+p_1\in m+n$. Thus we can define a function $f:m\bigsqcup n\to m+n$ such that for all $p\in m\bigsqcup n$, $f(p)=p_0m+p_1$. Given $p,q\in m\bigsqcup n$ such that $f(p)=f(q)$, we have $p_0m+p_1=f(p)=f(q)=q_0m+q_1$. If $p\in\{0\}\times m$ and $q\in\{1\}\times n$, then $f(p)=p_1\lt m\le m+q_1=f(q)$, a contradiction. Similarly, $p\in\{1\}\times n$ and $q\in\{0\}\times m$ also leads to a contradiction. Thus $p_0=q_0$, and we have $p_1=q_1$, implying $p=q$. Hence $f$ is injective. Let $k\in m+n$, then either $k\lt m$ or $k\ge m$.

• If $k\lt m$, then $(0,k)\in m\bigsqcup n$ and $f((0,k))=k$.
• If $k\ge m$, then $(1,k-m)\in m\bigsqcup n$ and $f((1,k-m))=k$.

In both cases there exists $p\in m\bigsqcup n$ such that $f(p)=k$. Hence $f$ is surjective. Therefore, $\abs{m\bigsqcup n}=\abs{m+n}=m+n$.

Let $m,n\in N$. Given $p\in m\times n$, $p_0n+p_1\lt p_0n+n=S(p_0)n\le mn$. Thus we can define a function $f:m\times n\to mn$ such that for all $p\in m\times n$, $f(p)=p_0n+p_1$. Given $p,q\in m\times n$ such that $f(p)=f(q)$, we have $p_0n+p_1=f(p)=f(q)=q_0n+q_1$. If $n=0$, then we immediately have a contradiction because $p_1\in n$. Thus $n\neq0$. Suppose $p_0\lt q_0$, then $S(p_0)\le q_0$, and we have $p_0n+(n+q_1)=S(p_0)n+q_1\le q_0n+q_1=p_0n+p_1$, implying $n+q_1=p_1\lt n$, a contradiction. Similarly, $p_0\gt q_0$ also leads to a contradiction, thus $p_0=q_0$, then we have $p_1=q_1$, implying $p=q$. Hence $f$ is injective. Let $k\in mn$, then $n\neq0$. Note that $k=n(k\divsymbol n)+(k\bmod n)$, thus $n(k\divsymbol n)\le n(k\divsymbol n)+(k\bmod n)\lt mn$, implying $k\divsymbol n\lt m$. Thus $(k\divsymbol n,k\bmod n)\in m\times n$, and we have $f((k\divsymbol n,k\bmod n))=k$. Hence $f$ is surjective. Therefore, $\abs{m\times n}=\abs{mn}=mn$.

Let $m,n\in N$. Suppose $g\in\{n\to m\}$. Then $\sum_{j=0}^{0-1}g(j)m^j=0\lt 1=m^0$. Let $k\in n$ and suppose $\sum_{j=0}^{k-1}g(j)m^j\lt m^k$. Then $\sum_{j=0}^{S(k)-1}g(j)m^j=\sum_{j=0}^{k-1}g(j)m^j+g(k)m^k\lt m^k+g(k)m^k=S(g(k))m^k\le mm^k=m^{S(k)}$. By induction, we have $\sum_{j=0}^{n-1}g(j)m^j\lt m^n$. Thus we can define a function $f:\{n\to m\}\to m^n$ such that for all $g\in\{n\to m\}$, $f(g)=\sum_{j=0}^{n-1}g(j)m^j$. Given $g,h\in\{n\to m\}$ such that $f(g)=f(h)$, we have $\sum_{j=0}^{n-1}g(j)m^j=f(g)=f(h)=\sum_{j=0}^{n-1}h(j)m^j$. Let $k\in n$ and suppose for all $l\in k$, $g(l)=h(l)$. Note that $$\p{\sum_{j=0}^{n-1}g(j)m^j}\bmod m^{k+1} =\p{\sum_{j=0}^{k}g(j)m^j+\sum_{j=k+1}^{n-1}g(j)m^j}\bmod m^{k+1} =\p{\p{\sum_{j=0}^{k}g(j)m^j}\bmod m^{k+1}+\p{\sum_{j=k+1}^{n-1}g(j)m^j}\bmod m^{k+1}}\bmod m^{k+1} =\p{\sum_{j=0}^{k}g(j)m^j}\bmod m^{k+1} $$ Similarly, we have $\p{\sum_{j=0}^{n-1}h(j)m^j}\bmod m^{k+1}=\p{\sum_{j=0}^{k}h(j)m^j}\bmod m^{k+1}$. Thus $\p{\sum_{j=0}^{k}g(j)m^j}\bmod m^{k+1}=\p{\sum_{j=0}^{k}h(j)m^j}\bmod m^{k+1}$. As we showed above, we have $\sum_{j=0}^{k}g(j)m^j\lt m^{k+1}$ and $\sum_{j=0}^{k}h(j)m^j\lt m^{k+1}$. Thus $\sum_{j=0}^{k}g(j)m^j=\sum_{j=0}^{k}h(j)m^j$. Since $g(l)=h(l)$ for all $l\in k$, we have $g(k)m^k=h(k)m^k$. If $m^k=0$, then $m=0$ and $k\neq0$, implying $0\in n$. But then $\{n\to m\}=\emptyset$, a contradiction. Thus $m^k\neq0$, and we have $g(k)=h(k)$, implying for all $l\in S(k)$, $g(l)=h(l)$. With the trivial base case, by induction, we have that for all $l\in n$, $g(l)=h(l)$, implying $g=h$. Hence $f$ is injective. Let $k\in m^n$. Let $j\in n$ and suppose for contradiction that $(k\bmod m^{j+1})\divsymbol m^j\ge m$, then $k\bmod m^{j+1}\ge m^j((k\bmod m^{j+1})\divsymbol m^j)\ge m^jm=m^{j+1}$, a contradiction. Thus for all $j\in n$, $(k\bmod m^{j+1})\divsymbol m^j\lt m$. And we can define a function $g:n\to m$ such that for all $j\in n$, $g(j)=(k\bmod m^{j+1})\divsymbol m^j$. Note that $\sum_{j=0}^{0-1}\p{(k\bmod m^{j+1})\divsymbol m^j}m^j=0=k\bmod m^0$. Let $l\in n$ and suppose $\sum_{j=0}^{l-1}\p{(k\bmod m^{j+1})\divsymbol m^j}m^j=k\bmod m^l$. Then $$\sum_{j=0}^{S(l)-1}\p{(k\bmod m^{j+1})\divsymbol m^j}m^j =\sum_{j=0}^{l-1}\p{(k\bmod m^{j+1})\divsymbol m^j}m^j+\p{(k\bmod m^{S(l)})\divsymbol m^{S(l)-1}}m^{S(l)-1} =k\bmod m^l+\p{(k\bmod m^{S(l)})\divsymbol m^l}m^l$$ Note that $$ k =m^{S(l)}(k\divsymbol m^{S(l)})+k\bmod m^{S(l)} =m^{S(l)}(k\divsymbol m^{S(l)})+m^l\p{(k\bmod m^{S(l)})\divsymbol m^l}+(k\bmod m^{S(l)})\bmod m^l =m^l\p{m(k\divsymbol m^{S(l)})+(k\bmod m^{S(l)})\divsymbol m^l}+(k\bmod m^{S(l)})\bmod m^l $$ and $(k\bmod m^{S(l)})\bmod m^l\lt m^l$. Thus $k\bmod m^l=(k\bmod m^{S(l)})\bmod m^l$, and we have $$k\bmod m^l+\p{(k\bmod m^{S(l)})\divsymbol m^l}m^l =(k\bmod m^{S(l)})\bmod m^l+\p{(k\bmod m^{S(l)})\divsymbol m^l}m^l =k\bmod m^{S(l)}$$ By induction, we have $$f(g)=\sum_{j=0}^{n-1}g(j)m^j=\sum_{j=0}^{n-1}\p{(k\bmod m^{j+1})\divsymbol m^j}m^j=k\bmod m^n=k$$ Hence $f$ is surjective. Therefore, $\abs{\{n\to m\}}=\abs{m^n}=m^n$. $\blacksquare$

Proof. The map $p\mapsto(1-p_0,p_1)$ from $\alpha\bigsqcup\beta$ to $\beta\bigsqcup\alpha$ is a bijection. Thus $\alpha+\beta=\beta+\alpha$.

