Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Reverse Implication

Theorem

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

Proof

 * align="right" | 3 ||
 * align="right" | 2
 * $p \implies q$
 * Sequent Introduction
 * 2
 * Implication Equivalent to Negation of Conjunction with Negative
 * Implication Equivalent to Negation of Conjunction with Negative


 * align="right" | 5 ||
 * align="right" | 1
 * $p \land \neg q$
 * Reductio Ad Absurdum
 * 2-4
 * 2-4