Composition of Ring Automorphisms is Ring Automorphism

Theorem
Let $R$ be a set.

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

Let: be (ring) automorphisms.
 * $\phi: \left({R, +_1, \circ_1}\right) \to \left({R, +_2, \circ_2}\right)$
 * $\psi: \left({R, +_2, \circ_2}\right) \to \left({R, +_3, \circ_3}\right)$

Then the composite of $\phi$ and $\psi$ is also a (ring) automorphism.

Proof
A ring automorphism is a ring isomorphism $f$ from a set to itself.

That is:
 * $\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\phi}\right)$
 * $\operatorname{Dom} \left({\psi}\right) = \operatorname{Cdm} \left({\psi}\right)$

From Composition of Ring Isomorphisms is Ring Isomorphism, $\psi \circ \phi$ is a ring isomorphism.

By definition of composition of mappings:
 * $\operatorname{Cdm} \left({\phi}\right) = \operatorname{Dom} \left({\psi}\right)$

Thus:
 * $\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\psi}\right) = R$

and so $\psi \circ \phi$ is a ring automorphism.