hi! can someone help me with the boolean laws used for this simplification? thank you

X = (A.B' + C)': ((A.C)' + B') (A.B' + C)' => (A.B')'C' #Demorgan's law (A.B').C' ((A.C)' + B') ===> (A.B').C'(A' + C') + (A.B')'.C.B' (A.B')'C'(A' + C') + (A.B').C.B' ===> (A.B').C'A' + (A.B')'C'C' + (A.B').C'.B' (A.B').C.A' + (A.B').C'C' + (A.B').C'.B' ===> (A'+B).A'C' + (A'+B).C' + (A' + B).C.B' (A'+B).A'C' + (A'+B).C' + (A' + B).C'.B' ===> A'C' + A'C'B + A.C' + BC + A'B'C' + 0 A'C' + A'C'B + A.C' + BC + A'B'C' + 0 ===> A'C' + BC' + A'B'C' ==> A'C' + BC' ==> (A+B')C (A+B')CH