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Prove that if $R$ is symmetric, then $R^{-1}$ is symmetric, $R$ being a relation over $A$, and $\lnot(A = \varnothing)$.

This came as an exercise in my book.

I couldn't do anything - there is no (evident) explanation about how to prove things like that in my book (actually it seemed like a bonus question). Some Google searches like "prove if a relation is symmetric then the inverse is symmetric" seem to return other topics.

All I could do (nothing) was begin with:

Our hypothesis is $\forall a,b \in A [aRb \implies bRa]$. We want to show, based on it, that $a,b\in A[aR^{-1}b \implies aR^{-1}b]$... and I'm stuck.

Note that, I don't really want the solution to this exercise. I just want an explanation of a "general procedure" I should be taking when working with this kind of exercises, as in, "observe that you can do this and that to reach this, etc".

What do I mean by this "kind" of exercises? Well, like these:

  • If $R$ is antisymmetric, then $R^{-1}$ is antisymmetric.
  • If $R$ is reflexive, then $R \cap R^{-1}$ is reflexive.
  • If $R$ is transitive, then $R \cap R^{-1}$ is transitive.
  • Etc...

2 Answers 2

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Think about the definition of $R^{-1}$

$aRb \implies bR^{-1}a$

If by symmetry of $R$ you know that $aRb \implies bRa$, what does $bRa$ imply about $R^{-1}$?

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    I'm guessing they want you to think of relations as being ordered pairs: if $aRb$ then $(a,b) \in R$. Likewise if $aR^{-1}b$ then $(a,b) \in R^{-1}$.2012-12-06
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You have a typo which may be preventing you from solving the problem. Instead of $a,b\in A[aR^{-1}b \implies aR^{-1}b]$ you want $a,b\in A[aR^{-1}b \implies bR^{-1}a]$ Now imagine you are given $c,d$ such that $cRd$. Now you know that $dRc$ because of symmetry. What does that tell you about $R^{-1}?$ What is your definition of $R^{-1}?$