Proof by Contradiction/Variant 1
Theorem
- $\left({p \vdash \left({q \land \neg q}\right)}\right) \vdash \neg p$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \vdash \left({q \land \neg q}\right)$ | Premise | (None) | ||
| 2 | 2 | $p$ | Assumption | (None) | ||
| 3 | 1, 2 | $q \land \neg q$ | Sequent Introduction | 1, 2 | by hypothesis | |
| 4 | 1, 2 | $q$ | Rule of Simplification: $\land \mathcal E_1$ | 3 | ||
| 5 | 1, 2 | $\neg q$ | Rule of Simplification: $\land \mathcal E_2$ | 3 | ||
| 6 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \mathcal E$ | 4, 5 | ||
| 7 | 1 | $\neg p$ | Proof by Contradiction: $\neg \mathcal I$ | 2 – 6 | Assumption 2 has been discharged |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): $\S 1.3$: Conjunction and Disjunction
