Boolean Expression simplify

Question

I have this expression: $$(x + y + z’)(x’ + y’ + z)$$ which I am trying to simplify. I decide to multiply it out in order to get, $${\color{red}{(xx')}}+(xy')+(xz)+(yx')+{\color{red}{(yy')}}+(yz)+(z'x')+(z'y')+{\color{red}{(z'z)}}.$$ I know that the $xx', yy'$ and $zz'$ would just be $0$, however, now I am stuck. I can't seem to find something to pull out and simplify further.

Answer 1

You can use the Consensus theorem to simplify:

Consensus theorem

$xy + y'z= xy + y'z + xz$

Applied to your expression, we can eliminate $xz$ given that we have $xy'$ and $yz$. We can also eliminate $yx'$ given $yz$ and $z'x'$. And finally we can eliminate $z'y'$ given $z'x'$ and $xy'$. So, that leaves $xy'+zy+z'x'$

Answer 2

Don't distribute everything. Begin by treating $(x+y)$ and $(x'+y')$ as blocks.

$\begin{split}&(x+y+z')(x'+y'+z)&\text{ is fairly simple.} \\=&(x+y)(x'+y')+(x+y)z+(x'+y')z'+zz'&\text{ by distribution} \\=&(x+y)(x'+y')+(x+y)z+(x'+y')z'&\text{ by contradiction and identity} \\=&(x+y)(x'+y')(z+z')+(x+y)z+(x'+y')z'&\text{ by identity and tautology} \\=&(x+y)(x'+y')z+(x+y)z+(x+y)(x'+y')z'+(x'+y')z'&\text{ distribution and association} \\=&(x+y)z+(x'+y')z'&\text{ by absorption AB+B=B} \\=&xz+yz+x'z'+y'z'&\text{ by distribution} \end{split}$