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

Theorem

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

Proof

 * align="right" | 3 ||
 * align="right" | 2
 * $p \lor q$
 * $\lor \mathcal I_1$
 * 2
 * align="right" | 4 ||
 * align="right" | 1, 3
 * $\bot$
 * $\neg \mathcal E$
 * 3, 1
 * align="right" | 5 ||
 * align="right" | 1
 * $\neg p$
 * Proof by Contradiction
 * 2 - 4
 * align="right" | 5 ||
 * align="right" | 1
 * $\neg p$
 * Proof by Contradiction
 * 2 - 4


 * align="right" | 7 ||
 * align="right" | 6
 * $p \lor q$
 * $\lor \mathcal I_2$
 * 6
 * align="right" | 8 ||
 * align="right" | 1, 7
 * $\bot$
 * $\neg \mathcal E$
 * 7, 1
 * align="right" | 9 ||
 * align="right" | 1
 * $\neg q$
 * Proof by Contradiction
 * 6 - 8
 * align="right" | 9 ||
 * align="right" | 1
 * $\neg q$
 * Proof by Contradiction
 * 6 - 8