Let $f$ be a bijection from $\alpha\bigsqcup\beta$ to $\alpha+\beta$ and $g$ be a bijection from $\beta\bigsqcup\gamma$ to $\beta+\gamma$. Then for all $p\in(\alpha+\beta)\bigsqcup\gamma$, $p_0\in\{0,1\}$ and when $p_0=0$, $f^{-1}(p_1)\in\alpha\bigsqcup\beta$, implying $f^{-1}(p_1)_0\in\{0,1\}$. Thus the map

• $p\to(0,f^{-1}(p_1)_1)$ if $p_0=0$ and $f^{-1}(p_1)_0=0$;
• $p\to(1,g((0,f^{-1}(p_1)_1)))$ if $p_0=0$ and $f^{-1}(p_1)_0=1$;
• $p\to(1,g((1,p_1)))$ if $p_0=1$;

from $(\alpha+\beta)\bigsqcup\gamma$ to $\alpha\bigsqcup(\beta+\gamma)$ is a bijection. Thus $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$.

If $\beta\le\gamma$ then $\beta\subseteq\gamma$, thus $\alpha\bigsqcup\beta\subseteq\alpha\bigsqcup\gamma$, implying $\alpha+\beta\le\alpha+\gamma$.

The map $p\mapsto(p_1,p_0)$ from $\alpha\times\beta$ to $\beta\times\alpha$ is a bijection. Thus $\alpha\beta=\beta\alpha$.

Let $f$ be a bijection from $\alpha\times\beta$ to $\alpha\beta$ and $g$ be a bijection from $\beta\times\gamma$ to $\beta\gamma$. The map $p\mapsto(f^{-1}(p_0)_0,g((f^{-1}(p_0)_1,p_1)))$ from $(\alpha\beta)\times\gamma$ to $\alpha\times(\beta\gamma)$ is a bijection. Thus $(\alpha\beta)\gamma=\alpha(\beta\gamma)$.

If $\beta\le\gamma$ then $\beta\subseteq\gamma$, thus $\alpha\times\beta\subseteq\alpha\times\gamma$, implying $\alpha\beta\le\alpha\gamma$.

Let $g_0$ denote the function $x\mapsto(0,x)$ from $\beta$ to $\{0\}\times\beta$, $g_1$ denote the function $x\mapsto(1,x)$ from $\gamma$ to $\{1\}\times\gamma$, and $h_a,h_b,h_c$ denote bijections from $\beta\bigsqcup\gamma$ to $\beta+\gamma$, from $\{\beta\to\alpha\}$ to $\alpha^\beta$, and from $\{\gamma\to\alpha\}$ to $\alpha^\gamma$, respectively. The map $f\mapsto(h_b((f\circ h_a)|_{\{0\}\times\beta}\circ g_0),h_c((f\circ h_a)|_{\{1\}\times\gamma}\circ g_1))$ from $\{\beta+\gamma\to\alpha\}$ to $\alpha^\beta\times\alpha^\gamma$ is a bijection. Thus $\alpha^{\beta+\gamma}=\alpha^\beta\alpha^\gamma$.

Let $g\in\{\gamma\to\{\beta\to\alpha\}\}$. Then there exists a unique function $h:\beta\times\gamma\to\alpha$ such that for all $p\in\beta\times\gamma$, $h(p)=g(p_1)(p_0)$. Then we can define a map $u$ from $\{\gamma\to\{\beta\to\alpha\}\}$ to $\{\beta\times\gamma\to\alpha\}$ such that for all $g\in\{\gamma\to\{\beta\to\alpha\}\}$, for all $p\in\beta\times\gamma$, $u(g)(p)=g(p_1)(p_0)$. Clearly, $u$ is a bijection. Let $h_a,h_b$ denote bijections from $\beta\times\gamma$ to $\beta\gamma$ and from $\{\beta\to\alpha\}$ to $\alpha^\beta$, respectively. The map $f\mapsto u(h_b^{-1}\circ f)\circ h_a^{-1}$ from $\{\gamma\to\alpha^\beta\}$ to $\{\beta\gamma\to\alpha\}$ is a bijection. Thus $\p{\alpha^\beta}^\gamma=\alpha^{\beta\gamma}$.

Let $g_a,g_b$ denote the functions from $\alpha\times\beta$ to $\alpha,\beta$ respectively, such that for all $p\in\alpha\times\beta$, $g_a(p)=p_0$ and $g_b(p)=p_1$. Let $h_a,h_b,h_c$ denote bijections from $\alpha\times\beta$ to $\alpha\beta$, from $\{\gamma\to\alpha\}$ to $\alpha^\gamma$, and from $\{\gamma\to\beta\}$ to $\beta^\gamma$, respectively. The map $f\mapsto(h_b(g_a\circ h_a^{-1}\circ f),h_c(g_b\circ h_a^{-1}\circ f))$ from $\{\gamma\to\alpha\beta\}$ to $\alpha^\gamma\times\beta^\gamma$ is a bijection. Thus $\p{\alpha\beta}^\gamma=\alpha^\gamma\beta^\gamma$.

If $\beta\le\gamma$ then $\beta\subseteq\gamma$. Since $\alpha\neq0$, we have $0\in\alpha$. Thus given $f\in\{\beta\to\alpha\}$, there exists a unique $g\in\{\gamma\to\alpha\}$ such that $g(x)=f(x)$ for all $x\in\beta$ and $g(x)=0$ for all $x\in\gamma\setminus\beta$. This defines an injective map from $\{\beta\to\alpha\}$ to $\{\gamma\to\alpha\}$, implying $\alpha^\beta\le\alpha^\gamma$.

If $\alpha\le\beta$ then $\alpha\subseteq\beta$, thus $\{\gamma\to\alpha\}\subseteq\{\gamma\to\beta\}$, implying $\alpha^\gamma\le\beta^\gamma$. $\blacksquare$

Proof. Let $g$ be a bijection from $X$ to $\abs{X}$. The map $f\mapsto (f\circ g)^{-1}(\{1\})$ from $\{\abs{X}\to2\}$ to $\mathcal P(X)$ is a bijection. Thus $\abs{\mathcal P(X)}=2^{\abs{X}}$. $\blacksquare$

Proof. By transfinite recursion, $\aleph$ is a class function from $\text{Ord}$ to $V$, and for all $\alpha\in\text{Ord}$, $\aleph_\alpha=G(\aleph\upharpoonright\alpha)$. Let $\alpha\in\text{Ord}$ and suppose that for all $\beta\in\alpha$, $\aleph_\beta\in\text{Car}$.

