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

Theorem

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

Proof

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 * $\neg p \lor \neg q$
 * Sequent Introduction
 * 2
 * De Morgan's Laws: Disjunction of Negations
 * De Morgan's Laws: Disjunction of Negations


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 * $p \land q$
 * Reductio Ad Absurdum
 * 2-4
 * }
 * }
 * }