If R and S are two equivalence relations on a set A, then

The correct answer is Option (3) → $(R∩S)^{-1}$ is also an equivalence relation.

• Let $R$ and $S$ be equivalence relations on a set $A$.
• Then both $R$ and $S$ are reflexive, symmetric, and transitive.
• The intersection $R \cap S$ consists of only those pairs that are in both $R$ and $S$.
• Reflexive: Since all $(a, a)$ are in both $R$ and $S$, they are also in $R \cap S$.
• Symmetric: If $(a, b) \in R \cap S$, then $(b, a) \in R$ and $(b, a) \in S$, so $(b, a) \in R \cap S$.
• Transitive: If $(a, b), (b, c) \in R \cap S$, then $(a, c) \in R$ and $(a, c) \in S$, so $(a, c) \in R \cap S$.

Hence, $R \cap S$ is always an equivalence relation.

Therefore, $(R ∩ S)^-1$ is also an equivalence relation.