Encyclopedia Cost Cost T5 Cost Uniqueness On Pos
ARTICLE 5 claims 4 theorems 1 model
Cost T5 Cost Uniqueness On Pos
A single formula for the cost of recognition is forced by five plain conditions, and a machine-checked theorem proves no other positive formula can work.
The uniqueness theorem
The cost function J(x) = (x + 1/x)/2 - 1 measures the price of a ratio x between two quantities. For x close to 1, meaning the quantities nearly match, the cost is tiny and grows quadratically: J(1 + ε) is about ε²/2. For x far from 1, the cost grows roughly linearly, so doubling a large ratio doubles the cost. The function is symmetric under inversion, J(x) = J(1/x), so swapping the two quantities leaves the cost unchanged. It vanishes only at x = 1, where the quantities are identical.
The theorem T5_cost_uniqueness_on_pos, proved in the machine-checked library of formal theorems, states that any function F on positive real numbers satisfying the framework's five plain conditions must equal J everywhere. The conditions are reciprocal symmetry, zero cost at unity, a forced composition law, calibration, and continuity. The proof works by showing F agrees with J on all exponentials, then extending that agreement to every positive number. The name "on pos" refers to the domain: all positive real numbers, excluding zero and negatives.
In Recognition Science, recognition (a discrete record of events) incurs a cost (a forced price, not a chosen one). The theorem says that any candidate cost function meeting the five conditions is identical to J. This is a uniqueness result, not an existence result: the framework's other theorems establish that J itself satisfies the conditions. The uniqueness is what makes J special: no alternative formula can pass the same tests.
What the theorem does not claim is as important as what it proves. It does not say that the five conditions are the only possible axioms for a cost function; other axiom sets could yield other functions. It does not claim that J is the unique function satisfying any subset of the conditions; dropping the composition law, for instance, would allow many functions. The theorem also does not assert that J is the actual cost used by any physical process; that is a modeling claim, not a theorem. The uniqueness is conditional on the five stated conditions, and the proof is airtight only within those conditions.
The practical consequence is that the framework's cost function is not an arbitrary choice. Any theory of recognition that accepts the five plain conditions is forced to use J, no matter how it is derived. This makes J a structural constant of the framework, like a fixed exchange rate between ratios and prices. The theorem also provides a check: if a proposed cost function differs from J anywhere on the positive reals, it must violate at least one of the five conditions.
THEOREM T5_cost_uniqueness_on_pos · IndisputableMonolith/Cost.lean
theorem T5_cost_uniqueness_on_pos {F : ℝ → ℝ} [JensenSketch F] :
