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

Theorem

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

Proofs

 * align="right" | 3 ||
 * align="right" | 2
 * $p \implies \neg q$
 * Sequent Introduction
 * 2
 * Modus Ponendo Tollens
 * Modus Ponendo Tollens


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