Question

Without using truth table prove that {[(pvq)^~p] →q}=q
$negation$
p​

Answer 1

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.

Answer 2

To prove,

(p∧q)∨(∼p∧∼q) = p↔q

Solution,

We can simply prove the mathematical equation as follows.

Taking the Right Hand Side,

⇒ p↔q

⇒ (p→q)∧(q→p)                                             (According to the definition)

⇒ (p∨∼q)∧(q∨∼p)                                        (By definition)

⇒((p∨∼q)∧q) ∨ ((p∨∼q)∧∼p)                        (By the distributive law)

⇒((p∧q) ∨ (∼q ∧q)) ∨((p∧∼p)∨(∼q∧∼p))       (Distributive law)

⇒(p∧q)∨(∼p∧∼q),                                         (Using the complement laws)

which is equal to the left-hand side of the equation.

L.H.S = R.H.S

Thus, we proved that (p∧q)∨(∼p∧∼q) = p↔q.