Question

simplify the boolean expession (x'y'z')+(x'yz')+(xy'z)

Answer 1

(x'y'z')+(x'yz')+(xy'z)
x'('y'z'+yz')+xy'z
x'z'(y'+y)+xy'z
x'z'(y'+y)+xy'z
x'z'+xy'z
{(x+z)(x'+y+z')}'
{(x+z)x'+(x+z)y+(x+z)z'}'
{x'z+(x+z)y+xz'}'
{x'z+(x+z)y}'
(x+z')(y+x'z')
(x+z')y+(x+z')x'z'
(x+z')y+x'z'
xy+(y+x)z'
Did this quite fact, check work.

Answer 2

Referring to work of @KenLJW
```
(x'y'z')+(x'yz')+(xy'z)
x'('y'z'+yz')+xy'z
x'z'(y'+y)+xy'z
x'z'(y'+y)+xy'z
x'z'+xy'z .............(1)
{(x+z)(x'+y+z')}'
{(x+z)x'+(x+z)y+(x+z)z'}'
{x'z+(x+z)y+xz'}' .................(2)
{x'z+(x+z)y}'
(x+z')(y+x'z')
(x+z')y+(x+z')x'z'
(x+z')y+x'z'
xy+(y+x)z'
```
Up to line (2), everything holds. xz' is, unfortunately, missing from the following line.
However, I think the result on line (1) is adequate as the result/answer.