# Boolean XOR conversion- rewriting to XOR in simplifed form

I have a Boolean Expression simplified to:

BC + AC + A'B'C'

and get to this be rewriting:

C(B + A) + A'B'C'

I need to rewrite to

(A + B) XOR C'

I can check with a Truth Table and can show they are equivalent. I am having issues with the boolean algebra to get the XOR solution. Some insight would be helpful.

• X XOR Y is X Y’+X’Y
– Eric
Commented May 28, 2023 at 16:21
• I know that but getting to the proper form is the issue. Commented May 28, 2023 at 16:56
• Just write it out till the point you get stuck. It’s literally just standard Boolean algebra simplifications at that point.
– Eric
Commented May 28, 2023 at 17:14
• So A XOR B = AB' + A'B; then subsitute A = C(B + A) an B = A'B'C' ? get started is where I am stuck. Commented May 28, 2023 at 17:50
• What? No, you substitute A+B and C’ and see if you can get it to match the other expression. Then, write it so that it’s rigorous, and you have your proof.
– Eric
Commented May 28, 2023 at 18:48

$$(A + B) XOR C' = (A+B)C'' + (A+B)'C' = (A+B)C + A'B'C'=...$$