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Simplifying simple logic

  1. Feb 9, 2006 #1
    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?
     
  2. jcsd
  3. Feb 9, 2006 #2
    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
     
    Last edited: Feb 9, 2006
  4. Feb 9, 2006 #3
    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
     
  5. Feb 9, 2006 #4
    yea

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

    (xy' + x'y)' = (x' + y)(x + y') = xy + x'y'
     
  6. Feb 9, 2006 #5
    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)
     
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