Reductio ad Absurdum/Variant 2/Proof by Truth Table

Theorem

 * $\neg p \implies \left({q \land \neg q}\right) \vdash p$

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

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