Saved Bookmarks
| 1. |
Without using truth table prove that:(p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q) ≡ p ∨ q |
|
Answer» L.H.S. = (p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q) ≡ [(p ∨ ∼ p) ∧ q] ∨ (p ∧ ∼ q) ..........(Distributive Law) ≡ (T ∧ q) ∨ (p ∧ ∼ q) .........(Complement Law) ≡ q ∨ (p ∧ ∼ q) ...........(Identity Law) ≡ (q ∨ p) ∧ (q ∨ ∼ q) .........(Distributive Law) ≡ (q ∨ p) ∧ T ..........(Complement Law) ≡ q ∨ p ..........(Identity Law) ≡ p ∨ q ......(Commutative Law) = R.H.S. |
|