Group has Latin Square Property/Proof 1

Proof
Thus, such a $g$ exists.

Suppose $g, g' \in G$ where $a \circ g = b = a \circ g'$.

Then:

Thus uniqueness holds.

To prove the second part of the theorem, let $h = b \circ a^{-1}$.

The remainder of the proof follows a similar procedure to the above.