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...