ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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.

Comparing the two quantitative density scales

Example

Fix $C_{1}, C_{2} > 0$ . For $0 < x < 1 / 2$ put $L = \log_{2} (1 / x)$ , $D_{1} = 2^{- C_{1} L^{2}}$ and $D_{2} = 2^{- C_{2} L^{2} / \log_{2} L}$ . The ratio of logarithmic losses is $\frac{\log_{2} (1 / D_{2})}{\log_{2} (1 / D_{1})} = \frac{C_{2}}{C_{1} \log_{2} L},$ and tends to zero as $x$ decreases to zero. Consequently $D_{2} > D_{1}$ for all sufficiently small $x$ . With $C_{1} = C_{2} = C$ and $x = 1 / 16$ , the fractions are $D_{1} = 2^{- 16 C}$ and $D_{2} = 2^{- 8 C}$ .

Facts & Assumptions

Given: $C_{1}, C_{2} > 0$ , $0 < x < 1 / 2$ , and $L, D_{1}, D_{2}$ as defined in the example.

[F1]

From Fox sudakov quantitative induced density bound: $δ = 2^{- C_{H} (\log_{2} (1 / x))^{2}}$

[F2]

From Loglog quantitative induced density bound: $δ = 2^{- C_{H} (\log_{2} (1 / x))^{2} / \log_{2} \log_{2} (1 / x)}$ .

[F3]

$\log_{b} x = \frac{\log x}{\log b}, b^{\log_{b} x} = x, \log_{b} (b^{u}) = u (u \in R)$ . (Change of base and inversion of the positive-base real exponential).

Verification

1.1

The two fractions have the forms in [F1] and [F2], with their constants allowed to differ. Since $L > 1$ , both losses are positive. Applying [F3] to their reciprocals yields losses $C_{1} L^{2}$ and $C_{2} L^{2} / \log_{2} L$ . Dividing and cancelling $L^{2} > 0$ gives $C_{2} / (C_{1} \log_{2} L)$ .

F1F2F3

2.1

For any $r > 0$ , take $x_{r} = 2^{- 2^{C_{2} / (C_{1} r) + 1}} \in (0, 1 / 2)$ . If $0 < x < x_{r}$ , then $L > 2^{C_{2} / (C_{1} r) + 1}$ and $\log_{2} L > C_{2} / (C_{1} r) + 1$ , so the ratio is less than $r$ . This proves the stated zero limit. Taking $r = 1$ makes the second loss smaller than the first; strict increase of base-two exponentiation gives $D_{2} > D_{1}$ after negating the losses.

step 1.1algebra

3.1

At $x = 1 / 16$ one has $L = 4$ and $\log_{2} L = 2$ . For equal constants $C > 0$ , the losses are $16 C$ and $8 C$ , so $D_{1} = 2^{- 16 C} < 2^{- 8 C} = D_{2}$ . The eventual comparison above does not assert dominance throughout the interval for unrelated constants.

step 1.1step 2.1algebra∎

Source notes

Proof/convention locator: Bucic, Nguyen, Scott and Seymour, Induced subgraph density I, 1.7–1.8, numerical comparison.

Depends on

Used by

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Dependency tree · two levels

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Sources

Bucic, Nguyen, Scott and Seymour, Induced subgraph density I (standard reference, not scraped)