Relation Isomorphism preserves Total Ordering

Theorem
Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.

Let $\struct {A, \RR}$ be a totally ordered set.

Then $\struct {B, \SS}$ is also a totally ordered set.

Proof
Let $\struct {A, \RR}$ be a totally ordered set.

Recall the definition:

From Relation Isomorphism preserves Ordering:
 * $\SS$ is an ordering.

Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.

Let $x, y \in A$.

Then either $x \mathrel \RR y$ or $y \mathrel \RR x$.

From the definition of relation isomorphism, either:
 * $\map \phi x \mathrel \SS \map \phi y$

or:
 * $\map \phi y \mathrel \SS \map \phi x$

and so by definition $\SS$ is also a total ordering.

So by definition $\struct {B, \SS}$ is also a totally ordered set.