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Question
Using the truth table, prove the following logical equivalence.
p ↔ q ≡ ~(p ∧ ~q) ∧ ~(q ∧ ~p)
Sum
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Solution
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| p | q | ~p | ~q | p↔q | p∧~q | ~(p∧~q) | (q∧~p) | ~(q∧~p) | ~(p∧~q)∧~(q ∧ ~p) |
| T | T | F | F | T | F | T | F | T | T |
| T | F | F | T | F | T | F | F | T | F |
| F | T | T | F | F | F | T | T | F | F |
| F | F | T | T | T | F | T | F | T | T |
In the above truth table, the entries in columns 5 and 10 are identical.
∴ p ↔ q ≡ ~(p ∧ ~q) ∧ ~(q ∧ ~p)
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