Lexicographically Ordered Pair of Ordered Semigroups with Cancellable Elements

Theorem
Let $\struct {S_1, \circ_1, \preccurlyeq_1}$ and $\struct {S_2, \circ_2, \preccurlyeq_2}$ be ordered semigroups.

Let $\struct {S_1 \times S_2, \odot} := \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}$ denote the external direct product of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.

Let $\struct {S_1 \times S_2, \preccurlyeq_l} := \struct {S_1, \preccurlyeq_1} \otimes^l \struct {S_2, \preccurlyeq_2}$ denote the lexicographic order of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$:
 * $\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \paren {x_1 \prec_1 y_1} \lor \paren {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$

Let every element of $\struct {S_1, \circ_1}$ be cancellable.

Then $\struct {S_1 \times S_2, \odot, \preccurlyeq_l}$ is also an ordered semigroup.

Proof
From Lexicographic Order is Ordering we have that $\struct {S_1 \times S_2, \preccurlyeq_l}$ is an ordered set.

From External Direct Product of Semigroups, $\struct {S_1 \times S_2, \odot}$ is a semigroup.

It remains to be shown that $\preccurlyeq_l$ is compatible with $\odot$.

Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \times S_2$ be arbitrary such that $\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2}$.


 * Case $1$: First suppose $x_1 = y_1$.


 * Case $2$: Now suppose $x_1 \prec_1 y_1$.

and the proof is complete.