I have this boolean equation:


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





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


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: $$\begin{split}X'Y'+XY+X'Y&=X'Y'+X'Y+XY\\ &=X'(Y'+Y)+XY\\ &=X'(1)+XY\\ &=X'+XY\\ &=(X'+X)(X'+Y)\\ &=(1)(X'+Y)\\ &=X'+Y.\end{split}$$

  • $\begingroup$ Are there any specific limitations to factoring in boolean? Can I factor out of 3+ statements or any part of the expression? $\endgroup$ – munchschair Feb 11 '14 at 3:39
  • $\begingroup$ 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. $\endgroup$ – Dave Wilding Feb 11 '14 at 9:15
  • $\begingroup$ 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 $X+Y$? $\endgroup$ – Dave Wilding Feb 11 '14 at 9:17
  • $\begingroup$ It's a proof that you can get to that. Factoring as in pulling alike terms out of the multiples. $\endgroup$ – munchschair Feb 11 '14 at 12:14
  • $\begingroup$ 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 $X_1,X_2,\dotsc,X_n$. $\endgroup$ – Dave Wilding Feb 11 '14 at 12:26

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