Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication

Theorem

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

Proof

 * align="right" | 4 ||
 * align="right" | 2, 3
 * $p \land q$
 * $\land \mathcal I$
 * 2, 3
 * align="right" | 5 ||
 * align="right" | 1, 2, 3
 * $\bot$
 * $\neg \mathcal E$
 * 4, 1
 * align="right" | 6 ||
 * align="right" | 1, 2
 * $\neg q$
 * Proof by Contradiction
 * 3-5
 * align="right" | 7 ||
 * align="right" | 1
 * $p \implies \neg q$
 * $\implies \mathcal I$
 * 2-6
 * }
 * align="right" | 7 ||
 * align="right" | 1
 * $p \implies \neg q$
 * $\implies \mathcal I$
 * 2-6
 * }
 * }
 * }