Modus Ponendo Tollens/Variant/Formulation 1/Reverse Implication

Theorem

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

Proof

 * align="right" | 3 ||
 * align="right" | 2
 * $p$
 * $\land \mathcal E_1$
 * 2
 * align="right" | 4 ||
 * align="right" | 2
 * $q$
 * $\land \mathcal E_2$
 * 2
 * align="right" | 5 ||
 * align="right" | 1, 2
 * $\neg q$
 * $\implies \mathcal E$
 * 1, 3
 * ... and show that the assumption leads to a contradiction ...
 * align="right" | 6 ||
 * align="right" | 1, 2
 * $\bot$
 * $\neg \mathcal E$
 * 5, 4
 * align="right" | 7 ||
 * align="right" | 1
 * $\neg \left({p \land q}\right)$
 * Proof by Contradiction
 * 2-6
 * }
 * align="right" | 7 ||
 * align="right" | 1
 * $\neg \left({p \land q}\right)$
 * Proof by Contradiction
 * 2-6
 * }
 * }
 * }