Maximal Finitely Satisfiable Theory with Witness Property is Satisfiable

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

Let $T$ be a set of $\LL$-sentences such that:
 * $T$ is finitely satisfiable
 * For every $\LL$-sentence $\phi$, either:
 * $\phi \in T$
 * or:
 * $\sqbrk {\neg \phi} \in T$


 * $T$ satisfies the witness property

Proof
Let $M$ be the set of all $\LL$-terms that contain no variables.

Let $\RR \subseteq M \times M$ be defined as:
 * $\map \RR {s, t} \iff \sqbrk {s = t} \in T$

By Equality in Finitely Satisfiable Theory is Equivalence Relation:
 * $\RR$ is an equivalence relation

Let $M' = M / \RR$.

For each $n$-ary function $f$ in $\LL$, define:
 * $\map {F_f} {\eqclass {t_1} \RR, \dotsc, \eqclass {t_n} \RR} = \eqclass {\map f {t_1, \dotsc, t_n}} \RR$

For each $n$-ary predicate $p$ in $\LL$, define:
 * $\map {P_p} {\eqclass {t_1} \RR, \dotsc, \eqclass {t_n} \RR} = \begin{cases}

\top & : \sqbrk {\map p {t_1, \dotsc, t_n}} \in T \\ \bot & : \sqbrk {\map p {t_1, \dotsc, t_n}} \notin T \end{cases}$

For these to be well-defined, we need to show that they are independent of which representatives $t_i$ are used.

Let $f$ be an arbitrary $n$-ary function in $\LL$.

Let $t_1, \dotsc, t_n, t'_1, \dotsc, t'_n \in M$.

Suppose that $\sqbrk {t_i = t'_i} \in T$ for each $1 \le i \le n$.

$\sqbrk {\map f {t_1, \dotsc, t_n} = \map f {t'_1, \dotsc, t'_n}} \notin T$.

Then, by assumption, $\sqbrk {\neg \paren {\map f {t_1, \dotsc, t_n} = \map f {t'_1, \dotsc, t'_n}}} \in T$.

As $T$ is finitely satisfiable, there is an $\LL$-structure $\AA$ such that:
 * $\AA \models_{\mathrm{PL}} \set {t_i = t'_i : 1 \le i \le n} \cup \set {\neg \paren {\map f {t_1, \dotsc, t_n} = \map f {t'_1, \dotsc, t'_n}}}$

But such a structure cannot exist, as $\map {\operatorname{val}_\AA} {t_i} = \map {\operatorname{val}_\AA} {t'_i}$ for every $1 \le i \le n$.

Thus, by Proof by Contradiction:
 * $\sqbrk {\map f {t_1, \dotsc, t_n} = \map f {t'_1, \dotsc, t'_n}} \in T$

Let $p$ be an arbitrary $n$-ary predicate in $\LL$.

Let $t_1, \dotsc, t_n, t'_1, \dotsc, t'_n \in M$.

Suppose that $\sqbrk {t_i = t'_i} \in T$ for each $1 \le i \le n$.

$\neg \paren {\sqbrk {\map p {t_1, \dotsc, t_n}} \in T \iff \sqbrk {\map p {t'_1, \dotsc, t'_n}} \in T}$.

As $\RR$ is an equivalence relation, it is symmetric, so if necessary, we can interchange the roles of $t_i$ and $t'_i$.

, suppose $\sqbrk {\map p {t_1, \dotsc, t_n}} \in T$ and $\sqbrk {\map p {t'_1, \dotsc, t'_n}} \notin T$.

By assumption, $\sqbrk {\neg \paren {\map p {t'_1, \dotsc, t'_n}}} \in T$.

As $T$ is finitely satisfiable, there is an $\LL$-structure $\AA$ such that:
 * $\AA \models_{\mathrm{PL}} \set {t_i = t'_i : 1 \le i \le n} \cup \set {\map p {t_1, \dotsc, t_n}, \neg \paren {\map p {t'_1, \dotsc, t'_n}}}$

But, again, such a structure cannot exist, as $\map {\operatorname{val}_\AA} {t_i} = \map {\operatorname{val}_\AA} {t'_i}$ for every $1 \le i \le n$.

Thus, by Proof by Contradiction:
 * $\sqbrk {\map p {t_1, \dotsc, t_n}} \in T \iff \sqbrk {\map p {t'_1, \dotsc, t'_n}} \in T$

The last step is to show that $\MM = \tuple {M', \set {F_f}, \set {P_p}}$ models $T$.