Can Boolean Logic Be Simplified to Use XOR Operations?

[electronic engineer]

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?

 

[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

 

[electronic engineer]

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

 

[waht]

yea

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

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

 

[electronic engineer]

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)