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 $\mathcal N$ :
 * 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 $\mathcal N \models \phi \left({m, \bar a}\right)$.

The condition on the right side of the 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 $\mathcal M$.

Generalization
Let $\mathcal N$ be an $\mathcal L$-structure, and $\mathcal M \subseteq \mathcal N$.

Then $\mathcal M$ is the universe of an elementary substructure of $\mathcal N$ :
 * 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 $\mathcal N \models \phi \left({m, \bar a}\right)$.

The condition on the right side of the 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 subset $\mathcal M$.

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.

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 Formulas are Preserved by Superstructures, 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:


 * $\mathcal M \models \psi_0 \left({\bar a}\right) \wedge \psi_1 \left({\bar a}\right)$


 * $\mathcal M \models\psi_0 \left({\bar a}\right)$
 * $\mathcal M \models\psi_0 \left({\bar a}\right)$

and:
 * $\mathcal M \models \psi_1 \left({\bar a}\right)$

By the inductive hypothesis:
 * $\mathcal M \models\psi_0 \left({\bar a}\right)$ and $\mathcal M \models \psi_1 \left({\bar a}\right)$


 * $\mathcal N \models\psi_0 \left({\bar a}\right)$ and $\mathcal N \models \psi_1 \left({\bar a}\right)$
 * $\mathcal N \models\psi_0 \left({\bar a}\right)$ 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.

Sufficient Condition
Let $\mathcal M$ be an elementary substructure of $\mathcal N$.

Then:
 * $\left({\mathcal N \models \exists x: \phi \left({x, \bar a}\right)}\right) \implies \left({\mathcal M \models \exists x \phi \left({x, \bar a}\right)}\right)$

Hence there exists $m \in \mathcal M$ such that:
 * $\mathcal M \models \phi \left({m, \bar a}\right)$

Passing back up to $\mathcal N$ yields the result.

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 easily shown, using formulae of the form $\phi \left({x, \bar v}\right) = f \left({\bar v}\right) = x$, that $\mathcal M$ is closed under functions, and hence the universe of a substructure.

It remains 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 $\psi$ is quantifier free.

From Quantifier Free Formulas are Preserved by Superstructures, quantifier free formulas with parameters from $\mathcal M$ are preserved when passing to and from superstructures.

Hence the result holds.

Let the result hold for $\psi$.

Consider $\neg \psi$:

We have:
 * $\left({\mathcal M \models \neg \psi \left({\bar a}\right)}\right) \iff \left({\mathcal M \not \models \psi \left({\bar a}\right)}\right)$

By the inductive hypothesis:
 * $\left({\mathcal M \not \models \psi \left({\bar a}\right)}\right) \iff \left({\mathcal N \not \models \psi \left({\bar a}\right)}\right)$

Hence the result follows for $\neg \psi$.

Suppose the result holds for $\psi_0$ and $\psi_1$.

Consider $\psi_0 \wedge \psi_1$:

We have that:
 * $\mathcal M \models \psi_0 \left({\bar a}\right) \wedge \psi_1 \left({\bar a}\right)$


 * $\mathcal M \models \psi_0 \left({\bar a}\right)$ and $\mathcal M \models \psi_1 \left({\bar a}\right)$
 * $\mathcal M \models \psi_0 \left({\bar a}\right)$ and $\mathcal M \models \psi_1 \left({\bar a}\right)$

By the inductive hypothesis:
 * $\mathcal M \models \psi_0 \left({\bar a}\right)$ and $\mathcal M \models \psi_1 \left({\bar a}\right)$


 * $\mathcal N \models \psi_0 \left({\bar a}\right)$ and $\mathcal N \models \psi_1 \left({\bar a}\right)$
 * $\mathcal N \models \psi_0 \left({\bar a}\right)$ and $\mathcal N \models \psi_1 \left({\bar a}\right)$

The result then follows for $\psi_0 \wedge \psi_1$.

Let the result hold 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:
 * $\mathcal N \models \exists x: \psi \left({x, \bar a}\right)$

Then 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.

Generalization
Another common statement of the theorem does not require $\mathcal M$ to be a substructure of $\mathcal N$, it is enough for it to be a subset of $\mathcal N$.

The ($\implies$) is shown just the same as above, while the other direction easily follows, since $\mathcal M$ satisfying the condition that for every $\mathcal L$-formula $\phi \left({x, \bar v}\right)$ and for every $\bar a$ in $\mathcal M$, if there is an $n$ in $\mathcal N$ such that $\mathcal N \models \phi \left({n, \bar a}\right)$, then there is an $m$ in $\mathcal M$ such that $\mathcal N \models \phi \left({m, \bar a}\right)$, is closed under functions (by directly applying the condition to formulae of the form $\phi \left({x, \bar y}\right) = \left({x = f \left({\bar y}\right)}\right)$), and hence the universe of a substructure, which reduces it to the statement above.