Encyclopedia Foundation Foundation Dalembert Inevitability Polynomial Form Forced

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Foundation Dalembert Inevitability Polynomial Form Forced

A single equation, not a choice: the d'Alembert form is the only polynomial rule that can govern a symmetric measure of deviation.

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The forced polynomial form

The d'Alembert functional equation, named for Jean le Rond d'Alembert's 1747 work on vibrating strings, is a classical object: it asks for functions F that satisfy F(xy) + F(x/y) = 2F(x)F(y). Its solutions include familiar functions like the cosine and hyperbolic cosine, and its theory was developed through the twentieth century by J. Aczél and others. The equation appears across mathematics because it encodes a deep consistency: the value at a product and the value at a quotient combine in a way that depends only on the values at the factors.

The Recognition Science framework asks a different question. Instead of assuming the equation, it asks what form a consistency rule must take if it is a polynomial. The framework's ledger, a discrete record of comparison events, assigns a cost to any positive number x, measuring how far x is from 1. The cost is symmetric, so F(x) = F(1/x), and normalized, so F(1) = 0. The framework then demands multiplicative consistency: the combined cost of a product and a quotient must be a symmetric quadratic polynomial in the individual costs. The declaration polynomial_form_forced proves that any such polynomial must have the form P(u, v) = 2u + 2v + c·u·v for some constant c. The d'Alembert form is not assumed; it is forced.

The proof proceeds by constraints. Normalization forces P(0, v) = 2v. Symmetry forces P(u, v) = P(v, u). Together with the polynomial condition, these narrow the possibilities to the bilinear family. A separate declaration, bilinear_family_forced, adds continuity and non-triviality to reach the same conclusion. The constant c remains free at this stage; choosing c = 2 is a normalization of units, not a further declaration. This is the honest boundary of what the declaration establishes.

In Recognition Science, this result closes a gap in a larger argument. The framework's axioms, including the specific form with c = 2, are not arbitrary postulates but the unique polynomial-compatible options. The framework's machine-checked library of formal declarations records this proof. What the declaration does not claim is that the constant c must be 2, that the equation applies to all functions, or that the framework's identification of the d'Alembert form with a physical law is itself proved. Those remain separate steps.

THEOREM polynomial_form_forced · IndisputableMonolith/Foundation/DAlembert/Inevitability.lean
polynomial_form_forced · IndisputableMonolith/Foundation/DAlembert/Inevitability.lean:141 
/-- For a symmetric polynomial P with P(0, v) = 2v, the only compatible form
    for a non-trivial F is P(u, v) = 2u + 2v + 2uv. -/
theorem polynomial_form_forced (P : ℝ → ℝ → ℝ)
    (hPoly : ∃ (a b c d e f : ℝ), ∀ u v, P u v = a + b*u + c*v + d*u*v + e*u^2 + f*v^2)
    (hSym : ∀ u v, P u v = P v u)  -- P is symmetric
    (hNorm0 : ∀ v, P 0 v = 2 * v)  -- From normalization
    (_hNonTriv : ∃ u₀ v₀, P u₀ v₀ ≠ 2 * u₀ + 2 * v₀)  -- Non-trivial (has uv term)
    (_hDeriv : P 0 0 = 0)  -- From F(1·1) + F(1/1) = 2F(1) = 0
    : ∃ (k : ℝ), ∀ u v, P u v = 2*u + 2*v + k*u*v := by
  obtain ⟨a, b, c, d, e, f, hP⟩ := hPoly
  -- From P(0, v) = 2v for all v:
  -- a + c*v + f*v^2 = 2*v for all v
  -- Comparing coefficients: a = 0, c = 2, f = 0
  have ha : a = 0 := by
    have := hNorm0 0
    simp only [mul_zero] at this
    have hP00 := hP 0 0
    simp at hP00
    rw [hP00] at this
    exact this
  have hc_f : c = 2 ∧ f = 0 := by
    -- From P(0, 1) = 2 and P(0, 2) = 4
    have h1 := hNorm0 1
    have h2 := hNorm0 2
    have hP01 := hP 0 1
    have hP02 := hP 0 2
    simp at hP01 hP02
    rw [hP01, ha] at h1
    rw [hP02, ha] at h2
    -- h1: c + f = 2
    -- h2: 2c + 4f = 4
    constructor <;> linarith
  have hc : c = 2 := hc_f.1
  have hf : f = 0 := hc_f.2
  -- From symmetry P(u, v) = P(v, u):
  -- Comparing P(1, 0) = P(0, 1): b + e = c + f = 2
  -- Comparing P(2, 0) = P(0, 2): 2b + 4e = 2c + 4f = 4
  -- So b = 2 and e = 0
  have hb_e : b = 2 ∧ e = 0 := by
    have h1 := hSym 1 0
    have h2 := hSym 2 0
    rw [hP 1 0, hP 0 1, ha, hc, hf] at h1
    rw [hP 2 0, hP 0 2, ha, hc, hf] at h2
    simp at h1 h2
    -- h1: b + e = 2
    -- h2: 2b + 4e = 4
    constructor <;> linarith
  have hb : b = 2 := hb_e.1
  have he : e = 0 := hb_e.2
  -- So P(u, v) = 2u + 2v + d*u*v
  use d
  intro u v
  rw [hP, ha, hb, hc, he, hf]
  ring
THEOREM bilinear_family_forced · IndisputableMonolith/Foundation/DAlembert/Inevitability.lean
bilinear_family_forced · IndisputableMonolith/Foundation/DAlembert/Inevitability.lean:275 
/-- **THEOREM: The consistency requirement forces the unique bilinear family.**

