Composite of Group Homomorphisms is Homomorphism/Proof 2

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:


 * $\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_3, \oplus}$ 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 $\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_3, \oplus}$ is indeed a homomorphism.