Theorem: Composite of Homomorphisms is Homomorphism/R-Algebraic Structure

Let $\struct {R, +_R, \times_R}$ be a ring.
Let:
:$\struct {S_1, \odot_1, \ldots, \odot_n}$
:$\struct {S_2, {\odot'}_1, \ldots, {\odot'}_n}$
:$\struct {S_3, {\odot' '}_1, \ldots, {\odot' '}_n}$
be algebraic structures each with $n$ operations.
Let:
:$\struct {S_1, *_1}_R$
:$\struct {S_2, *_2}_R$
:$\struct {S_3, *_3}_R$
be $R$-algebraic structures.
Let:
:$\phi: \struct {S_1, *_1}_R \to \struct {S_2, *_2}_R$
:$\psi: \struct {S_2, *_2}_R \to \struct {S_3, *_3}_R$
be homomorphisms.
Then the composite of $\phi$ and $\psi$ is also a homomorphism.

Proof:
What we are trying to prove is that $\paren {\psi \circ \phi}: \struct {S_1, *_1}_R \to \struct {S_3, *_3}_R$ is a homomorphism.
That is:
:$(1): \quad \forall k \in \closedint 1 n: \forall x, y \in S_1: \map {\paren {\psi \circ \phi} } {x \odot_k y} = \map {\paren {\psi \circ \phi} } x {\odot' '}_k \map {\paren {\psi \circ \phi} } y$
:$(2): \quad \forall x \in S: \forall \lambda \in R: \map {\paren {\psi \circ \phi} } {\lambda *_1 x} = \lambda *_3 \map \phi x$
where $\closedint 1 n = \set {1, 2, \ldots, n}$ denotes an integer interval.
To prove the above is the case, we need to demonstrate that the morphism property is held by each of the operations $\odot_1, \odot_2, \ldots, \odot_n$ under $\psi \circ \phi$.
Let $\odot_k$ be one of these operations.
We take two elements $x, y \in S_1$, and put them through the following wringer:
{{begin-eqn}}
{{eqn | l = \map {\paren {\psi \circ \phi} } {x \odot_k y}
      | r = \map \psi {\map \phi {x \odot_k y} }
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map \psi {\map \phi x {\odot'}_k \map \phi y}
      | c = {{Defof|Morphism Property}} applied to $\phi$
}}
{{eqn | r = \map \psi {\map \phi x} {\odot' '}_k \map \psi {\map \phi y}
      | c = {{Defof|Morphism Property}} applied to $\psi$
}}
{{eqn | r = \map {\paren {\psi \circ \phi} } x {\odot' '}_k \map {\paren {\psi \circ \phi} } y
      | c = {{Defof|Composition of Mappings}}
}}
{{end-eqn}}
Hence the morphism property is held by $\odot_k$ under $\psi \circ \phi$.
As $\odot_k$ is an arbitrary operation in $\struct {S_1, \odot_1, \ldots, \odot_n}$, it follows that morphism property is held by all of $\odot_1, \ldots, \odot_n$.
Thus $\paren {\psi \circ \phi}: \struct {S_1, *_1}_R \to \struct {S_3, *_3}_R$ satisfies $(1)$ above.
We take an element $x \in S_1$, and apply the following:
{{begin-eqn}}
{{eqn | l = \map {\paren {\psi \circ \phi} } {\lambda *_1 x}
      | r = \map \psi {\map \phi {\lambda *_1 x} }
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map \psi {\lambda *_2 \map \phi x}
      | c = {{Defof|Morphism Property}} applied to $\phi$
}}
{{eqn | r = \lambda *_3 \map \psi {\map \phi x}
      | c = {{Defof|Morphism Property}} applied to $\psi$
}}
{{eqn | r = \lambda *_3 \map {\paren {\psi \circ \phi} } x
      | c = {{Defof|Composition of Mappings}}
}}
{{end-eqn}}
Hence the morphism property is held by $\odot_k$ under $\psi \circ \phi$.
As $\odot_k$ is an arbitrary operation in $\struct {S_1, \odot_1, \ldots, \odot_n}$, it follows that morphism property is held by all of $\odot_1, \ldots, \odot_n$.
Thus $\paren {\psi \circ \phi}: \struct {S_1, *_1}_R \to \struct {S_3, *_3}_R$ satisfies $(2)$ above.
Hence the result.
{{qed}}