Given:
1. F : ℝ₊ → ℝ is a cost functional
2. F is symmetric: F(x) = F(1/x)
3. F is normalized: F(1) = 0
4. F has multiplicative consistency: F(xy) + F(x/y) = P(F(x), F(y)) for some **symmetric quadratic polynomial** P
5. F is non-trivial (not constant 0)

Then:
P(u, v) = 2u + 2v + c*u*v for some constant c.

This means F satisfies the generalized d'Alembert equation.
If we choose the canonical cost normalization c = 2, we recover the RCL. -/
theorem bilinear_family_forced (F : ℝ → ℝ) (P : ℝ → ℝ → ℝ)
    (hNorm : IsNormalized F)
    (hCons : HasMultiplicativeConsistency F P)
    (hPoly : ∃ (a b c d e f : ℝ), ∀ u v, P u v = a + b*u + c*v + d*u*v + e*u^2 + f*v^2)
    (hSymP : ∀ u v, P u v = P v u) -- Explicit symmetry of P
    (hNonTriv : ∃ x : ℝ, 0 < x ∧ F x ≠ 0)
    (hCont : ContinuousOn F (Set.Ioi 0)) -- Regularity: F is continuous on (0, ∞)
    : ∃ c : ℝ, (∀ u v, P u v = 2*u + 2*v + c*u*v) ∧
               (c = 2 → ∀ u v, P u v = 2*u + 2*v + 2*u*v) := by
  -- Derived reciprocity from symmetry of P
  have hSym : IsSymmetric F := F_symmetric_of_P_symmetric F P hCons hSymP
  -- Step 1: Normalization forces P(0, v) = 2v
  have hP0 : ∀ y : ℝ, 0 < y → P 0 (F y) = 2 * F y :=
    symmetry_and_normalization_constrain_P F P hSym hNorm hCons

  -- Use the polynomial form lemma
  -- We need to satisfy the hypotheses of `polynomial_form_forced`.
  -- `hNorm0`: ∀ v, P 0 v = 2 * v.
  -- We only have `P 0 (F y) = 2 F y`.
  -- However, since P is a polynomial and F is non-trivial (has range with at least 0 and some non-zero value),
  -- we can determine the coefficients.
  -- P(0, v) = a + c*v + f*v^2.
  -- P(0, 0) = a = 2*0 = 0 (from F(1)=0).
  -- P(0, F y) = c*(F y) + f*(F y)^2 = 2*(F y).
  -- This holds for y=1 (0=0) and some y where F y ≠ 0.
  -- If we only have two points, we can't uniquely determine a quadratic.
  -- But wait, `polynomial_form_forced` derived `a=0, c=2, f=0`.
  -- Let's reproduce that logic but being careful about the domain.

  obtain ⟨a, b, c, d, e, f, hP⟩ := hPoly

  -- 1. a = 0
  have ha : a = 0 := by
    have hCons1 := hCons 1 1 one_pos one_pos
    simp only [one_mul, one_div] at hCons1
    -- hCons1 : F 1 + F 1⁻¹ = P (F 1) (F 1)
    -- inv_one : 1⁻¹ = 1
    rw [inv_one, hNorm] at hCons1
    -- hCons1 : 0 + 0 = P 0 0
    simp only [add_zero] at hCons1
    -- hCons1 : 0 = P 0 0
    rw [hP 0 0] at hCons1
    simp at hCons1
    exact hCons1.symm

