If $R$ is an equivalence relation on a non-empty set $A$, then prove that $R^{-1}$ is also an equivalence relation on $A$.

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> [!TLDR]
> Each of reflexivity, symmetry and transitivity of $R^{-1}$ follows directly from the corresponding property of $R$, using $(x,y)\in R^{-1} \iff (y,x)\in R$.
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> By definition, $(x,y)\in R^{-1} \iff (y,x)\in R$.
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> **Reflexive:** Since $R$ is reflexive, $(a,a)\in R$ for all $a\in A$. As $(a,a)$ reversed is itself, $(a,a)\in R^{-1}$ for all $a\in A$. So $R^{-1}$ is reflexive.
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> **Symmetric:** Suppose $(a,b)\in R^{-1}$. By definition $(b,a)\in R$. Since $R$ is symmetric, $(b,a)\in R \Rightarrow (a,b)\in R$. But $(a,b)\in R$ means, by definition of the inverse relation, $(b,a)\in R^{-1}$. So $(a,b)\in R^{-1}\Rightarrow(b,a)\in R^{-1}$, i.e. $R^{-1}$ is symmetric.
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Q.If $R$ is an equivalence relation on a non-empty set $A$ , then prove that $R^{-1}$ is also an equivalence relation on $A$ .

Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

The three checks

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Q.If RRR is an equivalence relation on a non-empty set AAA, then prove that R−1R^{-1}R−1 is also an equivalence relation on AAA.

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Q.If $R$ is an equivalence relation on a non-empty set $A$ , then prove that $R^{-1}$ is also an equivalence relation on $A$ .

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