# Proof by Contradiction/Variant 3/Formulation 1

## Theorem

- $p \implies \neg p \vdash \neg p$

## Proof

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $p \implies \neg p$ | Premise | (None) | ||

2 | 2 | $p$ | Assumption | (None) | ||

3 | 1, 2 | $\neg p$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 2 | ||

4 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \mathcal E$ | 2, 3 | ||

5 | 1, 2 | $\neg p$ | Proof by Contradiction: $\neg \mathcal I$ | 2 – 4 | Assumption 2 has been discharged |

$\blacksquare$

## Sources

- 1965: E.J. Lemmon:
*Beginning Logic*... (previous) ... (next): $\S 1.3$: Conjunction and Disjunction: Theorem $23$ - 2000: Michael R.A. Huth and Mark D. Ryan:
*Logic in Computer Science: Modelling and reasoning about systems*... (previous) ... (next): $\S 1.2.1$: Rules for natural deduction: Example $1.21$