Coset Product is Well-Defined/Proof 1

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 $N \triangleleft G$ where $G$ is a group.

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

To show that the coset product is well-defined, we need to demonstrate that $\left({a \circ b}\right) \circ N = \left({a' \circ b'}\right) \circ N$.

So:

By Equal Cosets iff Product with Inverse in Coset‎:
 * $\left({a \circ b}\right)^{-1} \circ \left({a' \circ b'}\right) \in N \implies \left({a \circ b}\right) \circ N = \left({a' \circ b'}\right) \circ N$

and the job is finished.