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The relation $R$ is defined on the set $A$ such that $aRb$ if and only if both $a$ and $b$ tell lies. Let $L$ be the subset of $A$ consisting of all workers who tell lies. Then $aRb$ holds if and only if $a \in L$ and $b \in L$.
A relation $R$ is symmetric if $aRb \implies bRa$. If $aRb$ is true, then both $a$ and $b$ tell lies. Consequently, $b$ and $a$ also tell lies, which means $bRa$ is true. Thus, the relation is symmetric.
For $R$ to be an equivalence relation, it must be reflexive, symmetric, and transitive.
Check Reflexivity: $aRa$ requires $a$ to tell a lie. If there exists a worker $c \in A$ who does not tell lies (i.e., $c \notin L$), then $cRc$ is false. Since the relation is not reflexive for all $a \in A$, it is not an equivalence relation.
Final Answer: (i) Yes, it is symmetric. (ii) No, it is not an equivalence relation.
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