Proof by Contradiction/Variant 2/Formulation 1

Theorem

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

Proof

 * align="right" | 4 ||
 * align="right" | 1, 3
 * $q$
 * MPP
 * 1, 3
 * align="right" | 5 ||
 * align="right" | 2, 3
 * $\neg q$
 * MPP
 * 2, 3
 * align="right" | 6 ||
 * align="right" | 1, 2, 3
 * $\bot$
 * $\neg \mathcal E$
 * 4, 5
 * align="right" | 6 ||
 * align="right" | 1, 2, 3
 * $\bot$
 * $\neg \mathcal E$
 * 4, 5