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 and right identity are the same, and are therefore an identity.

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.