Rule of Material Implication/Formulation 1/Reverse Implication/Proof 1

Theorem

 * $\neg p \lor q \vdash p \implies q$

Proof

 * align="right" | 4 ||
 * align="right" | 2, 3
 * $\bot$
 * $\neg \mathcal E$
 * 3, 2
 * ... which of course is a direct contradiction ...
 * align="right" | 5 ||
 * align="right" | 2, 3
 * $q$
 * $\bot \mathcal E$
 * 4
 * ... from a falsehood, any statement can be derived - pick $q$ ...
 * align="right" | 6 ||
 * align="right" | 2
 * $p \implies q$
 * $\implies \mathcal I$
 * 3-5
 * $p \implies q$
 * $\implies \mathcal I$
 * 3-5


 * align="right" | 9 ||
 * align="right" | 7
 * $q$
 * Law of Identity
 * 7
 * The truth of $q$ still holds
 * align="right" | 10 ||
 * align="right" | 7
 * $p \implies q$
 * $\implies \mathcal I$
 * 8, 9
 * $\implies \mathcal I$
 * 8, 9