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.

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

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

Define:
 * $\MM = \tuple {M, \set {F_f}, \set {P_p}}$

We will now show that:
 * $\MM \models_{\mathrm{PL}} T$

Specifically, we will prove that, for every $\LL$-sentence $\phi$:
 * $\phi \in T \iff \MM \models_{\mathrm{PL}} \phi$

First, it can be easily seen that, for every term $t$ containing no variables:
 * $\map {\operatorname{val}_\MM} t = t$

where $\operatorname{val}$ is the value of a term.

Let $\map P n$ be defined as:
 * For every $\LL$-sentence $\phi$ with $n$ quantifiers, $\phi \in T \iff \MM \models_{\mathrm{PL}} \phi$

We proceed by induction on $\map P n$.

The cases for $\map P 0$ and $\map P {k + 1}$ are very similar, except that one of our cases will not apply in the former, so we will prove them simultaneously.

Let $n \in \N$ be arbitrary.

Let $\map Q m$ be defined as:
 * For every $\LL$-sentence $\phi$ with $n$ quantifiers and $m$ logical connectives, $\phi \in T \iff \MM \models_{\mathrm{PL}} \phi$

Now, perform complete induction on $\map Q m$, using the alternate formulation:
 * $\forall m \in \N, \paren {\forall \ell < m : \map Q \ell} \implies \map Q m$

Let $m \in \N$ be arbitrary.

Let $\phi$ be an arbitrary $\LL$-sentence with $n$ quantifiers and $m$ logical connectives.

By the definition of WFF, at least one of the following hold:
 * There exists an $k$-ary predicate $p$ and terms $t_1, \dotsc, t_k$ such that:
 * $\phi = \map p {t_1, \dotsc, t_k}$
 * There exists a WFF $\psi$ such that:
 * $\phi = \neg \psi$
 * There exist WFFs $\chi$ and $\psi$, as well as $\circ$ one of $\lor$, $\land$, $\implies$, or $\iff$ such that:
 * $\phi = \paren {\chi \circ \psi}$
 * Only if $n > 0$, there exists a variable $x$ and WFF $\psi$ such that:
 * $\phi = \paren {\forall x: \psi}$ or $\phi = \paren {\exists x: \psi}$

We will examine these case by case.

Predicate
Let $\phi = \map p {t_1, \dotsc, t_k}$.

We have:

Suppose $\sqbrk {\map p {t_1, \dotsc, t_k}} \in T$.

By definition of $P_p$:
 * $\map {\operatorname{val}_\MM} {\map p {t_1, \dotsc, t_k}} = \top$

Then, by definition of $\mathrm{PL}$:
 * $\MM \models_{\mathrm{PL}} \map p {t_1, \dotsc, t_k}$

Now, suppose $\sqbrk {\map p {t_1, \dotsc, t_k}} \notin T$.

For the same reason:
 * $\map {\operatorname{val}_\MM} {\map p {t_1, \dotsc, t_k}} = \bot$

Thus:
 * $\MM \not\models_{\mathrm{PL}} \map p {t_1, \dotsc, t_k}$

Negation
Let $\phi = \neg \psi$.

As $\psi$ has fewer logical connectives than $\phi$, we can apply the second induction hypothesis:
 * $\psi \in T \iff \MM \models_{\mathrm{PL}} \psi$

First, suppose $\phi \notin T$.

By assumption, either $\psi \in T$ or $\sqbrk {\neg \psi} \in T$.

Thus, by Modus Tollendo Ponens:
 * $\psi \in T$

Therefore, by the inductive statement above:
 * $\MM \models_{\mathrm{PL}} \psi$

Hence:

Therefore:
 * $\MM \not\models_{\mathrm{PL}} \phi$

Now, suppose:
 * $\phi \in T$

No $\mathrm{PL}$-structure models $\set {\psi, \neg \psi}$.

Thus, as $T$ is finitely satisfiable, and $\set {\psi, \phi}$ is finite:
 * $\psi \notin T$

Again, by the inductive statement:
 * $\MM \not\models_{\mathrm{PL}} \psi$

The same computation as above gives:
 * $\map {\operatorname{val}_\MM} \phi = \top$

Thus:
 * $\MM \models_{\mathrm{PL}} \phi$

Binary Connective
Let $\phi = \paren {\chi \circ \psi}$.

