Order Automorphism on Well-Ordered Class is Forward Moving

Theorem
Let $\struct {A, \preccurlyeq}$ be a well-ordered class.

Let $\phi$ be an order isomorphism on $\struct {A, \preccurlyeq}$.

Then:
 * $\forall a \in A: a \preccurlyeq \map \phi a$

Proof
Let us define an element $a$ of $A$ such that:
 * $\map \phi a \prec a$

as moving backwards.

there exists an element $a$ of $A$ that moves backwards:
 * $\map \phi a \prec a$

for some $a \in A$.

Then applying $\phi$ to both sides:
 * $\map \phi {\map \phi a} \prec \map \phi a$

That is:
 * $\map \phi a$ also moves backwards

Thus if some $a \in A$ moves backwards, there is another predecessor element that also moves backwards.

Hence there is no smallest element of the set of all elements of $A$ that move backwards.

But this contradicts the properties of a well-ordered class:
 * every non-empty subclass of $A$ has a smallest element under $\preccurlyeq$.

Hence there can be no elements of $A$ that move backwards.

That is:
 * $\forall a \in A: a \preccurlyeq \map \phi a$