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Let $V$ be the $V$ from $A$. It is clear $V$ satisfies exstensionality, seperation, and regularity. Pairing follows from reflection ($x=a\lor x=b$). Inductively, we can prove each $n\in V$, and so we have $\omega\in V$. If $X\in V$, then $\cup X\subseteq V$. Then define $\phi(x)\leftrightarrow \exists y(y\in X\land x\in y)$. By reflection $\cup X=\{x|\phi(x)\}$, and so $\cup X\in V$. A similar argument goes for powerset. For replacement, let $F=\{(x,y)|\phi(x,y)\}$ be a function. Then $F(X)\subseteq V$ (By definition). Then we can find some $V_\alpha$ that reflects $\phi(x,y)$ relative to $V$, and $x\in F(X)\leftrightarrow \exists y(y\in X\land\phi(x,y)^{V_\alpha})$, and so $F(X)\in V$. Therefore $V\vDash \phi$ for each axiom $\phi$ of $ZFC$. Now suppose an axiom of $ZFC\,\phi$, satisfies $\phi\vdash\psi$. Then every model of $\phi$ satisfies $\psi$, and so $V\vDash\psi$.

Proof the $V$ satisfies the reflection theorem: Note that the statement $x\in V_\alpha$ is $\Delta_1$, and so $V_\alpha^V=V_\alpha$. In addition, the mapping $\phi(x,\alpha)\leftrightarrow rank(x)<\alpha$ is $\Delta_1$, and so $\phi(x,\alpha)^V\leftrightarrow\phi(x,\alpha)$. Then for any $\alpha\in V$, $V_\alpha=\{x|rank(x)<\alpha\}$, and so $V\vDash Replacement\,for\,\alpha\mapsto V_\alpha$. As a consequence, for any $C\subseteq V$, the set $\hat C=\{x\in C|\forall y\in C(rank(x)\leq rank(y))\}$ is in $V$.

Let $H(u_0...u_n)=\hat C$ and $C=\{x|\phi^V(x_0...x_n)\}$. Then, for any set $M_0\in V$, define a sequence starting at $M_0$ with $M_{i+1}=M_i\cup(\cup\{H(u_0...u_n)|u_0...u_n\in M_i\})$. The function $F^V:{i\mapsto V_{ran(M_i)+1}}$ can be defined inductively, and it is easy to see that if $F(i)\in V$, $F(i+1)\in V$. Then $D=\cup \{H(u_0...u_n)|u_0...u_n\in\cup F(\omega)\}$ is in $V$. Then $x\in F(i)\leftrightarrow (i=0\land x\in M_0)\lor (\exists j(j+1=i\land x\in F(i)\cup ((D'\restriction F(i))(F(i))))$, where $D'$ is the union of the class of all sets such that each $x\in D'$ is $\hat C$ for some $C$ and $\cup D'=D$. And so $V_\alpha=\cup F(\omega)$ is in $V$ and $V_\alpha\vDash\phi\leftrightarrow V\vDash\phi$ (By induction on formula complexity).

Let $V$ be the $V$ from $A$. It is clear $V$ satisfies exstensionality, seperation, and regularity. Pairing follows from reflection ($x=a\lor x=b$). Inductively, we can prove each $n\in V$, and so we have $\omega\in V$. If $X\in V$, then $\cup X\subseteq V$. Then define $\phi(x)\leftrightarrow \exists y(y\in X\land x\in y)$. By reflection $\cup X=\{x|\phi(x)\}$, and so $\cup X\in V$. A similar argument goes for powerset. For replacement, let $F=\{(x,y)|\phi(x,y)\}$ be a function. Then $F(X)\subseteq V$ (By definition). Then we can find some $V_\alpha$ that reflects $\phi(x,y)$ relative to $V$, and $x\in F(X)\leftrightarrow \exists y(y\in X\land\phi(x,y)^{V_\alpha})$, and so $F(X)\in V$. Therefore $V\vDash \phi$ for each axiom $\phi$ of $ZFC$. Now suppose an axiom of $ZFC\,\phi$, satisfies $\phi\vdash\psi$. Then every model of $\phi$ satisfies $\psi$, and so $V\vDash\psi$.

