Lexicographic Order is Ordering

Theorem
Let $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ be ordered sets.

Let $\preccurlyeq$ be the lexicographic order on $S_1 \times S_2$''':
 * $\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right) \iff \left({x_1 \prec_1 y_1}\right) \lor \left({x_1 = y_1 \land x_2 \preceq_2 y_2}\right)$

Then $\preccurlyeq$ is an ordering on $S_1 \times S_2$.

Proof
In the following, $\left({x_1, x_2}\right), \left({y_1, y_2}\right), \left({z_1, z_2}\right) \in S_1 \times S_2$.

Checking in turn each of the criteria for an ordering:

Reflexivity
From Equality of Ordered Pairs:
 * $x_1 = x_2 \land y_1 = y_2 \iff \left({x_1, x_2}\right) = \left({y_1, y_2}\right)$

Thus:
 * $\left({x_1, x_2}\right) = \left({x_1, x_2}\right)$

and so:
 * $\left({x_1, x_2}\right) \preccurlyeq \left({x_1, x_2}\right)$

by definition of lexicographic order on $S_1 \times S_2$.

So $\preccurlyeq$ has been shown to be reflexive.

Transitivity
Let:
 * $\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right)$

and:
 * $\left({y_1, y_2}\right) \preccurlyeq \left({z_1, z_2}\right)$

$(1): \quad$ Let $x_1 = y_1 = z_1$.

Then by definition of lexicographic order on $S_1 \times S_2$:


 * $x_2 \preceq_2 y_2$

and:
 * $y_2 \preceq_2 z_2$

As $\preceq_2$ is an ordering, it is transitive.

Thus:
 * $x_2 \preceq z_2$

and it follows by definition of lexicographic order on $S_1 \times S_2$ that:
 * $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$

$(2): \quad$ Let $x_1 = y_1$ and $y_1 \ne z_1$.

Then by definition of lexicographic order on $S_1 \times S_2$:
 * $y_1 \prec z_1$

But as $x_1 = y_1$:
 * $x_1 \prec z_1$

and so by definition of lexicographic order on $S_1 \times S_2$:
 * $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$

$(3): \quad$ Let $x_1 \ne y_1$ and $y_1 = z_1$.

Then by definition of lexicographic order on $S_1 \times S_2$:
 * $x_1 \prec y_1$

But as $y_1 = z_1$:
 * $x_1 \prec z_1$

and so by definition of lexicographic order on $S_1 \times S_2$:
 * $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$

$(4): \quad$ Let $x_1 \ne y_1$ and $y_1 \ne z_1$.

Then by definition of lexicographic order on $S_1 \times S_2$:
 * $x_1 \prec y_1$

and:
 * $y_1 \prec z_1$

As $\preceq_1$ is an ordering, it is transitive.
 * $x_1 \prec z_1$

So by definition of lexicographic order on $S_1 \times S_2$:
 * $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$

Thus in all cases it can be seen that:
 * $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$

So $\preccurlyeq$ has been shown to be transitive.

Antisymmetry
Suppose that:
 * $\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right)$

and:
 * $\left({y_1, y_2}\right) \preccurlyeq \left({x_1, x_2}\right)$

Suppose $x_1 \ne y_1$.

Then by definition of lexicographic order on $S_1 \times S_2$:
 * $x_1 \preceq_1 y_1$

and:
 * $y_1 \preceq_1 x_1$

But $\preceq_1$ is an ordering and so $x_1 = y_1$.

From that contradiction it follows that $x_1 = y_1$.

Then by definition of lexicographic order on $S_1 \times S_2$:
 * $x_2 \preceq_2 y_2$

and:
 * $y_2 \preceq_2 x_2$

As $\preceq_2$ is an ordering, it is antisymmetric.

Therefore:
 * $x_2 = y_2$

and so:
 * $\left({x_1, x_2}\right) = \left({y_1, y_2}\right)$

So $\preccurlyeq$ has been shown to be antisymmetric.

$\preccurlyeq$ has been shown to be reflexive, transitive and antisymmetric.

Hence by definition it is an ordering.