A. ((x→y)∧x)→y = ((x'+y).x)'+y = (xy)'+y = x'+y'+y = x'+1=1 ( TRUE)
B. ((∼x→y)∧(∼x→∼y))→x = ((x+y).(x+y'))'+x = (x+yy')'+x = x'+x =1 ( TRUE )
C. (x→(x∨y)) = x'+x+y = 1+y =1 ( TRUE )
D. ((x∨y)↔(∼x→∼y)) = (x+y)↔(x+y') =((x+y)'.(x+y')).((x+y')'(x+y)) = (x'y'.(x+y')).((x'y.(x+y)) = x'y'.x'y= 0 ( FALSE)
so ans is D