Reductio ad Absurdum/Variant 2/Proof by Truth Table

Theorem

$\neg p \implies \paren {q \land \neg q} \vdash p$

Proof

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

$\begin{array}{|ccccccc||c|} \hline \neg & p & \implies & (q & \land & \neg & q) & p \\ \hline \T & \F & \F & \F & \F & \T & \F & \F \\ \T & \F & \F & \T & \F & \F & \T & \F \\ \F & \T & \T & \F & \F & \T & \F & \T \\ \F & \T & \T & \T & \F & \F & \T & \T \\ \hline \end{array}$

$\blacksquare$