# De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication

Jump to navigation
Jump to search

## Theorem

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

## Proof

By the tableau method of natural deduction:

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

1 | 1 | $\neg \paren {p \lor q}$ | Premise | (None) | ||

5 | 5 | $p$ | Assumption | (None) | ||

3 | 2 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 2 | ||

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

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

6 | 6 | $q$ | Assumption | (None) | ||

7 | 6 | $p \lor q$ | Rule of Addition: $\lor \II_2$ | 6 | ||

8 | 1, 7 | $\bot$ | Principle of Non-Contradiction: $\neg \mathcal E$ | 7, 1 | ||

9 | 1 | $\neg q$ | Proof by Contradiction: $\neg \mathcal I$ | 6 – 8 | Assumption 6 has been discharged | |

10 | 1 | $\neg p \land \neg q$ | Rule of Conjunction: $\land \mathcal I$ | 5, 9 |

$\blacksquare$