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?

Please explain? A way to absorb like terms? Am I going about this completely wrong? — rileypitcher, Sep 19 '19 at 16:05
No, you're pretty close to the target. Absorption says that $a + ab$ simplifies to $a$. — Fabio Somenzi, Sep 19 '19 at 16:12
No, that is incorrect. What is $a$ and what is $b$ in your case? — Fabio Somenzi, Sep 19 '19 at 16:18
Would it be my a is z' and my b is x' therefore it becomes z'+x'z'=z'? — rileypitcher, Sep 19 '19 at 16:24

score 0 · Accepted Answer · answered Sep 19 '19 at 16:30

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

answered Sep 19 '19 at 16:30

Vasily Mitch

10,129

score 0 · Answer 2 · answered Sep 19 '19 at 22:03

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A useful principle is:

Reduction

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

So:

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

answered Sep 19 '19 at 22:03

Bram28

100,612
6
70
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Boolean Algebra Simplification POS

2 Answers2