Semantic Consequence as Tautological Conditional

Theorem
Let $\FF$ be a finite set of WFFs of propositional logic.

Let $\mathbf A$ be another WFF.

Then the following are equivalent:

that is, $\mathbf A$ is a semantic consequence of $\FF$ $\ds \bigwedge \FF \implies \mathbf A$ is a tautology.

Here, $\ds \bigwedge \FF$ is the conjunction of $\FF$.

Necessary Condition
let:
 * $\FF \models_{\mathrm {BI} } \mathbf A$

$\ds \bigwedge \FF \implies \mathbf A$ were not a tautology.

Then there exists a boolean interpretation $v$ such that:


 * $\map v {\ds \bigwedge \FF} = \T$
 * $\map v {\mathbf A} = \F$

by definition of the boolean interpretation of $\implies$.

It now follows from the boolean interpretation of conjunction that:


 * $\map v {\mathbf B} = \T$

for every $\mathbf B \in \FF$.

Hence, by definition of model:


 * $v \models_{\mathrm {BI} } \FF$

It now follows from our assumption that:
 * $\map v {\mathbf A} = \T$

This is a contradiction, hence $\ds \bigwedge \FF \implies \mathbf A$ is a tautology.

Sufficient Condition
Let $\ds \bigwedge \FF \implies \mathbf A$ be a tautology.

Let $v$ be an arbitrary model of $\FF$.

Then:


 * $\map v {\mathbf B} = \T$

for every $\mathbf B \in \FF$, whence by the boolean interpretation of conjunction:


 * $\map v \FF = \T$

Since $\ds \bigwedge \FF \implies \mathbf A$ is a tautology, it must be the case that:


 * $\map v {\mathbf A} = \T$

Hence, since $v$ was arbitrary:


 * $\FF \models_{\mathrm {BI} } \mathbf A$