Saved Bookmarks
| 1. |
Prove that :~ (p ∨ q) ∨ (~ p ∧ q) ≡ ~ pWithout using truth table. |
|
Answer» L.H.S. =p↔q =(p→q)∧(q→p) =(p∨∼q)∧(q∨∼p) =((p∨∼q)∧q)∨((p∨∼q)∧∼p) =((p∧q)∨(∼q∧q))∨((p∧∼p)∨(∼q∧∼p)) =(p∧q)∨(∼p∧∼q) = R.H.S. |
|