ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 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.

Recurrence to a dyadic interval under doubling

Example

Assume countable choice. For Lebesgue doubling and $E = [0, 1 / 4)$ , almost every $x \in E$ has $D^{n} x \in E$ for infinitely many positive integers $n$ . The points of $E$ returning at time one form $[0, 1 / 8)$ , of measure $1 / 8$ .

Facts & Assumptions

[F1]

Finite measure preservation implies infinitely many positive returns for almost every point of a measurable set. Poincare recurrence for finite measure-preserving systems.

[F2]

Doubling preserves Lebesgue probability on the circle and its completion. Doubling preserves Lebesgue measure.

Verification

Given: Assume countable choice. For Lebesgue doubling and $E = [0, 1 / 4)$ , almost every $x \in E$ has $D^{n} x \in E$ for infinitely many positive integers $n$ . The points of $E$ returning at time one form $[0, 1 / 8)$ , of measure $1 / 8$ .

1.1

The set $E$ is Borel, with $λ (E) = 1 / 4$ , and [F2] gives a measure-preserving system of total mass one. Applying [F1] with exactly this $E$ proves the stated almost-everywhere infinitely-many-positive-returns conclusion. Countable choice enters through [F2]; the recurrence theorem itself needs no choice axiom.

F1F2

2.1

The two inverse branches give $D^{- 1} E = [0, 1 / 8) \cup [1 / 2, 5 / 8)$ . Intersecting with $E$ leaves $[0, 1 / 8)$ , whose measure is $1 / 8$ . This is the first-return-one set because there is no smaller positive time. Return times need not all be one: $1 / 5 \in E$ has successive images $2 / 5, 4 / 5, 3 / 5, 1 / 5$ , so its first positive return time is four. Zero is fixed and returns at every positive time. These calculations are compatible with recurrence; recurrence alone supplies no return-frequency value or every-point assertion.

1.1F2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

E–W Theorem 2.11 and Example 2.4 (standard reference, not scraped)