Equivalence of Definitions of Equivalence Relation

Theorem
Let $\RR$ be a relation on a set $S$.

Definition 1 implies Definition 2
Let $\RR$ be an equivalence relation by definition 1.

By definition, $\RR$ is reflexive, symmetric and transitive.

From Relation Contains Diagonal Relation iff Reflexive:
 * $\Delta_S \subseteq \RR$

From Relation equals Inverse iff Symmetric:
 * $\RR^{-1} = \RR$

and so by definition of set equality:
 * $\RR^{-1} \subseteq \RR$

From Relation contains Composite with Self iff Transitive:
 * $\RR \circ \RR \subseteq \RR$

From Union is Smallest Superset:


 * $\Delta_S \cup \RR^{-1} \cup \RR \circ \RR \subseteq \RR$

Thus $\RR$ is an equivalence relation by definition 2.

Definition 2 implies Definition 1
Let $\RR$ be an equivalence relation by definition 2.

That is:
 * $\Delta_S \cup \RR^{-1} \cup \RR \circ \RR \subseteq \RR$

By Union is Smallest Superset:


 * $(1): \quad \Delta_S \subseteq \RR$
 * $(2): \quad \RR^{-1} \subseteq \RR$
 * $(3): \quad \RR \circ \RR \subseteq \RR$

From Relation Contains Diagonal Relation iff Reflexive, $(1)$ gives directly that $\RR$ is reflexive.

From Inverse Relation Equal iff Subset, $(2)$ gives that $\RR^{-1} = \RR$.

So from Relation equals Inverse iff Symmetric $\RR$ is symmetric.

From Relation contains Composite with Self iff Transitive, $(3)$ gives that $\RR$ is transitive.

So $\RR$ has been shown to be reflexive, symmetric and transitive.

Thus $\RR$ is an equivalence relation by definition 1.