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$\overline{(\bar{X} + Z\bar{Y} + Y\bar{Z})(\bar{X}+Y\bar{Z})}$

The farthest I've gotten is DeMorgans:

$X(\bar{Z} + Y)(\bar{Y} + Z) + X(\bar{Y} + Z)$

The answer should be $X\bar{Y} + XZ$, but I'm unable to get anywhere close to it. I tried foiling the binomial but it results in a single X value when the whole expression is evaluated, and that's not correct.

S. Coughing
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1 Answers1

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Continuing from where you stopped using $1$ for $True$

\begin{eqnarray*} X(\bar{Z} + Y)(\bar{Y} + Z) + X(\bar{Y} + Z) & = & X(\bar{Y} + Z)((\bar{Z} + Y) + 1) \\ & \stackrel{x+1=1}{=} & X(\bar{Y} + Z)\\ & = & X\bar{Y} + XZ \end{eqnarray*}

trancelocation
  • 32,243
  • Can I get an explanation of the first step? Is X factored out? And where does the one come from? – S. Coughing Mar 12 '21 at 12:57
  • @S.Coughing When you do Boolean algebra, you need some basic rules like $1b=b, b+1=1$ and distributivity $a(b+c)=ab+ac$. In the first step I use distributivity and $1b=b$ together. You will see such "tricks" quite often when transforming Boolean expressions. So, it's good to get accustomed to such rules. – trancelocation Mar 12 '21 at 14:06