# Proof by Contradiction/Variant 2/Formulation 2

## Theorem

- $\vdash \paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$

## Proof

By the tableau method of natural deduction:

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

1 | 1 | $\paren {p \implies q} \land \paren {p \implies \neg q}$ | Assumption | (None) | ||

2 | 1 | $p \implies q$ | Rule of Simplification: $\land \mathcal E_1$ | 1 | ||

3 | 1 | $p \implies \neg q$ | Rule of Simplification: $\land \mathcal E_2$ | 1 | ||

4 | 1 | $\neg p$ | Sequent Introduction | 2, 3 | Proof by Contradiction: Formulation 1 | |

5 | $\paren {\paren {p \implies q} \land \paren {p \implies \neg q} } \implies \neg p$ | Rule of Implication: $\implies \mathcal I$ | 1 – 4 | Assumption 1 has been discharged |

$\blacksquare$

## Sources

- 1964: Donald Kalish and Richard Montague:
*Logic: Techniques of Formal Reasoning*... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T34}$