As both $\chi$ and $\psi$ have fewer logical connectives than $\phi$:
 * $\chi \in T \iff \MM \models_{\mathrm{PL}} \chi$
 * $\psi \in T \iff \MM \models_{\mathrm{PL}} \psi$

For the same reasons as the previous case, $T$ contains exactly one of:
 * $\set {\chi, \psi}$
 * $\set {\neg \chi, \psi}$
 * $\set {\chi, \neg \psi}$
 * $\set {\neg \chi, \neg \psi}$

Additionally, $T$ contains exactly one of:
 * $\phi$
 * $\neg \phi$

As $T$ is finitely satisfiable, they must agree according to $f^\circ$, the truth function of $\circ$.

If $\phi \in T$, then the values of $\chi$ and $\psi$ in $\MM$ must be such that $\map {\operatorname{val}_\MM} \phi = \top$.

If $\phi \notin T$, then for the same reason, $\map {\operatorname{val}_\MM} \phi = \bot$.

In both cases, we find that:
 * $\phi \in T \iff \MM \models_{\mathrm{PL}} \phi$

Quantifier
Again, this case does not apply if $n = 0$.

Let:
 * $\phi_a = \forall x : \map \psi x$

and:
 * $\phi_e = \exists x : \map \psi x$

First, suppose $\phi_a \in T$.

Then, for every term $t$ containing no variables:
 * $\map \psi {x := t} \in T$

As $\psi$ has exactly $n - 1$ quantifiers, we can apply the first induction hypothesis.

Thus, for every $t \in M$:
 * $\MM \models_{\mathrm{PL}} \map \psi {x := t}$

But then, as $\map {\operatorname{val}_\MM} t = t$:
 * $\map {\operatorname{val}_\MM} {\map \psi x} \sqbrk {\paren {x/t}} = \top$

Therefore, by definition of value of a formula:
 * $\map {\operatorname{val}_\MM} {\forall x: \map \psi x} = \top$

Thus:
 * $\MM \models_{\mathrm{PL}} \forall x: \map \psi x$

Second, suppose $\phi_e \notin T$.

Since:
 * $\set {\neg \exists x : \map \psi x, \neg \forall x : \neg \map \psi x}$

is finite and not satisfiable, we have:
 * $\sqbrk {\forall x: \neg \map \psi x} \in T$

Then, as the number of quantifiers in $\neg \map \psi x$ is the same as in $\map \psi x$, we can apply the first case above.

This gives us:
 * $\MM \models_{\mathrm{PL}} \forall x: \neg \map \psi x$

which states that, for every $t \in M$:
 * $\map {\operatorname{val}_\MM} {\neg \map \psi x} \sqbrk {\paren {x/t}} = \top$

That is, for every $t \in M$:
 * $\map {\operatorname{val}_\MM} {\map \psi x} \sqbrk {\paren {x/t}} = \bot$

Thus, there is no $t \in M$ for which the above results in $\top$.

By definition:
 * $\map {\operatorname{val}_\MM} {\exists x: \map \psi x} = \bot$

Which gives:
 * $\MM \not\models_\mathrm{PL} \exists x: \map \psi x$

For the last two cases, those being:
 * $\phi_a \notin T$

and:
 * $\phi_e \in T$

observe that the first reduces to the second in exactly the same manner as before.

Thus, we will only consider:
 * $\sqbrk {\exists x: \map \psi x} \in T$

By the witness property, there exists some term $t_\psi$ containing no variables such that:
 * $\map \psi {x := t_\psi} \in T$

By the first induction hypothesis:
 * $\MM \models_{\mathrm{PL}} \map \psi {x := t_\psi}$

Since $\map {\operatorname{val}_\MM} {t_\psi} = t_\psi$:
 * $\map {\operatorname{val}_\MM} {\map \psi x} \sqbrk {\paren {x/t_\psi}} = \top$

Thus, by definition:
 * $\MM \models_{\mathrm{PL}} \exists x: \map \psi x$