• If $\alpha$ is a zero ordinal, then $\aleph_\alpha=G(\aleph\upharpoonright\emptyset)=\omega\in\text{Car}$.
• If $\alpha$ is a successor ordinal, then $\aleph_\alpha=G(\aleph\upharpoonright\alpha)=\bigcup\{S(p_1):p\in\aleph\upharpoonright\alpha\}$. Note that for all $p\in \aleph\upharpoonright\alpha$, there exist $\beta,\gamma$ such that $\beta\in\alpha$ and $p=(\beta,\gamma)$. Thus $p_1=\gamma=\aleph\upharpoonright\alpha(\beta)=\aleph_\beta\in\text{Car}$, and hence $S(p_1)\in\text{Car}$. Then for all $\gamma\in\{S(p_1):p\in\aleph\upharpoonright\alpha\}$, $\gamma\in\text{Car}$, and thus $\bigcup\{S(p_1):p\in\aleph\upharpoonright\alpha\}\in\text{Car}$. Hence $\aleph_\alpha\in\text{Car}$.
• If $\alpha$ is a limit ordinal, then $\aleph_\alpha=G(\aleph\upharpoonright\alpha)=\bigcup\{p_1:p\in\aleph\upharpoonright\alpha\}$. By the same reasoning as above, for all $p\in \aleph\upharpoonright\alpha$, $p_1\in\text{Car}$. Then for all $\gamma\in\{p_1:p\in\aleph\upharpoonright\alpha\}$, $\gamma\in\text{Car}$, and thus $\bigcup\{p_1:p\in\aleph\upharpoonright\alpha\}\in\text{Car}$. Hence $\aleph_\alpha\in\text{Car}$.

In all cases, $\aleph_\alpha\in\text{Car}$. Thus by transfinite induction, for all $\alpha\in\text{Ord}$, $\aleph_\alpha\in\text{Car}$. Hence $\aleph$ is a class function from $\text{Ord}$ to $\text{Car}$.

Suppose $\alpha\in\text{Ord}$ and for all $\beta\in\alpha$, for all $\gamma\in\beta$, $\aleph_\gamma\lt\aleph_\beta$.

• If $\alpha$ is a zero ordinal, then trivially, for all $\gamma\in\alpha$, $\aleph_\gamma\lt\aleph_\alpha$.
• If $\alpha$ is a successor ordinal, then for some $\beta\in\alpha$ we have $\beta^+=\alpha$. If $\gamma\in\alpha$, then $\gamma\le\beta$, thus $\aleph_\gamma\le\aleph_\beta\lt S(\aleph_\beta)\le\bigcup\{S(p_1):p\in\aleph\upharpoonright\alpha\}=\aleph_\alpha$.
• If $\alpha$ is a limit ordinal, then given $\gamma\in\alpha$, there exists $\beta\in\alpha$ such that $\gamma\in\beta$, implying $\aleph_\gamma\lt\aleph_\beta\le\bigcup\{p_1:p\in\aleph\upharpoonright\alpha\}=\aleph_\alpha$.

In all cases, we have that for all $\gamma\in\alpha$, $\aleph_\gamma\lt\aleph_\alpha$. By transfinite induction, for all $\alpha\in\text{Ord}$, for all $\gamma\in\alpha$, $\aleph_\gamma\lt\aleph_\alpha$. We have shown that $\aleph$ is strictly increasing. $\blacksquare$

Proof. $\aleph_0=G(\aleph\upharpoonright0)=\omega$.

Note that $\aleph_{\alpha^+}=\bigcup\{S(p_1):p\in\aleph\upharpoonright\alpha^+\}$. If $u\in\bigcup\{S(p_1):p\in\aleph\upharpoonright\alpha^+\}$, then for some $p\in\aleph\upharpoonright\alpha^+$, $u\in S(p_1)$. And for some $\beta\in\alpha^+$, $p=(\beta,\aleph\upharpoonright\alpha^+(\beta))=(\beta,\aleph_\beta)$, thus $u\in S(\aleph_\beta)\le S(\aleph_\alpha)$. For the other direction, suppose $u\in S(\aleph_\alpha)$. Note that $(\alpha,\aleph_\alpha)\in\aleph\upharpoonright\alpha^+$, thus $u\in\bigcup\{S(p_1):p\in\aleph\upharpoonright\alpha^+\}$. Hence $\bigcup\{S(p_1):p\in\aleph\upharpoonright\alpha^+\}=S(\aleph_\alpha)$.

Suppose $\alpha$ is a limit ordinal, then $\aleph_\alpha=\bigcup\{p_1:p\in\aleph\upharpoonright\alpha\}$. If $u\in\bigcup\{p_1:p\in\aleph\upharpoonright\alpha\}$, then for some $p\in\aleph\upharpoonright\alpha$, $u\in p_1$. And for some $\beta\in\alpha$, $p=(\beta,\aleph\upharpoonright\alpha(\beta))=(\beta,\aleph_\beta)$, thus $u\in\aleph_\beta\le\bigcup_{\beta\in\alpha}\aleph_\beta$. For the other direction, suppose $u\in\bigcup_{\beta\in\alpha}\aleph_\beta$. Then for some $\beta\in\alpha$, $u\in\aleph_\beta$. Note that $(\beta,\aleph_\beta)\in\aleph\upharpoonright\alpha$, thus $u\in\bigcup\{p_1:p\in\aleph\upharpoonright\alpha\}$. Hence $\bigcup\{p_1:p\in\aleph\upharpoonright\alpha\}=\bigcup_{\beta\in\alpha}\aleph_\beta$. $\blacksquare$

Proof. Suppose such set $X$ exists. Let $\alpha$ be an ordinal, then $\aleph\upharpoonright\alpha$ is an injection from $\alpha$ to $X$, implying $\abs{\alpha}\le\abs{X}$. Since $\alpha$ is an arbitrary ordinal, we also have $S(\abs{X})=\abs{S(\abs{X})}\le\abs{X}$, a contradiction. $\blacksquare$

Proof. Since $X$ is non-empty and has no maximal element, we have some $\mu\in X$, and some $\nu\in X$ such that $\mu\lt\nu$. Note that $\mu\in\nu\subseteq\bigcup X$, implying $\bigcup X$ is non-empty. If $\bigcup X$ has a maximal element, then for some ordinal $\alpha$, $\alpha^+=\bigcup X$. Then $\alpha$ is not an upper bound of $X$, implying for some $\mu\in X$, $\mu\gt\alpha$. Then we have $\mu\ge\alpha^+=\bigcup X$. But then for some $\nu\in X$, we have $\nu\gt\mu\ge\bigcup X\ge\nu$, a contradiction. Thus $\bigcup X$ is a limit ordinal. $\blacksquare$

Proof. Suppose there exists a limit cardinal that does not belong to the range of $\aleph$. Then there exists a least limit cardinal $\kappa$ with such property. Let $X$ denote the subset of $\kappa$ consisting of limit cardinals less than $\kappa$. Then $\bigcup X$ is a cardinal and is the least upper bound of $X$.

If $X$ is empty, then $\kappa$ cannot be greater than $\omega$. But then $\kappa$ also cannot be smaller than $\omega$ because $\omega$ is the least limit ordinal. Thus $\aleph_0=\omega=\kappa$, a contradiction.

If $X$ has a maximal element $\mu$, then $\mu$ must fall in the range of $\aleph$, implying for some ordinal $\alpha$, $\aleph_\alpha=\mu$. If $S(\mu)\lt\kappa$, then $S(\mu)\in X$, implying $\mu$ is not a maximal element of $X$, a contradiction. If $S(\mu)\gt\kappa$, then $S(\mu)$ is not the least cardinal greater than $\mu$, a contradiction. Hence $\aleph_{\alpha^+}=S(\aleph_\alpha)=S(\mu)=\kappa$, a contradiction.

