# Boolean algebra Simplification of “xy'+xz+x'y+yz+z" [closed]

F = x'y'z' + xyz'

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

F = x'y + xy' + z

How can this display?

## closed as off-topic by José Carlos Santos, John B, Paul Frost, Peter Foreman, mrtaurhoApr 6 at 13:17

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E.g. for $$F = x'y'z' + xyz'$$ make one for $$\left(X^{\complement}\cap Y^{\complement}\cap Z^{\complement}\right)\cup\left(X\cap Y\cap Z^{\complement}\right)$$.