Encyclopedia Foundation Foundation Dalembert Factorization Forcing Rcl Combiner Satisfies Gate

ARTICLE 3 claims 3 theorems

Foundation Dalembert Factorization Forcing Rcl Combiner Satisfies Gate

A single algebraic rule, checked by machine, pins down the exact formula that combines two recognition costs.

The combiner gate

The recognition framework keeps a discrete record of events, and it assigns a cost to recognizing any two of them together. The question is what form that combined cost can take. The answer, proved in the framework's machine-checked library of formal theorems, is a single formula: P(u,v) = 2uv + 2u + 2v. The declaration rclCombiner_satisfies_gate establishes that this formula satisfies four plain algebraic conditions, and a companion theorem shows those conditions force the formula uniquely.

The four conditions are the gate. First, the combiner is symmetric: recognizing u then v costs the same as recognizing v then u. Second, for any fixed first argument, the cost responds linearly to the second argument; this is the affine-response step that the framework's B2 closure program treats as the hard analytic core. Third, the boundary law holds: combining u with zero costs exactly 2u. Fourth, the normalization fixes the unit diagonal at P(1,1) = 6. The theorem rclCombiner_satisfies_gate checks that the displayed formula meets all four conditions.

The force runs the other way too. A separate theorem, gate_forces_rcl, proves that any combiner satisfying the same four conditions must equal the formula above. Symmetry and the boundary law already narrow the field to a bilinear family, and the normalization at (1,1) selects the coefficient 2. The result is an exact equivalence: satisfying the gate is the same as being the canonical combiner. This is a theorem in the machine-checked library, not a heuristic or a fitted choice.

In Recognition Science, this gate is the algebraic core that later steps use to force the golden ratio and the eight-tick cycle. The formula itself is the payoff: it is the unique way to combine two recognition costs that respects symmetry, linear response, the boundary law, and the unit normalization. What the theorem does not claim is that the gate explains why recognition has these properties; it only proves that if they hold, the formula follows. The analytic passage from factorization to affine response is assumed, not derived here, and the physical bridge from recognition to ordinary space remains open.

THEOREM rclCombiner_satisfies_gate · IndisputableMonolith/Foundation/DAlembert/FactorizationForcing.lean
/-- The canonical RCL polynomial satisfies the full factorization gate. -/
theorem rclCombiner_satisfies_gate :
    FactorizationAssociativityGate rclCombiner where
  symmetric := by
    intro u v
    unfold rclCombiner
    ring
  rightAffine := by
    intro u
    refine ⟨2 * u + 2, 2 * u, ?_⟩
    intro v
    unfold rclCombiner
    ring
  zeroBoundary := by
    intro u
    unfold rclCombiner
    ring
  unitDiagonal := by
    unfold rclCombiner
    norm_num
THEOREM gate_forces_rcl · IndisputableMonolith/Foundation/DAlembert/FactorizationForcing.lean
/-- Canonical normalization selects the RCL member of the bilinear family. -/
theorem gate_forces_rcl (P : ℝ → ℝ → ℝ)
    (hGate : FactorizationAssociativityGate P) :
    ∀ u v, P u v = 2 * u * v + 2 * u + 2 * v := by
  obtain ⟨c, hc⟩ := gate_forces_bilinear_family P hGate
  have hc_two : c = 2 := by
    have h11 : P 1 1 = c * 1 * 1 + 2 * 1 + 2 * 1 := by
      simpa using hc 1 1
    linarith [hGate.unitDiagonal, h11]
  intro u v
  calc
    P u v = c * u * v + 2 * u + 2 * v := hc u v
    _ = 2 * u * v + 2 * u + 2 * v := by rw [hc_two]
THEOREM gate_forces_bilinear_family · IndisputableMonolith/Foundation/DAlembert/FactorizationForcing.lean
/-- Once the affine-response step is known, symmetry and the boundary law force
    the entire bilinear family. -/
theorem gate_forces_bilinear_family (P : ℝ → ℝ → ℝ)
    (hGate : FactorizationAssociativityGate P) :
    ∃ c : ℝ, ∀ u v, P u v = c * u * v + 2 * u + 2 * v := by
  classical
  choose α β hAffine using hGate.rightAffine
  have hβ : ∀ u, β u = 2 * u := by
    intro u
    have h0 : P u 0 = α u * 0 + β u := hAffine u 0
    rw [hGate.zeroBoundary u] at h0
    linarith
  let c : ℝ := α 1 - 2
  refine ⟨c, ?_⟩
  intro u v
  have hsym1 : P u 1 = P 1 u := hGate.symmetric u 1
  have hαu : α u = c * u + 2 := by
    dsimp [c]
    have hcalc : α u * 1 + β u = α 1 * u + β 1 := by
      calc
        α u * 1 + β u = P u 1 := by symm; exact hAffine u 1
        _ = P 1 u := hGate.symmetric u 1
        _ = α 1 * u + β 1 := hAffine 1 u
    rw [hβ u, hβ 1] at hcalc
    linarith
  calc
    P u v = α u * v + β u := hAffine u v
    _ = (c * u + 2) * v + 2 * u := by rw [hαu, hβ u]
    _ = c * u * v + 2 * u + 2 * v := by ring

What this page does not claim

The theorem does not derive the affine-response step; it assumes that step as a premise. The gate does not explain why recognition has these four properties, only that they force the formula. The physical bridge from recognition to three spatial dimensions is not established by this declaration.

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

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