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Proof on symmetry

  • Thread starter Kolmin
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  • #1
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Homework Statement



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

Homework Equations



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


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.
 

Answers and Replies

  • #2
Bacle2
Science Advisor
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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?
 
  • #3
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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?
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:
 
  • #4
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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].

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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