TAKE OPTION A
((x→y)∧x)→y((x→y)∧x)→y
Truth value : T ,T ,T ,T
Take option B :
((∼x→y)∧(∼x∧∼y))→x
Truth value :T ,T, T,T
Take option C
(x→(x∨y))
Truth value : T ,T,T ,T
taking option d
((x∨y)↔(∼x∨∼y))
Truth value :
F , T, T, F
Hence option d is correct .