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I have this boolean equation:

X'Y'+XY+X'Y=X'+Y

I want to prove it.

Now I was wondering if I can rearrange this equation, if I could, so I can factor out the other side; tell me if this is allowed. I haven't seen anything to say I could in my textbook:

X'Y'+XY+X'Y

X'Y'+X'Y+XY see now I move the X'Y to the left

X'(Y+Y')+XY

X'+X'Y+XY

X'+Y(X'+X)

X'+Y

Am I doing it right? I've been trying this equation in other ways and haven't been able to prove it otherwise.

munchschair
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1 Answers1

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Almost all of your rearrangements are correct, except it is not clear how you get from $X'(Y+Y')+XY$ to $X'+X'Y+XY$. I would write your argument like this: $$\begin{split}X'Y'+XY+X'Y&=X'Y'+X'Y+XY\\ &=X'(Y'+Y)+XY\\ &=X'(1)+XY\\ &=X'+XY\\ &=(X'+X)(X'+Y)\\ &=(1)(X'+Y)\\ &=X'+Y.\end{split}$$

Dave Wilding
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  • Are there any specific limitations to factoring in boolean? Can I factor out of 3+ statements or any part of the expression? – munchschair Feb 11 '14 at 03:39
  • What specifically do you mean by "factoring"? Because I would call what you have done "simplifying". You can certainly simplify boolean expressions that have 3+ terms. – Dave Wilding Feb 11 '14 at 09:15
  • To help me understand: could you tell me what you see in $X'Y'+XY+X'Y$ that makes you want to get $X'+Y$ rather than $X+Y$? – Dave Wilding Feb 11 '14 at 09:17
  • It's a proof that you can get to that. Factoring as in pulling alike terms out of the multiples. – munchschair Feb 11 '14 at 12:14
  • OK, I see. When you are working in boolean you can factor in two different ways: $X_1Y+X_2Y+\dotsb +X_nY=(X_1+\dotsb +X_n)Y$ and $(X_1X_2\dotsm X_n)+Y=(X_1+Y)(X_2+Y)\dotsm(X_n+Y)$ for any number of $X_1,X_2,\dotsc,X_n$. – Dave Wilding Feb 11 '14 at 12:26
  • I'm not sure how you got from X' + XY to (X'+X)(X'+Y) As for me, there's a formula where C'+CD = C'+D. It's in the book I am using, you can expand the C' and then use that to eliminate the A. – munchschair Feb 11 '14 at 15:56
  • To get from $X'+XY$ to $(X'+X)(X'+Y)$ I am using distributive law (b) with $A=X'$, $B=X$ and $C=Y$. The formula in your book is a rule that you can most certainly use, I am just breaking down the steps of how you prove that formula. – Dave Wilding Feb 11 '14 at 21:22