Proof the $V$ satisfies the reflection theorem: Note that the statement $x\in V_\alpha$ is $\Delta_1$, and so $V_\alpha^V=V_\alpha$. In addition, the mapping $\phi(x,\alpha)\leftrightarrow rank(x)<\alpha$ is $\Delta_1$, and so $\phi(x,\alpha)^V\leftrightarrow\phi(x,\alpha)$. Then for any $\alpha\in V$, $V_\alpha=\{x|rank(x)<\alpha\}$, and so $V\vDash Replacement\,for\,\alpha\mapsto V_\alpha$. As a consequence, for any $C\subseteq V$, the set $\hat C=\{x\in C|\forall y\in C(rank(x)\leq rank(y))\}$ is in $V$.

Let $H(u_0...u_n)=\hat C$ and $C=\{x|\phi^V(x_0...x_n)\}$. Then, for any set $M_0\in V$, define a sequence starting at $M_0$ with $M_{i+1}=M_i\cup(\cup\{H(u_0...u_n)|u_0...u_n\in M_i\})$. The function $F^V:{i\mapsto V_{ran(M_i)+1}}$ can be defined inductively, and it is easy to see that if $F(i)\in V$, $F(i+1)\in V$. Then $D=\cup \{H(u_0...u_n)|u_0...u_n\in\cup F(\omega)\}$ is in $V$. Then $x\in F(i)\leftrightarrow (i=0\land x\in M_0)\lor (\exists j(j+1=i\land x\in F(i)\cup ((D'\restriction F(i))(F(i))))$, where $D'$ is the union of the class of all sets such that each $x\in D'$ is $\hat C$ for some $C$ and $\cup D'=D$. What I mean by $(D'\restriction F(i))(F(i))$, is really $(G\restriction F(i))(F(i))$, where $G=\{(x,u_0...u_n)|x=H(u_0...u_n)\land u_0...u_n\in F(i)\}$. And so $V_\alpha=\cup F(\omega)$ is in $V$ and $V_\alpha\vDash\phi\leftrightarrow V\vDash\phi$ (By induction on formula complexity).

Let $V$ be the $V$ from $A$. It is clear $V$ satisfies exstensionality, seperation, and regularity. Pairing follows from reflection ($x=a\lor x=b$). Inductively, we can prove each $n\in V$, and so we have $\omega\in V$. If $X\in V$, then $\cup X\subseteq V$. Then define $\phi(x)\leftrightarrow \exists y(y\in X\land x\in y)$. By reflection $\cup X=\{x|\phi(x)\}$, and so $\cup X\in V$. A similar argument goes for powerset. For replacement, let $F=\{(x,y)|\phi(x,y)\}$ be a function. Then $F(X)\subseteq V$ (By definition). Then we can find some $V_\alpha$ that reflects $\phi(x,y)$ relative to $V$, and $x\in F(X)\leftrightarrow \exists y(y\in X\land\phi(x,y)^{V_\alpha})$, and so $F(X)\in V$. Therefore $V\vDash \phi$ for each axiom $\phi$ of $ZFC$. Now suppose an axiom of $ZFC\,\phi$, satisfies $\phi\vdash\psi$. Then every model of $\phi$ satisfies $\psi$, and so $V\vDash\psi$.

Proof the $V$ satisfies the reflection theorem: Note that the statement $x\in V_\alpha$ is $\Delta_1$, and so $V_\alpha^V=V_\alpha$. In addition, the mapping $\phi(x,\alpha)\leftrightarrow rank(x)<\alpha$ is $\Delta_1$, and so $\phi(x,\alpha)^V\leftrightarrow\phi(x,\alpha)$. Then for any $\alpha\in V$, $V_\alpha=\{x|rank(x)<\alpha\}$, and so $V\vDash Replacement\,for\,\alpha\mapsto V_\alpha$. As a consequence, for any $C\subseteq V$, the set $\hat C=\{x\in C|\forall y\in C(rank(x)\leq rank(y))\}$ is in $V$.

Let $H(u_0...u_n)=\hat C$ and $C=\{x|\phi^V(x_0...x_n)\}$. Then, for any set $M_0\in V$, define a sequence starting at $M_0$ with $M_{i+1}=M_i\cup(\cup\{H(u_0...u_n)|u_0...u_n\in M_i\})$. The function $F^V:{i\mapsto V_{ran(M_i)+1}}$ can be defined inductively, and it is easy to see that if $F(i)\in V$, $F(i+1)\in V$. Then $D=\cup \{H(u_0...u_n)|u_0...u_n\in\cup F(\omega)\}$ is in $V$. Then $x\in F(i)\leftrightarrow (i=0\land x\in M_0)\lor (\exists j(j+1=i\land x\in F(i)\cup (D'\restriction F(i)))$, where $D'$ is the union of the class of all sets such that each $x\in D'$ is $\hat C$ for some $C$ and $\cup D'=D$. And so $V_\alpha=\cup F(\omega)$ is in $V$ and $V_\alpha\vDash\phi\leftrightarrow V\vDash\phi$ (By induction on formula complexity).

