Question

iv) No parallelogram is a rectangle,
(3) Without using truth table prove that {[(pvg)a-p]+q=gi;​

Answer 1

Answer:

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.

Hence proved.

Step-by-step explanation:

ok