Encyclopedia Cosmology Cosmology Polarized Birth Interface Cost Interface Cost Card
ARTICLE 5 claims 5 theorems
Cosmology Polarized Birth Interface Cost Interface Cost Card
In a polarized birth field, carrying the bulk is free; only the boundary between charge regions is paid for.
The interface cost
The declaration interfaceCost_card establishes a precise accounting rule for a discrete spatial structure called a polarized birth field. The field assigns each cell a charge of +1, 0, or -1, and the framework's ledger, a discrete record of recognition events, posts a cost for every adjacency between two cells. The cost is not arbitrary: it is the forced function J(x) = (x + x⁻¹)/2 - 1, evaluated at the golden ratio φ raised to the difference in charges between the two cells.
Two facts make the accounting simple. A carried edge connects cells of equal charge, so the charge difference is zero, and J(φ⁰) = J(1) = 0 exactly. The bulk is carried for literally zero recognition cost. An interface edge connects a cell on the x = 0 spine to a cell at x = ±1, so the charges differ by exactly ±1, and the cost is J(φ) = (√5 - 2)/2 > 0, a genuine positive number. The theorem proves that the total cost of the entire field equals the interface cost alone, with the carried cost exactly zero.
In two dimensions, the diamond lattice at time step t has total cost (8t - 4) · J(φ). In three dimensions, the octahedron lattice has total cost (8t² - 8t + 4) · J(φ). The cost scales with the number of interface edges, which grows with the boundary area, not with the bulk volume. The framework reads this as a literal cost-unit statement of a compute-watch law: recognition activity is paid at the codimension-1 interface, while the interior is carried coarse for free.
THEOREM Jpow_zero · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- A zero-rung gap (a carried, monochromatic edge) costs nothing: `J(φ^0) = J(1) = 0`. -/
lemma Jpow_zero : Jpow 0 = 0 := by
rw [Jpow, zpow_zero, Cost.Jcost_unit0]
THEOREM Jpow_of_abs_one · Jcost_phi_pos · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- Either a single rung up or a single rung down costs exactly `J(φ)`: the recognition cost of a
forced interface distinction. -/
lemma Jpow_of_abs_one {d : ℤ} (h : d = 1 ∨ d = -1) : Jpow d = Cost.Jcost Constants.phi := by
rcases h with h | h <;> subst h
· exact Jpow_one
· exact Jpow_neg_one
/-- `J(φ) > 0`: a forced interface distinction has a genuine, strictly positive recognition cost.
`J(φ) = (φ - 1)²/(2φ) = (√5 - 2)/2`. -/
lemma Jcost_phi_pos : 0 < Cost.Jcost Constants.phi := by
rw [Cost.Jcost_eq_sq Constants.phi_ne_zero]
apply div_pos
· have hne : Constants.phi - 1 ≠ 0 := sub_ne_zero.mpr Constants.phi_ne_one
positivity
· have := Constants.phi_pos; linarith
THEOREM totalCost_eq_interfaceCost · totalCost_eq_interfaceCost · carriedCost_eq_zero · carriedCost_eq_zero · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- **The total cost equals the interface cost**, because the carried bulk contributes nothing. -/
theorem totalCost_eq_interfaceCost (t : ℕ) : totalCost t = interfaceCost t := by
have hsplit := Finset.sum_filter_add_sum_filter_not (E t)
(fun p => polarized t p.1 ≠ polarized t p.2) (edgeCost t)
have hBeq : (E t).filter (fun p => polarized t p.1 ≠ polarized t p.2) = B t := by
rw [E, B, Finset.filter_filter]
have hMeq : (E t).filter (fun p => ¬ (polarized t p.1 ≠ polarized t p.2)) = carried t := by
rw [carried]
apply Finset.filter_congr
intro p _
simp
rw [hBeq, hMeq] at hsplit
have hzero : ∑ p ∈ carried t, edgeCost t p = 0 := by
apply Finset.sum_eq_zero
intro p hp
exact edgeCost_carried_zero t p hp
rw [hzero, add_zero] at hsplit
simp only [totalCost, interfaceCost]
exact hsplit.symm
/-- **Carried cost is exactly zero.** The whole bulk the engine carries coarse costs no recognition. -/
theorem carriedCost_eq_zero (t : ℕ) : carriedCost t = 0 := by
simp only [carriedCost]
apply Finset.sum_eq_zero
intro p hp
exact edgeCost_carried_zero t p hp
THEOREM totalCost_card · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- **The total recognition cost of the polarized diamond field is `(8t - 4) • J(φ)`** (`t ≥ 1`):
the carried bulk is free, and the whole cost sits on the `8t - 4` interface edges. -/
theorem totalCost_card (t : ℕ) (ht : 1 ≤ t) :
totalCost t = (8 * t - 4) • Cost.Jcost Constants.phi := by
rw [totalCost_eq_interfaceCost, interfaceCost_card t ht]
THEOREM totalCost_card · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- **The total recognition cost of the polarized diamond field is `(8t - 4) • J(φ)`** (`t ≥ 1`):
the carried bulk is free, and the whole cost sits on the `8t - 4` interface edges. -/
theorem totalCost_card (t : ℕ) (ht : 1 ≤ t) :
totalCost t = (8 * t - 4) • Cost.Jcost Constants.phi := by
rw [totalCost_eq_interfaceCost, interfaceCost_card t ht]
What this page does not claim
The declaration does not claim that the polarized birth field is the actual physical universe. It does not claim that the cost function J is derived from the interface structure; J is a separate theorem. It does not claim that the bulk is physically free, only that its recognition cost is zero within this model.
