Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-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.

A nonstationary independent source with no entropy-rate limit

Example

Constructs independent binary coordinates whose block entropy per symbol has subsequences tending to zero and one by alternating super-exponentially long deterministic and fair-coin blocks.

Facts & Assumptions

Given: Independent binary variables (Xj)j≥0 and finite-block entropy as in Finite-block information and entropy in bits.

[F1]

Finite-block information and entropy in bits: entropy is computed from the finite joint law.

[F2]

Information rate of a general finite-alphabet process: a general process need not be stationary, and its block information rate is defined by the limit of Hn/n when it exists.

Verification

Given: The independent coordinates and block convention below.

1.1F1givenalgebra

Partition the coordinates into consecutive blocks of lengths Lj=22j for j≥1. In odd-numbered blocks set Xi=0 deterministically; in even-numbered blocks let the Xi be fair independent bits. Independence gives the block law p0:N−1(x0N−1)=∏i=0N−1pi(xi). Expanding its finite entropy sum over positive-probability words and using log⁡2∏ipi(xi)=∑ilog⁡2pi(xi) yields H(X0N−1)=∑i=0N−1H(Xi), since each other coordinate probability sums to one. Thus the entropy contribution is zero on odd blocks and one bit per coordinate on even blocks.

2.1F2step 1.1algebra∎

Let Nj=∑r=1jLr be the endpoint of block j. For j≥2, ∑r<jLr≤(j−1)22j−1, so Lj∑r<jLr≥22j−1j−1⟶∞. Thus the previous blocks contribute a vanishing fraction of Nj. Therefore H(X0Nj−1)/Nj→0 along odd j and tends to 1 along even j. The block entropy per symbol has no limit, so the process has no entropy rate in the sense of [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources