Composite of Homomorphisms is Homomorphism/Algebraic Structure

Theorem
Let:
 * $\struct {S_1, \circ_1, \circ_2, \ldots, \circ_n}$
 * $\struct {S_2, *_1, *_2, \ldots, *_n}$
 * $\struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$

be algebraic structures.

Let:
 * $\phi: \struct {S_1, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {S_2, *_1, *_2, \ldots, *_n}$
 * $\psi: \struct {S_2, *_1, *_2, \ldots, *_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$

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:


 * $\paren {\psi \bullet \phi}: \struct {S_1, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$ is a homomorphism.

To prove the above is the case, 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 $\struct {S_1, \circ_1, \circ_2, \ldots, \circ_n}$, it follows that it holds for all of them.

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