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A smal problem with boolean algebra

  1. Sep 23, 2010 #1
    There's something I can't seem to figure out.

    say we have the function
    f=xz +x'z'

    why is this equals (x xor z)' ?
    i thought using De-Morgan's law we shall get

    (x xor z)'=(x'z+z'x)'=(xz'zx')=0

    then why does (x xor z)'=xz+x'z' ?


    and one more small thing,
    say we have
    (x xor y) xor (x xor z)'
    why does (x xor y) xor (x xor z)' = (x xor y xor x xor z)'

    how did they come to this term?
     
  2. jcsd
  3. Sep 23, 2010 #2

    vela

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    De Morgan's law also tells you (ab)' = a' + b', so

    (x xor z)' = (x'z+z'x)' = (x'z)'(z'x)' = ...

    I'll let you finish it.
     
  4. Sep 23, 2010 #3
    thanks so much i feel so stupid for missing the obvious.

    one more thing though,
    why does the following equation stand?

    (x XOR y) XOR (x XOR z)' = (x XOR y XOR x XOR z)'
     
  5. Sep 25, 2010 #4

    vela

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    That's kind of like asking why a trig identity is true. It's true because it's a consequence of how things are defined. You just have to prove it by using algebraic manipulations or just listing all possible values of x, y, and z and showing the two sides give the same results.
     
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