Proof by Contradiction/Variant 2/Formulation 2

Theorem

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

Proof

 * align="right" | 4 ||
 * align="right" | 1
 * $\neg p$
 * Sequent Introduction
 * 2, 3
 * Proof by Contradiction: Formulation 1
 * Proof by Contradiction: Formulation 1