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.