Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


$$\begin{align*} (xy’+z)’\cdot ((xz)’+y’) &=(x'+yz’)\cdot (x’+z’+y’)\\ &=x’x’ + x’z’ + x’y’ + yz’x’ + yz’z’ + yz’y’\\ &=x’ + x’z’ + x’y’ + yz’x’ + yz’ + z’ \end{align*}$$

Can it be further simplified?

share|cite|improve this question
>Can it be further simplified? Perhaps, using $x' + x'y' = x'$ which you may have seen as $x+xy=x$. But you should check your work at the very first step. If $x=0,y=0,z=1$, we have $xy'+z=1, (xy'+z)'=0$ and so the expression you have to simplify has value $0$. But $x'+yz'=1$ and $x'+z'+y'=1$ also, and so $$0=(xy'+z)'.((xz)'+y') \neq (x'+yz').(x'+z'+y') = 1$$ when $x=0,y=0,z=1$. – Dilip Sarwate Jan 2 '12 at 18:08
By De Morgan's Laws, $(ab)' = a'+b'$ and $(a+b)' = a'b'$. So $(xy'+z)' = ((xy')')z' = (x'+y)z' = x'z' + yz'$, but you have $x'+yz'$. – Arturo Magidin Jan 2 '12 at 18:26

$$\begin{align*} (xy’+z)’\cdot ((xz)’+y’) &= (xy')'z'((xz)'+y')\\ &= (x'+y)z'(x'+z'+y')\\ &= (x'z' + yz')(x'+z'+y')\\ &= (x'z' + yz')x' + (x'z'+yz')z' + (x'z'+yz')y'\\ &= x'x'z' + x'yz' + x'z'z' + yz'z' + x'y'z' + yy'z'\\ &= x'z' + x'yz' + x'z' + yz' + x'y'z' + 0\\ &= x'z'(1+y+y') + yz'\\ &= x'z' + yz'\\ &= (x'+y)z'. \end{align*}$$

share|cite|improve this answer
Alternatively, setting $z^\prime(x^\prime+z^\prime+y^\prime)=z^\prime$ in the second line (using $a(a+b)=a$) gets us to the end result much more quickly. – Dilip Sarwate Jan 2 '12 at 21:53
@DilipSarwate That z'(x'+z'+y')=z' (whatever it means exactly, which doesn't come as clear, but it doesn't matter) uses more than that (a(a+b))=a (and substitution). You have to use that (a+b)=(b+a) and possibly that (a+(b+c))=((a+b)+c) also. Of course, this isn't a problem for Boolean Algebra, but what you did relies on more than just that (a(a+b))=a (and substitution). – Doug Spoonwood Jan 6 '12 at 2:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.