Answer : option A) -> B

Solution :

Put given C= AB' +A'B

Hence, AC'+A'C= A(AB'+A'B)' + A'(AB'+A'B)

->Now by rule : (X+Y)'= X'Y'

Hence = A( (AB')'.(A'B)' ) + A'(AB'+A'B)

-> Again similar rule : (XY)' = X'+Y' (DeMorgan's)

Hence = A((A'+B).(A+B')) + A'(AB'+A'B)

-> Rule: X (Y+Z)= XY + XZ

Hence = A(A'B'+AB)+A'(AB'+A'B')

= AA'B+AAB+A'AB'+A'A'B

-> Since AA' =0

Hence = AB+A'B ,Taking B out= B(A+A') = B.1 =B