Rule of Transposition/Formulation 1/Reverse Implication/Proof

Theorem

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

Proof

 * align="right" | 3 ||
 * align="right" | 2
 * $\neg \neg p$
 * $\neg \neg \mathcal I$
 * 2
 * align="right" | 4 ||
 * align="right" | 1, 2
 * $\neg \neg q$
 * MTT
 * 1, 3
 * align="right" | 5 ||
 * align="right" | 1, 2
 * $q$
 * $\neg \neg \mathcal E$
 * 4
 * align="right" | 6 ||
 * align="right" | 1
 * $p \implies q$
 * $\implies \mathcal I$
 * 2, 5
 * }
 * align="right" | 6 ||
 * align="right" | 1
 * $p \implies q$
 * $\implies \mathcal I$
 * 2, 5
 * }
 * }
 * }