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:

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

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Sep 24 '19 at 07:01
For this type of question it would be more useful if you share what you have tried so far. You surely have tried something as you arrived at a different result than the supposed correct result. When doing boolean algebra truth tables are good to check correctness. — skyking, Sep 24 '19 at 07:01
@skyking has suggested a good approach. There are only 8 possibilities to look at. — almagest, Sep 24 '19 at 07:49
Both $xz+yz'+x'y'$ and $xy+x'z'+y'z$ are correct, though they look different. If you've got one of them, it is correct. — user376343, Sep 24 '19 at 07:58
the result I have tried so far is 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' No idea how to continue. As I checked on the truth table, the result is 1. — Ganchimeg, Sep 24 '19 at 08:00
Please move your try from a comment to the question itself. Use $ to delimit mathematical expressions. — Robert Shore, Sep 24 '19 at 09:13

score 2 · Answer 1 · answered Sep 24 '19 at 09:59

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).

score 1 · Answer 2 · answered Sep 25 '19 at 03:00

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:

$x'y'+ xz + yz'$

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

2 Answers2