  -- 2. From hSymP: P(u,v) = P(v,u)
  -- a + bu + cv + duv + eu^2 + fv^2 = a + bv + cu + duv + ev^2 + fu^2
  -- (b-c)u + (c-b)v + (e-f)u^2 + (f-e)v^2 = 0
  -- This implies b=c and e=f.
  have hb_c : b = c := by
    have h1 := hSymP 1 0
    rw [hP 1 0, hP 0 1] at h1
    -- h1 : a + b*1 + c*0 + d*0 + e*1 + f*0 = a + b*0 + c*1 + d*0 + e*0 + f*1
    -- i.e., a + b + e = a + c + f
    -- Using ha: a = 0, we get b + e = c + f
    simp only [ha, mul_zero, mul_one, add_zero, zero_add] at h1
    -- We need another equation to separate b, e, c, f
    have h2 := hSymP 2 0
    rw [hP 2 0, hP 0 2] at h2
    simp only [ha, mul_zero, add_zero, zero_add] at h2
    -- h1: b + e = c + f
    -- h2: 2b + 4e = 2c + 4f
    -- From h2: b + 2e = c + 2f
    -- Subtracting h1: e = f
    -- So b = c
    linarith
  have he_f : e = f := by
    have h1 := hSymP 1 0
    have h2 := hSymP 2 0
    rw [hP 1 0, hP 0 1] at h1
    rw [hP 2 0, hP 0 2] at h2
    simp only [ha, mul_zero, mul_one, add_zero, zero_add] at h1 h2
    linarith

  -- Now P(0, v) = c*v + f*v^2 (using a=0, b=c, e=f).
  -- And P(0, F y) = 2 * F y.
  -- So c*(F y) + f*(F y)^2 = 2*(F y).
  -- (c - 2)*(F y) + f*(F y)^2 = 0.
  -- This must hold for all y > 0.
  -- Since F is non-trivial, there exists y such that F y ≠ 0.
  obtain ⟨y0, hy0_pos, hy0_ne⟩ := hNonTriv
  have hc_2 : c = 2 ∧ f = 0 := by
    -- Let k = F y0 (a nonzero value in the range).
    let k : ℝ := F y0
    have hk_ne : k ≠ 0 := by
      -- hy0_ne : F y0 ≠ 0
      simpa [k] using hy0_ne

    -- The polynomial identity on the range: (c - 2) * F(y) + f * (F(y))^2 = 0.
    have poly_identity : ∀ y : ℝ, 0 < y → (c - 2) * (F y) + f * (F y)^2 = 0 := by
      intro y hy
      have h := hP0 y hy
      rw [hP 0 (F y)] at h
      simp [ha, hb_c, he_f] at h
      linarith

    -- Use IVT to find y1 with F y1 = k/2 (since F(1)=0 and F(y0)=k).
    have hF1 : F 1 = 0 := hNorm
    have hInterval_pos : Set.uIcc 1 y0 ⊆ Set.Ioi 0 := by
      intro x hx
      rcases hx with ⟨hx_lo, _hx_hi⟩
      have hmin_pos : 0 < min 1 y0 := lt_min one_pos hy0_pos
      exact lt_of_lt_of_le hmin_pos hx_lo
    have hContInterval : ContinuousOn F (Set.uIcc 1 y0) :=
      hCont.mono hInterval_pos
    have h1_mem : 1 ∈ Set.uIcc 1 y0 := Set.left_mem_uIcc
    have hy0_mem : y0 ∈ Set.uIcc 1 y0 := Set.right_mem_uIcc
    have hk2_in_image : k / 2 ∈ F '' Set.uIcc 1 y0 := by
      have hPreconn := isPreconnected_uIcc (a := 1) (b := y0)
      by_cases hk : 0 ≤ k
      · -- monotone direction: 0 ≤ k, so k/2 ∈ Icc 0 k
        have hIVT : Set.Icc 0 k ⊆ F '' Set.uIcc 1 y0 := by
          simpa [hF1, k] using hPreconn.intermediate_value h1_mem hy0_mem hContInterval
        have hk2_between : k / 2 ∈ Set.Icc 0 k := by
          constructor <;> linarith
        exact hIVT hk2_between
      · -- reverse direction: k < 0, so k/2 ∈ Icc k 0
        have hk_lt : k < 0 := lt_of_not_ge hk
        have hIVT : Set.Icc k 0 ⊆ F '' Set.uIcc 1 y0 := by
          simpa [hF1, k] using hPreconn.intermediate_value hy0_mem h1_mem hContInterval
        have hk2_between : k / 2 ∈ Set.Icc k 0 := by
          constructor <;> linarith
        exact hIVT hk2_between
    obtain ⟨y1, hy1_mem, hFy1⟩ := hk2_in_image
    have hy1_pos : 0 < y1 := hInterval_pos hy1_mem

    -- Evaluate the polynomial identity at y0 and y1, then solve for c and f.
    have h_y0 : (c - 2) * k + f * k^2 = 0 := by
      have h := poly_identity y0 hy0_pos
      simpa [k] using h
    have h_y1 : (c - 2) * (k / 2) + f * (k / 2)^2 = 0 := by
      have h := poly_identity y1 hy1_pos
      -- rewrite F y1 = k/2
      simpa [hFy1, k] using h

