Coset Product is Well-Defined/Proof 3

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

Let $N$ be a normal subgroup of $G$.

Let $a, b \in G$.

Then the coset product:
 * $\left({a \circ N}\right) \circ \left({b \circ N}\right) = \left({a \circ b}\right) \circ N$

is well-defined.

Proof
Let $N \lhd G$ where $G$ is a group.

Let $a, a', b, b' \in G: N \circ a = N \circ a', N \circ b = N \circ b'$.

We need to show that $N \circ a \circ b = N \circ a' \circ b'$.

So: