Inverse in Group is Unique/Proof 3

Theorem
Let $\left({G, \circ}\right)$ be a group.

Then every element $x \in G$ has exactly one inverse:
 * $\forall x \in G: \exists_1 x^{-1} \in G: x \circ x^{-1} = e^{-1} = x \circ x$

where $e$ is the identity element of $\left({G, \circ}\right)$.

Proof
Let $x, y \in G$.

We already have, from the definition of inverse element, that:
 * $\forall x \in G: \exists x^{-1} \in G: x \circ x^{-1} = e = x^{-1} \circ x$

By the Latin Square Property, there exists exactly one $a \in G$ such that $a \circ x = y$.

Similarly, there exists exactly one $b \in G$ such that $x \circ b = y$.

Substituting $e$ for $y$, it follows that $x^{-1}$ as defined above is unique.