Encyclopedia Foundation Foundation Dalembert Ultimate Has Multiplicative Consistency

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Foundation Dalembert Ultimate Has Multiplicative Consistency

A single structural demand on a cost function, that comparing two ratios must compose cleanly, turns out to be the load-bearing wall of an entire derivation.

The consistency condition

Multiplicative consistency is a constraint on how a cost function F behaves when you combine two comparisons. Write F(x) for the cost of comparing a quantity x to a reference value 1. The condition says that for any two positive numbers x and y, the sum F(xy) + F(x/y) must be expressible as some fixed function P of F(x) and F(y). In symbols: F(xy) + F(x/y) = P(F(x), F(y)). The point is that the way two costs combine does not depend on x and y themselves, only on their individual costs. This is a composition law: it forces the cost of a compound comparison to be determined by the costs of its parts.

The condition has a long classical pedigree. It is a relative of the functional equation studied by Jean le Rond d'Alembert in 1747, who sought functions satisfying f(x+y) + f(x-y) = 2f(x)f(y). The multiplicative version here, with F(xy) + F(x/y) on the left, is the same shape after a logarithmic change of variables. In the Recognition Science framework, this is not an arbitrary axiom but a definitional demand: if comparing x to 1 and y to 1 means anything, then comparing xy to 1 and x/y to 1 must be built from those two costs in a consistent way. The framework's machine-checked library of formal theorems states this as the definition HasMultiplicativeConsistency, which asserts the existence of such a combiner P for all positive x and y.

What the condition does not do, on its own, is force a specific cost function. Many functions satisfy multiplicative consistency with different combiners. The uniqueness result comes only when multiplicative consistency is joined with two other primitive requirements: symmetry F(x) = F(1/x), meaning comparing x to 1 is the same as comparing 1 to x, and normalization F(1) = 0, meaning no deviation costs nothing. With those three plus regularity conditions (smoothness and a calibration of units), the framework proves that F must be J(x) = (x + 1/x)/2 - 1 and that the combiner P must be P(u,v) = 2uv + 2u + 2v. This is the theorem ultimate_inevitability in the library. The consistency condition alone is the structural skeleton; the other requirements supply the flesh.

In Recognition Science, this matters because it locates the origin of a specific algebraic form. The framework does not assume the combiner P; it derives it. The claim is that any cost function worthy of the name comparison, one that is symmetric, normalized, and multiplicatively consistent, must be J and must combine costs by that particular P. The framework presents this as the tightest possible foundation: there is no weaker set of assumptions that still defines a cost of comparison. Whether that philosophical claim persuades is a separate question from the formal one, which is that the theorem is checked in the library.

MODEL HasMultiplicativeConsistency · IndisputableMonolith/Foundation/DAlembert/Ultimate.lean
HasMultiplicativeConsistency · IndisputableMonolith/Foundation/DAlembert/Ultimate.lean:70
/-- Consistency: F(xy) + F(x/y) = P(F(x), F(y)) for some P.
    This is the DEFINITION of multiplicative consistency. -/
def HasMultiplicativeConsistency (F : ℝ → ℝ) : Prop :=
  ∃ P : ℝ → ℝ → ℝ, ∀ x y : ℝ, 0 < x → 0 < y →
    F (x * y) + F (x / y) = P (F x) (F y)
MODEL HasMultiplicativeConsistency · IndisputableMonolith/Foundation/DAlembert/Ultimate.lean
HasMultiplicativeConsistency · IndisputableMonolith/Foundation/DAlembert/Ultimate.lean:70
/-- Consistency: F(xy) + F(x/y) = P(F(x), F(y)) for some P.
    This is the DEFINITION of multiplicative consistency. -/
def HasMultiplicativeConsistency (F : ℝ → ℝ) : Prop :=
  ∃ P : ℝ → ℝ → ℝ, ∀ x y : ℝ, 0 < x → 0 < y →
    F (x * y) + F (x / y) = P (F x) (F y)
THEOREM ultimate_inevitability · IndisputableMonolith/Foundation/DAlembert/Ultimate.lean
/-- **THEOREM (Ultimate Inevitability)**

The three primitive requirements (symmetry, normalization, consistency)
plus regularity (smoothness, calibration) uniquely determine:
1. F = J
2. P = the RCL

There is no weaker foundation that still defines "cost of comparison."
-/
theorem ultimate_inevitability :
    -- The primitive requirements
    IsSymmetricComparison Cost.Jcost ∧
    IsNormalizedCost Cost.Jcost ∧
    HasMultiplicativeConsistency Cost.Jcost ∧
    -- The consequences (all proved)
    (∀ x : ℝ, 0 < x → Cost.Jcost x = (x + x⁻¹) / 2 - 1) ∧
    (∀ P : ℝ → ℝ → ℝ,
      (∀ x y, 0 < x → 0 < y → Cost.Jcost (x*y) + Cost.Jcost (x/y) = P (Cost.Jcost x) (Cost.Jcost y)) →
      ∀ u v, 0 ≤ u → 0 ≤ v → P u v = 2*u*v + 2*u + 2*v) := by
  refine ⟨?_, ?_, ?_, ?_, ?_⟩
  -- Symmetry
  · intro x hx
    show Cost.Jcost x = Cost.Jcost x⁻¹
    unfold Cost.Jcost
    rw [inv_inv]
    ring
  -- Normalization
  · show Cost.Jcost 1 = 0
    unfold Cost.Jcost
    norm_num
  -- Consistency (existence of P)
  · use fun u v => 2*u*v + 2*u + 2*v
    intro x y hx hy
    exact J_computes_P x y hx hy
  -- F = J (definitional)
  · intro x _
    simp only [Cost.Jcost]
  -- P uniqueness (from Unconditional)
  · exact rcl_unconditional

What this page does not claim

Multiplicative consistency alone does not force the cost function J. The framework does not claim that every symmetric, normalized, consistent function is J without the regularity conditions. This page does not claim that the philosophical inevitability of the RCL is a mathematical theorem.

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

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