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 [Definition:Logical Formula|$\mathcal L$-[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 $\mathcal M$.

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 $\mathcal N$ 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 subset $\mathcal M$.

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.

Proof of Generalization
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 $\mathcal N \models \phi(m, \bar a)$.

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

Then, we only 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.

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 $\phi(x, \bar v)$ 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 $\mathcal N \models \phi(m, \bar a)$, is closed under functions (by directly applying the conditon to formulae of the form $\phi(x,\bar y)=(x=f(\bar y))$), and hence the universe of a substructure, which reduces it to the statement above.