Lexicographic Order forms Well-Ordering on Ordered Pairs of Ordinals

Theorem
The Lexicographic Order $\operatorname{Le}$ is a strict well-ordering on $\left({\operatorname{On} \times \operatorname{On}}\right)$.

Proof 1
This is an instance of Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering.

Total Ordering
Suppose $\left({x, y}\right) \operatorname{Le} \left({x, y}\right)$.

Then:
 * $x < x \lor \left({x = x \land y < y}\right)$.

Both are contradictory, so $\operatorname{Le}$ is irreflexive.

Let:
 * $\left({\alpha, \beta}\right) \operatorname{Le} \left({\gamma, \delta}\right)$

and:
 * $\left({\gamma, \delta}\right) \operatorname{Le} \left({\epsilon, \zeta}\right)$

There are two cases:

Let $\alpha < \gamma$.

Then:
 * $\alpha < \epsilon$

so:
 * $\left({\alpha, \beta}\right) \operatorname{Le} \left({\epsilon, \zeta}\right)$

Let $\alpha = \gamma$.

Then:
 * $\alpha < \epsilon \lor \left({\alpha = \epsilon \land \delta < \zeta}\right)$

But also, if $\alpha = \gamma$, then $\beta < \delta$.

Therefore:
 * $\left({\alpha = \epsilon \land \beta < \zeta}\right)$

Therefore:
 * $\left({\alpha, \beta}\right) \operatorname{Le} \left({\epsilon, \zeta}\right)$

In either case, $\operatorname{Le}$ is transitive.

So $\operatorname{Le}$ is a strict ordering.

Strict Total Ordering
Let:
 * $\neg \left({\alpha, \beta}\right) \operatorname{Le} \left({\gamma, \delta}\right)$

and:
 * $\neg \left({\gamma, \delta}\right) \operatorname{Le} \left({\alpha, \beta}\right)$

Then:
 * $\neg \alpha < \gamma$

and:
 * $\neg \gamma < \alpha$

so:
 * $\alpha = \gamma$

Similarly:
 * $\neg \beta < \delta$

and:
 * $\neg \delta < \beta$

so:
 * $\beta = \delta$

By Equality of Ordered Pairs:
 * $\left({\alpha, \beta}\right) = \left({\gamma, \delta}\right)$

Therefore $\operatorname{Le}$ is a strict total ordering.

Well-Ordering
Let $A$ be a nonempty subset $A$ of $\left({\operatorname{On} \times \operatorname{On}}\right)$

Let $A$ be any class.

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

Let the mapping $1^{st}$ send each ordered pair $\left({x, y}\right)$ to its first member $x$.


 * $\displaystyle 1^{st} = \left\{ {\left({\left({x, y}\right) z}\right): z = x}\right\}$

Then $1^{st}: A \to \operatorname{On}$ is a mapping.

Take $\operatorname{Im}\left({A}\right)$, the image of $A$ under $1^{st}$.


 * $\operatorname{Im}\left({A}\right) \subseteq \operatorname{On}$

so by Subset of Ordinals has Minimal Element, $\operatorname{Im}\left(A\right)$ has a minimal element.

Let this minimal element be $\alpha$.

Let $B = \left\{ {y \in \operatorname{On} : \left({\alpha, y}\right) \in A}\right\}$.

$\alpha$ is a minimal element of $\operatorname{Im}\left({A}\right)$.

So:
 * $\left({\alpha, y}\right) \in \operatorname{Im} \left({A}\right)$

for some $y \in \operatorname{On}$.

Therefore $B$ is nonempty.

Furthermore, $B$ is some subset of the ordinals.

By Subset of Ordinals has Minimal Element it follows that $B$ has a minimal element.

Let this minimal element be $\beta$.

Therefore:
 * $\left({\alpha, \beta}\right) \in A$

Suppose there is some element $\left({\gamma, \delta}\right)$ of $A$ such that:
 * $\left({\gamma, \delta}\right) \operatorname{Le} \left({\alpha, \beta}\right)$

Then:
 * $\gamma \le \alpha$

But for all ordered pairs in $A$, $\alpha$ is a minimal first element.

Therefore $\gamma = \alpha$

But this implies that $\delta < \beta$ and $\left({\alpha, \delta}\right) \in A$.

This contradicts the fact that $\beta$ is the minimal element satisfying $\left({\alpha, \beta}\right) \in A$.

From this contradiction it follows that $\left({\alpha, \beta}\right)$ is the $\operatorname{Le}$-minimal element of $A$.