Logical consequence using Replacement Axiom Let $\mathfrak{A}=<A,E>$ be a countable model of $\mathrm{ZF}$. And $\mathfrak{A}_{A}=(\mathfrak{A},a)_{a\in A}$ is defined to be the model for the language $\{\in\}\cup \{c_{a}:a\in A\}$ by interpreting each new constant $c_{a}$ by the element $a$.
How can we show that for every $\phi$, if $\mathfrak{A}_{A}\vDash \forall y\exists z\exists x[z\not\in y\land\phi(x,z)\land x\in c_{a}]$, then $\exists x\forall y\exists z[z\not\in y\land\phi(x,z)\land x\in c_{a}]$ is also true? In Chang and Keisler's model theory, they says it can be shown by using the axiom of replacement in $\mathrm{ZF}$. But there is no further argument. I have no idea..
 A: Here's the idea of what's going on: The first sentence essentially says "the collection of all $z$ which are related to some $x\in a$ by $\phi$ is a proper class". The second sentence essentially says "there is some $x\in a$ such that the collection of all $z$ which are related to $x$ by $\phi$ is a proper class". Prove the contrapositive: If for every $x\in a$, the collection of all $z$ which are related to $x$ by $\phi$ is a set, then we can gather up all these $z$'s into a single set by Replacement and Union, i.e., the collection of all $z$ which are related to some $x\in a$ by $\phi$ is a set.
More formally: Suppose $$\mathfrak{A}_A\models \forall x\exists y\forall z[(x\in c_a\land \phi(x,z))\rightarrow z\in y].$$ Then for each $x\in a$, there is some set $b_x$ which contains all $z$ such that $\phi(x,z)$. Let $b_x' = \{z\in b_x\mid \phi(x,z)\}$. Then $b_x'$ is a set by Separation, and we can define the function $x\mapsto b_x'$ by $\forall z\,(z\in y \leftrightarrow \phi(x,z))$. By Replacement, there is a set $c = \{b_x'\mid x\in a\}$. Let $d = \bigcup c$. Then $d$ witnesses $$\mathfrak{A}_A\models \exists y \forall z\forall x[(x\in c_a\land \phi(x,z))\rightarrow z\in y].$$
