Advertisements
Advertisements
Question
Using truth table prove that:
`p → (q ∨ r) ≡ (p → q) ∨ (p → r)`
Chart
Solution
To prove: `p → (q ∨ r) ≡ (p → q) ∨ (p → r)`
Truth table:
| p | q | r | q ∨ r | `bb(p → q)` | `bb(p → r)` | `bb((p → q) ∨ (p → r))` | `bb(p → (q ∨ r))` |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| T | T | T | T | T | T | T | T |
| T | T | F | T | T | F | T | T |
| T | F | T | T | F | T | T | T |
| T | F | F | F | F | F | F | F |
| F | T | T | T | T | T | T | T |
| F | T | F | T | T | T | T | T |
| F | F | T | T | T | T | T | T |
| F | F | F | F | T | T | T | T |
From column 7 and 8
`p → (q ∨ r) ≡ (p → q) ∨ (p → r)`
shaalaa.com
Is there an error in this question or solution?
