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theorem proved term proof high

C_competing_violates_budget

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Physicists deriving the ILG spatial kernel amplitude cite this result to show that the competing value φ^{-3/2} together with the J-penalty exceeds the half-rung budget. The declaration rules out C = φ^{-3/2} in favor of the forced value φ^{-2}. Its proof combines the half-rung equality with a strict inequality between the two amplitudes and applies linear arithmetic.

claim$J(φ) + φ^{-3/2} > 1/2$, where $J(φ) = φ - 3/2$ is the first-rung J-cost penalty.

background

The module derives the ILG spatial-kernel amplitude C = φ^{-2} from the half-rung budget identity J(φ) + C = 1/2, which follows from φ² = φ + 1 alone and identifies C with 2 - φ. J(φ) = φ - 3/2 is the first-rung cost penalty, while the competing amplitude is defined as C' = φ^{-3/2}. The local setting contrasts the structurally forced C with the alternative value drawn from an earlier three-channel argument.

proof idea

The proof recalls the equality Jphi_penalty + C_kernel = 1/2 from the half-rung budget lemma, invokes the strict inequality C_kernel < C_kernel_competing, and concludes via linarith.

why it matters in Recognition Science

The theorem eliminates the competing amplitude φ^{-3/2} by budget violation and feeds the master certificate ilgSpatialKernelCert together with the consolidated one-statement theorem ilg_spatial_kernel_one_statement. It enforces the half-rung identity that selects C = φ^{-2} over φ^{-3/2}, consistent with the three-channel factorization C = φ^{-1} · φ^{-1} in D = 3 and the Recognition Science forcing chain.

scope and limits

formal statement (Lean)

 213theorem C_competing_violates_budget :
 214    Jphi_penalty + C_kernel_competing > 1 / 2 := by

proof body

Term-mode proof.

 215  have h1 : Jphi_penalty + C_kernel = 1 / 2 := half_rung_budget
 216  have h2 : C_kernel < C_kernel_competing := C_competing_gt_C_kernel
 217  linarith
 218
 219/-! ## §6. Structural three-channel factorization sketch
 220
 221The amplitude `C = φ⁻²` admits the structural decomposition
 222  `C = (longitudinal weight) × (transverse-collective weight) = φ⁻¹ · φ⁻¹`
 223in `D = 3` spatial dimensions. We record the per-channel weight and
 224the product as named definitions for transparency.
 225-/
 226
 227/-- The per-channel ladder weight: `φ⁻¹`. One step on the φ-ladder. -/

used by (2)

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depends on (11)

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