Analysis of properties of logic operators.

Discussion in 'Homework Help' started by Tacticus, Oct 9, 2010.

  1. Tacticus

    Thread Starter New Member

    Sep 21, 2010
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    0
    The question is attached to the post.

    I'm not sure how I should start in this question.



    Commutative

    xy = yx

    Associative

    x + y = y + x


    so, are they asking me to prove:

    (x -> y) = ( y -> x)

    and

    (x -> y) + (y -> x) = (y -> x) + (x -> y)


    This question is really confusing me.

    Thanks for help in advance.
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    First of all, get your thoughts right. You have misunderstood the associative property. Look, here for an explanation: http://www.onlinemathlearning.com/number-properties.html (a random find among many).

    In order to test the commutative property of the \rightarrow operation, put in the first column of the truth table the operand y and x on the second. Apply \rightarrow as the truth table shows you and compare the results you got with the third column of the initial table. Are they the same?
    For example: 0\rightarrow1=1. Does 1\rightarrow0=1? If not, then \rightarrow is not commutative.

    Same for associative. Take permutations of 3 variables x,y and z, for the operation (x\rightarrowy)\rightarrowz. If all the results of this expression are the same with the results of the expression x\rightarrow(y\rightarrowz), then the \rightarrow operator is associative.
     
  3. Tacticus

    Thread Starter New Member

    Sep 21, 2010
    2
    0
    Thanks for the help :)
     
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