Let $V$ be the $V$ from $A$. It is clear $V$ satisfies exstensionality, seperation, and regularity. Pairing follows from reflection ($x=a\lor x=b$). Inductively, we can prove each $n\in V$, and so we have $\omega\in V$. If $X\in V$, then $\cup X\subseteq V$. Then define $\phi(x)\leftrightarrow \exists y(y\in X\land x\in y)$. By reflection $\cup X=\{x|\phi(x)\}$, and so $\cup X\in V$. A similar argument goes for powerset. For replacement, let $F=\{(x,y)|\phi(x,y)\}$ be a function. Then $F(X)\subseteq V$ (By definition). Then we can find some $V_\alpha$ that reflects $\phi(x,y)$ relative to $V$, and $x\in F(X)\leftrightarrow \exists y(y\in X\land\phi(x,y)^{V_\alpha})$, and so $F(X)\in V$. Therefore $V\vDash \phi$ for each axiom $\phi$ of $ZFC$. Now suppose an axiom of $ZFC\,\phi$, satisfies $\phi\vdash\psi$. Then every model of $\phi$ satisfies $\psi$, and so $V\vDash\psi$.

Proof the $V$ satisfies the reflection theorem: Note that the statement $x\in V_\alpha$ is $\Delta_1$, and so $V_\alpha^V=V_\alpha$. In addition, the mapping $\phi(x,\alpha)\leftrightarrow rank(x)<\alpha$ is $\Delta_1$, and so $\phi(x,\alpha)^V\leftrightarrow\phi(x,\alpha)$. Then for any $\alpha\in V$, $V_\alpha=\{x|rank(x)<\alpha\}$, and so $V\vDash Replacement\,for\,\alpha\mapsto V_\alpha$. As a consequence, for any $C\subseteq V$, the set $\hat C=\{x\in C|\forall y\in C(rank(x)\leq rank(y))\}$ is in $V$.

Let $H(u_0...u_n)=\hat C$ and $C=\{x|\phi^V(x_0...x_n)\}$. Then, for any set $M_0\in V$, define a sequence starting at $M_0$ with $M_{i+1}=M_i\cup(\cup\{H(u_0...u_n)|u_0...u_n\in M_i\})$. The function $F^V:{i\mapsto V_{ran(M_i)+1}}$ can be defined inductively, and it is easy to see that if $F(i)\in V$, $F(i+1)\in V$. Then $D=\cup \{H(u_0...u_n)|u_0...u_n\in\cup F(\omega)\}$ is in $V$. Then $x\in F(i)\leftrightarrow (i=0\land x\in M_0)\lor (\exists j(j+1=i\land x\in F(i)\cup ((D'\restriction F(i))(F(i))))$, where $D'$ is the union of the class of all sets such that each $x\in D'$ is $\hat C$ for some $C$ and $\cup D'=D$. And so $V_\alpha=\cup F(\omega)$ is in $V$ and $V_\alpha\vDash\phi\leftrightarrow V\vDash\phi$ (By induction on formula complexity).

Let $V$ be the $V$ from $A$. It is clear $V$ satisfies exstensionality, seperation, and regularity. Pairing follows from reflection ($x=a\lor x=b$). Inductively, we can prove each $n\in V$, and so we have $\omega\in V$. If $X\in V$, then $\cup X\subseteq V$. Then define $\phi(x)\leftrightarrow \exists y(y\in X\land x\in y)$. By reflection $\cup X=\{x|\phi(x)\}$, and so $\cup X\in V$. A similar argument goes for powerset. For replacement, let $F=\{(x,y)|\phi(x,y)\}$ be a function. Then $F(X)\subseteq V$ (By definition). Then we can find some $V_\alpha$ that reflects $\phi(x,y)$ relative to $V$, and $x\in F(X)\leftrightarrow \exists y(y\in X\land\phi(x,y)^{V_\alpha})$, and so $F(X)\in V$. Therefore $V\vDash \phi$ for each axiom $\phi$ of $ZFC$. Now suppose an axiom of $ZFC\,\phi$, satisfies $\phi\vdash\psi$. Then every model of $\phi$ satisfies $\psi$, and so $V\vDash\psi$.

