Proof by Contradiction/Variant 1

Theorem

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

Proof

 * align="right" | 3 ||
 * align="right" | 1, 2
 * $q \land \neg q$
 * Sequent Introduction
 * 1, 2
 * by hypothesis
 * align="right" | 4 ||
 * align="right" | 1, 2
 * $\bot$
 * Rule of Not-Elimination
 * 3
 * Rule of Not-Elimination
 * 3