ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge 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.

Half-mass cutoff and exact coarse entropy export

Example

In the fixed-reference effective-area model of Exact ideal emitter mass ODE and entropy accounting, with its stated positive constants (in particular $G > 0$ ), choose the positive cutoff $M_{\min} = M_{0} / 2$ . Then the model stops at $τ = (7 / 8) t_{*}$ , where $t_{*} = M_{0}^{3} / (3 α)$ , exports energy $c^{2} M_{0} / 2$ , and has $S_{B} (M_{0}) - S_{B} (M_{0} / 2) = \frac{3 π k_{B} G M_{0}^{2}}{ℏ c}, S_{r a d} (τ) = \frac{4 π k_{B} G M_{0}^{2}}{ℏ c}, Δ (S_{B} + S_{r a d}) = \frac{π k_{B} G M_{0}^{2}}{ℏ c} .$ The entropy offset cancels. This calculation stops at positive mass and asserts no endpoint at $M = 0$ or physical generalized second law.

Facts & Assumptions

Given: The definitions and exact cutoff solution in Exact ideal emitter mass ODE and entropy accounting, including $G, c, ℏ, k_{B} > 0$ , with $M_{0} > 0$ and $M_{\min} = M_{0} / 2$ .

[F1]

Exact ideal emitter mass ODE and entropy accounting, Statement: $τ = (M_{0}^{3} - M_{\min}^{}) / (3 α)$ , emitted energy is $c^{2} (M_{0} - M_{\min})$ , $S_{B} (M) = 4 π k_{B} G M^{2} / (ℏ c) + S_{0}$ , and the entropy export and total change are $4 / 3$ and $1 / 3$ times the B-entropy decrease.

Verification

1.1givenF1algebra

Substituting $M_{\min} = M_{0} / 2$ into the cutoff time gives $τ = (1 - 1 / 8) M_{0}^{3} / (3 α) = (7 / 8) t_{*}$ . The integrated energy is $c^{2} (M_{0} - M_{0} / 2) = c^{2} M_{0} / 2$ .

2.1F1step 1.1algebra

The entropy decrease is $[4 π k_{B} G / (ℏ c)] (M_{0}^{2} - M_{0}^{2} / 4) = 3 π k_{B} G M_{0}^{2} / (ℏ c)$ , with the same $S_{0}$ at both endpoints cancelling.

3.1F1step 2.1algebra∎

Multiplying the decrease in step 2.1 by $4 / 3$ and $1 / 3$ , respectively, gives $S_{r a d} (τ) = 4 π k_{B} G M_{0}^{2} / (ℏ c)$ and total change $π k_{B} G M_{0}^{2} / (ℏ c)$ . These are exact outputs of the defined positive-cutoff model.

Depends on

Exact ideal emitter mass ODE and entropy accounting

Half-mass cutoff and exact coarse entropy export

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels