Answer to Question #283385 in Discrete Mathematics for Mahesh

Question

Show ((p ∨ q) ∧ ¬(¬p ∧ (¬q ∨ ¬r))) ∨ (¬p ∧ ¬q) ∨ (¬p ∧ ¬r) is tautology, by using replacement process.

Answer 1

Solution:

((p ∨ q) ∧ ¬(¬p ∧ (¬q ∨ ¬r))) ∨ (¬p ∧ ¬q) ∨ (¬p ∧ ¬r)

= [(p ∨ q) ∧ ¬(¬p ∧ ¬(q ∧ r))] ∨ [¬(p ∨ q)] ∨ [¬(p ∨ r)]

= [(p ∨ q) ∧ (p ∨ (q ∧ r))] ∨ ¬[(p ∨ q) ∧ (p ∨ r)]

= [(p ∨ q) ∧ (p ∨ (q ∧ r))] ∨ ¬[p ∨ (q ∧ r)]

= (p ∨ q) ∨ T

= T

Hence, it is a Tautology.