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
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 * align="right" |
 * $\left({p \implies \neg p}\right) \implies \neg p$
 * $\implies \mathcal I$
 * 1, 2
 * }
 * 1, 2
 * }
 * }