Consistency of Logical Formulas has Finite Character/Proof 2

Theorem

Let $P$ be the property of collections of logical formulas defined as:

$\forall \FF: \map P \FF$ denotes that $\FF$ is consistent.

Then $P$ is of finite character.

That is:

$\FF$ is a consistent set of formulas if and only if every finite subset of $\FF$ is also a consistent set of formulas.

Proof

Sufficient Condition

Let $\FF$ be a consistent set of formulas.

Aiming for a contradiction, suppose there exists $\GG \subseteq \FF$ where $\GG$ is finite such that $\GG$ is inconsistent.

But then from Set of Logical Formulas is Inconsistent iff it has Finite Inconsistent Subset, $\FF$ is itself inconsistent.

From this contradiction it follows that $\GG$ must be consistent.

As $\GG$ is arbitrary, it follows that every finite subset of $\FF$ is consistent.

$\Box$

Necessary Condition

Let every finite subset of $\FF$ is consistent.

Aiming for a contradiction, suppose $\FF$ is inconsistent.

Then by Set of Logical Formulas is Inconsistent iff it has Finite Inconsistent Subset there exists a finite subset $\GG \subseteq \FF$ such that $\GG$ is inconsistent.

But by hypothesis $\GG$ is consistent.

From this contradiction it follows that $\FF$ must also be consistent.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles