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