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f(x+y , y) = (x+y)’ + y

               = (x’.y’) + y

               = (x’+ y) . (y’+ y)

               = (x’+ y)                    (because,  y’+ y = 1)

 

f(f(x+y, y), z)   =  f((x’+ y), z)

                       =  (x’+ y)’+ z

                       =  (x.y’)+ z

 

Answer C is correct

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