Verify this page
Every tagged claim above names its theorem. To check one yourself rather than trust this page, elaborate the source module with Lean 4 and audit its axiom basis:
$ lake env lean IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)
A page whose claims cannot be reproduced this way does not ship. In production, every anchor links to the exact declaration in the public source release, and this block carries the build receipt for the page itself.
Derived articles
This page is generated by a question-recursion engine: the questions its answers raise become the next pages. The current agenda, with open targets marked red:
- What physical process, if any, corresponds to the discrete time step t in the diamond and octahedron lattices?
- How does the interface cost relate to the boundary area of the lattice as t grows?
- Does the zero carried cost hold for lattices other than the diamond and octahedron?
- What is the recognition cost of a single interface edge in physical units, if the framework's constants are applied?
MACHINE LAYER · GROUNDED CLAIM TABLE · CLICK TO EXPAND
THEOREM Jpow_zero · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- A zero-rung gap (a carried, monochromatic edge) costs nothing: `J(φ^0) = J(1) = 0`. -/ lemma Jpow_zero : Jpow 0 = 0 := by rw [Jpow, zpow_zero, Cost.Jcost_unit0]A carried edge connects cells of equal charge, so the charge difference is zero, and J(φ⁰) = J(1) = 0 exactly. Jpow_zero · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.leanTHEOREM Jpow_of_abs_one · Jcost_phi_pos · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- Either a single rung up or a single rung down costs exactly `J(φ)`: the recognition cost of a forced interface distinction. -/ lemma Jpow_of_abs_one {d : ℤ} (h : d = 1 ∨ d = -1) : Jpow d = Cost.Jcost Constants.phi := by rcases h with h | h <;> subst h · exact Jpow_one · exact Jpow_neg_one/-- `J(φ) > 0`: a forced interface distinction has a genuine, strictly positive recognition cost. `J(φ) = (φ - 1)²/(2φ) = (√5 - 2)/2`. -/ lemma Jcost_phi_pos : 0 < Cost.Jcost Constants.phi := by rw [Cost.Jcost_eq_sq Constants.phi_ne_zero] apply div_pos · have hne : Constants.phi - 1 ≠ 0 := sub_ne_zero.mpr Constants.phi_ne_one positivity · have := Constants.phi_pos; linarithAn interface edge connects a cell on the x = 0 spine to a cell at x = ±1, so the charges differ by exactly ±1, and the cost is J(φ) = (√5 - 2)/2 > 0. Jpow_of_abs_one · Jcost_phi_pos · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.leanTHEOREM totalCost_eq_interfaceCost · totalCost_eq_interfaceCost · carriedCost_eq_zero · carriedCost_eq_zero · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- **The total cost equals the interface cost**, because the carried bulk contributes nothing. -/ theorem totalCost_eq_interfaceCost (t : ℕ) : totalCost t = interfaceCost t := by have hsplit := Finset.sum_filter_add_sum_filter_not (E t) (fun p => polarized t p.1 ≠ polarized t p.2) (edgeCost t) have hBeq : (E t).filter (fun p => polarized t p.1 ≠ polarized t p.2) = B t := by rw [E, B, Finset.filter_filter] have hMeq : (E t).filter (fun p => ¬ (polarized t p.1 ≠ polarized t p.2)) = carried t := by rw [carried] apply Finset.filter_congr intro p _ simp rw [hBeq, hMeq] at hsplit have hzero : ∑ p ∈ carried t, edgeCost t p = 0 := by apply Finset.sum_eq_zero intro p hp exact edgeCost_carried_zero t p hp rw [hzero, add_zero] at hsplit simp only [totalCost, interfaceCost] exact hsplit.symm/-- **Carried cost is exactly zero.** The whole bulk the engine carries coarse costs no recognition. -/ theorem carriedCost_eq_zero (t : ℕ) : carriedCost t = 0 := by simp only [carriedCost] apply Finset.sum_eq_zero intro p hp exact edgeCost_carried_zero t p hpThe theorem proves that the total cost of the entire field equals the interface cost alone, with the carried cost exactly zero. totalCost_eq_interfaceCost · totalCost_eq_interfaceCost · carriedCost_eq_zero · carriedCost_eq_zero · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.leanTHEOREM totalCost_card · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- **The total recognition cost of the polarized diamond field is `(8t - 4) • J(φ)`** (`t ≥ 1`): the carried bulk is free, and the whole cost sits on the `8t - 4` interface edges. -/ theorem totalCost_card (t : ℕ) (ht : 1 ≤ t) : totalCost t = (8 * t - 4) • Cost.Jcost Constants.phi := by rw [totalCost_eq_interfaceCost, interfaceCost_card t ht]In two dimensions, the diamond lattice at time step t has total cost (8t - 4) · J(φ). totalCost_card · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.leanTHEOREM totalCost_card · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean
/-- **The total recognition cost of the polarized diamond field is `(8t - 4) • J(φ)`** (`t ≥ 1`): the carried bulk is free, and the whole cost sits on the `8t - 4` interface edges. -/ theorem totalCost_card (t : ℕ) (ht : 1 ≤ t) : totalCost t = (8 * t - 4) • Cost.Jcost Constants.phi := by rw [totalCost_eq_interfaceCost, interfaceCost_card t ht]In three dimensions, the octahedron lattice has total cost (8t² - 8t + 4) · J(φ). totalCost_card · IndisputableMonolith/Cosmology/PolarizedBirthInterfaceCost.lean