Semantic Consequence as Tautological Conditional

Theorem
Let $\mathcal F$ 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 $\mathcal F$ iff $\displaystyle \bigwedge \mathcal F \implies \mathbf A$ is a tautology.

Here, $\displaystyle \bigwedge \mathcal F$ is the conjunction of $\mathcal F$.

Necessary Condition
Assume that $\mathcal F \models_{\mathrm{BI}} \mathbf A$.

Suppose $\displaystyle \bigwedge \mathcal F \implies \mathbf A$ were not a tautology.

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


 * $v \left({\displaystyle \bigwedge \mathcal F}\right) = T$
 * $v \left({\mathbf A}\right) = F$

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

It now follows from the boolean interpretation of conjunction that:


 * $v \left({\mathbf B}\right) = T$

for every $\mathbf B \in \mathcal F$.

Hence, by definition of model:


 * $v \models_{\mathrm{BI}} \mathcal F$.

It now follows from our assumption that $v \left({\mathbf A}\right) = T$.

This is a contradiction, hence $\displaystyle \bigwedge \mathcal F \implies \mathbf A$ is a tautology.

Sufficient Condition
Assume that $\displaystyle \bigwedge \mathcal F \implies \mathbf A$ is a tautology.

Let $v$ be an arbitrary model of $\mathcal F$.

Then:


 * $v \left({\mathbf B}\right) = T$

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


 * $v \left({\displaystyle \mathcal F}\right) = T$

Since $\displaystyle \bigwedge \mathcal F \implies \mathbf A$ is a tautology, it must be that:


 * $v \left({\mathbf A}\right) = T$

Hence, since $v$ was arbitrary:


 * $\mathcal F \models_{\mathrm{BI}} \mathbf A$