If $X$ is non-empty and has no maximal element, then $\bigcup X$ is a limit cardinal. If $\bigcup X\lt\kappa$, then $\bigcup X\in X$, and there exists $\mu\in X$ such that $\mu\gt\bigcup X\ge\mu$, a contradiction. If $\bigcup X\gt\kappa$, then $\bigcup X$ is not the least upper bound of $X$, a contradiction. Thus $\bigcup X=\kappa$. Note that for each $\mu\in X$, there exists an ordinal $\alpha_\mu$ such that $\aleph_{\alpha_\mu}=\mu$. Also note that $\{\alpha_\mu:\mu\in X\}$ is a non-empty set of ordinals with no maximal element. Let $\alpha$ denote $\bigcup_{\mu\in X}\alpha_\mu$, then $\alpha$ is a limit ordinal. If $u\in\aleph_\alpha$, then for some $\beta\in\alpha$, $u\in\aleph_\beta$, and for some $\mu\in X$, $\beta\in\alpha_\mu$, thus $\aleph_\beta\lt\aleph_{\alpha_\mu}=\mu$, implying $u\in\mu$. Hence $u\in\bigcup X$. For the other direction, if $u\in\bigcup X$, then for some $\mu\in X$, $u\in\mu=\aleph_{\alpha_\mu}$. Since $\alpha_\mu\in\{\alpha_\mu:\mu\in X\}$, $\alpha_\mu\le\alpha$, implying $\aleph_{\alpha_\mu}\le\aleph_\alpha$. Hence $u\in\aleph_\alpha$. Therefore, we have $\aleph_\alpha=\bigcup X=\kappa$, a contradiction.

We have shown that every limit cardinal is in the range of $\aleph$. Uniqueness is trivial. $\blacksquare$

Proof. Given a function $f$ from an ordinal $\alpha$ to $X$, if $f$ is not a surjection, then there exists a unique least element of $X$ not in the range of $f$. Let $A$ denote the class of non-surjective functions from an ordinal to $X$, then this defines a class function $F:A\to X$. Define a class function $G:V\to V$ by:

• $G(x)=F(x)$ if $x\in A$;
• $G(x)=\emptyset$ if otherwise.

Then $G$ recursively defines a class function $H:\text{Ord}\to V$ such that for all $\alpha\in\text{Ord}$, $H(\alpha)=G(H\upharpoonright\alpha)$. Assume for contradiction that for all ordinal $\alpha$, $H\upharpoonright\alpha\in A$. Let $\alpha$ be an ordinal, then $H\upharpoonright\alpha$ is a function from $\alpha$ to $X$. Given $\gamma,\delta\in\alpha$ such that $H\upharpoonright\alpha(\gamma)=H\upharpoonright\alpha(\delta)$, suppose for contradiction that $\gamma\lt\delta$. Note that $H\upharpoonright\delta\in A$, thus $H\upharpoonright\alpha(\delta)=H(\delta)=G(H\upharpoonright\delta)=F(H\upharpoonright\delta)$, which is the least element of $X$ not in the range of $H\upharpoonright\delta$. But then $H\upharpoonright\alpha(\gamma)=H(\gamma)=H\upharpoonright\delta(\gamma)$, which is in the range of $H\upharpoonright\delta$, a contradiction. With a similar argument, $\gamma\gt\delta$ also leads to a contradiction. Hence $\gamma=\delta$. We have shown that $H\upharpoonright\alpha$ is injective from $\alpha$ to $X$. Since $\alpha$ is an arbitrary ordinal, we have $S(\abs{X})=\abs{S(\abs{X})}\le\abs{X}$, a contradiction. Hence there exists an ordinal $\alpha$ such that $H\upharpoonright\alpha\notin A$. Let $\epsilon$ be the least ordinal such that $H\upharpoonright\epsilon\notin A$. Let $\alpha$ be an ordinal and suppose for all $\beta\in\alpha$, if $\beta\le\epsilon$, then $H\upharpoonright\beta$ is an order-embedding from $\beta$ to $X$ with property $\varphi(H\upharpoonright\beta)$, where $\varphi(f)$ represents that for all $x,y\in X$, if $x$ is in range of $f$ and $y$ is not in range of $f$, then $x\lt y$ (meaning $x\le y$ and $x\neq y$). Let $\alpha\le\epsilon$.

• If $\alpha$ is a zero ordinal, then $H\upharpoonright\alpha=\emptyset$, which is trivially an order-embedding from $\alpha$ to $X$ with property $\varphi(H\upharpoonright\alpha)$.
• If $\alpha$ is a successor ordinal, then some $\beta\in\alpha$ has $\beta^+=\alpha$. Since $\beta\lt\epsilon$, $H\upharpoonright\beta$ is an order-embedding from $\beta$ to $X$ with property $\varphi(H\upharpoonright\beta)$ and $H\upharpoonright\beta\in A$. Then $H\upharpoonright\alpha(\beta)=H(\beta)=G(H\upharpoonright\beta)=F(H\upharpoonright\beta)$, which is the least element of $X$ not in the range of $H\upharpoonright\beta$. Then $H\upharpoonright\alpha$ is clearly an order-embedding from $\alpha$ to $X$ with property $\varphi(H\upharpoonright\alpha)$.
• If $\alpha$ is a limit ordinal, then given $\gamma,\delta\in\alpha$, there exists $\beta\in\alpha$ such that $\gamma,\delta\in\beta$. Since for any $\beta\in\alpha$, $H\upharpoonright\beta$ is an order-embedding from $\beta$ to $X$ with property $\varphi(H\upharpoonright\beta)$, $H\upharpoonright\alpha$ is clearly an order-embedding from $\alpha$ to $X$ with property $\varphi(H\upharpoonright\alpha)$.

By transfinite induction, for every ordinal $\alpha$, if $\alpha\le\epsilon$, then $H\upharpoonright\alpha$ is an order-embedding from $\alpha$ to $X$. Thus $H\upharpoonright\epsilon$ is an order-embedding from $\epsilon$ to $X$. Note that $H\upharpoonright\epsilon\notin A$, implying $H\upharpoonright\epsilon$ is also a surjection from $\epsilon$ to $X$. Denote $H\upharpoonright\epsilon$ as $f$ and let $x,y\in X$ such that $x\le y$. Suppose $f^{-1}(x)\gt f^{-1}(y)$, then $y\le x$ and $x\neq y$, a contradiction. We have shown that $f$ is an order-isomorphism from $\epsilon$ to $X$. $\blacksquare$

Proof. Suppose for contradiction that there exists a limit cardinal $\kappa$ such that $\kappa^2\gt\kappa$. Then there exists a least limit cardinal $\kappa$ such that $\kappa^2\gt\kappa$.

Define the relation $\le$ on $\kappa\times\kappa$ such that for all $p,q\in\kappa\times\kappa$, $p\le q$ if and only if

• $p_0\cup p_1\lt q_0\cup q_1$, or
• $p_0\cup p_1=q_0\cup q_1$ and $p_0\lt q_0$, or
• $p_0\cup p_1=q_0\cup q_1$, $p_0=q_0$, and $p_1\le q_1$.