∀ {x : ℝ}, 0 < x → F x = Jcost x :=
by
intro x hx
have hAgree : AgreesOnExp F := by
intro t
exact le_antisymm (JensenSketch.axis_upper (F:=F) t) (JensenSketch.axis_lower (F:=F) t)
exact (agree_on_exp_extends (F:=F) hAgree) hx
MODEL Jcost · IndisputableMonolith/Cost.lean
noncomputable def Jcost (x : ℝ) : ℝ := (x + x⁻¹) / 2 - 1
THEOREM Jcost_symm · IndisputableMonolith/Cost.lean
lemma Jcost_symm {x : ℝ} (hx : 0 < x) : Jcost x = Jcost x⁻¹ := by
have hx0 : x ≠ 0 := ne_of_gt hx
rw [Jcost_eq_sq hx0, Jcost_eq_sq (inv_ne_zero hx0)]
field_simp [hx0]
ring
THEOREM Jcost_eq_zero_iff · IndisputableMonolith/Cost.lean
/-- J(x) = 0 iff x = 1, for positive x. -/
lemma Jcost_eq_zero_iff (x : ℝ) (hx : 0 < x) : Jcost x = 0 ↔ x = 1 := by
constructor
· intro h
by_contra h1
exact absurd h (ne_of_gt (Jcost_pos_of_ne_one x hx h1))
· intro h
rw [h]
exact Jcost_unit0
THEOREM Jcost_one_plus_eps_quadratic · IndisputableMonolith/Cost.lean
lemma Jcost_one_plus_eps_quadratic (ε : ℝ) (hε : |ε| ≤ (1 : ℝ) / 2) :
∃ (c : ℝ), Jcost (1 + ε) = ε ^ 2 / 2 + c * ε ^ 3 ∧ |c| ≤ 2 := by
classical
have hbounds := abs_le.mp hε
have hpos : 0 < 1 + ε := by
have : -(1 : ℝ) / 2 ≤ ε := by simpa [neg_div] using hbounds.1
linarith
have hne : 1 + ε ≠ 0 := ne_of_gt hpos
have hcalc : Jcost (1 + ε) = ε ^ 2 / (2 * (1 + ε)) := by
simpa [pow_two, add_comm, add_left_comm, add_assoc, sub_eq_add_neg]
using (Jcost_eq_sq hne)
let c : ℝ := -1 / (2 * (1 + ε))
have h_eq :
Jcost (1 + ε) = ε ^ 2 / 2 + c * ε ^ 3 := by
have : ε ^ 2 / (2 * (1 + ε)) = ε ^ 2 / 2 + (-1 / (2 * (1 + ε))) * ε ^ 3 := by
field_simp [hne]
ring
simpa [hcalc, c] using this
have hden_pos : 0 < 2 * (1 + ε) := by nlinarith [hpos]
have habs : |c| = 1 / (2 * (1 + ε)) := by
simp [c, div_eq_mul_inv, abs_mul, abs_inv, abs_of_pos hpos]
-- Use 1/(2(1+ε)) ≤ 1 from (1+ε) ≥ 1/2
have hone_le : (1 : ℝ) ≤ 2 * (1 + ε) := by
have : (1 / 2 : ℝ) ≤ 1 + ε := by linarith
simpa [two_mul] using mul_le_mul_of_nonneg_left this (by norm_num : (0 : ℝ) ≤ 2)
have hdiv_le_one : 1 / (2 * (1 + ε)) ≤ 1 := by
have hpos1 : 0 < (1 : ℝ) := by norm_num
simpa [one_div] using one_div_le_one_div_of_le hpos1 hone_le
have hbound : |c| ≤ 2 := by
have h1 : |c| ≤ 1 := by simpa [habs] using hdiv_le_one
have h12 : (1 : ℝ) ≤ 2 := by norm_num
exact le_trans h1 h12
exact ⟨c, h_eq, hbound⟩
What this page does not claim
The five conditions are the only possible axioms for a cost function. J is the unique function satisfying any subset of the five conditions. J is the actual cost used by any physical process. The theorem proves existence of a cost function satisfying the conditions.
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/Cost.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 are the five plain conditions stated in full detail?
- How does the proof extend agreement from exponentials to all positive reals?
- What would a counterexample to one of the five conditions look like?
- How does this uniqueness theorem connect to the golden ratio and the forcing chain?
