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Using the truth table prove the following logical equivalence.
∼ (p ∨ q) ∨ (∼ p ∧ q) ≡ ∼ p
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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
Concept: Statement Patterns and Logical Equivalence
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