Composite of Group Homomorphisms is Homomorphism/Proof 2

Theorem
Let:
 * $\left({G_1, \circ}\right)$
 * $\left({G_2, *}\right)$
 * $\left({G_3, \oplus}\right)$

be groups.

Let:
 * $\phi: \left({G_1, \circ}\right) \to \left({G_2, *}\right)$
 * $\psi: \left({G_2, *}\right) \to \left({G_3, \oplus}\right)$

be homomorphisms.

Then the composite of $\phi$ and $\psi$ is also a homomorphism.

Proof
So as to alleviate possible confusion over notation, let the composite of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$.

Then what we are trying to prove is denoted:


 * $\left({\psi \bullet \phi}\right): \left({G_1, \circ}\right) \to \left({G_3, \oplus}\right)$ is a homomorphism.

To prove the above is the case, we need to demonstrate that the morphism property is held by $\circ$ under $\psi \bullet \phi$.

We take two elements $x, y \in G_1$, and put them through the following wringer:

Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by $\circ$ under $\psi \bullet \phi$.

Thus $\left({\psi \bullet \phi}\right): \left({G_1, \circ}\right) \to \left({G_3, \oplus}\right)$ is indeed a homomorphism.