Antilexicographic Product of Totally Ordered Sets is Totally Ordered

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

Let $S_1 \otimes^a S_2 = \struct {S_1 \times S_2, \preccurlyeq_a}$ be the antilexicographic product of $S_1$ and $S_2$.

Then $\struct {S_1 \times S_2, \preccurlyeq_a}$ is itself a totally ordered set.

Proof
From Antilexicographic Order is Ordering, we have that $\preccurlyeq_a$ is an ordering.

It remains to be shown that $\preccurlyeq_a$ is connected.

By definition of antilexicographic product, we have that:
 * $\tuple {a, b} \preccurlyeq_a \tuple {c, d} \iff \tuple {c \prec_2 d} \lor \paren {c = d \land a \preccurlyeq_1 b}$

We note that as $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ are both totally ordered sets, then $\preccurlyeq_1$ and $\preccurlyeq_2$ are both connected.

Hence either:
 * $c \prec_2 d$ or $d \prec_2 c$ or $c = d$

and also either
 * $a \preccurlyeq_2 b$ or $b \preccurlyeq_2 a$

If $c \prec_2 d$ or $d \prec_2 c$, it follows directly that:
 * $\tuple {a, b} \preccurlyeq_a \tuple {c, d}$ or $\tuple {c, d} \preccurlyeq_a \tuple {a, b}$

Suppose $c = d$.

We have that either:
 * $a \preccurlyeq_2 b$ or $b \preccurlyeq_2 a$

and so it follows that:
 * $\tuple {a, b} \preccurlyeq_a \tuple {c, d}$ or $\tuple {c, d} \preccurlyeq_a \tuple {a, b}$

Hence, by definition, $\preccurlyeq_a$ is connected.

So we have shown that:
 * $\preccurlyeq_a$ is connected
 * $\preccurlyeq_a$ is reflexive, transitive and antisymmetric.

Thus by definition, $\preccurlyeq_a$ is a total ordering and so $\struct {S_1 \times S_2, \preccurlyeq_a}$ is a totally ordered set.