Prove this proposition concerning a theory with ∀∃-axiomatization Setting
A theory $\pmb{T}$ has a $\forall\exists$-axiomatization if it can be axiomatized by sentences of the form $$\forall v_1\ldots \forall v_n \exists w_1 \ldots \exists w_n ~~ \phi(\bar{v},\bar{w})$$ where $\phi$ is a quantifier free $\mathcal{L}$-formula. 
Furthermore, suppose whenever $(\mathcal{M}_i : i \in \mathbb{I})$ is a chain of models of $\pmb{T}$, then $$\mathcal{M} = \bigcup \mathcal{M}_i \models \pmb{T}.$$ Let $\Gamma = \{ \phi : \phi \text{ is a $\forall\exists$-sentence and $\pmb{T} \models \phi$}\}$. Let $\mathcal{M} \models \Gamma$. 
Finally, suppose there is $\mathcal{N} \models \pmb{T}$ such that if $\psi$ is an $\exists\forall$-sentence and $\mathcal{M} \models \psi$, then $\mathcal{N} \models \psi$.
Now I would like to show that there is $\mathcal{N}' \supseteq \mathcal{M} $ with $\mathcal{N}' \equiv \mathcal{N}$.
Problem


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*First, I am not sure if the assumption is that $\mathcal{M} \subseteq \mathcal{N}'$? I am guessing since you can always create an extension, then $\mathcal{N}'$ is assumed to exist?

*Suppose my presumption is correct so that $\mathcal{N}'$ exist, but as an extension of $\mathcal{M}$, it does not need to satisfy the same sentences correct? So why is it true that $\mathcal{N}' \equiv \mathcal{N}$?

*Finally, if I am reading the proposition totally wrong, and instead I just have to show/construct some $\mathcal{N}'$ so that $\mathcal{N}' \equiv \mathcal{N}$, how would I go about doing such a construction?
 A: Please excuse my barbaric notation and terminology; it has been years since I sat in a logic class. I will sketch a proof of the fact that, if $\mathcal M,\mathcal N$ are models such that every existential sentence which holds in $\mathcal M$ also holds in $\mathcal N$, then there is a model $\mathcal N'$ such that $\mathcal M$ is isomorphic to a submodel of $\mathcal N'$ and $\mathcal N$ is isomorphic to an elementary submodel of $\mathcal N'$.
Let $U$ be the first-order theory of $\mathcal N$ in an expanded language which has a constant symbol for each element of $\mathcal N$. Thus any model of $U$ (more precisely, its reduct to the original language) will contain an isomorphic copy of $\mathcal N$ as an elementary submodel.
Let $Q$ be the quantifier-free theory of $\mathcal M$ in an expanded language which has a constant symbol for each element of $\mathcal M$. Thus any model of $Q$ will contain a submodel isomorphic to $\mathcal M$.
Since any model $\mathcal N'$ of $U\cup Q$ will do what we want, all we have left to show is that $U\cup Q$ is consistent. By the compactness theorem, it will suffice to show that $U\cup Q'$ is consistent for each finite set $Q'\subseteq Q$. This follows from the hypothesis that every existential sentence which holds in $\mathcal M$ also holds in $\mathcal N$. [*]
P.S. By virtue of the Keisler-Shelah theorem (elementarily equivalent models have isomorphic ultrapowers) the statement can be strengthened to read: if every existential sentence which holds in $\mathcal M$ also holds in $\mathcal N$, then $\mathcal M$ is isomorphic to a submodel of an ultrapower of $\mathcal N$.
[*] Consider a finite set $Q'\subseteq Q$. By forming a conjunction, we may assume that $Q'$ consists of a single quantifier-free sentence $\varphi(c_1,\dots,c_n)$, where $c_1,\dots,c_n$ are constants in the expanded language for $\mathcal M$. Since $\varphi(c_1,\dots,c_n)$ holds in $\mathcal M$, so does the existential semtemce $\exists x_1,\dots,x_n)\varphi(x_1,\dots,x_n)$ of the original language. Hence the sentence $\exists x_1,\dots,x_n)\varphi(x_1,\dots,x_n)$ also holds in $\mathcal N$. Therefore, the constants $c_,\dots,c_n$ can be assigned values in $\mathcal N$ so as to satisfy $\varphi(c_1,\dots,c_n)$, showing that $U\cup Q'$ is consistent.
