Alphabeta Math
Pipeline-generated
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.

Actual photon box trace and Planck energy/entropy flux — Examples

1 · Prerequisites

2 · Summary

The counterexample shows why the discrete zero-energy oscillator cannot be retained in the occupation space merely because the zero wavevector has Lebesgue measure zero in a continuum integral: its infinitely many zero-energy occupations make the actual Gibbs multiplier noncompact and non-trace-class. The half-mass example specializes the positive-cutoff effective-area model to Mmin⁡=M0/2, giving an exact 7/8 lifetime fraction and a finite entropy budget. It stops at positive mass and makes no claim about a zero-mass endpoint or a physical generalized second law.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Zero-energy bosonic mode trace obstruction

Statement refuted

For a periodic bosonic box, deleting the zero wavevector only in the continuum mode integral is sufficient to leave the actual finite-box Gibbs trace finite even if the zero-energy mode remains in the discrete occupation space.

Facts & Assumptions

Given: Assume Countable Choice The Axiom of Countable Choice (ACω) for the Fock basis and Lebesgue-null singleton suppliers. Let I0=(Z3∖{0})×{1,2} be the enumerated base mode set of Actual photon box trace and Planck energy/entropy flux, with its stated max-norm shell and lexicographic ordering.

[A1]

Form the ordered disjoint union I′={0⋆}⊔I0, placing the new label 0⋆ first. Give 0⋆ energy ϵ0⋆=0 and each i=(n,s)∈I0 its energy from Actual photon box trace and Planck energy/entropy flux. Let ΩB′ be the set of finite-support maps m:I′→N0, let E′(m)=∑i∈I′ϵimi, and set H′=ℓ2(ΩB′,C) with its occupation basis. Define H′ by (H′a)m=E′(m)am on D(H′)={a∈H′:∑mE′(m)2∣am∣2<∞}, and define the bounded positive diagonal multiplier Gβ by Gβ∣m⟩=e−βE′(m)∣m⟩. Its diagonal partition sum is Zβ′:=∑m∈ΩB′e−βE′(m)∈[0,+∞]. This is the stipulated extension of the finite-box diagonal model by one zero-energy bosonic mode, without a zero-point term.

[F1]

Occupation bases and actual CCR/CAR domains, Statement: for a specified enumerated mode set, finite-support bosonic occupation configurations give an orthonormal basis of the bosonic Fock space. The current batch 45 evidence records this draft carrier as Step-1-ready; that is not independent review or proof acceptance, so the use remains conditional on its inspected interface.

[F2]

Trace class operator, Definition: a trace-class operator is compact and its singular values have finite sum. The compactness condition uses the exact definition in def-compact-linear-operator: the image of every bounded set must have compact closure; compactness of that closure is the finite-open-subcover property in def-metric-compactness.

[F3]

Actual photon box trace and Planck energy/entropy flux, Statement: the base model has nonzero wavevectors only and its corresponding diagonal Gibbs trace is finite under the explicitly stated summability conditions.

[F4]

The Axiom of Countable Choice (ACω): the Fock-basis supplier assumes ACω.

[F5]

Every affine hyperplane of Rn, and hence every proper linear subspace, is Lebesgue null, Statement: assuming ACω, every proper linear subspace of R3 is Lebesgue measurable and null; in particular, {0} is null.

Counterexample

Given: The explicit extended mode and occupation model in [A1].

1.1givenA1F1F3F4

For each integer r≥0, let m(r)∈ΩB′ have occupation r at 0⋆ and occupation zero at every i∈I0. These are distinct finite-support configurations, so [F1] makes ∣m(r)⟩ an orthonormal sequence in H′.

2.1givenA1F2F5step 1.1algebra∎

The diagonal multiplier satisfies E′(m(r))=rϵ0⋆=0, hence Gβ∣m(r)⟩=∣m(r)⟩. Let B={a∈H′:∥a∥≤1} and let K=Gβ(B)‾. If Gβ were compact, its image of the bounded set B would have compact closure K by the compact-operator definition recorded in [F2]. All vectors ∣m(r)⟩ belong to K. The sets K∩{x∈H′:∥x−y∥<1/2}, for y∈K, form an open cover of K. Distinct vectors ∣m(r)⟩ have distance 2, while every such ball has diameter at most 1; no finite subfamily covers all these vectors. This contradicts compactness, so Gβ is not compact and therefore is not trace class by [F2]. The diagonal partition sum Zβ′ contains ∑r=0∞e−βrϵ0⋆=∑r=0∞1=∞. The continuum zero-mode set {0}⊂R3 has Lebesgue measure zero by [F5], but that fact does not remove the infinitely many discrete zero-mode occupation states.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

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 Mmin⁡=M0/2. Then the model stops at τ=(7/8)t∗, where t∗=M03/(3α), exports energy c2M0/2, and has SB(M0)−SB(M0/2)=3πkBGM02ℏc,Srad(τ)=4πkBGM02ℏc,Δ(SB+Srad)=πkBGM02ℏc. 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,kB>0, with M0>0 and Mmin⁡=M0/2.

[F1]

Exact ideal emitter mass ODE and entropy accounting, Statement: τ=(M03−Mmin⁡3)/(3α), emitted energy is c2(M0−Mmin⁡), SB(M)=4πkBGM2/(ℏc)+S0, and the entropy export and total change are 4/3 and 1/3 times the B-entropy decrease.

Verification

1.1givenF1algebra

Substituting Mmin⁡=M0/2 into the cutoff time gives τ=(1−1/8)M03/(3α)=(7/8)t∗. The integrated energy is c2(M0−M0/2)=c2M0/2.

2.1F1step 1.1algebra

The entropy decrease is [4πkBG/(ℏc)](M02−M02/4)=3πkBGM02/(ℏc), with the same S0 at both endpoints cancelling.

3.1F1step 2.1algebra∎

Multiplying the decrease in step 2.1 by 4/3 and 1/3, respectively, gives Srad(τ)=4πkBGM02/(ℏc) and total change πkBGM02/(ℏc). These are exact outputs of the defined positive-cutoff model.