False Statement implies Every Statement/Formulation 1/Proof by Truth Table

Theorem

 * $\neg p \vdash p \implies q$

Proof
We apply the Method of Truth Tables.

As can be seen by inspection, where the truth value in the relevant column on the LHS is $T$, that under the one on the RHS is also $T$:

$\begin{array}{|cc||ccc|} \hline \neg & p & p & \implies & q \\ \hline T & F & F & T & F \\ T & F & F & T & T \\ F & F & T & F & F \\ F & F & T & T & T \\ \hline \end{array}$