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Switching Algebra proof question

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove with algebraic manipulation the following equality: X Y' + Y Z' + X' Z = X' Y + Y' Z + X Z'

    2. Relevant equations

    All you need to know to prove it are the switching axioms and theorems listed on the second slide of http://meseec.ce.rit.edu/eecc341-winter2001/341-12-13-2001.pdf" [Broken].

    3. The attempt at a solution

    xy' + yz' + x'z = (xy' + yz' + x'z)' ' involution
    = ((x'+y)(y'+z)(x+z'))' DeMorgan
    = (xyz + x'y'z')' distribute and use aa' = 0
    = (x' + y' + z')(x + y + z) DeMorgan
    = (xy' + yz' + x'z) + (x'y + y'z + xz') distribute and use aa' = 0

    since we have a = a + b then b = a or b = 0 by idempotency or identities, but I think we can show b ~= 0 if x, y ,or z are not all 0 or 1.

    I feel like there might be an easier way. What do you think?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 3, 2010 #2
    all you need are distribution ,aa=a and a+a'=1

    there is no need for DeMorgan or anything else
     
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