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

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

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(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.
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.

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