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Boolean Algebra Proof (Distribution and XOR)

  1. Sep 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Use the definition of exclusive or (XOR), the facts that XOR commutes and
    associates (if you need this) and all the non-XOR axioms and theorems you know
    from Boolean algebra to prove this distributive rule:

    A*(B (XOR) C) = (A*B) (XOR) (A*C)

    2. Relevant equations
    All the boolean algebra theorems

    3. The attempt at a solution
    So here's what I've tried

    A*(B (XOR) C)
    =A*B*C' + A*B'*C
    =A*X+A*X'
    =A*1
    =A*(X (XOR) X')

    And

    (A*B) (XOR) (A*C)
    =A*B*(A*C)' + (A*B)'*B*C
    =A*B*(A' + C') + (A' + B')*B*C
    =A*B*A' + A*B*C' + A'*B*C + B'*B*C
    =A*B*C' + A'*B*C
    =B*(A (XOR) C)
    I feel like I go in circles no matter what I do.
    Any clues?
     
  2. jcsd
  3. Sep 24, 2016 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    This looks wrong. :))
    Mistake here, too. :H
     
  4. Sep 24, 2016 #3
    Oops, that's not how demorgans works...brb with corrections.
     
  5. Sep 24, 2016 #4
    Ok, the answer came out once I realized how to use DeMorgan's Rule for the right hand side:

    (A*B) (XOR) (A*C)
    =(A*B) * (A*C)' + (A*B)' * (A*C)
    =A*B * (A' + C') + (A' + B') * A*C
    =A*A'*B + A*B*C' + A'*A*C + B'*A*C
    =A*B*C' + A*B'*C
    =A*(B*C' + B'*C)
    =A* [B (XOR) C]

    Thanks for the help!
     
    Last edited: Sep 24, 2016
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