Then $(\kappa\times\kappa,\le)$ is clearly a well-ordered set, and thus order-isomorphic to some ordinal $\alpha$. Then $\alpha\ge\abs{\alpha}=\abs{\kappa\times\kappa}=\kappa^2\gt\kappa$. Let $f$ be an order-isomorphism from $\alpha$ to $\kappa\times\kappa$, then there exists $p\in\kappa\times\kappa$ not in range of $f|_\kappa$, implying $f^{-1}(p)\in\alpha\setminus\kappa$. Then for all $\beta\in\kappa$, $\beta\lt f^{-1}(p)$, thus $f(\beta)\lt p$, implying $f(\beta)_0\cup f(\beta)_1\le p_0\cup p_1$. Since $\kappa$ is a limit ordinal, $(p_0\cup p_1)^+\in\kappa$. Denote $(p_0\cup p_1)^+$ as $\gamma$, then $f(\beta)_0\cup f(\beta)_1\lt\gamma$, thus $f(\beta)\in\gamma\times\gamma$. Since $f|_\kappa$ is an injection from $\kappa$ to $\gamma\times\gamma$, $\kappa\le\abs{\gamma\times\gamma}$. If $\gamma\in\omega$, then $\gamma$ is a cardinal and $\kappa\le\gamma\gamma\lt\omega$, a contradiction. Thus $\gamma\ge\omega$, implying $\abs{\gamma}$ is a limit cardinal. Since $\abs{\gamma}\le\gamma\lt\kappa$, $\abs{\gamma}^2\le\abs{\gamma}$, and thus $\abs{\gamma}^2=\abs{\gamma}$. Since $\gamma\times\gamma$ is bijective to $\abs{\gamma}\times\abs{\gamma}$, we have $\abs{\gamma\times\gamma}=\abs{\gamma}^2$. Then we have $\kappa\le\abs{\gamma\times\gamma}=\abs{\gamma}^2=\abs{\gamma}\le\gamma\lt\kappa$, a contradiction.

We have shown that for all limit cardinal $\kappa$, $\kappa^2\le\kappa$, and thus $\kappa^2=\kappa$. $\blacksquare$

Proof.

• $\kappa+\mu\le\kappa+\kappa=2\kappa\le\kappa\kappa=\kappa^2=\kappa$.
• $\kappa\mu\le\kappa\kappa=\kappa^2=\kappa$.
• For base case $\kappa^0=1\le\kappa$. Suppose $n\in\omega$ and $\kappa^n\le\kappa$, then $\kappa^{S(n)}=\kappa^n\kappa\le\kappa\kappa=\kappa^2=\kappa$. Thus by induction, for all $n\in\omega$, $\kappa^n\le\kappa$. If $n\in\omega\setminus\{0\}$, then $n\ge1$, thus $\kappa^n\ge\kappa$, and hence $\kappa^n=\kappa$.

$\blacksquare$

Proof. Let $f\in\{N\to2\}$, then $$\sup_{i\in N}\sum_{j\in i}2f(j)3^{-j}$$ is a real number. This defines a function $F:\{N\to2\}\to R$. Suppose $f,g\in\{N\to2\}$ and $F(f)=F(g)$. Suppose $n\in N$ and $f(k)=g(k)$ for all $k\in n$. Suppose for contradiction that $f(n)=0$ and $g(n)=1$. Define $f',g'\in\{N\to2\}$ such that $f'(j)=f(j)$ and $g'(j)=g(j)$ if $j\le n$, and $f'(j)=1$ and $g'(j)=0$ if $j\gt n$. Then clearly, $F(f)\le F(f')$ and $F(g)\ge F(g')$. Note that $\sum_{j\in n}2f(j)3^{-j}+3^{-n}$ is an upper bound of $\{\sum_{j\in i}2f'(j)3^{-j}:i\in N\}$, and $F(g')\gt\sum_{j\in n}2f(j)3^{-j}+3^{-n}$. Thus $F(g')\gt F(f')$. Hence $F(f)\lt F(g)$, a contradiction. By a symmetric argument, the case when $f(n)=1$ and $g(n)=0$ also leads to a contradiction. Hence $F(n)=g(n)$, implying $f(k)=g(k)$ for all $k\in S(n)$. With the trivial base case, by induction, for all $n\in N$, $f(k)=g(k)$ for all $k\in n$. Thus $f=g$. We have shown that $F$ is an injection, thus $2^{\aleph_0}=\abs{\{N\to2\}}\le\abs{R}$. Since $R\subseteq\mathcal P(Q)$, we have $\abs{R}\le\abs{\mathcal P(Q)}=2^{\abs{Q}}=2^{\aleph_0}$. Therefore, $2^{\aleph_0}=\abs{R}$.

Since the set $C$ is just $R\times R$ which is naturally bijective to $R^2$, it is sufficient to show that given non-zero natural number $n$, $2^{\aleph_0}=\abs{R^n}$. Suppose $n\in N$ and if $n\neq0$ then $2^{\aleph_0}=\abs{R^n}$.

• If $n=0$, then $\abs{R^{S(n)}}=\abs{R^1}=\abs{R}=2^{\aleph_0}$.
• If $n\neq0$, then $\abs{R^{S(n)}}=\abs{R^n\times R}=\abs{R^n}\abs{R}=2^{\aleph_0}2^{\aleph_0}=2^{\aleph_0}$.

By induction, for all $n\in N$, if $n\neq0$ then $2^{\aleph_0}=\abs{R^n}$. $\blacksquare$

Proof. Suppose otherwise, then for some $\alpha\in\kappa$, $\kappa\in\alpha^+$ or $\kappa=\alpha^+$. If $\kappa\in\alpha^+$, then $\kappa\in\alpha$ or $\kappa=\alpha$, a contradiction. If $\kappa=\alpha^+$, note that $f:1\to\alpha^+$ with $f(0)=\alpha$ is a map such that for all $\beta\in\alpha^+$, $\beta\in f(0)$ or $\beta=f(0)$. Clearly, there is no such map from $0$ to $\alpha^+$. Hence $N\in\kappa=\text{cf}(\kappa)=\text{cf}(\alpha^+)=1$, a contradiction. $\blacksquare$

Proof. Let $\alpha\in\kappa$ and suppose for inductive hypothesis that for all $\beta\in\alpha$, $\abs{V_\beta}\lt\kappa$. Then $\abs{\mathcal P(V_\beta)}\lt\kappa$. Let $A=\{(\beta,x)\in\alpha\times V_\alpha:x\subseteq V_\beta\}$. Then $\abs{V_\alpha}\le\abs{A}$. Suppose for contradiction that $\abs{A}\ge\kappa$, then there exists a surjection $F$ from $A$ to $\kappa$. Let $\beta\in\alpha$, then $\{F(\beta,x):x\subseteq V_\beta\}$ has a strict upper bound in $\kappa$, because otherwise $\kappa=\text{cf}(\kappa)\le\abs{\{F(\beta,x):x\subseteq V_\beta\}}\le\abs{\mathcal P(V_\beta)}\lt\kappa$, a contradiction. Hence $\{F(\beta,x):x\subseteq V_\beta\}$ has a least strict upper bound $\beta^*$ in $\kappa$, and so $\beta\mapsto\beta^*$ defines a function $G$ from $\alpha$ to $\kappa$. Now for all $\gamma\in\kappa$, there exists $(\beta,x)\in A$, where $\beta\in\alpha$, such that $\gamma=F(\beta,x)\lt\beta^*=G(\beta)$. Thus $\kappa=\text{cf}(\kappa)\le\abs{\alpha}\le\alpha\lt\kappa$, a contradiction. Therefore, $\abs{A}\lt\kappa$, and hence $\abs{V_\alpha}\lt\kappa$. By transfinite induction, for all $\alpha\in\kappa$, $\abs{V_\alpha}\lt\kappa$. Now suppose $x\in V_\kappa$, then for some $\alpha\in\kappa$, $x\subseteq V_\alpha$, hence $\abs{x}\le\abs{V_\alpha}\lt\kappa$. $\blacksquare$

