De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Reverse Implication

Theorem

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

Proof

 * align="right" | 6 ||
 * align="right" | 2
 * $p$
 * Reductio Ad Absurdum
 * 3-5
 * 3-5


 * align="right" | 10 ||
 * align="right" | 2
 * $q$
 * Reductio Ad Absurdum
 * 7-9
 * 7-9


 * align="right" | 13 ||
 * align="right" | 1
 * $\neg p \lor \neg q$
 * Reductio Ad Absurdum
 * 2-12
 * 2-12