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Proof on symmetry

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]R[/itex] is simmetric iff [itex]R=R^{-1}[/itex]

    2. Relevant equations

    [itex]( \forall x \forall y ((x,y) \in R \rightarrow (y,x) \in R)) \leftrightarrow R=R^{-1}[/itex]


    3. The attempt at a solution

    My problem is with my formulation in [2.] of the statement I have to prove.

    Is that formulation right or the right one is [itex]( \forall x \in A \forall y \in A ((x,y) \in R \rightarrow (y,x) \in R)) \leftrightarrow R=R^{-1}[/itex]?

    The difference is significative, at least for my purpose. In the first case, I can prove it, in the second one, I cannot (or I am not able), so I would like to know if the second one is redundant.
     
  2. jcsd
  3. Aug 25, 2012 #2

    Bacle2

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    What is A? And, just to makesure, what is the definition of symmetric you're starting with? I know this is a standard term, but maybe your definition is slightly different?
     
  4. Aug 25, 2012 #3
    Well, that's the problem. :smile:

    Given a set [itex]A[/itex], a relation [itex]R[/itex] on [itex]A[/itex] is symmetric if [itex]\forall x \in A \forall y \in A(xRy \rightarrow yRx)[/itex].

    So, here we are.The formulation in [3.] should be the correct one to translate the statement in [1.] in logical terms, but - at the same time - it is problematic cause I am free to derive the [itex][ \rightarrow ][/itex] part of the proof only if there is not [itex]A[/itex] in the definition. Then, is it redundant or not?

    PS: Not sure you read my previous thread, but the all problem arises cause I am trying to use the software Proof Designer, which is a tool that can be use to implement the learning system presented in the book How to prove it: a structured approach to develop proof-skills. This software (and the books) are great, but the (positive!) side-effect is that human flexibility is not the point here. :smile:
     
  5. Aug 26, 2012 #4

    Just for the curious reader, who is studying How to prove it and who is frustrated by the lack of results obtained with Proof Designer, I think I found a way to fix the problem.

    We define the theorem that has to be proved in the following way: [tex] \forall x \in A \forall y \in A((x,y) \in A \times A \rightarrow (y,x) \in A \times A) \leftrightarrow A \times A = (A \times A)^{-1} [/tex].
    Then we define [itex]R=A \times A[/itex] and the problem is solved.

    PS: [itex]A[/itex] obviously is not redundant, so my question was simply wrong!
     
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