$(xy’+z)’\cdot((xz)’+y')$
$$\begin{align*} (xy’+z)’\cdot ((xz)’+y’) &=(x'+yz’)\cdot (x’+z’+y’)\\ &=x’x’ + x’z’ + x’y’ + yz’x’ + yz’z’ + yz’y’\\ &=x’ + x’z’ + x’y’ + yz’x’ + yz’ + z’ \end{align*}$$
Can it be further simplified?
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ >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 $x=0,y=0,z=1$. $\endgroup$– Dilip SarwateCommented Jan 2, 2012 at 18:08
-
1$\begingroup$ 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 $x'+yz'$. $\endgroup$– Arturo MagidinCommented Jan 2, 2012 at 18:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
$$\begin{align*} (xy’+z)’\cdot ((xz)’+y’) &= (xy')'z'((xz)'+y')\\ &= (x'+y)z'(x'+z'+y')\\ &= (x'z' + yz')(x'+z'+y')\\ &= (x'z' + yz')x' + (x'z'+yz')z' + (x'z'+yz')y'\\ &= x'x'z' + x'yz' + x'z'z' + yz'z' + x'y'z' + yy'z'\\ &= x'z' + x'yz' + x'z' + yz' + x'y'z' + 0\\ &= x'z'(1+y+y') + yz'\\ &= x'z' + yz'\\ &= (x'+y)z'. \end{align*}$$
answered Jan 2, 2012 at 18:47
-
2$\begingroup$ 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. $\endgroup$ Commented Jan 2, 2012 at 21:53
-
$\begingroup$ @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). $\endgroup$ Commented Jan 6, 2012 at 2:34