Isomorphisms between Lower Sections of Well-Ordered Classes are Nested

Theorem
Let $\struct {A, \preccurlyeq_A}$ and $\struct {B, \preccurlyeq_B}$ be well-ordered classes.

Let $\phi_1$ and $\phi_2$ be order isomorphisms from a lower section of $\struct {A, \preccurlyeq_A}$ to a lower section of $\struct {B, \preccurlyeq_B}$.

Then either:
 * $\phi_1 \subseteq \phi_2$

or:
 * $\phi_2 \subseteq \phi_1$

where $\subseteq$ denotes the subset relation on mappings:
 * $f \subseteq g \iff \forall x \in \Dom f: \map f x = \map g x$

Proof
Let us label the domains of $\phi_1$ and $\phi_2$:
 * $D_1 = \Dom {\phi_1}$
 * $D_2 = \Dom {\phi_2}$

Because $D_1$ and $D_2$ are both lower sections of $\struct {A, \preccurlyeq_A}$, they are comparable under the subset relation.

, suppose $D_1 \subseteq D_2$.

Then the restriction $\phi_2 {\restriction_{D_1} }$ is an order isomorphisms from $D_1$ to a lower section of $\struct {B, \preccurlyeq_B}$.

From Lower Sections of Well-Ordered Classes are Order Isomorphic at most Uniquely:
 * $\phi_2 {\restriction_{D_1} } = \phi_1$

That is:
 * $\phi_1 \subseteq \phi_2$

Similarly if $D_2 \subseteq D_1$ then:
 * $\phi_2 \subseteq \phi_1$