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electronic engineer
Feb9-06, 07:16 AM
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
Feb9-06, 09:37 AM
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
Feb9-06, 12:24 PM
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
Feb9-06, 02:28 PM
yea

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

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

electronic engineer
Feb9-06, 02:48 PM
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)