Ordinals Isomorphic to the Same Well-Ordered Set

Theorem
Let $A$ and $B$ be ordinals.

Let $(\prec,S)$ be a strict well-ordering.

Let $( \in, A )$ and $(\prec,S)$ be order isomorphic.

Let $( \in, B )$ and $(\prec,S)$ be order isomorphic.

Then, $A = B$.

Proof
Let $\phi_1$ denote the function creating the order isomorphism between $( \in, A )$ and $(\prec,S)$.

Let $\phi_2$ denote the function creating the order isomorphism between $( \in, B )$ and $(\prec,S)$.

Then, $\phi_2^{-1} : S \to B$ is an order isomorphism by Inverse of Order Isomorphism.

$\phi_1 \circ \phi_2^{-1} : A \to B$ is an order isomorphism by Composite of Order Isomorphisms

So $( \in,A$ and $(\in,B)$ are order-isomorphic ordinals. By Isomorphic Ordinals are Equal, $A = B$