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 posets.

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, and by Well-Ordering is Total Ordering a total ordering.

From Order Isomorphism on Totally Ordered Sets, 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:
 * $$\exists m_1 \in T_1: \forall t \in T_1: t = m_1 \or t \not \preccurlyeq_1 m_1$$

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, 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 $$\left({S_1, \preccurlyeq_1}\right)$$.