Proof by Contradiction/Variant 3/Formulation 2

Theorem

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

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $\neg p$
 * Sequent Introduction
 * 1
 * Proof by Contradiction: Variant 3, Formulation 1
 * Proof by Contradiction: Variant 3, Formulation 1