Proof the $V$ satisfies the reflection theorem: Note that the statement $x\in V_\alpha$ is $\Delta_1$, and so $V_\alpha^V=V_\alpha$. In addition, the mapping $\phi(x,\alpha)\leftrightarrow rank(x)<\alpha$ is $\Delta_1$, and so $\phi(x,\alpha)^V\leftrightarrow\phi(x,\alpha)$. Then for any $\alpha\in V$, $V_\alpha=\{x|rank(x)<\alpha\}$, and so $V\vDash Replacement\,for\,\alpha\mapsto V_\alpha$. As a consequence, for any $C\subseteq V$, the set $\hat C=\{x\in C|\forall y\in C(rank(x)\leq rank(y))\}$ is in $V$.

Let $H(u_0...u_n)=\hat C$ and $C=\{x|\phi^V(x_0...x_n)\}$. Then, for any set $M_0\in V$, define a sequence starting at $M_0$ with $M_{i+1}=M_i\cup(\cup\{H(u_0...u_n)|u_0...u_n\in M_i\})$. The function $F^V:{i\mapsto V_{ran(M_i)+1}}$ can be defined inductively, and it is easy to see that if $F(i)\in V$, $F(i+1)\in V$. Then $D=\cup \{H(u_0...u_n)|u_0...u_n\in\cup F(\omega)\}$ is in $V$. Then $x\in F(i)\leftrightarrow (i=0\land x\in M_0)\lor (\exists j(j+1=i\land x\in F(i)\cup (D\restriction F(i)))$. And so $V_\alpha=\cup F(\omega)$ is in $V$ and $V_\alpha\vDash\phi\leftrightarrow V\vDash\phi$ (By induction on formula complexity).

Let $V$ be the $V$ from $A$. It is clear $V$ satisfies exstensionality, seperation, and regularity. Pairing follows from reflection ($x=a\lor x=b$). Inductively, we can prove each $n\in V$, and so we have $\omega\in V$. If $X\in V$, then $\cup X\subseteq V$. Then define $\phi(x)\leftrightarrow \exists y(y\in X\land x\in y)$. By reflection $\cup X=\{x|\phi(x)\}$, and so $\cup X\in V$. A similar argument goes for powerset. For replacement, let $F=\{(x,y)|\phi(x,y)\}$ be a function. Then $F(X)\subseteq V$ (By definition). Then we can find some $V_\alpha$ that reflects $\phi(x,y)$ relative to $V$, and $x\in F(X)\leftrightarrow \exists y(y\in X\land\phi(x,y)^{V_\alpha})$, and so $F(X)\in V$. Therefore $V\vDash \phi$ for each axiom $\phi$ of $ZFC$. Now suppose an axiom of $ZFC\,\phi$, satisfies $\phi\vdash\psi$. Then every model of $\phi$ satisfies $\psi$, and so $V\vDash\psi$.

Proof the $V$ satisfies the reflection theorem: Note that the statement $x\in V_\alpha$ is $\Delta_1$, and so $V_\alpha^V=V_\alpha$. In addition, the mapping $\phi(x,\alpha)\leftrightarrow rank(x)<\alpha$ is $\Delta_1$, and so $\phi(x,\alpha)^V\leftrightarrow\phi(x,\alpha)$. Then for any $\alpha\in V$, $V_\alpha=\{x|rank(x)<\alpha\}$, and so $V\vDash Replacement\,for\,\alpha\mapsto V_\alpha$. As a consequence, for any $C\subseteq V$, the set $\hat C=\{x\in C|\forall y\in C(rank(x)\leq rank(y))\}$ is in $V$.

Let $H(u_0...u_n)=\hat C$ and $C=\{x|\phi^V(x_0...x_n)\}$. Then, for any set $M_0\in V$, define a sequence starting at $M_0$ with $M_{i+1}=M_i\cup(\cup\{H(u_0...u_n)|u_0...u_n\in M_i\})$. The function $F^V:{i\mapsto V_{ran(M_i)+1}}$ can be defined inductively, and it is easy to see that if $F(i)\in V$, $F(i+1)\in V$. Then $D=\cup \{H(u_0...u_n)|u_0...u_n\in\cup F(\omega)\}$ is in $V$. Then $x\in F(i)\leftrightarrow (i=0\land x\in M_0)\lor (\exists j(j+1=i\land x\in F(i)\cup (D'\restriction F(i)))$, where $D'$ is the union of the class of all sets such that each $x\in D'$ is $\hat C$ for some $C$ and $\cup D'=D$. And so $V_\alpha=\cup F(\omega)$ is in $V$ and $V_\alpha\vDash\phi\leftrightarrow V\vDash\phi$ (By induction on formula complexity).