Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication

Theorem

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

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $\neg p \lor \neg \neg q$
 * Sequent Introduction
 * 1
 * De Morgan's Laws: Disjunction of Negations
 * align="right" | 3 ||
 * align="right" | 1
 * $p \implies \neg \neg q$
 * Sequent Introduction
 * 1
 * Rule of Material Implication
 * 1
 * Rule of Material Implication


 * align="right" | 6 ||
 * align="right" | 1, 4
 * $q$
 * $\neg \neg \mathcal E$
 * 5
 * 5