Cosmology Graded Rung Cost Interface Cost Eq Card

The cost of a distinction

In the Recognition Science framework, the universe keeps a discrete record of events, a ledger, where each event is a recognition that something has changed. The framework models this change as a field of integer-valued rungs, like floors in a building, assigned to regions of space. The cost of a recognition event is a fixed, forced quantity, not a free parameter. The theorem interfaceCost_eq_card establishes exactly what that cost is for any such field.

The theorem states a simple accounting rule. Consider any finite set of adjacencies, or edges, between regions. The rule applies when the rung values across every edge differ by at most one step, a condition the framework calls UnitStep. Under that condition, the total cost of the entire field equals the number of edges where the rung actually changes, multiplied by a single constant, J(phi), where phi is the golden ratio. Edges where the rung is the same cost nothing. The proof shows this for any rung profile, not just a simple binary one, and it is verified in a machine-checked library of formal theorems with no unproved assumptions.

The constant J(phi) is not arbitrary. It comes from a proved functional equation for the cost function J, which forces its value at the golden ratio to be positive and fixed. The theorem therefore says that each forced unit-rung distinction carries the same price tag, and the entire same-rung bulk is carried for free. This is a structural result about the ledger itself, independent of any particular physical configuration.

What the theorem does not claim is that any particular physical system must obey this rule. It proves a conditional statement: if a field satisfies the UnitStep condition, then the cost formula holds. Whether a real physical field actually satisfies that condition is a separate question, one the framework addresses elsewhere. The theorem also does not say what the numerical value of J(phi) is, only that it is positive and the same for every distinction.

/-- **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]

/-- **Carried cost is exactly zero.** The whole same-rung bulk the engine carries coarse is free. -/
theorem carriedCost_eq_zero (k : V → ℤ) (E : Finset (V × V)) :
    carriedCost k E = 0 := by
  simp only [carriedCost]
  apply Finset.sum_eq_zero
  intro p hp
  rw [Finset.mem_filter] at hp
  exact edgeCost_carried k hp.2

/-- **Graded-rung cost ledger headline (Phase 56).** For any finite ordered edge set `E` and any
integer rung field `k` with the forced minimal-distinction property `UnitStep k E`: the carried
same-rung bulk costs exactly zero, the total cost equals the interface cost, the total cost is the
interface edge count times the one-rung cost `J(phi)`, and `J(phi) > 0` is a genuine positive cost.
The cost of carrying any rung profile is `J(phi)` per forced unit-rung distinction, bulk free. -/
theorem t56_graded_cost_ledger (k : V → ℤ) (E : Finset (V × V)) (hk : UnitStep k E) :
    carriedCost k E = 0
    ∧ totalCost k E = interfaceCost k E
    ∧ totalCost k E = (E.filter (fun p => k p.1 ≠ k p.2)).card • Cost.Jcost Constants.phi
    ∧ 0 < Cost.Jcost Constants.phi :=
  ⟨carriedCost_eq_zero k E, totalCost_eq_interfaceCost k E,
   totalCost_eq_card k E hk, Jcost_phi_pos⟩

What this page does not claim

The theorem does not prove that any particular physical system obeys the UnitStep condition. The theorem does not derive the numerical value of J(phi), only that it is positive and constant across distinctions. The theorem does not apply to fields where rung values can jump by more than one step between adjacent regions.

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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