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

Theorem

\(\ds \neg p\)

\(\)

\(\ds \)

\(\ds \vdash \ \ \)

\(\ds p \implies q\)

\(\)

\(\ds \)

Proof

We apply the Method of Truth Tables.

As can be seen by inspection, where the truth value in the relevant column on the left hand side is $\T$, that under the one on the right hand side is also $\T$:

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

$\blacksquare$