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Sum
Using the truth table, prove the following logical equivalence.
~p ∧ q ≡ [(p ∨ q)] ∧ ~p
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Solution
| 1 | 2 | 3 | 4 | 5 | 6 |
| p | q | ~p | ~p∧q | (p∨q) | (p∨q)∧~p |
| T | T | F | F | T | F |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | F | F | F |
In the above truth table, the entries in columns 4 and 6 are identical.
∴ ~p ∧ q ≡ [(p ∨ q)] ∧ ~p
Is there an error in this question or solution?
