# Order Type Multiplication is Well-Defined Operation

## 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:

 $\ds \alpha \cdot \beta$ $=$ $\ds \map \ot {\struct {S_1, \preccurlyeq_1} \otimes^a \struct {S_2, \preccurlyeq_2} }$ $\ds$ $=$ $\ds \map \ot {\struct {T_1, \preccurlyeq_{1'} } \otimes^a \struct {T_2, \preccurlyeq_{2'} } }$ Order Isomorphism is Preserved by Antilexicographic Order $\ds$ $=$ $\ds \alpha' \cdot \beta'$

$\blacksquare$