Rule of Material Implication/Formulation 1/Reverse Implication/Proof 2

Theorem

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

Proof

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 * align="right" | 1
 * $\neg \left({\neg \neg p \land \neg q}\right)$
 * Sequent Introduction
 * 1
 * De Morgan's Laws: $p \lor q \vdash \neg \left({\neg p \land \neg q}\right)$
 * align="right" | 3 ||
 * align="right" | 1
 * $\neg \neg p \implies q$
 * Sequent Introduction
 * 2
 * Implication Equivalent to Negation of Conjunction with Negative: $\neg \left({p \land \neg q}\right) \vdash p \implies q$
 * 2
 * Implication Equivalent to Negation of Conjunction with Negative: $\neg \left({p \land \neg q}\right) \vdash p \implies q$


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 * align="right" | 4
 * $\neg \neg p$
 * $\neg \neg \mathcal I$
 * 4
 * 4


 * align="right" | 7 ||
 * align="right" | 1
 * $p \implies q$
 * $\implies \mathcal I$
 * 4, 6
 * 4, 6