    -- Multiply the y1 equation by 4 to align it with the y0 equation.
    have h_y1_4 : 2 * (c - 2) * k + f * k^2 = 0 := by
      have h' := congrArg (fun z => 4 * z) h_y1
      -- simplify 4*(...) and 4*0
      ring_nf at h'
      -- `ring_nf` chooses its own normal form; bridge to our preferred one.
      have hrew : c * k * 2 - k * 4 + k ^ 2 * f = 2 * (c - 2) * k + f * k ^ 2 := by ring
      -- h' : c*k*2 - k*4 + k^2*f = 0
      calc
        2 * (c - 2) * k + f * k ^ 2
            = c * k * 2 - k * 4 + k ^ 2 * f := by simpa [hrew] using (Eq.symm hrew)
        _ = 0 := h'

    -- Subtract to get (c - 2) * k = 0, hence c = 2 (since k ≠ 0).
    have hk_mul : (c - 2) * k = 0 := by
      linarith [h_y0, h_y1_4]
    have hc : c = 2 := by
      rcases mul_eq_zero.mp hk_mul with hc0 | hk0
      · linarith
      · exact False.elim (hk_ne hk0)

    -- Plug back to get f = 0.
    have hf : f = 0 := by
      have hk2_ne : k^2 ≠ 0 := pow_ne_zero 2 hk_ne
      have hfk2 : f * k^2 = 0 := by
        -- from h_y0 with c=2
        simpa [hc] using h_y0
      rcases mul_eq_zero.mp hfk2 with hf0 | hk20
      · exact hf0
      · exact False.elim (hk2_ne hk20)

    exact ⟨hc, hf⟩

  have hc : c = 2 := hc_2.1
  have hf : f = 0 := hc_2.2
  have hb : b = 2 := by rw [hb_c, hc]
  have he : e = 0 := by rw [he_f, hf]

  -- So P(u, v) = 2u + 2v + d*u*v.
  use d
  constructor
  · intro u v
    rw [hP, ha, hb, hc, he, hf]
    ring
  · intro hd u v
    rw [hP, ha, hb, hc, he, hf, hd]
    ring
MODEL axiom_bundle_necessary · IndisputableMonolith/Foundation/DAlembert/Inevitability.lean
axiom_bundle_necessary · IndisputableMonolith/Foundation/DAlembert/Inevitability.lean:474 
/-- **COROLLARY: The Recognition Science axiom bundle (A1, A2, A3) is transcendentally necessary.**

- A1 (Normalization): F(1) = 0
  → Definitional for "cost of deviation from unity"

- A2 (RCL): F(xy) + F(x/y) = 2F(x)F(y) + 2F(x) + 2F(y)
  → PROVED: The unique polynomial form for multiplicative consistency (up to scale)

- A3 (Calibration): F''(1) = 1
  → Sets the natural scale (removes family degeneracy)

Therefore: The entire axiom bundle is not arbitrary but forced by the structure of comparison. -/
theorem axiom_bundle_necessary :
    -- A1: Normalization is definitional
    (∀ F : ℝ → ℝ, (∀ x : ℝ, 0 < x → F x = Cost.Jcost x) → F 1 = 0) ∧
    -- A2: RCL is the unique polynomial form (proven above)
    (∀ F P, IsNormalized F → HasMultiplicativeConsistency F P →
      (∃ a b c d e f, ∀ u v, P u v = a + b*u + c*v + d*u*v + e*u^2 + f*v^2) →
      (∀ u v, P u v = P v u) → -- Symmetry requirement
      (∃ x, 0 < x ∧ F x ≠ 0) → -- Non-triviality
      ContinuousOn F (Set.Ioi 0) → -- Regularity
      ∃ c, ∀ u v, P u v = 2*u + 2*v + c*u*v) ∧
    -- A3: Calibration pins down the scale (J''(1) = 1)
    (deriv (deriv (fun x => Cost.Jcost x)) 1 = 1) := by
  constructor
  · intro F hF
    have h := hF 1 one_pos
    simp only [Cost.Jcost, inv_one] at h
    linarith
  constructor
  · intro F P hNorm hCons hPoly hSymP hNonTriv hCont
    -- Use bilinear_family_forced and extract the first conjunct
    obtain ⟨c, hc, _⟩ := bilinear_family_forced F P hNorm hCons hPoly hSymP hNonTriv hCont
    exact ⟨c, hc⟩
  · -- Prove J''(1) = 1 (calibration)
    -- J(x) = x/2 + 1/(2x) - 1, so J''(x) = x⁻³, thus J''(1) = 1.
    exact Cost.deriv2_Jcost_one

What this page does not claim

The constant c is not forced to be 2 by this declaration; that requires a separate normalization choice. The declaration does not claim the d'Alembert equation holds for all functions, only for those satisfying the stated symmetry, normalization, polynomial consistency, and regularity conditions. The framework's identification of this mathematical form with a physical law is not a proved declaration.

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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/Inevitability.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.

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