Order Type Multiplication is Well-Defined Operation

Theorem
The multiplication operation on order types is well-defined.

Proof
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.

Let $\struct {T_1, \preccurlyeq_{1'} }$ and $\struct {T_2, \preccurlyeq_{2'} }$ be ordered sets such that:


 * $\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {T_1, \preccurlyeq_{1'} }$


 * $\struct {S_2, \preccurlyeq_2}$ is isomorphic to $\struct {T_2, \preccurlyeq_{2'} }$

Let $\alpha := \map \ot {S_1, \preccurlyeq_1}$ and $\beta := \map \ot {S_2, \preccurlyeq_2}$ denote the order types of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ respectively.

Let $\alpha' := \map \ot {T_1, \preccurlyeq_{1'} }$ and $\beta' := \map \ot {T_2, \preccurlyeq_{2'} }$ denote the order types of $\struct {T_1, \preccurlyeq_{1'} }$ and $\struct {T_2, \preccurlyeq_{2'} }$ respectively.

It is required to show that:
 * $\alpha \cdot \beta$

is the same as:
 * $\alpha' \cdot \beta'$

We have that: