0
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ X XOR Y is X Y’+X’Y $\endgroup$
    – Eric
    Commented May 28, 2023 at 16:21
  • $\begingroup$ I know that but getting to the proper form is the issue. $\endgroup$ Commented May 28, 2023 at 16:56
  • $\begingroup$ Just write it out till the point you get stuck. It’s literally just standard Boolean algebra simplifications at that point. $\endgroup$
    – Eric
    Commented May 28, 2023 at 17:14
  • $\begingroup$ 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. $\endgroup$ Commented May 28, 2023 at 17:50
  • $\begingroup$ 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. $\endgroup$
    – Eric
    Commented May 28, 2023 at 18:48

1 Answer 1

1
$\begingroup$

Since it is easier to work with and, or, and not (or at leas: we have more, and more familiar, principles for them), I would definitely start with the RHS (the XOR expression), and work your way back to the LHS.

So:

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

... Oh, and LOL! I was going to say: "And you can take it from here" but you're there already! Or at least: you are now at the spot where you ended up by starting with the LHS :P

Moral: when proving an equivalence between LHS and RHS, you can work from both sides .. while it is intutive to go from LHS to RHS, sometimes it's easier to go from RHS to LHS ... and sometimes you work from both sides and try to meet somewhere in the middle

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .