Finitely Satisfiable Set of Sentences has Maximal Finitely Satisfiable Extension

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

Let $\FF$ be a finitely satisfiable set of $\LL$-sentences.

Then there exists a finitely satisfiable set of $\LL$-sentences $\FF' \supseteq \FF$ such that:
 * For all $\LL$-sentences $\phi$, either $\phi \in \FF'$ or $\sqbrk {\neg \phi} \in \FF'$

Proof
Let $S$ be the set of all $\LL$-sentences.

For every finite subset $P \subseteq S$, define:
 * $M_P = \set {t \in \Bbb B^P : \exists \MM : \paren {\MM \models_{\mathrm {PL}} \FF \cap P} \land \paren {\forall \phi \in P : \map t \phi = T \iff \MM \models_{\mathrm {PL}} \phi}}$

Define:
 * $M = \bigcup_{P \mathop \in I} M_K$

where $I$ is the set of all finite subsets of $S$.

We want to show that $M$ is a binary mess over $S$.

For axiom $\text M 1$:
 * Let $P \in I$ be arbitrary.
 * As Intersection is Subset and Subset of Finite Set is Finite, $\FF \cap P$ is a finite subset of $\FF$.
 * As $\FF$ is finitely satisfiable, there is some $\LL$-structure $\MM_P$ such that $\MM_P \models_{\mathrm {PL}} \FF \cap P$.
 * Define $t_P : P \to \Bbb B$ as:
 * $\map {t_P} \phi = \begin{cases}

T & : \MM_P \models_{\mathrm {PL}} \phi \\ F & : \MM_P \not\models_{\mathrm {PL}} \phi \end{cases}$
 * As it satisfies the rule, $t_P \in M_P \subseteq M$, proving axiom $\text M 1$.

For axiom $\text M 2$:
 * Let $P \in I$ and $t \in M$ be arbitrary.
 * By definition, $t \in M_Q$ for some $Q \in I$.
 * Thus, there is some $\LL$-structure $\MM_Q$ such that:
 * $\MM_Q \models_{\mathrm {PL}} \FF \cap Q$
 * $\map t \phi = T \iff \MM_Q \models_{\mathrm {PL}} \phi$ for all $\phi \in Q$
 * As $\FF \cap Q \cap P \subseteq \FF \cap Q$, it follows that $\MM_Q \models \FF \cap Q \cap P$
 * As $Q \cap P \subseteq Q$, it follows that $\map t \phi = T \iff \MM_Q \models_{\mathrm {PL}} \phi$ for all $\phi \in Q \cap P$
 * Thus, $t \restriction_P$ satisfies the criteria for $M_{Q \cap P}$, and $t \restriction_P \in M$, proving axiom $\text M 2$.

By the Consistency Principle for Binary Mess:
 * There exists a mapping $f : S \to \Bbb B$ that is consistent with $M$.

Define:
 * $\FF' = \set {\phi \in S : \map f \phi = T}$

We now need to verify that the following properties hold for $\FF'$:
 * $\FF' \supseteq \FF$.
 * $\FF'$ is finitely satisfiable.
 * $\forall \phi \in S: \phi \in \FF' \lor \sqbrk {\neg \phi} \in \FF'$.

Let $\phi \in \FF$ be arbitrary.

We need to show that $\map f \phi = T$.

By definition of consistent, $f\restriction_{\set \phi} \in M$.

Thus, there is an $\LL$-structure $\MM_{\set \phi}$ such that:
 * $\MM_{\set \phi} \models \set \phi$
 * $\map {f\restriction_{\set \phi}} \phi = T \iff \MM_{\set \phi} \models \set \phi$

It immediately follows that $\map {f\restriction_{\set \phi}} \phi = T$, and thus that $\map f \phi = T$.

Let $P \subseteq \FF'$ be an arbitrary finite subset.

By definition, for every $\phi \in P$, $\map f \phi = T$.

By definition of consistent, $f\restriction_P \in M$.

Thus, there is an $\LL$-structure $\MM_P$ such that:
 * $\forall \phi \in P: \map {f\restriction_P} \phi = T \iff \MM_P \models \phi$

It follows immediately that:
 * $\forall \phi \in P: \MM_P \models \phi$

As $P$ was arbitrary, $\FF'$ is finitely satisfiable by definition.

, suppose $\phi \notin \FF'$ and $\sqbrk {\neg \phi} \notin \FF'$, where $\phi \in S$.

Then, $\map f \phi = \map f {\neg \phi} = F$, by definition of $\FF'$.

By definition of consistent, we have that $f\restriction_{\set {\phi, \neg \phi}} \in M$.

Thus, there is an $\LL$-structure $\MM_{\set {\phi, \neg \phi}}$ such that:
 * $\map {f\restriction_{\set {\phi, \neg \phi}}} \phi = T \iff \MM_{\set {\phi, \neg \phi}} \models \phi$
 * $\map {f\restriction_{\set {\phi, \neg \phi}}} {\neg \phi} = T \iff \MM_{\set {\phi, \neg \phi}} \models \neg \phi$

It follows immediately that:
 * $\MM_{\set {\phi, \neg \phi}} \not\models \phi$
 * $\MM_{\set {\phi, \neg \phi}} \not\models \neg \phi$

By definition of $\mathrm{PL}$:
 * $\map {\operatorname{val}_{\MM_{\set {\phi, \neg \phi}}}} \phi = F$
 * $\map {\operatorname{val}_{\MM_{\set {\phi, \neg \phi}}}} {\neg \phi} = F$

As $F \ne T$, that is a contradiction.

Thus, at least one of $\phi$ and $\neg \phi$ is in $\FF'$