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