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 $C\in V$.  the set $\{x\in C|\forall y\in C(rank(x)\leq  rank(y))\}\in V$. From here, repeat the standard proof of the reflection theorem.