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Sum
Using the truth table, prove the following logical equivalence.
p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
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Solution
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| p | q | r | q∨r | p∧(q∨r) | p∧q | p∧r | (p∧q)∨(p∧r) |
| T | T | T | T | T | T | T | T |
| T | T | F | T | T | T | F | T |
| T | F | T | T | T | F | T | T |
| T | F | F | F | F | F | F | F |
| F | T | T | T | F | F | F | F |
| F | T | F | T | F | F | F | F |
| F | F | T | T | F | F | F | F |
| F | F | F | F | F | F | F | F |
In the above truth table, the entries in columns 5 and 8 are identical.
∴ p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Is there an error in this question or solution?
