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.