There is a boolean expression (A+B)(AB)'.
The answer's truth table compared to this initial have the matching combinations that are resembling one of a XOR gate. I'm wondering if there is an identity or it is possible to further simplify the finished answer into something that resembles a XOR gate. My working out is as follows
(A+B)(AB)'
= (A+B)(A'+B')
= AA' + AB' + BA' + BB'
= 0 + AB' + BA' + 0
= AB' + BA'
Im not sure if this is fully simplified or if there was another path to follow that would give the correct answer. (The initial converted to a circuit will have 4 gates while the "simplified" version will have 5 unless I miscounted)
Thank you in advance
The Identity for the XOR gate is (A+B)(AB)'. Although it is not displayed in most sites that iterate DeMorgans laws, their truth tables have common outputs and when XOR is googled, the formula (A+B)(AB)' is shown. Therefore, (A+B)(AB)' = A xor B or the simplified version AB' + BA' = A xor B.