# Boolean Algebra Simplification POS

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?

• Do you know what absorption is? – Fabio Somenzi Sep 19 at 16:00
• No, you're pretty close to the target. Absorption says that $a + ab$ simplifies to $a$. – Fabio Somenzi Sep 19 at 16:12
• So could I say z'+x'z'= x'? – coder12345 Sep 19 at 16:17
• No, that is incorrect. What is $a$ and what is $b$ in your case? – Fabio Somenzi Sep 19 at 16:18

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