ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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 lower bound of size $2^{c \sqrt{\log_{2} n}}$ is still subpolynomial in $n$

Example

Fix $c, ε > 0$ . Then the function $2^{c \sqrt{\log_{2} n}}$ grows more slowly than $n^{ε}$ .

Facts & Assumptions

Given: Positive reals $c$ and $ε$ .

[L1]

For $n > 1$ , $\log_{2} n$ is defined (The logarithm to a positive base other than one).

Verification

technique · direct

1.1

Write $n = 2^{L}$ with $L : = \log_{2} n$ . Then $2^{c \sqrt{L}} / n^{ε} = 2^{c \sqrt{L} - ε L}$ .

L1algebra

2.1

If $L \geq (2 c / ε)^{2}$ , then $c \sqrt{L} \leq (ε / 2) L$ , so $c \sqrt{L} - ε L \leq - (ε / 2) L$ . Hence for all sufficiently large $n$ , $2^{c \sqrt{\log_{2} n}} / n^{ε} \leq 2^{- (ε / 2) \log_{2} n} = 1 / n^{ε / 2}$ , which tends to $0$ .

step 1.1algebra

3.1

Therefore $2^{c \sqrt{\log_{2} n}} = o (n^{ε})$ .

step 2.1∎

Depends on

The logarithm to a positive base other than one

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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

Maria Chudnovsky, The Erdős-Hajnal Conjecture: A Survey (standard reference, not scraped)

A lower bound of size 2clog⁡2n2clog2​n​ is still subpolynomial in nn