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: $\neg \EE$ | 3, 1 | ||
5 | 1 | $\neg p$ | Proof by Contradiction: $\neg \II$ | 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: $\neg \EE$ | 7, 1 | ||
9 | 1 | $\neg q$ | Proof by Contradiction: $\neg \II$ | 6 – 8 | Assumption 6 has been discharged | |
10 | 1 | $\neg p \land \neg q$ | Rule of Conjunction: $\land \II$ | 5, 9 |
$\blacksquare$