=== Sufficient Condition === Let $\mathcal M$ be an elementary substructure of $\mathcal N$. Then $\mathcal N \models \exists x \phi \left({x, \bar a}\right)$ implies that $\mathcal M \models \exists x \phi \left({x, \bar a}\right)$. Hence there exists some $m$ in $\mathcal M$ such that: :$\mathcal M \models\phi \left({m, \bar a}\right)$. Passing back up to $\mathcal N$ yields the result. {{explain|Worth making it explicit exactly what that result is, so as to clarify what is being sought here.}} {{qed|lemma}} === Necessary Condition === Let $\mathcal M$ be such that: :For every $\mathcal L$-formula $\phi \left({x, \bar v}\right)$ and for every $\bar a$ in $\mathcal M$: ::if there exists an $n$ in $\mathcal N$ such that $\mathcal N \models \phi \left({n, \bar a}\right)$ ::then there exists an $m$ in $\mathcal M$ such that $\mathcal N \models \phi \left({m, \bar a}\right)$. It is to be shown that $\mathcal M$ is an elementary substructure of $\mathcal N$. That is: :for every $\mathcal L$-formula $\psi \left({\bar v}\right)$ and: :for every $\bar a$ in $\mathcal M$: ::$\left({\mathcal M \models \phi \left({\bar a}\right)}\right) \iff \left({\mathcal N \models\phi \left({\bar a}\right)}\right)$ The proof proceeds by induction on complexity of formulas. Let $\psi$ be quantifier free. From Quantifier Free Formula is Preserved by Superstructure, quantifier free formulas with parameters from $\mathcal M$ are preserved when passing to and from superstructures. Hence the result holds. Let the result holds for $\psi$. Consider $\neg \psi$: We have that: :$\left({\mathcal M \models \neg \psi \left({\bar a}\right)}\right) \iff \left({\mathcal M \not \models \psi \left({\bar a}\right)}\right)$ By the inductive hypothesis: :$\left({\mathcal M \not\models\psi \left({\bar a}\right)}\right) \iff \left({\mathcal N \not\models\psi \left({\bar a}\right)}\right)$ Thus the result follows for $\neg \psi$. Let the result hold for $\psi_0$ and $\psi_1$. Consider $\psi_0 \wedge \psi_1$: We have: :$\mathcal M \models \psi_0 \left({\bar a}\right) \wedge \psi_1 \left({\bar a}\right)$ {{iff}}: :$\mathcal M \models\psi_0 \left({\bar a}\right)$ and: :$\mathcal M \models \psi_1 \left({\bar a}\right)$ By the inductive hypothesis: :$\mathcal M \models\psi_0 \left({\bar a}\right)$ and $\mathcal M \models \psi_1 \left({\bar a}\right)$ {{iff}}: :$\mathcal N \models\psi_0 \left({\bar a}\right)$ and $\mathcal N \models \psi_1 \left({\bar a}\right)$ Thus the result follows for $\psi_0 \wedge \psi_1$. Let the result holds for $\psi$. Consider: :$\exists x: \psi \left({x}\right)$ We prove the two directions of this case separately. First let: :$\mathcal M \models \exists x: \psi \left({x, \bar a}\right)$ Then there exists $m \in \mathcal M$ such that: :$\mathcal M \models \psi \left({m, \bar a}\right)$ By the inductive hypothesis: :$\mathcal N \models \psi \left({m, \bar a}\right)$ and so: :$\mathcal N \models \exists x: \psi \left({x, \bar a}\right)$ Conversely, let: :$\mathcal N \models \exists x: \psi \left({x, \bar a}\right)$ By assumption, there exists $m \in \mathcal M$ such that: :$\mathcal N \models \psi \left({m, \bar a}\right)$ By the inductive hypothesis: :$\mathcal M \models \psi \left({m, \bar a}\right)$ and hence: :$\mathcal M \models \exists x: \psi \left({x, \bar a}\right)$ completing the proof. {{qed}}