Canonical Order Well-Orders Ordered Pairs of Ordinals

Theorem
The canonical order, $R_0$ strictly well-orders the ordered pairs of ordinal numbers.

Strict Ordering
Let $\tuple {x, y} \mathrel {R_0} \tuple {x, y}$.

Then:
 * $\map \max {x, y} < \map \max {x, y} \lor \tuple {x, y} \mathrel {\operatorname {Le} } \tuple {x, y}$

Both lead to contradictions, so:
 * $\neg \tuple {x, y} \mathrel {R_0} \tuple {x, y}$

and $R_0$ is antireflexive.

Let:
 * $(1): \quad \tuple {\alpha, \beta} \mathrel {R_0} \tuple {\gamma, \delta}$

and:
 * $(2): \quad \tuple {\gamma, \delta} \mathrel {R_0} \tuple {\epsilon, \zeta}$

There are two cases:
 * $\map \max {\alpha, \beta} < \map \max {\gamma, \delta}$

or:
 * $\map \max {\alpha, \beta} = \map \max {\gamma, \delta}$

In either case:
 * $\tuple {\alpha, \beta} \mathrel {R_0} \tuple {\epsilon, \zeta}$

and $R_0$ is transitive.

Strict Total Ordering
Suppose:
 * $\neg \tuple {\alpha, \beta} \mathrel {R_0} \tuple {\gamma, \delta}$

and:
 * $\neg \tuple {\gamma, \delta} \mathrel {R_0} \tuple {\alpha, \beta}$

Then:
 * $\map \max {\alpha, \beta} \le \map \max {\gamma, \delta}$

and:
 * $\map \max {\gamma, \delta} \le \map \max {\alpha, \beta}$

So:
 * $\map \max {\alpha, \beta} = \map \max {\gamma, \delta}$

Therefore:
 * $\neg \tuple {\alpha, \beta} \mathrel {\operatorname {Le} } \tuple {\gamma, \delta}$

and:
 * $\neg \tuple {\gamma, \delta} \mathrel {\operatorname {Le} } \tuple {\alpha, \beta}$

From Lexicographic Order forms Well-Ordering on Ordered Pairs of Ordinals:
 * $\tuple {\alpha, \beta} = \tuple {\gamma, \delta}$

Well-Ordering
Take any nonempty subset $A$ of $\paren {\On \times \On}$.

We shall allow $A$ to be any class.

This isn't strictly necessary, but it will not alter the proof.

The $\max$ mapping sends each element in $A$ to an element of $\On$.

Therefor the image of $\max$ has a minimal element, $N$.

Take $B$ to be the class of all ordered pairs $\tuple {x, y}$, such that $\map \max {x, y} = N$.

Let the $\operatorname {Le}$-minimal element of $B$ be denoted $C$.

Then:
 * $\map \max C = N$

and $C$ is seen to be $\operatorname {Le}$-minimal.

Therefore $C$ is the $R_0$-minimal element of $A$.