Simplifying Boolean algebra... What am I missing?

Question

I'm reviewing boolean algebra, but I'm having trouble with a basic simplification: $$\begin{equation}\begin{aligned} &x'z'+ xyz +xz'\\ &= z'(x+x')+xyz\\ &= z'+xyz\\ &= ??? \\ &= z'+xy \end{aligned}\end{equation}\tag{2}\label{eq2}$$

I can't seem to find a theorem or postulate that satisfies the final step in any table. What is it that I'm not seeing?

score 4 · Accepted Answer · answered Feb 17 '21 at 06:53

4

You have

\begin{eqnarray*}z'+xyz & = & z'(1+xy) + xyz \\ & = & (z'+z)xy + z'\\ & = & z' + xy \end{eqnarray*}

answered Feb 17 '21 at 06:53

trancelocation

32,243

Thanks for the reply, but where does the 1+xy come from? – LosOnTheGo Feb 17 '21 at 10:06
1

$1$ stands for $True$ and for any Boolean variable $b$ you have $1+b = 1$ and $1b=b$. The use as presented in my answer is a standard trick in Boolean algebra. – trancelocation Feb 17 '21 at 10:12

Simplifying Boolean algebra... What am I missing?

1 Answers1