Left and Right Identity are the Same

Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure.

Let $e_L \in S$ be a left identity, and $e_R \in S$ be a right identity.

Then:
 * $e_L = e_R$

that is, both the left identity and right identity are the same, and are therefore an identity $e$.

Furthermore, $e$ is the only left identity and right identity for $\circ$.

Proof
Let $\left({S, \circ}\right)$ be an algebraic structure such that:


 * $\exists e_L \in S: \forall x \in S: e_L \circ x = x$
 * $\exists e_R \in S: \forall x \in S: x \circ e_R = x$

Then $e_L = e_L \circ e_R = e_R$ by both the above, hence the result.

The uniqueness of the left and right identity is a direct result of Identity is Unique.