LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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.

Finite termination of the dyadic estimator parameter searches

Statement

Let $a > 0$ and $C > 0$ be rational and let $k \geq 1$ be an integer. Each of the following exact rational searches terminates:

increase an integer $j \geq 0$ until $2^{j} a \geq 1$ ;
after fixing $k$ , increase an integer $m \geq 0$ until $2^{2 m} a^{2} \geq C k$ .

The first successful exponent is the least one satisfying its comparison. The second returns the least power of two $M = 2^{m}$ with $M^{2} a^{2} \geq C k$ .

Facts & Assumptions

Given: Positive rationals $a, C$ and integer $k \geq 1$ .

[F1]

Every complete ordered field is Archimedean: for every real $x$ there is a natural number $n$ with $x < n$ .

[F2]

Bernoulli's inequality $(1 + x)^{n} \geq 1 + n x$ : if $x \geq - 1$ and $n$ is natural, then $(1 + x)^{n} \geq 1 + n x$ .

Proof

1.1givenF1F2algebra

Apply [F1] to $1 / a$ and obtain a natural $n > 1 / a$ . Then [F2] at $x = 1$ gives $2^{n} = (1 + 1)^{n} \geq 1 + n > 1 / a$ , so $2^{n} a > 1$ . Thus the first search has a successful exponent no larger than $n$ .

1.2givenF1F2algebra

After $k$ is fixed, apply [F1] to $C k / (2 a^{2})$ and obtain a natural $m > C k / (2 a^{2})$ . By [F2] at exponent $2 m$ , $2^{2 m} = (1 + 1)^{2 m} \geq 1 + 2 m > C k / a^{2}$ , hence $(2^{m})^{2} a^{2} > C k$ . Thus the second search has a successful exponent no larger than $m$ .

2.1givenstep 1.1step 1.2construct∎

Each tested comparison is between rationals and is decided by finite integer arithmetic. Since step 1.1 or step 1.2 supplies a successful exponent, incrementing from zero reaches one after finitely many tests; returning at the first successful test returns the least such exponent.

Depends on

Used by

Variance-aware quantum Monte Carlo Algorithm

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

J. Lebl, Basic Analysis I (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed. (standard reference, not scraped)