Group has Latin Square Property/Proof 2

Proof
We shall prove that this is true for the first equation:

Because the statements:
 * $a \circ g = b$

and
 * $g = a^{-1} \circ b$

are equivalent, we may conclude that $g$ is indeed the only solution of the equation.

The proof that the unique solution of $h$ is $b \circ a^{-1}$ in the second equation follows similarly.