Proof. Since $x\subseteq V_\kappa$, for all $u\in x$, $u\in V_\kappa$, thus for some $\alpha_u\in\kappa$, $u\subseteq V_{\alpha_u}$, and so $u\in V_{\alpha_u^+}$, where $\alpha_u^+\in\kappa$. Hence $\rank(u)\le\alpha_u^+$ and $\rank$ defines a map from $x$ to $\kappa$. Suppose for contradiction that for all $\alpha\in\kappa$, there exists $u\in x$ such that $\alpha\le\rank(u)$, then $\kappa=\text{cf}(\kappa)\le\abs{x}$, a contradiction. Hence there exists $\alpha\in\kappa$ such that for all $u\in x$, $\rank(u)\lt\alpha$, and thus $u\in V_\alpha$. So $x\subseteq V_\alpha$, and thus $x\in V_{\alpha^+}$, implying $x\in V_\kappa$. $\blacksquare$

Proof. The only thing to prove is that axiom schema of specification and axiom schema of replacement are satisfied by the structure. We will first make clear the definitions of the second order forms of these axiom schemas:

• axiom of specification: given any natural number $n$, for every collection $\mathcal C$ of $(n+2)$-tuples of the universe, for every $n$-tuple $w$ of the universe, for every $X$ in the universe, there exists $Y$ in the universe that contains exactly the objects $u$ in $X$ such that $(w_1,\ldots,w_n,X,u)\in\mathcal C$.
• axiom of replacement: given any natural number $n$, for every collection $\mathcal C$ of $(n+3)$-tuples of the universe, for every $n$-tuple $w$ of the universe, for every $X$ in the universe, if for all $x\in X$ there exists a unique $y$ in the universe such that $(w_1,\ldots,w_n,X,x,y)\in\mathcal C$, then there exists $Y$ in the universe, such that for all $x\in X$, there exists $y\in Y$ such that $(w_1,\ldots,w_n,X,x,y)\in\mathcal C$.

By the hypothesis, we have $V$ as the universe of the structure, which satisfies the above axioms.

For any natural number $n$, let $X,Y,u,w_1,\ldots,w_n$ be distinct variable symbols, and let $\varphi$ be a formula with free variables among $X,u,w_1,\ldots,w_n$. Let $\mathcal C$ be the collection of $(n+2)$-tuples $T$ of the universe such that $\varphi$ is true in the structure with respect to the assignment function $\sigma$ defined by

• $\sigma(w_j)=T_j$ if $w_j\in\text{free}(\varphi)$ for all $j$ in $1,\ldots,n$,
• $\sigma(X)=T_{n+1}$ if $X\in\text{free}(\varphi)$, and
• $\sigma(u)=T_{n+2}$ if $u\in\text{free}(\varphi)$.

Assign arbitrary $w^*_j\in V$ to $w_j$ for all $j$ in $1,\ldots,n$, then $(w^*_1,\ldots,w^*_n)$ is an $n$-tuple of $V$. Assign arbitrary $X^*\in V$ to $X$, then there exists $Y^*\in V$ that contains exactly the objects $u^*$ in $X^*$ such that $(w^*_1,\ldots,w^*_n,X^*,u^*)\in\mathcal C$. We assign such $Y^*$ to $Y$. Now assign arbitrary $u^*\in V$ to $u$. If $u^*\in Y^*$, then $u^*\in X^*$ and $(w^*_1,\ldots,w^*_n,X^*,u^*)\in\mathcal C$, implying that $\varphi$ is true in the structure with respect to the assignment function $\sigma$ defined by

• $\sigma(w_j)=w^*_j$ if $w_j\in\text{free}(\varphi)$ for all $j$ in $1,\ldots,n$,
• $\sigma(X)=X^*$ if $X\in\text{free}(\varphi)$, and
• $\sigma(u)=u^*$ if $u\in\text{free}(\varphi)$.

Note that $\sigma$ agrees with our assignments. Thus $(u\in Y)\to((u\in X)\land\varphi)$ is satisfied. If $u^*\in X^*$ and $\varphi$ is true in the structure with respect to the assignment function $\sigma$ described above, then $(w^*_1,\ldots,w^*_n,X^*,u^*)\in\mathcal C$, thus $u^*\in Y^*$. Hence $((u\in X)\land\varphi)\to(u\in Y)$ is satisfied. Therefore, $(u\in Y)\leftrightarrow((u\in X)\land\varphi)$ is satisfied. Since $u^*$ is arbitrary, $\forall u((u\in Y)\leftrightarrow((u\in X)\land\varphi))$ is satisfied. Since we have $Y^*$ as an instance, $\exists Y\forall u((u\in Y)\leftrightarrow((u\in X)\land\varphi))$ is satisfied. Since the rest of the assignments are arbitrary, $\forall w_1\ldots\forall w_n\forall X\exists Y\forall u((u\in Y)\leftrightarrow((u\in X)\land\varphi))$ is satisfied.

For any natural number $n$, let $X,Y,x,y,w_1,\ldots,w_n$ be distinct variable symbols, and let $\varphi$ be a formula with free variables among $X,x,y,w_1,\ldots,w_n$. Let $\mathcal C$ be the collection of $(n+3)$-tuples $T$ of the universe such that $\varphi$ is true in the structure with respect to the assignment function $\sigma$ defined by

• $\sigma(w_j)=T_j$ if $w_j\in\text{free}(\varphi)$ for all $j$ in $1,\ldots,n$,
• $\sigma(X)=T_{n+1}$ if $X\in\text{free}(\varphi)$,
• $\sigma(x)=T_{n+2}$ if $x\in\text{free}(\varphi)$, and
• $\sigma(y)=T_{n+3}$ if $y\in\text{free}(\varphi)$.

