Composite of Relation Isomorphisms is Relation Isomorphism

Theorem
Let $\left({S_1, \mathcal R_1}\right)$, $\left({S_2, \mathcal R_2}\right)$ and $\left({S_3, \mathcal R_3}\right)$ be relational structures.

Let: and: be relation isomorphisms.
 * $\phi: \left({S_1, \mathcal R_1}\right) \to \left({S_2, \mathcal R_2}\right)$
 * $\psi: \left({S_2, \mathcal R_2}\right) \to \left({S_3, \mathcal R_3}\right)$

Then $\psi \circ \phi: \left({S_1, \mathcal R_1}\right) \to \left({S_3, \mathcal R_3}\right)$ is also a relation isomorphism.

Proof
From Composite of Bijections, $\psi \circ \phi$ is a bijection, as, by definition, an relation isomorphism is also a bijection.

By definition of composition of mappings, $\psi \circ \phi \left({x}\right) = \psi \left({\phi \left({x}\right)}\right)$.

As $\phi$ is a relation isomorphism, we have:
 * $\forall x_1, y_1 \in S_1: x_1 \,\mathcal R_1\, y_1 \implies \phi \left({x_1}\right) \,\mathcal R_2\, \phi \left({y_1}\right)$

As $\psi$ is a relation isomorphism, we have:
 * $\forall x_2, y_2 \in S_2: x_2 \,\mathcal R_2\, y_2 \implies \psi \left({x_2}\right) \,\mathcal R_3\, \psi \left({y_2}\right)$

By setting $x_2 = \phi \left({x_1}\right), y_2 = \phi \left({y_1}\right)$, it follows that:
 * $\forall x_1, y_1 \in S_1: x_1 \,\mathcal R_1\, y_1 \implies \psi \left({\phi \left({x_1}\right)}\right) \,\mathcal R_3\, \psi \left({\phi \left({y_1}\right)}\right)$

Similarly we can show that:
 * $\forall x_3, y_3 \in S_3: x_3 \,\mathcal R_3\, y_3 \implies \phi^{-1} \left({\psi^{-1} \left({x_3}\right)}\right) \,\mathcal R_1\, \phi^{-1} \left({\psi^{-1} \left({y_3}\right)}\right)$

Hence the result, by definition of a relation isomorphism.