Proof of Lemma 4.2, p. 555 — existence of the refined mapping π on each
ProvedLocalSearchFL.UFL.exists_refined_piLet be any two assignments of the clients to facilities, with , and . Then there is a permutation of such that
- maps every onto itself;
- (Property 3.1) whenever , ;
- whenever , every with satisfies .
This is the mapping on which the facility cost bound of Lemma 4.2 is built: a client whose serving facility in is closed is reassigned through , and the fixed points of are exactly the clients of a captured block that cannot be moved to another block.
Formalization Note The family of bijections , one for each , is encoded as a single permutation of all clients with . The assignments are arbitrary maps: the statement is purely combinatorial.
import Mathlib import Definitions.Def_LocalSearchFL_UFL_captures
namespace LocalSearchFL.UFL
/-- The mapping π of the proof of Lemma 4.2 (p. 555, first paragraph of the proof): for any
assignments `σS`, `σO` of the clients, there is a permutation `π` of the clients that maps each
`N_O(o)` onto itself, satisfies Property 3.1 (if `s` does not capture `o` then
`π(N^o_s) ∩ N^o_s = ∅`), and, when `s` captures `o`, fixes every `j ∈ N^o_s` with
`π(j) ∈ N^o_s`. -/
theorem exists_refined_pi {Cl Fa : Type} [Fintype Cl] [DecidableEq Cl] [DecidableEq Fa]
(σS σO : Cl → Fa) :
∃ π : Equiv.Perm Cl, IsRefinedPi σS σO π := by sorry
end LocalSearchFL.UFL
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What the Lean code literally says, in plain math · claude-opus-5-5
Setting. Let be a finite type of clients with decidable equality. Let be any type of facilities with decidable equality; it need not be finite. Let be arbitrary maps. Write:
- and ;
- .
Say that captures when , as a strict inequality of natural numbers.
Conclusion. For every such pair of maps, there exists a bijection satisfying all three of:
- (i) for every client ;
- (ii) for all facilities such that does not capture , and every : ;
- (iii) for all facilities such that captures , and every with : .
No distances, costs or sets of open facilities appear in the statement. The maps are completely unrestricted.
Degenerate cases.
- If , the empty permutation witnesses the claim.
- If has one client , then is captured by , and the identity satisfies (i)–(iii).
- In general, the claim includes every client whose own pair is uncaptured. For each such , it asserts that moves to a client with the same -value but a different -value.
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.