Simplifying x'y'+y'z+xz+xy+yz'

Question

Someone help me to simplify this boolean expression.

$x'y'+y'z+xz+xy+yz'$

I understand the laws used but still not getting the exact answer. I would appreciate if someone solved this for me.

My attempt so far:

$$x'y'+y'z+xz+xy+yz'= x'y'+y'z+(y+y')xz+xy+yz'$$ $$= x'y'+y'z+yxz+y'xz+xy+yz'$$ $$= x'y'+y'z(x+1)+xy(z+1)+yz'$$ $$= x'y'+y'z+xy+yz'$$

Answer 1

Continuing your attempt. You reached $$x'y'+y'z+xy+yz'$$

Now $xy=xy(z+z')=xyz+xyz'$, but $xyz'+yz'=yz'$, so we get $$x'y'+y'z+xyz+yz'$$

Similarly, $y'z=x'y'z+xy'z$, but $x'y'+x'y'z=x'y'$, so we get $$x'y'+xy'z+xyz+yz'=x'y'+x(y+y')z+yz'=x'y'+xz+yz'$$

Another way of expressing it is $$(xy'z'+x'yx)'$$ so you are not going to get any simpler (6 of the 8 expressions in the truth table are true and 2 are false).

Answer 2

This is a job for the

Consensus Theorem

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

Ok, applying that to your original statement, we see that between $x'y'$ and $xz$ we can get rid of $y'z$, and between $xz$ and $yz'$ we can get rid of $xy$, leaving us with:

$x'y'+ xz + yz'$