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" | 4 ||
 * align="right" | 3
 * $\neg p \lor \neg q$
 * $\lor \mathcal I_1$
 * 3
 * align="right" | 5 ||
 * align="right" | 2, 3
 * $\bot$
 * $\neg \mathcal E$
 * 4, 2
 * align="right" | 6 ||
 * align="right" | 2
 * $p$
 * Reductio Ad Absurdum
 * 3-5
 * align="right" | 6 ||
 * align="right" | 2
 * $p$
 * Reductio Ad Absurdum
 * 3-5


 * align="right" | 8 ||
 * align="right" | 7
 * $\neg p \lor \neg q$
 * $\lor \mathcal I_2$
 * 3
 * align="right" | 9 ||
 * align="right" | 2, 7
 * $\bot$
 * $\neg \mathcal E$
 * 8, 2
 * align="right" | 10 ||
 * align="right" | 2
 * $q$
 * Reductio Ad Absurdum
 * 7-9
 * align="right" | 10 ||
 * align="right" | 2
 * $q$
 * Reductio Ad Absurdum
 * 7-9


 * align="right" | 12 ||
 * align="right" | 1, 2
 * $\bot$
 * $\neg \mathcal E$
 * 11, 1
 * align="right" | 13 ||
 * align="right" | 1
 * $\neg p \lor \neg q$
 * Reductio Ad Absurdum
 * 2-12
 * $\neg p \lor \neg q$
 * Reductio Ad Absurdum
 * 2-12