Boolean Algebra Manipulation #2

Question

Ok, so I know I just asked a similar question recently, but I have one more. I don't even know how to start this one: (De Morgan's maybe?) (x+y+z)(x'+y+z')(x+y')(y'+z) = x'y'z + xy'z' + xyz

Any suggestions on how to approach this?

Answer 1

You could simply multiply the whole thing out, simplifying as you go. For instance,

$$(x+y')(y'+z)=xy'+xz+y'y'+y'z\;;$$

we can simplify $y'y'$ to $y'$, and absorption lets us reduce $xy'+y'$ to $y'$ and then $y'+y'z$ to $y'$, leaving us with just $xz+y'$. Then we can multiply that by $x'+y+z'$, simplify, and finally multiply the result by $x+y+z$; it’s tedious, but it works.

I’d try to be a bit more efficient by starting with the distributive laws:

$$(x+y')(y'+z)=y'+xz$$

(as we already saw the hard way!), and

$$(x+y+z)(x'+y+z')=y+(x+z)(x'+z')\;,$$

so the lefthand side can be reduced to

$$\Big(y+(x+z)(x'+z')\Big)(y'+xz)\;.$$

Now multiply out $(x+z)(x'+z')$ and simplify to some (farily nice) expression $e$, and multiply out the resulting $(y+e)(y'+xz)$; the remaining simplifications turn out to be easy.