Assign arbitrary $w^*_j\in V$ to $w_j$ for all $j$ in $1,\ldots,n$, then $(w^*_1,\ldots,w^*_n)$ is an $n$-tuple of $V$. Assign arbitrary $X^*\in V$ to $X$. If for all $x^*\in X^*$ there exists a unique $y^*\in V$ such that $(w^*_1,\ldots,w^*_n,X^*,x^*,y^*)\in\mathcal C$, then there exists $Y^*\in V$, such that for all $x^*\in X^*$, there exists $y^*\in Y^*$ such that $(w^*_1,\ldots,w^*_n,X^*,x^*,y^*)\in\mathcal C$. Suppose $\forall x((x\in X)\to\exists!y\varphi)$ is satisfied. Let $x^*\in X^*$, if we assigned $x^*$ to $x$, then both $x\in X$ and $(x\in X)\to\exists!y\varphi$ are satisfied, thus $\exists!y\varphi$ is satisfied. Hence there exists a unique $y^*\in V$ such that when $y^*$ is assigned to $y$, $\varphi$ is satisfied. Note that given $y^*\in V$, $(w^*_1,\ldots,w^*_n,X^*,x^*,y^*)\in\mathcal C$ if and only if $\varphi$ is satisfied with $y$ assigned $y^*$. Thus there exists a unique $y^*\in V$ such that $(w^*_1,\ldots,w^*_n,X^*,x^*,y^*)\in\mathcal C$. Hence there exists $Y^*\in V$, such that for all $x^*\in X^*$, there exists $y^*\in Y^*$ such that $(w^*_1,\ldots,w^*_n,X^*,x^*,y^*)\in\mathcal C$. We assign such $Y^*$ to $Y$. Suppose we have arbitrary $x^*\in V$. If $x^*\in X^*$, then there exists $y^*\in Y^*$ such that $(w^*_1,\ldots,w^*_n,X^*,x^*,y^*)\in\mathcal C$. Again, note that given $y^*\in V$, $(w^*_1,\ldots,w^*_n,X^*,x^*,y^*)\in\mathcal C$ if and only if $\varphi$ is satisfied with $y$ assigned $y^*$. Thus if we assign such $y^*$ to $y$, then $(y\in Y)\land\varphi$ is satisfied. Since we have $y^*$ as an instance, $\exists y((y\in Y)\land\varphi)$ is satisfied, thus $(x\in X)\to\exists y((y\in Y)\land\varphi)$ is satisfied. If not $x^*\in X^*$, then $(x\in X)\to\exists y((y\in Y)\land\varphi)$ is trivially satisfied. Since $x^*$ is arbitrary, $\forall x((x\in X)\to\exists y((y\in Y)\land\varphi))$ is satisfied. Since we have $Y^*$ as an instance, $\exists Y\forall x((x\in X)\to\exists y((y\in Y)\land\varphi))$ is satisfied. Therefore $(\forall x((x\in X)\to\exists!y\varphi)\to\exists Y\forall x((x\in X)\to\exists y((y\in Y)\land\varphi)))$ is satisfied. Since the rest of the assignments are arbitrary, $\forall w_1\ldots\forall w_n\forall X(\forall x((x\in X)\to\exists!y\varphi)\to\exists Y\forall x((x\in X)\to\exists y((y\in Y)\land\varphi)))$ is satisfied. $\blacksquare$

Proof. We will show below that for all $X\in V$, for all $Y\subseteq X$, $Y\in V$. Let $X\in V$ and $Y\subseteq X$. Let $\mathcal C$ be the collection of $2$-tuples $w$ of $V$ such that $w_2\in Y$. Then there exists $Y'\in V$ that contains exactly the objects $u$ in $X$ such that $(X,u)\in\mathcal C$. Note that $u\in Y$ if and only if $u\in X$ and $(X,u)\in\mathcal C$, if and only if $u\in Y'$. Thus $Y=Y'$, implying that $Y\in V$.

We will show below that for all $X\in V$, for all $x\in X$, $x\in V$. Let $X\in V$ and $x\in X$. Suppose for contradiction that $x\notin V$. Let $X'$ be the collection of objects $u$ in $X$ such that $u\neq x$. Then $X'\subseteq X$, thus $X'\in V$. Since $V$ satisfies axiom of extensionality, and given $u\in V$, $u\in X$ if and only if $u\in X'$, we have $X=X'$, a contradiction. Hence $x\in V$.

Suppose object $X\in V_\kappa$. Then $X\in V$. Note that the formula $\forall X((X\in V_\kappa)\to(X\subseteq V_\kappa))$ is satisfied by the model $V$. Thus $\forall x((x\in X)\to(x\in V_\kappa))$ is true when $X$ is taken to be the object $X$. That means for all object $x\in V$, we have $(x\in X)\to(x\in V_\kappa)$. Given object $x\in X$, we have $x\in V$, thus $(x\in X)\to(x\in V_\kappa)$, and since $x\in X$, we have $x\in V_\kappa$. We have shown that for all $x\in X$, $x\in V_\kappa$.

Suppose object $X\in V_\kappa$ and $Y$ is a sub-collection of $X$. Then $X\in V$, thus $Y\in V$. And $X\subseteq V_\kappa$, thus $Y\subseteq V_\kappa$. By a lemma above, we have $\abs{X}\lt\kappa$ in $V$. Since $Y\subseteq X$, we have $\abs{Y}\le\abs{X}$ in $V$. Thus $\abs{Y}\lt\kappa$ in $V$. Hence by a lemma above, we have $Y\in V_\kappa$ in $V$.

Since $\omega\in\kappa$, $\omega^+\in\kappa$. Trivially, $0\subseteq V_0$. Let $n\in\omega$ and suppose $n\subseteq V_n$. Then $n\in V_{n^+}$ and $n\subseteq V_{n^+}$. Thus $S(n)\subseteq V_{n^+}=V_{S(n)}$. By induction, given $n\in\omega$, we have $n\subseteq V_n$, and thus $n\in V_{n^+}\subseteq V_\omega$. Hence $\omega\subseteq V_\omega$, implying $\omega\in V_{\omega^+}\subseteq V_\kappa$. Therefore, $\omega\in V_\kappa$. And the object $\omega$ is in the object $V_\kappa$. $\blacksquare$

Proof.

