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Question
Using truth table prove that:
~ (p `leftrightarrow` q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p)
Chart
Solution
| p | q | (p `leftrightarrow` q) | ~(p `leftrightarrow` q) | ~p | ~q | L = (p ∧ ~ q) |
S = (q ∧ ~ p) |
L ∨ S |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| T | T | T | F | F | F | F | F | F |
| T | F | F | T | F | T | T | F | T |
| F | T | F | T | T | F | F | T | T |
| F | F | T | F | T | T | F | F | F |
From columns 4 and 9
~ (p `leftrightarrow` q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p)
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