# Proof by Contradiction/Variant 1

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## 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