Proof by Contradiction/Variant 3/Formulation 1

Theorem

 * $p \implies \neg p \vdash \neg p$

Proof

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