Kenneth Rosen Edition 6th Exercise 1.2 Question 30 (Page No. 29)

Question

Show that (p ∨ q) ∧ (¬p ∨ r) → (q ∨ r) is a tautology. Prove it without Truth tables.

Answer 1

    (p v q) ∧ (~p v r) -> (q v r)

= (p + q)(~p + r ) - > (q + r)

= ~((p +q)(~p + r)) + (q + r)

= (p + q)' + (p' + r)' + (q + r)

= p'q' + pr' + q + r

= (p'q' + q ) + ( pr' + r )

= (p' + q ) + ( p+ r)

= (p' + p ) + q + r

= T + q + r

= T

Answer 2