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$(xy’+z)’\cdot((xz)’+y')$

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

Can it be further simplified?

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    >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 $x=0,y=0,z=1$.2012-01-02
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    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 $x'+yz'$.2012-01-02

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