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Quick Boolean Agebra Question

  1. Sep 30, 2013 #1
    1. The problem statement, all variables and given/known data
    xy + compliment(xy) = 1


    2. Relevant equations



    3. The attempt at a solution

    is it true? because x+compliment(x) = 1

    maybe its not true...
     
  2. jcsd
  3. Sep 30, 2013 #2

    CompuChip

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    Yes. Take ##x + \overline{x} = 1## and rename x to a: ##a + \overline{a} = 1##.
    Now set ##a = xy##.

    Note, however, that ##\overline{xy} \neq \overline{x} \overline{y}##.
     
  4. Oct 1, 2013 #3
    what if x=1 and y=0 the xy=0 and compliment(xy)=0 then 0+0 != 1
     
  5. Oct 1, 2013 #4

    tiny-tim

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    Hi ZeroPivot! :smile:

    (guys, thanks for the compliments, but it's complements! :wink:)
    No, complement(xy) = 1 :wink:
     
  6. Oct 3, 2013 #5
    i meant compliment(x)compliment(y) = 0

    but thanks.
     
  7. Oct 3, 2013 #6

    tiny-tim

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    Hi ZeroPivot! :smile:
    So you meant, is ##xy + \bar{x}\bar{y} = 1## ?

    No.
     
  8. Oct 4, 2013 #7

    CompuChip

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    Because if it were, ##\overline{xy}## would equal ##\bar x \bar y##. In fact it equals ##\bar x + \bar y##.
     
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