Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 4 / 5E / iv / 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 givesTheir common relation is the equivalence relation generated by $R\cup S$, and under these conditions it is already obtained by the two-step composite $R\circ S$.