Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma

Theorem
Let $\LL$ be a language of predicate logic.

Let $T$ be a finitely satisfiable $\LL$-theory.

Let $\phi$ be an $\LL$-sentence.

Then either:
 * $T \cup \set \phi$

or:
 * $T \cup \set {\neg \phi}$

is finitely satisfiable.

Proof
, suppose that $T \not\models \phi$ and $T \not\models \neg \phi$.

that $T \cup \set \phi$ is not finitely satisfiable.

Then by definition there must be a finite subset $K$ of $T \cup \set \phi$ which is not satisfiable.

Since $T$ is finitely satisfiable, $\phi \in K$.

Therefore $\Delta = K \setminus \set \phi$ is a finite subset of $T$.

Thus, for every $\LL$-structure $\MM$ that models $\Delta$, $\phi$ is not valid.

Hence, for every $\LL$-structure $\MM$ that models $\Delta$, $\neg \phi$ is valid.

By definition, then, $\Delta \models \neg \phi$
 * where $\models$ denotes semantic consequence.

Let $\Sigma$ be a finite subset of $T$.

Since $\Delta \cup \Sigma$ is a finite subset of $T$, it is satisfiable.

But $\Delta \models \neg \phi$.

Hence $\set {\neg \phi} \cup \Sigma$ is satisfiable.

But any finite subset of $T \cup \set {\neg \phi}$ is either in $T$ or of the form $\Sigma \cup \set {\neg \phi}$.

This demonstrates that $T \cup \set {\neg \phi}$ is finitely satisfiable, contradicting our assumption.