Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-13
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.

Lyapunov condition for nonidentical summands

Example

Assume AC. Take independent symmetric signs ϵn,k∈{−1,1}, 1≤k≤n, and set Xn,k=kϵn,k. With sn2=∑k=1nk2, the Lyapunov condition holds for δ=1, and sn−1∑k=1nXn,k⇒N(0,1). In every row of length at least two the summand laws are distinct.

Facts & Assumptions

[F1]

The normalized third-moment condition implies the CLT with delta=1. Lyapunov central limit theorem.

[F2]

Under DC and countable choice independent copies of a two-point law exist. Countably many independent copies of a prescribed law exist.

[F3]

Verification

Given: Assume AC. Take independent symmetric signs ϵn,k∈{−1,1}, 1≤k≤n, and set Xn,k=kϵn,k. With sn2=∑k=1nk2, the Lyapunov condition holds for δ=1, and sn−1∑k=1nXn,k⇒N(0,1). In every row of length at least two the summand laws are distinct.

1.1F2F3

Use [F2]–[F3] on the law assigning mass 1/2 to each sign, and index its coordinates by n(n−1)/2+k for 1≤k≤n. These indices are distinct across the array and exhaust the positive integers. Thus the required signs exist and are independent. Direct two-point integration gives EXn,k=(k−k)/2=0, EXn,k2=k2 and E∣Xn,k∣3=k3. Different k have disjoint supports {−k,k}, so their laws differ.

2.1step 1.1F1∎

At least n/2 integers k in the row satisfy k≥n/2, so sn2≥(n/2)(n/2)2=n3/8. Also ∑k=1nk3≤n n3=n4. Hence sn−3∑kE∣Xn,k∣3≤83/2/n→0. These bounds remain valid for n=1. The rows are centered, independent and have finite moments and positive s_n, so [F1] with delta=1 proves the assertion. AC is used by the copy construction and inherited in [F1].

Depends on

Used by

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Dependency tree · two levels

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Sources