• Logical axioms: For LA1, use induction to show that there exists a tuple of steps $(\varphi_1,\ldots,\varphi_n)$ to obtain the tautology $\varphi$ such that the variable symbols of each $\varphi_i$ is a sub-collection of those of $\varphi$. Then use induction to show that if the propositional variable symbols are assigned the same truth values as the corresponding first-order formulas given a first-order assignment function, then the original propositional formula is evaluated to the same truth value as the first-order formula formed by replacing variable symbols with formulas. Since the original formula is a tautology, we can show that LA1 is satisfied. For LA2, use induction to show that there exists a tuple of steps $(\varphi_1,\ldots,\varphi_n)$ to obtain $\varphi$ such that $t$ is substitutable for $x$ in each $\varphi_i$. Then use induction to show that given an assignment function $\sigma$ for the free variables of $\varphi[t/x]$ and $t$, if we define an assignment function $\tau$ for the free variables of $\varphi$ and $x$ that matches $\sigma$ for variables other than $x$, and maps $x$ to the object that $\sigma$ sends $t$ to, then $\delta_\tau(\varphi)=\delta_\sigma(\varphi[t/x])$. Using this, we can show that LA2 is satisfied. For LA6, use induction to show that when $x$ and $y$ are assigned the same object, $\varphi$ and $\varphi'$ are evaluated to the same truth value. The rest are trivial.
• Axiom of extensionality: Let $X,Y\in V_\kappa$. Then $X,Y\subseteq V_\kappa$. Hence if for all $u\in V_\kappa$, $u\in X$ if and only if $u\in Y$, then given $u\in X$, we have $u\in V_\kappa$ and $u\in X$, thus $u\in Y$; given $u\in Y$, we have $u\in V_\kappa$ and $u\in Y$, thus $u\in X$. Hence $X=Y$.
• Axiom of pairing: Let $a,b\in V_\kappa$. Then $a,b\in V$. Thus there exists $c\in V$ such that for all $u\in V$, $u\in c$ if and only if $u=a$ or $u=b$. Also, for some $\alpha\in\kappa$, $a\subseteq V_\alpha$ and for some $\beta\in\kappa$, $b\subseteq V_\beta$. Thus $a\in V_{\alpha^+}$ and $b\in V_{\beta^+}$.
  • if $\alpha^+=\beta^+$, then $a,b\in V_{\alpha^+}$.
  • If $\alpha^+\in\beta^+$, then $V_{\alpha^+}\subseteq V_{\beta^+}$, thus $a,b\in V_{\beta^+}$.
  • If $\beta^+\in\alpha^+$, then $V_{\beta^+}\subseteq V_{\alpha^+}$, thus $a,b\in V_{\alpha^+}$.
  Note that $\alpha^+,\beta^+\in\kappa$, thus there exists $\gamma\in\kappa$ such that $a,b\in V_\gamma$. Let $u\in V$, if $u\in c$, then $u=a$ or $u=b$, and in both cases we have $u\in V_\gamma$. Thus $c\subseteq V_\gamma$ and hence $c\in V_{\gamma^+}$. Since $\gamma^+\in\kappa$, $c\in V_\kappa$, and thus $c\subseteq V_\kappa$. Let $x\in V_\kappa$. If $x\in c$, then $x\in V$ and $x\in c$, thus $x=a$ or $x=b$. If $x=a$ or $x=b$, then $x\in V$ and $(x=a)\lor(x=b)$, thus $x\in c$.
• Axiom schema of specification: Let $n$ be a natural number and $\varphi$ be a formula as required. Fix an assignment function for the variable symbols $w_1,\ldots,w_n$. Let $X\in V_\kappa$, then there exists $Y\in V$ such that for all $u\in V$, $u\in Y$ if and only if $u\in X$ and $\varphi$. Thus $Y\subseteq X$. Note that there exists $\alpha\in\kappa$ such that $X\subseteq V_\alpha$. Thus $Y\subseteq V_\alpha$ and hence $Y\in V_{\alpha^+}\subseteq V_\kappa$. Let $u\in V_\kappa$. If $u\in Y$, then $u\in V$ and $u\in Y$, thus $u\in X$ and $\varphi$. If $u\in X$ and $\varphi$, then $u\in V$ and $(u\in X)\land\varphi$, thus $u\in Y$.
• Axiom of union: Let $X\in V_\kappa$. Then there exists $Y\in V$ such that for all $u\in V$, $u\in Y$ if and only if there exists $x\in V$ such that $x\in X$ and $u\in x$. Also, for some $\alpha\in\kappa$, $X\subseteq V_\alpha$. Let $u\in V$. If $u\in Y$ then there exists $x\in V$ such that $x\in X$ and $u\in x$. Thus $x\in V_\alpha$ and hence for some $\beta\in\alpha$, $x\subseteq V_\beta$, and we have $u\in V_\beta$. Therefore, we have $Y\subseteq V_\beta$, and thus $Y\in V_{\beta^+}\subseteq V_\kappa$. Let $u\in V_\kappa$. If $u\in Y$, then $u\in V$ and $u\in Y$, thus there exists $x\in V$ such that $x\in X$ and $u\in x$. Since $X\in V_\kappa$ and $x\in X$, we have $x\in V_\kappa$. If there exists $x\in V_\kappa$ such that $x\in X$ and $u\in x$, then $x\in V$, and $u\in V$, thus $u\in Y$.
• Axiom of power set: Let $X\in V_\kappa$. Then there exists $Y\in V$ such that for all $y\in V$, $y\in Y$ if and only if for all $x\in V$, $x\in y$ implies $x\in X$. Also, for some $\alpha\in\kappa$, $X\subseteq V_\alpha$. Let $y\in V$. If $y\in Y$ then for all $x\in V$, $x\in y$ implies $x\in X$. Thus $y\subseteq X\subseteq V_\alpha$, and we have $y\in V_{\alpha^+}$. Therefore, we have $Y\subseteq V_{\alpha^+}$, and thus $Y\in V_{\alpha^{++}}\subseteq V_\kappa$. Let $y\in V_\kappa$. If $y\in Y$, then $y\in V$ and $y\in Y$, thus for all $x\in V$, $x\in y$ implies $x\in X$, hence for all $x\in V_\kappa$, $x\in y$ implies $x\in X$. If for all $x\in V_\kappa$, $x\in y$ implies $x\in X$, then let $x\in V$, if $x\in y$, then $x\in V_\kappa$, thus $x\in y$ implies $x\in X$. Since we also have $y\in V$, we have $y\in Y$.
• Axiom of infinity: We showed above that $N\in V_\kappa$. Let $z\in V_\kappa$, if for all $y\in V_\kappa$, not $y\in z$, then given $u\in V$, either $u\in V_\kappa$, then $u\notin z$, or $u\notin V_\kappa$, then also $u\notin z$, hence $z=\emptyset$. Since $\emptyset\in N$, we have $z\in N$. Let $x\in V_\kappa$. Suppose $x\in N$. Let $X\in V_\kappa$. Suppose for all $u\in V_\kappa$, $u\in X$ if and only if $u\in x$ or $u=x$. Let $u\in V$. If $u\in X$, then $u\in V_\kappa$ and $u\in X$, thus $u\in x$ or $u=x$, implying $u\in S(x)$. If $u\in S(x)$, then $u\in x$ or $u=x$, in both cases we also have $u\in V_\kappa$, thus $u\in X$. Hence $X=S(x)$. Since $x\in N$, we have $S(x)\in N$, and thus $X\in N$.
• Axiom schema of replacement: Let $n$ be a natural number and $\varphi$ be a formula as required. Fix an assignment function for the variable symbols $w_1,\ldots,w_n$. Let $X\in V_\kappa$. Suppose for all $x\in V_\kappa$, if $x\in X$ then there exists a unique $y\in V_\kappa$ such that we have $\varphi$. Let $x\in V$. If $x\in X$, then $x\in V_\kappa$ and $x\in X$, thus there exists a unique $y\in V_\kappa$ such that we have $\varphi$. Now consider the formula $(y\in V_\kappa)\land\varphi$, denoted $\varphi'$. We can see that the unique $y\in V_\kappa$ that satisfies $\varphi$ also uniquely satisfies $\varphi'$ in $V$. Thus there exists $Y\in V$ such that for all $x\in V$, if $x\in X$ then there exists $y\in V$ such that $y\in Y$ and $\varphi'$. Then we can see that for some $Y\in V$, for all $x\in X$, there exists a unique $y\in Y$ such that $\varphi'$. Note that $\varphi'(x,y)$ defines a function $f:X\to Y$. Let $Y'$ denote the range of $f$, then $f$ is a surjection from $X$ to $Y'$, thus $\abs{X}\ge\abs{Y'}$. Since $X\in V_\kappa$, we have $\abs{X}\lt\kappa$, thus $\abs{Y'}\lt\kappa$. If $y'\in Y'$, then there exists $x'\in V$ such that $x'\in X$ and $f(x')=y'$. And we have $\varphi'(x',f(x'))$ and thus $\varphi'(x',y')$, implying $y'\in V_\kappa$. Therefore, $Y'\subseteq V_\kappa$. Since $Y'\subseteq V_\kappa$ and $\abs{Y'}\lt\kappa$, by a lemma above, we have $Y'\in V_\kappa$. Let $x\in V_\kappa$. If $x\in X$, then $x\in V$ and $x\in X$, thus there exists $y\in V$ such that $y\in Y$ and $\varphi'$, implying $y\in V_\kappa$ and $\varphi$.
• Axiom of foundation: Let $S\in V_\kappa$. Suppose there exists $x\in V_\kappa$ such that $x\in S$, then there exists $x\in V$ such that $x\in S$. Thus there exists $s\in V$ such that $s\in S$ and there does not exist $y\in V$ such that $y\in S$ and $y\in s$. Hence $s\in V_\kappa$ and there does not exist $y\in V_\kappa$ such that $y\in S$ and $y\in s$.
• Witness axioms: These are trivial, since witness functions are interpreted to satisfy witness axioms. Just use the fact that the witness term is substitutable for the substituted variable symbol in $\varphi$ and the results we had for LA2.

$\blacksquare$