Negation implies Negation of Conjunction/Case 1
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Theorem
- $\neg p \implies \neg \paren {p \land q}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg p$ | Assumption | (None) | ||
| 2 | 2 | $p \land q$ | Assumption | (None) | ||
| 3 | 2 | $p$ | Rule of Simplification: $\land \EE_1$ | 2 | ||
| 4 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 3, 1 | ||
| 5 | 1 | $\neg \paren {p \land q}$ | Proof by Contradiction: $\neg \II$ | 2 – 4 | Assumption 2 has been discharged | |
| 6 | $\neg p \implies \neg \paren {p \land q}$ | Rule of Implication: $\implies \II$ | 1 – 5 | 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 $\text{T43}$