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Boolean Algebra simplication

  1. Apr 19, 2015 #1
    1. Simplify the expression:

    F = xyz' + xy'z' + x'yz + xyz


    2. Postulates and theorems


    3. The attempt at a solution

    F = xyz' + xy'z' + x'yz + xyz

    = x(yz' + y'z') + yz(x' + x) (Distributive)

    = x(yz' + y'z') + yz.1 (Complement)


    = x(yz' + y'z') + yz (identity)

    This is where I need help.
     
  2. jcsd
  3. Apr 19, 2015 #2
    Notice that you can factor the z' out of the first term.
     
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