Composite of Homomorphisms is Homomorphism

Theorem
Let: be algebraic structures.
 * $$\left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right)$$
 * $$\left({S_2, *_1, *_2, \ldots, *_n}\right)$$
 * $$\left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$$

Let: be homomorphisms.
 * $$\phi: \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_2, *_1, *_2, \ldots, *_n}\right)$$
 * $$\psi: \left({S_2, *_1, *_2, \ldots, *_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$$

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:

$$\left({\psi \bullet \phi}\right): \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$$ is a homomorphism.

To prove homomomorphism, we need to demonstrate that the morphism property is held by each of the operations $$\circ_1, \circ_2, \ldots, \circ_n$$ under $$\psi \bullet \phi$$.

Let $$\circ_k$$ be one of these operations.


 * We take two elements $$x, y \in S_1$$, 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 $$\circ_k$$ under $$\psi \bullet \phi$$.

As this holds for any arbitrary operation $$\circ_k$$ in $$\left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right)$$, it follows that it holds for all of them.

Thus $$\left({\psi \bullet \phi}\right): \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$$ is indeed a homomorphism.