LemmaStatement: AI-adaptedProof: AI-generatedPipeline-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.

Uniqueness of denominator-bounded phase candidates

Statement

Let $N \geq 1$ , $Q > 4 N^{2}$ be integers and $0 \leq y < Q$ . There is at most one reduced fraction $p / q$ with $1 \leq q \leq N$ satisfying $2 ∣ y q - p Q ∣ \leq q$ .

Facts & Assumptions

Given: Integers $N \geq 1$ , $Q > 4 N^{2}$ , an integer $y$ with $0 \leq y < Q$ , and reduced fractions with positive denominators.

[F1]

A fraction is reduced when its integer numerator and positive denominator are coprime; coprime means gcd equal to $1$ (Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ , Coprime integers: $\gcd (a, b) = 1$ ).

[F2]

Rational subtraction is computed by cross multiplication: $p_{1} / q_{1} - p_{2} / q_{2} = (p_{1} q_{2} - p_{2} q_{1}) / (q_{1} q_{2})$ when $q_{1}, q_{2} > 0$ (Arithmetic on the rationals).

[F3]

Rational absolute value is multiplicative and satisfies $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (Absolute value and the triangle inequality).

Proof

1.1givenF2F3algebra

For any fraction $p / q$ passing the test, division by the positive integer $2 Q q$ gives $∣ y / Q - p / q ∣ \leq 1 / (2 Q)$ . Thus two passing fractions have distance at most $1 / Q$ by the triangle inequality.

2.1givenF1F2step 1.1algebra∎

If $p_{1} / q_{1}$ and $p_{2} / q_{2}$ are distinct reduced fractions, [F2] makes their difference a nonzero integer divided by $q_{1} q_{2}$ , so its absolute value is at least $1 / (q_{1} q_{2}) \geq 1 / N^{2}$ . Since $Q > 4 N^{2}$ , $1 / Q < 1 / (4 N^{2}) < 1 / N^{2}$ , contradicting the bound from step 1.1. Therefore the two fractions coincide.

Depends on

Used by

Rational reconstruction from a phase sample Algorithm
Denominator-bounded rational phase reconstruction Definition
Denominator-bounded reconstruction at the strict boundary Example
Correctness of phase-sample rational reconstruction Theorem

Dependency tree · two levels

16 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

Andrew M. Childs, Lecture Notes on Quantum Algorithms, §9 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)