Relation Isomorphism preserves Well-Ordering

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

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

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

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

Recall the definition:

By Well-Ordering is Total Ordering, $\RR$ is a total ordering.

From Relation Isomorphism preserves Total Ordering:
 * $\SS$ is a total ordering on $B$.

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

Let $A' \subseteq A$.

As $\struct {A, \RR}$ is a well-ordered set:
 * $\exists m_1 \in A': \forall a \in A': a = m_1 \lor \neg \paren {a \mathrel \RR m_1}$

Now consider the image $m_2$ of $m_1$ in $B$:
 * $m_2 = \map \phi {m_1}$

$\struct {B, \SS}$ is not a well-ordered set.

Let there exist $B' \subseteq B$ such that:
 * $\exists x \in B': x \ne m_2, x \mathrel \SS m_2$

By Inverse of Relation Isomorphism is Relation Isomorphism, $\phi^{-1}$ is a relation isomorphism.

Hence:
 * $\map {\phi^{-1} } x \ne \map {\phi^{-1} } {m_2}, \map {\phi^{-1} } x \mathrel \SS \map {\phi^{-1} } {m_2}$

But this contradicts the fact that $\struct {A, \RR}$ is well-ordered.

Hence by Proof by Contradiction $\struct {B, \SS}$ must be a well-ordered set.