De Morgan's Laws (Logic)/Conjunction/Formulation 1/Reverse Implication

Theorem

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

Proof

 * align="right" | 3 ||
 * align="right" | 2
 * $\neg p \lor \neg q$
 * Sequent Introduction
 * 2
 * De Morgan's Laws: Disjunction of Negations
 * align="right" | 4 ||
 * align="right" | 1, 2
 * $\bot$
 * $\neg \mathcal E$
 * 3, 1
 * align="right" | 5 ||
 * align="right" | 1
 * $p \land q$
 * Reductio Ad Absurdum
 * 2-4
 * }
 * $p \land q$
 * Reductio Ad Absurdum
 * 2-4
 * }
 * }