English
For s and H as in the statement of preimage_find_eq_disjointed, the preimage of {n} under x ↦ Nat.find (H x) equals disjointed s n.
Русский
Для заданных s и H, предобраз множества {n} под x ↦ Nat.find(H x) равен disjointed s n.
LaTeX
$$$(\forall x, \exists n, x \in s n) \Rightarrow (\{x : α \mid \operatorname{Nat.find}(H x) = n\} = \operatorname{disjointed} s n)$$$
Lean4
/-- Any well-founded relation can be extended to a well-ordering on that type. -/
theorem exists_well_order_ge : ∃ s, r ≤ s ∧ IsWellOrder α s :=
⟨(wellOrderExtension r).lt, fun _ _ h => Prod.Lex.left _ _ (rank_lt_of_rel h), ⟨⟩⟩