Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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.

Two-level entropy first variation

Example

On H=C2 with its standard inner product and standard basis, fix p∈(0,1) and set σ:=diag⁡(p,1−p). For ∣ε∣<min⁡{p,1−p} define ρ(ε):=diag⁡(p+ε,1−p−ε) and s(ρ(ε)):=−tr⁡(ρ(ε)log⁡ρ(ε)). Then every ρ(ε) is a positive-definite density matrix, ρ(0)=σ, and

ddεs(ρ(ε))∣ε=0=log⁡ ⁣(1−pp).

Facts & Assumptions

Given: H=C2 with its standard inner product and basis, p∈(0,1), and ∣ε∣<min⁡{p,1−p}.

[F1]

Finite-dimensional entropy first variation at a faithful state: for a differentiable density path through a faithful finite-dimensional reference, s′(0)=tr⁡(ρ′(0)(−log⁡σ)).

[F2]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: log⁡ is differentiable on (0,∞) with log⁡′(x)=1/x.

[F3]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: for a,b>0, log⁡(a)−log⁡(b)=log⁡(a/b).

[F5]

The natural logarithm as the inverse of the exponential function: log⁡ is the real inverse of the exponential on (0,∞).

Verification

1.1givenF1algebra

The bounds on ε imply p+ε>0 and 1−p−ε>0, and their sum is 1. Thus ρ(ε) is a positive diagonal trace-one matrix, σ=ρ(0) is strictly positive, and the path satisfies the finite-dimensional hypotheses in [F1].

1.2givenF2F3F4F5algebra

The eigenvalues give s(ρ(ε))=−(p+ε)log⁡(p+ε)−(1−p−ε)log⁡(1−p−ε). By the difference-quotient formula in [F2], ddεlog⁡(p+ε)=1/(p+ε) and, after replacing the increment by its negative, ddεlog⁡(1−p−ε)=−1/(1−p−ε). The product rule [F4] therefore gives ddεs(ρ(ε))=log⁡(1−p−ε)−log⁡(p+ε); at 0, [F3] turns this into log⁡((1−p)/p).

2.1F1step 1.2algebra∎

Directly, ρ′(0)=diag⁡(1,−1) and −log⁡σ=diag⁡(−log⁡p,−log⁡(1−p)), so tr⁡(ρ′(0)(−log⁡σ))=log⁡(1−p)−log⁡p=log⁡((1−p)/p). This agrees with the local calculation and the general result [F1].

Depends on

Dependency tree · 2 levels

Sources