Order Isomorphism on Well-Ordered Set preserves Well-Ordering

Theorem
Let $\left({S_1, \preccurlyeq_1}\right)$ and $\left({S_2, \preccurlyeq_2}\right)$ be ordered sets.

Let $\phi: \left({S_1, \preccurlyeq_1}\right) \to \left({S_2, \preccurlyeq_2}\right)$ be an order isomorphism.

Then $\left({S_1, \preccurlyeq_1}\right)$ is a well-ordered set iff $\left({S_2, \preccurlyeq_2}\right)$ is also a well-ordered set.

Proof
Let $\left({S, \preccurlyeq_1}\right)$ be a well-ordered set.

Then by definition $\preccurlyeq_1$ is a well-ordering.

By Well-Ordering is Total Ordering, $\preccurlyeq_1$ is a total ordering.

From Order Isomorphism on Totally Ordered Set preserves Total Ordering, it follows that $\left({S_2, \preccurlyeq_2}\right)$ is a totally ordered set.

All that remains is to show that $\preccurlyeq$ is well-founded.

Let $T_1 \subseteq S_1$.

As $\left({S_1, \preccurlyeq_1}\right)$ is a well-ordered set:
 * $\exists m_1 \in T_1: \forall t \in T_1: t = m_1 \lor t \not \preccurlyeq_1 m_1$

Now consider the image $m_2$ of $m_1$ in $S_2$:
 * $m_2 = \phi \left({m_1}\right)$

Suppose $\left({S_2, \preccurlyeq_2}\right)$ is not a well-ordered set.

Let there exist $T_2 \subseteq S_2$ such that:
 * $\exists x \in T_2: x \ne m_2, x \preccurlyeq_2 m_2$

By Inverse of Order Isomorphism is Order Isomorphism, if $\phi$ is an order isomorphism then so is $\phi^{-1}$.

Hence:
 * $\phi^{-1} \left({x}\right) \ne \phi^{-1} \left({m_2}\right), \phi^{-1} \left({x}\right) \preccurlyeq_2 \phi^{-1} \left({m_2}\right)$

But this contradicts the fact that $\left({S_1, \preccurlyeq_1}\right)$ is well-ordered.

Hence $\left({S_2, \preccurlyeq_2}\right)$ must be a well-ordered set.

The same technique is used to show that if $\left({S_2, \preccurlyeq_2}\right)$ is a well-ordered set then so is $\left({S_1, \preccurlyeq_1}\right)$.