Encyclopedia Foundation Foundation Cost Axioms Composition Normalization Implies Symmetry
Foundation Cost Axioms Composition Normalization Implies Symmetry
Two simple assumptions about a cost function force a hidden symmetry: the cost of a ratio equals the cost of its reciprocal.
The symmetry of cost
In mathematics, a function's symmetry often hides a deeper structure. The declaration Composition_Normalization_implies_symmetry in the Recognition Science framework's machine-checked library of formal theorems proves exactly such a hidden symmetry. It shows that any cost function satisfying two plain conditions must treat a ratio and its reciprocal as equally costly. Formally, for any positive real number x, the cost at x equals the cost at 1/x.
The first condition is normalization: the cost of unity, the ratio 1, is zero. Perfect balance costs nothing. The second is a composition law, a d'Alembert functional equation: F(xy) + F(x/y) = 2F(x)F(y) + 2F(x) + 2F(y) for all positive x and y. This equation forces multiplicative consistency, meaning the cost of a product and a quotient must relate to the costs of the parts in a fixed way. The theorem states that these two assumptions alone, without any continuity or smoothness requirement, imply the reciprocal symmetry F(x) = F(1/x).
This result is a stepping stone, not the destination. The full Recognition Science framework derives a unique cost function, J(x) = (x + 1/x)/2 - 1, from three axioms: normalization, the composition law, and a calibration condition on the second derivative. The symmetry theorem is a consequence of the first two axioms alone. It does not require the third, and it does not by itself identify the cost function. The uniqueness of J requires additional regularity hypotheses, such as continuity and convexity, which are not part of this declaration.
The practical meaning is that the framework's cost function, whatever its exact form, cannot favor one direction over its inverse. A ratio of 2 and a ratio of 1/2 carry the same cost. This symmetry is a structural constraint that any viable cost function must obey, and it emerges from the most basic assumptions about how recognition works.
THEOREM Composition_Normalization_implies_symmetry · IndisputableMonolith/Foundation/CostAxioms.lean
/-- Composition + Normalization implies symmetry: F(x) = F(1/x).
Proof: Apply Composition with x = 1:
F(1 * y) + F(1 / y) = 2F(1)F(y) + 2F(1) + 2F(y)
F(y) + F(1/y) = 2 * 0 * F(y) + 2 * 0 + 2F(y) (by Normalization: F(1) = 0)
F(y) + F(1/y) = 2F(y)
F(1/y) = F(y)
Therefore F(y) = F(1/y) for all y > 0, which is symmetry. -/
theorem Composition_Normalization_implies_symmetry (F : ℝ → ℝ) [Composition F] [Normalization F] :
∀ {x : ℝ}, 0 < x → F x = F x⁻¹ := by
intro x hx
-- Apply Composition with x = 1, y = x
have h := Composition.dAlembert (F := F) 1 x one_pos hx
-- F(1 * x) + F(1 / x) = 2F(1)F(x) + 2F(1) + 2F(x)
-- Simplify: F(1) = 0, 1 * x = x, 1 / x = x⁻¹
simp only [one_mul, one_div, Normalization.unit_zero, zero_mul, add_zero, mul_zero] at h
-- h is now: F(x) + F(x⁻¹) = 2F(x)
-- Subtracting F(x) from both sides: F(x⁻¹) = F(x)
have h_symm : F x⁻¹ = F x := by
have h_sub : F x⁻¹ = (F x + F x⁻¹) - F x := by ring
rw [h_sub, h]
ring
exact h_symm.symm
What this page does not claim
The declaration does not identify the cost function J(x) = (x + 1/x)/2 - 1. The declaration does not require or use continuity, convexity, or any smoothness hypothesis. The declaration does not establish the calibration condition or any other third axiom.
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/CostAxioms.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)
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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 does the full uniqueness theorem require beyond normalization and composition?
- How does the calibration condition on the second derivative select the unique cost function?
- What are the classical d'Alembert functional equation solutions without regularity assumptions?
MACHINE LAYER · GROUNDED CLAIM TABLE · CLICK TO EXPAND
THEOREM Composition_Normalization_implies_symmetry · IndisputableMonolith/Foundation/CostAxioms.lean
/-- Composition + Normalization implies symmetry: F(x) = F(1/x). Proof: Apply Composition with x = 1: F(1 * y) + F(1 / y) = 2F(1)F(y) + 2F(1) + 2F(y) F(y) + F(1/y) = 2 * 0 * F(y) + 2 * 0 + 2F(y) (by Normalization: F(1) = 0) F(y) + F(1/y) = 2F(y) F(1/y) = F(y) Therefore F(y) = F(1/y) for all y > 0, which is symmetry. -/ theorem Composition_Normalization_implies_symmetry (F : ℝ → ℝ) [Composition F] [Normalization F] : ∀ {x : ℝ}, 0 < x → F x = F x⁻¹ := by intro x hx -- Apply Composition with x = 1, y = x have h := Composition.dAlembert (F := F) 1 x one_pos hx -- F(1 * x) + F(1 / x) = 2F(1)F(x) + 2F(1) + 2F(x) -- Simplify: F(1) = 0, 1 * x = x, 1 / x = x⁻¹ simp only [one_mul, one_div, Normalization.unit_zero, zero_mul, add_zero, mul_zero] at h -- h is now: F(x) + F(x⁻¹) = 2F(x) -- Subtracting F(x) from both sides: F(x⁻¹) = F(x) have h_symm : F x⁻¹ = F x := by have h_sub : F x⁻¹ = (F x + F x⁻¹) - F x := by ring rw [h_sub, h] ring exact h_symm.symmThe declaration proves that any cost function satisfying normalization and the composition law must treat a ratio and its reciprocal as equally costly. Composition_Normalization_implies_symmetry · IndisputableMonolith/Foundation/CostAxioms.lean