C = A XOR B = A'B + AB' You can use the bubble pushing method... add inversion bubbles at the inputs of AND/NAND gates. This also inverts the inputs. But since inverts of A' and B and A and B' result in the same set of inputs, nothing else is needed. An AND gate with two inverted inputs is a NOR gate. So you basically get (A'+B)' + (A+B')'. Another way of doing this is to use demorgans law and change the equation to POS form. ((A'B+AB')')' = ((A+B')(A'+B))' = (A+B')'+(A'+B)'