Tarski-Vaught Test

Theorem
Let $\mathcal{M},\mathcal{N}$ be $\mathcal{L}$-structures such that $\mathcal{M}$ is a substructure of $\mathcal{N}$.

Then $\mathcal{M}$ is an elementary substructure of $\mathcal{N}$ if and only if for every $\mathcal{L}$-formula $\phi(x, \bar{v})$ and for every $\bar{a}$ in $\mathcal{M}$, if there is an $n$ in $\mathcal{N}$ such that $\mathcal{N}\models \phi(n, \bar{a})$, then there is an $m$ in $\mathcal{M}$ such that $\mathcal{N}\models \phi(m, \bar{a})$.

The condition on the right side of the iff statement above can be rephrased as: every existential statement with parameters from $\mathcal{M}$ which is satisfied in $\mathcal{N}$ can be witnessed by an element from the substructure $\mathcal{M}$.

Proof
The direction ($\implies$) is easy: if $\mathcal{M}$ is an elementary substructure of $\mathcal{N}$, then $\mathcal{N}\models\exists x \phi(x, \bar{a})$ implies that $\mathcal{M}\models \exists x \phi(x, \bar{a})$. Hence, there is some $m$ in $\mathcal{M}$ such that $\mathcal{M}\models\phi(m, \bar{a})$. Passing back up to $\mathcal{N}$ yields the result.

For the converse, suppose that for every $\mathcal{L}$-formula $\phi(x, \bar{v})$ and for every $\bar{a}$ in $\mathcal{M}$, if there is an $n$ in $\mathcal{N}$ such that $\mathcal{N}\models \phi(n, \bar{a})$, then there is an $m$ in $\mathcal{M}$ such that $\mathcal{N}\models \phi(m, \bar{a})$.

We need to show that $\mathcal{M}$ is an elementary substructure of $\mathcal{N}$, that is, for every $\mathcal{L}$-formula $\psi(\bar{v})$ and for every $\bar{a}$ in $\mathcal{M}$, $\mathcal{M}\models\phi(\bar{a})$ iff $\mathcal{N}\models\phi(\bar{a})$.

The proof proceeds by induction on complexity of formulas.

If $\psi$ is quantifier free, then since quantifier free formulas with parameters from $\mathcal{M}$ are preserved when passing to and from superstructures, the result holds.

Suppose the result holds for $\psi$, and consider $\neg\psi$:

Since $\mathcal{M}\models\neg\psi(\bar{a})$ iff $\mathcal{M}\not\models\psi(\bar{a})$, and by the inductive hypothesis $\mathcal{M}\not\models\psi(\bar{a})$ iff $\mathcal{N}\not\models\psi(\bar{a})$, the result follows for $\neg\psi$.

Suppose the result holds for $\psi_0$ and $\psi_1$, and consider $\psi_0 \wedge \psi_1$:

Since $\mathcal{M}\models\psi_0 (\bar{a}) \wedge \psi_1 (\bar{a})$ iff $\mathcal{M}\models\psi_0 (\bar{a})$ and $\mathcal{M}\models \psi_1 (\bar{a})$, and by the inductive hypothesis $\mathcal{M}\models\psi_0 (\bar{a})$ and $\mathcal{M}\models \psi_1 (\bar{a})$ iff $\mathcal{N}\models\psi_0 (\bar{a})$ and $\mathcal{N}\models \psi_1 (\bar{a})$, the result follows for $\psi_0 \wedge \psi_1$.

Suppose the result holds for $\psi$, and consider $\exists x \psi(x)$:

We prove the two directions of this case separately.

First, suppose $\mathcal{M}\models\exists x \psi(x,\bar{a})$. Then there is some $m$ in $\mathcal{M}$ for which $\mathcal{M}\models\psi(m,\bar{a})$, and by the inductive hypothesis, this gives us $\mathcal{N}\models\psi(m,\bar{a})$ and so $\mathcal{N}\models\exists x \psi(x,\bar{a})$.

Conversely, suppose $\mathcal{N}\models\exists x \psi(x,\bar{a})$, then by assumption, there is an $m$ in $\mathcal{M}$ such that $\mathcal{N}\models\psi(m,\bar{a})$. By the inductive hypothesis, this gives us $\mathcal{M}\models\psi(m,\bar{a})$ and hence $\mathcal{M}\models\exists x \psi(x,\bar{a})$, completing the proof.