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Question
Using the truth table proves the following logical equivalence.
∼ (p ↔ q) ≡ (p ∧ ∼ q) ∨ (q ∧ ∼ p)
Chart
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Solution
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| p | q | ∼ p | ∼ q | p ↔ q | ∼ (p ↔ q) | p ∧ ∼ q | q ∧ ∼ p | (p ∧ ∼ q) ∨ (q ∧ ∼ p) |
| T | T | F | F | T | F | F | F | F |
| T | F | F | T | F | T | T | F | T |
| F | T | T | F | F | T | F | T | T |
| F | F | T | T | T | F | F | F | F |
The entries in columns 6 and 9 are identical.
∴ ∼ (p ↔ q) ≡ (p ∧ ∼ q) ∨ (q ∧ ∼ p)
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