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Question
Using the truth table prove the following logical equivalence.
∼ (p ∨ q) ∨ (∼ p ∧ q) ≡ ∼ p
Solution
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| p | q | ∼ p | p ∨ q | ∼ (p ∨ q) | ∼ p ∧ q | ∼ (p ∨ q) ∨ (∼ p ∧ q) |
| T | T | F | T | F | F | F |
| T | F | F | T | F | F | F |
| F | T | T | T | F | T | T |
| F | F | T | F | T | F | T |
The entries in columns 3 and 7 are identical.
∴ ∼ (p ∨ q) ∨ (∼ p ∧ q) ≡ ∼ p
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