No Boolean Interpretation Models a WFF and its Negation

Theorem
Let $$\mathcal M$$ be a model for propositional calculus.

Let $$\mathbf A$$ be a propositional WFF.

Then $$\mathcal M$$ can not satisfy both $$\mathbf A$$ and $$\neg \mathbf A$$.

Proof
Let $$\mathcal M \models \mathbf A$$.

Then $$\mathcal M \left({\mathbf A}\right) = T$$ by definition.

By definition of Logical Negation, $$\mathcal M \left({\neg \mathbf A}\right) = F$$.

Now suppose $$\mathcal M \models \neg \mathbf A$$.

Then $$\mathcal M \left({\neg \mathbf A}\right) = T$$, again by definition.

From this contradiction comes the result.