= Solution
Condition (iv) says in particular that $T$ is an <equivalence relation>, so it is symmetric and transitive. It therefore implies (i) and (ii), closing the chain of implications.
Uniqueness is literal: if two <equivalence relations> both contain $R,S$ and are each contained in every such relation, each is contained in the other, so they are equal. Combining the preceding proofs gives
$$
\boxed{(\mathrm i)\Longleftrightarrow(\mathrm{ii})\Longleftrightarrow
(\mathrm{iii})\Longleftrightarrow(\mathrm{iv}).}
$$
Their common relation is the <equivalence relation> generated by $R\cup S$, and under these conditions it is already obtained by the two-step composite $R\circ S$.
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