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