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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...
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up vote 1 down vote accepted

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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Hm..., so I could do something like... $$(aRb \implies bRa) \implies (bR^{-1}a \implies aR^{-1}b)$$ and then... I'm done since this proves that $R^{-1}$ is symmetric? – Zol Tun Kul Dec 6 '12 at 5:08
I do believe so. Not too bad, huh? – cheepychappy Dec 6 '12 at 5:11
Well that was silly... thanks a lot! I guess the other exercises are similar to this. – Zol Tun Kul Dec 6 '12 at 5:12
Indeed. Out of curiosity, what book are you working from? – cheepychappy Dec 6 '12 at 5:17
Well, it is in spanish, "Introduction to Discrete Mathematics" ("Introduccion a la Matematica Discreta") by Manuel Murillo Tsiji – Zol Tun Kul Dec 6 '12 at 5:20

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 $R^{-1}?$

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