Relativisation is Standard Model

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

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


 * $\displaystyle A \subseteq B \implies ( B \models P \iff P^B )$

Proof
The proof shall proceed by induction on the well-formed parts of $P$. $P$ must be of the form $x \in y$, $( Q \land R )$, $\neg Q$, or $\forall x: Q$ for some propositions $Q$ and $R$.

Atoms
Suppose $P$ is of the form $x \in y$. Then, $A = \{x,y\}$.


 * $\displaystyle B \models P \iff ( x \in y \land x \in B \land y \in B )$ by Definition:Standard Structure


 * $\displaystyle P^B \iff x \in y$ by Definition:Relativisation

If $A \subseteq B$, then $x \in B \land y \in B$ and the two statements are equivalent.

Inductive Step for $\neg$
Suppose $P$ is 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 $( Q \implies R )$.

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 $\forall x:$
Suppose that $P$ is of the form $\forall x: Q$. Suppose that the statement holds for $Q$.

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

Furthermore, $x \notin A$ because:


 * $\displaystyle x \in A \implies x$ is a free variable in $\displaystyle \forall x: Q$, which is a contradiction.