ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

One-dimensional finite Gaussian matching and shell composition

Physical scope

A one-coordinate finite Gaussian regulator example and a separate normalized shell calculation; no empirical claim, continuum limit, or generic matching theorem.

Example

Assume AC. Use one positive mass unit $m_{0}$ and a one-site finite regulator with $M = 2 m_{0}$ , $A = (m_{0}^{2})$ , $E = m_{0}$ , $μ_{W} = \frac{3}{2} m_{0}$ , $Λ = 3 m_{0}$ , source $J = (m_{0})$ , and truncation order $N = 0$ . Then $K = M^{2} I + A = (5 m_{0}^{2})$ . Let $γ_{K^{- 1}} = N_{1} (0, K^{- 1})$ be the centered normalized Gaussian with covariance $K^{- 1}$ . For $Z_{H} (J) = \int e^{J^{T} χ} d γ_{K^{- 1}} (χ)$ and the action increment $Δ S_{W} = - \log Z_{H} (J)$ , the exact increment is $- 1 / 10$ , the order-zero approximation is $- 1 / 8$ , the vector remainder is $1 / (20 m_{0})$ , and the action remainder is $1 / 40 \leq 1 / 32$ .

For a separate dimensionless shell coordinate, let $γ_{C} = N_{1} (0, C)$ be the centered normalized Gaussian of variance $C$ , define $T_{C} f (x) = \int f (x + z) d γ_{C} (z)$ , and take $f (x) = x^{2}$ , $C_{1} = 1$ , and $C_{2} = 3$ . The successive normalized shell value at zero and the combined-shell value are both $4$ .

This is a model-specific finite calculation. It reports no observation and asserts no continuum limit or general interacting matching rule.

Example references

Finite-regulator EFT model and scale data

Finite-regulator Wilsonian action

Finite-regulator derivative expansion and truncation

Finite heavy-scalar Gaussian matching

Successive finite Gaussian shells compose

Multivariate normal law, including singular covariance

Normalized finite Gaussian shell transform

Facts & Assumptions

Given: The finite model and units in the Example, AC, and the normalized Gaussian conventions.

[A1]

The Axiom of Choice: AC is the assumption used by the exact finite Gaussian suppliers.

[F1]

Finite heavy-scalar Gaussian matching: in the finite positive Gaussian sector, $Δ S_{W} (J) = - \frac{1}{2} J^{T} K^{- 1} J$ , with the stated spectral-band truncation bounds.

[F2]

Successive finite Gaussian shells compose: two normalized Gaussian shell transforms compose by adding their covariance matrices.

[F3]

Multivariate normal law, including singular covariance: $N_{1} (0, C)$ has mean zero and covariance $C$ , including $C = 0$ .

[F4]

Normalized finite Gaussian shell transform: $T_{C} f (x) = \int f (x + z) d γ_{C} (z)$ for a nonnegative Borel function $f$ .

Verification

technique · exact one-dimensional arithmetic and covariance moments

Given: The data in the Example.

1.1A1F1givenalgebra

Here $K = 5 m_{0}^{2}$ is positive, $J$ is an eigenvector of $A$ with eigenvalue $E^{2} = m_{0}^{2}$ , and $E < M$ . Thus $K^{- 1} J = 1 / (5 m_{0})$ and $S_{0} J = M^{- 2} J = 1 / (4 m_{0})$ . By [F1], $Z_{H} (J) = e^{1 / 10}$ and $Δ S_{W} = - 1 / 10$ , while the order-zero action is $- \frac{1}{2} J^{T} S_{0} J = - 1 / 8$ .

2.1F1step 1.1algebra

The vector remainder is $∣ 1 / (5 m_{0}) - 1 / (4 m_{0}) ∣ = 1 / (20 m_{0}) \leq 1 / (16 m_{0})$ , the bound in [F1] for $N = 0$ . The action remainder is $∣ - 1 / 10 + 1 / 8 ∣ = 1 / 40$ , while [F1] bounds it by $\frac{1}{2} (m_{0}^{2} / (4 m_{0}^{2})) (m_{0}^{2} / (4 m_{0}^{2})) = 1 / 32$ . Both inequalities are exact rational comparisons.

3.1A1F2F3F4givenalgebra∎

For $Z_{C} \sim N_{1} (0, C)$ , [F3] gives $E Z_{C} = 0$ and $E Z_{C}^{2} = C$ . Hence [F4] gives $T_{C} f (x) = E [(x + Z_{C})^{2}] = x^{2} + C$ . Therefore $T_{3} (T_{1} f) (0) = T_{3} (x^{2} + 1) (0) = 4 = T_{4} f (0)$ , in agreement with [F2]. The shell measures are normalized, so no determinant factor is introduced.

One-dimensional finite Gaussian matching and shell composition

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels