Extend Theory to Satisfy Witness Property

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

Let $T$ be a set of $\LL$-sentences.

Then there exists a language $\LL^*$ and a set of $\LL^*$-sentences $T^*$ satisfying the following properties:
 * $T^*$ is finitely satisfiable $T$ is finitely satisfiable.
 * If $T^*$ is satisfiable, then $T$ is also.
 * For every $\LL^*$-WFF of $1$ free variable $\map \phi x$, there exists some constant $c_\phi$ such that:
 * $T^* \models \paren {\exists x: \map \phi x} \implies \map \phi {x := c_\phi}$


 * $T^*$ satisfies the witness property.

Proof
We will recursively define a sequence of languages.

Define:
 * $\LL_0 := \LL$

Let the set of predicates in $\LL$ be $\PP$.

Let the set of functions in $\LL$ be $\FF = \set {\KK, \FF_1, \FF_2, \dotsc}$.

For every $i \in \N$, define:
 * $\Phi_i$ as the set of $\LL_i$-WFFs of $1$ free variable.
 * $\KK_i$ as the set of constants in $\LL_i$.
 * $\LL_{i + 1}$ as the language of predicate logic with:
 * Predicates $\PP$.
 * Functions $\set {\KK_i \cup \Phi_i, \FF_1, \dotsc}$.

Define:
 * $\ds \LL^* = \bigcup_{i \mathop \in \N} \LL_i$
 * $T^* = T \cup \set {\sqbrk {\paren {\exists x : \map \phi x} \implies \map \phi {x := \phi}} : \phi \in \Phi}$

where $\Phi$ is the set of $\LL^*$-WFFs of $1$ free variable.

It remains to show that all of the stated properties hold.

Satisfiability
As $T \subseteq T^*$, it is trivial that:
 * $T^*$ finitely satisfiable implies $T$ finitely satisfiable.
 * $T^*$ satisfiable implies $T$ satisfiable.

We now show that $T$ finitely satisfiable implies $T^*$ finitely satisfiable.

Proceed by mathematical induction to show that:
 * $T_i = T^* \cap \LL_i$

is finitely satisfiable for every $i \in \N$.

By assumption, $T_0 = T$ is finitely satisfiable.

Suppose that, for $i \in \N$, $T_i$ is finitely satisfiable.

Let $P$ be an arbitrary finite subset of $T_{i + 1}$.

Let $P_L = P \cap T_i$ and $P_K = P \setminus T_i$.

As $P_L$ is a finite subset of $T_i$, then by assumption there is some $\LL_i$-structure $\MM = \tuple {M, F, P}$ such that:
 * $\MM \models_{\mathrm{PL}} P_L$

By definition of $T_{i + 1}$, each $\psi \in P_K$ is of the form:
 * $\paren {\exists x : \map \phi x} \implies \map \phi {x := \phi}$

for some $\LL_i$-WFF $\phi$ of $1$ free variable.

Let $Q$ be the set of all of those $\phi$ where the expression above appears in $P_K$.

Let $Q' = \set {\phi \in Q : \MM \models_{\mathrm{PL}} \exists x : \map \phi x}$

For each $\phi \in Q'$, by definition of $\mathrm{PL}$:
 * There exists some $a \in M$ such that $\map {\operatorname{val}_\MM} \phi \sqbrk {\paren {x / a}} = \top$

As $Q'$ is finite, by the Principle of Finite Choice, there exists a mapping $g : Q' \to M$ such that:
 * $\forall \phi \in Q': \map {\operatorname{val}_\MM} \phi \sqbrk {\paren {x / \map g \phi}} = \top$

As $M$ is non-empty, let $z \in M$ be arbitrary.

Define $\MM' = \tuple {M, F \cup \set {G_\phi : \phi \in Q'} \cup \set {Z_\phi : \phi \in \Phi_i \setminus Q'}, P}$, where:
 * $\map {G_\phi} {} = \map g \phi$
 * $\map {Z_\phi} {} = z$

Each $\psi \in P_L$ is modeled by $\MM'$, since the value depends only on symbols contained in $L_i$.

For each $\phi \in Q'$, $\map \phi {x := \phi}$ is modeled by $\MM'$, as follows from the definition of $g$.

Thus, for each $\psi \in P_K$ either:
 * The corresponding $\phi$ is in $Q'$, in which case the right hand side of the conditional is valid.
 * The corresponding $\phi$ is not in $Q'$, in which case the left hand side of the conditional is not valid.

In either case, based on the truth function of conditional, the complete $\psi$ is valid.

Thus,
 * $\MM' \models_{\mathrm{PL}} P$

Every finite subset of $T^*$ is contained in some $T_i$, so $T^*$ is finitely satisfiable.

Witness Property
By definition of $T^*$, that:
 * $T^* \models \paren {\exists x: \map \phi x} \implies \map \phi {x := \phi}$

holds for every $\LL^*$-WFF of $1$ free variable $\map \phi x$.

Because of how $\LL^*$ is defined, $\phi$ is also a constant in $\LL^*$, and plays the role of $c_\phi$ in the theorem statement.

That $T^*$ satisfies the witness property follows immediately from the lemma.