Model Defined by Implication

Theorem

Let $P$ and $Q$ be logical formulas.

Let $P \models Q$ denote that $Q$ is a logical consequence of $P$.

Then $P \models Q$ iff:

$\models P \implies Q$

where $\models$ in this context means equals true in all interpretations.

That is, $P \models Q$ iff $P \implies Q$ is a tautology.

Proof

Sufficient Condition

Let $P \models Q$.

Then $Q$ is true in every model of $P$.

So let $\mathcal M$ be an arbitrary model of $P$.

From Implication Properties, we have $q \vdash p \implies q$:

If something is true, anything implies it.

So as $Q$ is true in every model of $P$, then $P \implies Q$ is likewise true in every model of $P$.

Hence $\models P \implies Q$.

Necessary Condition

Now let $P \not \models Q$.

That is, there exists a model for $P$ for which $Q$ is false.

Let $\mathcal M \left({P}\right)$ be such a model.

That is: $\mathcal M \left({P}\right) = T$ while $\mathcal M \left({Q}\right) = F$.

From Tautology and Contradiction it follows that $\mathcal M \left({\neg Q}\right) = T$.

Thus $\mathcal M \left({P}\right) = T$ and $\mathcal M \left({\neg Q}\right) = T$.

Hence from the definition of conjunction, $\mathcal M \left({P \land \neg Q}\right) = T$.

So from Conjunction and Implication, we have that $\mathcal M \left({\neg \left({P \implies Q}\right)}\right) = T$ and hence $\mathcal M \left({P \implies Q}\right) = F$.

So $\not \models P \implies Q$.

Hence the result.

$\blacksquare$