Cosmology Graded Rung Cost Polarized Unit Step

The unit-step property

The recognition cost ledger, a discrete record of what reality charges for distinguishing one state from another, assigns a price to every adjacency between two regions. The central question is how that price behaves when the regions differ. The declaration polarized_unitStep establishes that for a specific family of fields, the difference between neighboring regions is always exactly zero or exactly one step, never more. This is the forced minimal-distinction property: a unit recognition step resolves at most one phi-rung, so the integer-valued rung field k satisfies k p - k q ∈ {0, +1, -1} for every edge (p, q).

This property matters because it makes the cost law tractable. The framework proves that for any finite edge set and any integer rung field satisfying UnitStep, a carried adjacency (equal rungs) costs exactly zero, while an interface adjacency (different rungs) costs exactly J(phi), the recognition cost of the golden ratio. The total cost therefore equals the number of interface edges times J(phi). The polarized_unitStep theorem shows that the specific polarized birth field, recovered as a special case, satisfies this condition. This is a theorem in the machine-checked library of formal theorems, proved with zero sorry and only the three standard axioms.

The declaration does not claim that every possible field satisfies UnitStep, nor that the rung difference is always exactly one. It establishes the property for the polarized field and, by extension, for any field that maintains the single-rung-step invariant. The engine's T-3 refiner descends one rung at a time, so the invariant is maintained in practice. What the declaration does not do is prove that the cost law holds without the UnitStep hypothesis, or that the polarized field is the only field with this property. It also does not claim that the rung difference can never be zero; equal rungs are explicitly allowed.

/-- **The polarized birth field satisfies the forced minimal-distinction property.** Every adjacency
of the diamond either keeps the charge (carried, gap `0`) or flips it across the spine, and the
Phase-55 `level_diff` shows a flip is exactly `±1` rung. So `UnitStep (polarized t) (E t)`. -/
theorem polarized_unitStep (t : ℕ) : UnitStep (polarized t) (E t) := by
  intro p hp
  rw [E, Finset.mem_filter] at hp
  by_cases h : polarized t p.1 = polarized t p.2
  · exact Or.inl (sub_eq_zero.mpr h)
  · rcases PolarizedBirthInterfaceCost.Diamond.level_diff t p.1 p.2 hp.2 h with h1 | hm1
    · exact Or.inr (Or.inl h1)
    · exact Or.inr (Or.inr hm1)

/-- A carried (equal-rung) adjacency costs exactly zero: `J(phi^0) = J(1) = 0`. -/
theorem edgeCost_carried (k : V → ℤ) {p : V × V} (h : k p.1 = k p.2) :
    edgeCost k p = 0 := by
  have hz : k p.1 - k p.2 = 0 := sub_eq_zero.mpr h
  rw [edgeCost, hz, Jpow_zero]

/-- Under the unit-step law, an interface (different-rung) adjacency costs exactly `J(phi)`: the gap
is forced to `±1` rung, and `J(phi^(±1)) = J(phi)` by reciprocal symmetry. -/
theorem edgeCost_interface (k : V → ℤ) {E : Finset (V × V)} (hk : UnitStep k E)
    {p : V × V} (hp : p ∈ E) (hne : k p.1 ≠ k p.2) :
    edgeCost k p = Cost.Jcost Constants.phi := by
  have hd : k p.1 - k p.2 = 1 ∨ k p.1 - k p.2 = -1 := by
    rcases hk p hp with h0 | h1 | hm1
    · exact absurd (sub_eq_zero.mp h0) hne
    · exact Or.inl h1
    · exact Or.inr hm1
  rw [edgeCost]
  exact Jpow_of_abs_one hd

/-- **The total recognition cost of any unit-step rung field is `(interface edge count) * J(phi)`.**
Carried bulk is free; the whole cost sits on the forced unit-rung distinctions. -/
theorem totalCost_eq_card (k : V → ℤ) (E : Finset (V × V)) (hk : UnitStep k E) :
    totalCost k E = (E.filter (fun p => k p.1 ≠ k p.2)).card • Cost.Jcost Constants.phi := by
  rw [totalCost_eq_interfaceCost, interfaceCost_eq_card k E hk]

/-- **The polarized birth field satisfies the forced minimal-distinction property.** Every adjacency
of the diamond either keeps the charge (carried, gap `0`) or flips it across the spine, and the
Phase-55 `level_diff` shows a flip is exactly `±1` rung. So `UnitStep (polarized t) (E t)`. -/
theorem polarized_unitStep (t : ℕ) : UnitStep (polarized t) (E t) := by
  intro p hp
  rw [E, Finset.mem_filter] at hp
  by_cases h : polarized t p.1 = polarized t p.2
  · exact Or.inl (sub_eq_zero.mpr h)
  · rcases PolarizedBirthInterfaceCost.Diamond.level_diff t p.1 p.2 hp.2 h with h1 | hm1
    · exact Or.inr (Or.inl h1)
    · exact Or.inr (Or.inr hm1)

What this page does not claim

The declaration does not prove that every possible field satisfies UnitStep. The declaration does not claim that the rung difference is always exactly one; zero is explicitly allowed. The declaration does not prove that the cost law holds without the UnitStep hypothesis.

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/GradedRungCost.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)

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