Group has Latin Square Property/Proof 3

Proof
Suppose that $\exists x, y \in G: a \circ x = b = a \circ y$.

So such an element, if it exists, is unique.

Now it is demonstrated that $g = a^{-1} b$ satisfies the requirement for $a \circ g = b$

Since $a \in G$, it follows by group axiom $G3$: existence of inverses that $a^{-1} \in G$.

Then:

Thus, such a $g$ exists.

The properties of $h$ are proved similarly.