Which of the following Boolean expressions is not a tautology?

Step-by-step solution

1   ( p ⇒ q ) ∨ ( ~ q ⇒ p ) By applying a ⇒ b ≡ ~ a ∨ b   &   ~ ~ a = a , we get = ( ~ p ∨ q ) ∨ ( q ∨ p ) Since, the operation disjunction is commutative, hence, we have = ( ~ p ∨ q ) ∨ ( p ∨ q ) Now, applying the distributive law i.e. a ∨ b ∨ c ≡ a ∨ b ∨ a ∨ c , we get = ( ~ p ∨ p ) ∨ q Also, we know that ~ a ∨ a ≡ t   &   t ∨ a ≡ t , wher