Reflexive Euclidean Relation is Equivalence

Theorem
A relation is an equivalence it is either left-Euclidean or right-Euclidean, and also reflexive.

Sufficient Condition
Let $\mathcal R$ be an equivalence relation.

Then by definition it is reflexive.

It is also transitive and symmetric.

So, let $x, y, z \in \mathcal R$ such that $x \mathcal R y$ and $x \mathcal R z$.

From symmetry, we have $y \mathcal R x$, and by transitivity it follows that $y \mathcal R z$.

Hence $\mathcal R$ is right-Euclidean.

Now let $x, y, z \in \mathcal R$ such that $x \mathcal R z$ and $y \mathcal R z$.

From symmetry, we have $z \mathcal R y$, and by transitivity it follows that $x \mathcal R y$.

Hence $\mathcal R$ is left-Euclidean.

So $\mathcal R$ is both left-Euclidean and right-Euclidean.

We know by definition that it is reflexive.

Thus an equivalence relation is both left-Euclidean and right-Euclidean and also reflexive.

Necessary Condition
Let $\mathcal R$ be a relation which is both left-Euclidean and reflexive.

Checking in turn each of the criteria for equivalence:

Reflexive
By definition: $\mathcal R$ has been defined as a relation which is reflexive.

Symmetric
Suppose $x \mathcal R y$.

We have that $y \mathcal R y$ from the reflexivity of $\mathcal R$.

So we have $y \mathcal R y$ and $x \mathcal R y$, and so $y \mathcal R x$ from the fact that $\mathcal R$ is left-Euclidean.

Hence $\mathcal R$ is symmetric.

Transitive
Suppose $x \mathcal R y$ and $y \mathcal R z$.

As $\mathcal R$ is symmetric (from above), we have that $z \mathcal R y$.

So we have $x \mathcal R y$ and $z \mathcal R y$, and by the fact that $\mathcal R$ is left-Euclidean, we have that $x \mathcal R z$.

So, if $\mathcal R$ be a relation which is both left-Euclidean and reflexive, it is also an equivalence relation.

A similar argument shows that if $\mathcal R$ be a relation which is both right-Euclidean and reflexive, it is also an equivalence relation.