Composition of Ring Homomorphisms is Ring Homomorphism/Proof 2

Theorem
Let: be rings.
 * $\left({R_1, +_1, \circ_1}\right)$
 * $\left({R_2, +_2, \circ_2}\right)$
 * $\left({R_3, +_3, \circ_3}\right)$

Let: be homomorphisms.
 * $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$
 * $\psi: \left({R_2, +_2, \circ_2}\right) \to \left({R_3, +_3, \circ_3}\right)$

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:


 * $\left({\psi \bullet \phi}\right): \left({R_1, +_1, \circ_1}\right) \to \left({R_3, +_3, \circ_3}\right)$ is a homomorphism.

To prove the above is the case, we need to demonstrate that the morphism property is held by $+_1$ and $\circ_1$ under $\psi \bullet \phi$.

We take two elements $x, y \in R_1$, and put them through the following wringer with respect to $+_1$:

The same applies to $\circ_1$:

Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by both $+_1$ and $\circ_1$ under $\psi \bullet \phi$.

Thus $\left({\psi \bullet \phi}\right): \left({R_1, +_1, \circ_1}\right) \to \left({R_3, +_3, \circ_3}\right)$ is a homomorphism.