MACHINE LAYER · GROUNDED CLAIM TABLE · CLICK TO EXPAND
THEOREM T5_cost_uniqueness_on_pos · IndisputableMonolith/Cost.lean
theorem T5_cost_uniqueness_on_pos {F : ℝ → ℝ} [JensenSketch F] : ∀ {x : ℝ}, 0 < x → F x = Jcost x := by intro x hx have hAgree : AgreesOnExp F := by intro t exact le_antisymm (JensenSketch.axis_upper (F:=F) t) (JensenSketch.axis_lower (F:=F) t) exact (agree_on_exp_extends (F:=F) hAgree) hxThe theorem T5_cost_uniqueness_on_pos, proved in the machine-checked library of formal theorems, states that any function F on positive real numbers satisfying the framework's five plain conditions must equal J everywhere. T5_cost_uniqueness_on_pos · IndisputableMonolith/Cost.leanMODEL Jcost · IndisputableMonolith/Cost.lean
noncomputable def Jcost (x : ℝ) : ℝ := (x + x⁻¹) / 2 - 1The cost function J(x) = (x + 1/x)/2 - 1 measures the price of a ratio x between two quantities. Jcost · IndisputableMonolith/Cost.leanTHEOREM Jcost_symm · IndisputableMonolith/Cost.lean
lemma Jcost_symm {x : ℝ} (hx : 0 < x) : Jcost x = Jcost x⁻¹ := by have hx0 : x ≠ 0 := ne_of_gt hx rw [Jcost_eq_sq hx0, Jcost_eq_sq (inv_ne_zero hx0)] field_simp [hx0] ringThe function is symmetric under inversion, J(x) = J(1/x), so swapping the two quantities leaves the cost unchanged. Jcost_symm · IndisputableMonolith/Cost.leanTHEOREM Jcost_eq_zero_iff · IndisputableMonolith/Cost.lean
/-- J(x) = 0 iff x = 1, for positive x. -/ lemma Jcost_eq_zero_iff (x : ℝ) (hx : 0 < x) : Jcost x = 0 ↔ x = 1 := by constructor · intro h by_contra h1 exact absurd h (ne_of_gt (Jcost_pos_of_ne_one x hx h1)) · intro h rw [h] exact Jcost_unit0It vanishes only at x = 1, where the quantities are identical. Jcost_eq_zero_iff · IndisputableMonolith/Cost.leanTHEOREM Jcost_one_plus_eps_quadratic · IndisputableMonolith/Cost.lean
lemma Jcost_one_plus_eps_quadratic (ε : ℝ) (hε : |ε| ≤ (1 : ℝ) / 2) : ∃ (c : ℝ), Jcost (1 + ε) = ε ^ 2 / 2 + c * ε ^ 3 ∧ |c| ≤ 2 := by classical have hbounds := abs_le.mp hε have hpos : 0 < 1 + ε := by have : -(1 : ℝ) / 2 ≤ ε := by simpa [neg_div] using hbounds.1 linarith have hne : 1 + ε ≠ 0 := ne_of_gt hpos have hcalc : Jcost (1 + ε) = ε ^ 2 / (2 * (1 + ε)) := by simpa [pow_two, add_comm, add_left_comm, add_assoc, sub_eq_add_neg] using (Jcost_eq_sq hne) let c : ℝ := -1 / (2 * (1 + ε)) have h_eq : Jcost (1 + ε) = ε ^ 2 / 2 + c * ε ^ 3 := by have : ε ^ 2 / (2 * (1 + ε)) = ε ^ 2 / 2 + (-1 / (2 * (1 + ε))) * ε ^ 3 := by field_simp [hne] ring simpa [hcalc, c] using this have hden_pos : 0 < 2 * (1 + ε) := by nlinarith [hpos] have habs : |c| = 1 / (2 * (1 + ε)) := by simp [c, div_eq_mul_inv, abs_mul, abs_inv, abs_of_pos hpos] -- Use 1/(2(1+ε)) ≤ 1 from (1+ε) ≥ 1/2 have hone_le : (1 : ℝ) ≤ 2 * (1 + ε) := by have : (1 / 2 : ℝ) ≤ 1 + ε := by linarith simpa [two_mul] using mul_le_mul_of_nonneg_left this (by norm_num : (0 : ℝ) ≤ 2) have hdiv_le_one : 1 / (2 * (1 + ε)) ≤ 1 := by have hpos1 : 0 < (1 : ℝ) := by norm_num simpa [one_div] using one_div_le_one_div_of_le hpos1 hone_le have hbound : |c| ≤ 2 := by have h1 : |c| ≤ 1 := by simpa [habs] using hdiv_le_one have h12 : (1 : ℝ) ≤ 2 := by norm_num exact le_trans h1 h12 exact ⟨c, h_eq, hbound⟩For x close to 1, meaning the quantities nearly match, the cost is tiny and grows quadratically: J(1 + ε) is about ε²/2. Jcost_one_plus_eps_quadratic · IndisputableMonolith/Cost.lean