Boolean Algebra Simplification POS

Question

I'm given this expression $$ (x+y'+z')(x'+z') $$ the $'$ meaning not. I have to simplify this to 3 literals and show my answer as a product of sums.

Every calculator I check says the answer is $(x'y')+z'$. So far all I can think of to do as the first step is to expand the given expression using distribution giving me $$ xx'+xz'+x'y'+y'z'+x'z'+z'z'. $$ From there I know $xx'$ is $1$ and $z'z'$ is $z'$, giving me $xz'+x'y'+y'z'+x'z'+z'$ and this is where I get stuck. Any suggestions?

Answer 1

Expression $xx'$ means “$x$ is true and not $x$ is true”, so $xx'=0$. Thus the expression is: $$ xz'+x'y'+y'z'+x'z'+z'z'=(x+x')z'+x'y'+y'z'+z'= x'y' + (y'+1)z'=x'y'+z' $$

Answer 2

A useful principle is:

Reduction

$x(x'+z) = x + z$ (given $x$, the $x'+z$ term reduces to just $z$)

So:

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