## simplifying simple logic

let's consider this boolean logic experession:

s=x'y'z+x'yz'+xy'z'+xyz

can i simplify it to:

s=x'y'z+x'yz'+xy'z'

as xyz=1 where x,y,z in high logic(1)

what's the simplest expression?

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 You have 2 xor's s=x'y'z+x'yz'+xy'z'+xyz x'y'z + xyz = z(x'y' + xy) = z(x xor y) x'yz' + xy'z' = z'(x'y + xy') = z'(x xor y) add these up and go from there

 Quote by waht You have 2 xor's s=x'y'z+x'yz'+xy'z'+xyz x'y'z + xyz = z(x'y' + xy) = z(x xor y) x'yz' + xy'z' = z'(x'y + xy') = z'(x xor y) add these up and go from there
x xor y= x'y+xy' that's right, but i don't think x'y'+xy=x xor y

are you sure from that?

thanks

## simplifying simple logic

yea

xy +x'y' is just the complement of (x xor y)

(xy' + x'y)' = (x' + y)(x + y') = xy + x'y'

 s=x'y'z+x'yz'+xy'z'+xyz x'y'z + xyz = z(x'y' + xy) = z(x xor y)' x'yz' + xy'z' = z'(x'y + xy') = z'(x xor y) so s=(z) xor (y) xor (z)