OAI/Analysis/CoulombTransport/CounterexampleAssembly.lean
6 declarations, 0 contain sorry, 0 declared as axioms
- lemma
OAI.Problem356.branchMarginal_eq_splitMeasureline 18Source documentation
The four-map marginal is the generic finite split measure, with central weight
1/3and four equal outer weights1/6.lemma branchMarginal_eq_splitMeasure (nu : Measure E3) (H : Fin 4 → E3 → E3) : branchMarginal nu (H 0) (H 1) (H 2) (H 3) = Splitting.splitMeasure nu (1 / 3) (fun _ : Fin 4 => 1 / 6) H := by …Mentions: OAI.Problem356.E3, OAI.Problem356.branchMarginal, OAI.Problem356.Splitting.splitMeasure
- lemma
OAI.Problem356.ae_branchMarginal_memline 25Source documentation
Support of the central measure and its four images gives support of the five-component marginal.
lemma ae_branchMarginal_mem (nu : Measure E3) {H : Fin 4 → E3 → E3} (hH : ∀ i, Measurable (H i)) {B U : Set E3} (hU : MeasurableSet U) (hfull : ∀ᵐ x ∂nu, x ∈ B) (hBU : B ⊆ U) (hHU : ∀ i, H i '' B ⊆ U) : ∀ᵐ x ∂branchMarginal nu (H 0) (H 1) (H 2) (H 3), x ∈ U := by …Mentions: OAI.Problem356.E3, OAI.Problem356.branchMarginal
- lemma
OAI.Problem356.coulombCost_cycle_assemblyline 42 - lemma
OAI.Problem356.triplePotential_cycle_assemblyline 48 - theorem
OAI.Problem356.hasCoulombCounterexample_of_branchesline 53 - theorem
OAI.Problem356.branchCoupling_optimal_of_certificateline 135Source documentation
The concrete six-graph coupling is itself the finite optimizer. This exposes the witness used by the counterexample assembly to the separated support approximation argument.
theorem branchCoupling_optimal_of_certificate (nu : Measure E3) [IsProbabilityMeasure nu] (H : Fin 4 → E3 → E3) (hH : ∀ i, Measurable (H i)) {U : Set E3} (hU : MeasurableSet U) (hmuU : ∀ᵐ x ∂branchMarginal nu (H 0) (H 1) (H 2) (H 3), x ∈ U) {p : E3 → ℝ≥0∞} (hp : Measurable p) (hpfinite : (∫⁻ x, p x ∂branchMarginal nu (H 0) (H 1) (H 2) (H 3)) ≠ ⊤) {C : ℝ≥0∞} (hC : C ≠ ⊤) (hbound : ∀ t, tripleFst t ∈ U → tripleSnd t ∈ U → tripleThd t ∈ U → triplePotential p t ≤ coulombCost t + C) (hcontact01 : ∀ᵐ x ∂nu, triplePotential p (x, (H 0 x, H 1 x)) = coulombCost (x, (H 0 x, H 1 x)) + C) …Mentions: OAI.Problem356.E3, OAI.Problem356.branchMarginal, OAI.Problem356.tripleFst, OAI.Problem356.tripleSnd, OAI.Problem356.tripleThd, OAI.Problem356.triplePotential, OAI.Problem356.coulombCost