Rule of Explosion/Variant 2
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Theorem
- $\vdash \paren {p \land \neg p} \implies q$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \land \neg p$ | Assumption | (None) | ||
| 2 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
| 3 | 1 | $\neg p$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
| 4 | 1 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 2 | ||
| 5 | 1 | $q$ | Modus Tollendo Ponens $\mathrm {MTP}_{{{6}}}$ | 4, 3 |
$\blacksquare$