Relativisation is Standard Model

Theorem
Let $P$ be a well-formed formula.

Let $A$ be a finite set such that $x \in A$ $x$ is a free variable in $P$.

Then:
 * $A \subseteq B \implies \paren {B \models P \iff P^B}$

Proof
The proof proceeds by Induction on Well-Formed Formulas of $P$.

$P$ must be of the form:
 * $x \in y$
 * $\paren {Q \land R}$
 * $\neg Q$

or:
 * $\forall x: Q$

for some propositions $Q$ and $R$.

Atoms
Let $P$ be of the form $x \in y$.

Then:
 * $A = \set {x, y}$

By definition of standard structure:
 * $B \models P \iff \paren {x \in y \land x \in B \land y \in B}$

By definition of relativisation:
 * $P^B \iff x \in y$

If $A \subseteq B$, then:
 * $x \in B \land y \in B$

and the two statements are equivalent.

Inductive Step for $\neg$
Let $P$ be of the form $\neg Q$ and that the statement holds for $Q$.

Then the free variables in $Q$ are precisely those in $P$.

Inductive Step for $\implies$
Suppose $P$ is of the form $\paren {Q \implies R}$.

Suppose, further, that the statement holds for $Q$ and $R$.

The free variables of $Q$ and $R$ have to all be members of $A$, and thus are members of $B$ since $A \subseteq B$.

Inductive Step for $\forall x:$
Let $P$ be of the form:
 * $\forall x: Q$

Let the statement hold for $Q$.

Then the free variables of $Q$ are either in $A$ or $x$.

Furthermore, $x \notin A$ because:
 * $x \in A \implies x$ is a free variable in $\forall x: Q$

which is a contradiction.