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Boolean algebra - XOR

  1. Sep 23, 2016 #1

    Rectifier

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    The problem
    This is not a complete homework problem. I am at the last step of the solution to a long problem and only interested to know whether these following expressions are equivalent.

    My answer:
    ## a \oplus ab \oplus ac ##

    Answer in my book:
    ## a \oplus b \oplus c ##

    The attempt
    I have tried rewriting that expression and get
    ## a \oplus ab \oplus ac = a \oplus a(b \oplus c) ##

    I am not advancing any longer from there. I am surely missing some algebraic law for boolean expressions. Could you please help me?
     
  2. jcsd
  3. Sep 23, 2016 #2
    Did you try building a truth table?
    If you do you'll see that your answer is different from the book's answer.
    For example ABC = 0,1,0
    The book answer gives 1
    Your answer results in 0
     
  4. Sep 23, 2016 #3

    jedishrfu

    Staff: Mentor

    Part of your problem may be your understanding of operator precedence.

    It seems to vary slightly across programming languages. For C, Java and Javascript it is:

    NOT > AND > XOR > OR

    a XOR ( a AND b) XOR ( a AND c)

    with XOR equivalent to:

    x XOR y = (x OR y) AND (NOT ( x AND y) // mentor note: fixed the expression (thx cpscdave)

    reworking the second and third terms should get you to the goal.
     
    Last edited: Sep 23, 2016
  5. Sep 23, 2016 #4

    Rectifier

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    Well, I guess I am doing something wrong then. :/ Thanks for the help.
     
  6. Sep 23, 2016 #5
    Think you mean
    (x OR y) AND [NOT ( x AND y)]

    or alternatively

    (x OR y) AND ( NOT x or NOT y)
     
  7. Sep 23, 2016 #6

    jedishrfu

    Staff: Mentor

    Yes you are right my apologies.

    I got my up and down arrows confused in my conversion to ands and ors.

    I fixed my post.
     
  8. Sep 23, 2016 #7

    SammyS

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    a ⊕ b ⊕ c does have a rather interesting truth table.
     
  9. Sep 25, 2016 #8

    NascentOxygen

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    Staff: Mentor

    I would proceed by converting one XOR at a time to its equivalent in AND and OR, starting with this:

    ## \mathrm {a \oplus ab \oplus ac = a (\overline {ab \oplus ac }) + \overline a (ab \oplus ac)}##
     
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