Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication
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Theorem
- $\neg \paren {p \land q} \vdash p \implies \neg q$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg \paren {p \land q}$ | Premise | (None) | ||
| 2 | 2 | $p$ | Assumption | (None) | ||
| 3 | 3 | $q$ | Assumption | (None) | ||
| 4 | 2, 3 | $p \land q$ | Rule of Conjunction: $\land \II$ | 2, 3 | ||
| 5 | 1, 2, 5 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 4, 1 | ||
| 6 | 1, 2 | $\neg q$ | Proof by Contradiction: $\neg \II$ | 3 – 5 | Assumption 3 has been discharged | |
| 7 | 1 | $p \implies \neg q$ | Rule of Implication: $\implies \II$ | 2 – 6 | Assumption 2 has been discharged |
$\blacksquare$