Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 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.

Qid logarithmic and constant divisibility

Statement

Every nonempty finite graph H is ℓ-divisive for each of ℓ(x)=log⁡2(1/x) and ℓ(x)=2. Both functions are subreciprocal on (0,1/2). The divisibility constants may depend on H.

Facts & Assumptions

Given: A nonempty finite graph H and the two functions log⁡2(1/x) and 2 on (0,1/2).

[F1]

For each nonempty H, constants k1,k2>0 make a strict bound ind⁡H(G)<xk1∣G∣∣H∣ yield a QID x-restricted sequence of length at least 2log⁡2(1/x) and width at least ⌊xk2∣G∣⌋ when G is nonempty and 0<x≤1/(8∣H∣). At least half its indices form a subsequence uniformly x-sparse in G or in G‾, so that subsequence has length at least log⁡2(1/x) and the same width lower bound. (Special copy trichotomy produces a restricted blockade).

[F2]

From Subreciprocal function and ell divisibility: A function ℓ:(0,1/2)→(0,∞) is subreciprocal when it is nonincreasing and satisfies 1<ℓ(x)≤1/x throughout its domain. A nonempty finite H is ℓ-divisive if fixed witnesses 0<c<1/2 and d>1 ensure that for every 0<x<c and nonempty finite G, the bound ind⁡H(G)≤xd∣G∣∣H∣ yields a QID sequence uniformly x-sparse in one of G,G‾, with length at least ℓ(x) and width at least ⌊xd∣G∣⌋.

[F3]

For every real y, its unique integer part satisfies ⌊y⌋≤y<⌊y⌋+1. (Integer part: for every real x there is exactly one integer m with m≤x<m+1).

[F4]

For b>0, b≠1, and x>0, log⁡bx=log⁡xlog⁡b,blog⁡bx=x,log⁡b(bu)=u(u∈R). (Change of base and inversion of the positive-base real exponential).

[F5]

The natural logarithm log⁡:(0,∞)→R is strictly increasing and log⁡1=0. (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

Proof

1.1F4F5algebra

By [F5], log⁡2>0, so [F4] implies that log⁡2 is strictly increasing and log⁡22=1. Its inverse u↦2u is also strictly increasing: if u<v but 2u≥2v, applying log⁡2 would give u≥v. Likewise, for any fixed 0<a<1, [F5] gives log⁡a<0, so log⁡a is strictly decreasing by [F4]. If u<v but au≤av, applying log⁡a would give u≥v; thus au>av.

2.1F3F4step 1.1algebra

For y≥2, let N=⌊log⁡2y⌋ by [F3]. Then N≥1 and N≤log⁡2y<N+1. The integer inequality 2N≥N+1 follows by induction: it is equality at N=1, and 2N+1≥2N+2≥N+2. Thus [F4] and step 1.1 give log⁡2y<N+1≤2N≤y. In particular log⁡2y≤y without an asymptotic restriction.

3.1F2step 1.1step 2.1algebra

If 0<x<1/2, then y=1/x>2, so 1<log⁡2(1/x)≤1/x by the preceding bound. As x increases, 1/x decreases and the increasing logarithm makes log⁡2(1/x) nonincreasing. The constant function 2 is nonincreasing and 1<2<1/x. Both satisfy [F2].

4.1F1F2F3step 1.1step 3.1

For fixed H choose k1,k2 from [F1] and any d>max⁡(1,k1,k2). Let c=1/(16∣H∣). For 0<x<c and nonempty G, the premise ind⁡H(G)≤xd∣G∣∣H∣ implies ind⁡H(G)<xk1∣G∣∣H∣ because 0<x<1 and d>k1. The uniformly x-sparse subsequence supplied by [F1] has length at least log⁡2(1/x) and width at least ⌊xk2∣G∣⌋≥⌊xd∣G∣⌋ by floor monotonicity: if u≤v but ⌊u⌋>⌊v⌋, integrality and [F3] give u≥⌊u⌋≥⌊v⌋+1>v, a contradiction. These are the required witnesses for logarithmic divisibility.

5.1F2step 1.1step 4.1algebra∎

The same witnesses have c≤1/16<1/4, hence x<c gives log⁡2(1/x)>2. The same uniformly x-sparse subsequence therefore has length at least 2 and the same width. This witnesses constant divisibility, including every zero-floor-width case.

Source notes

Proof/convention locator: Bucic, Nguyen, Scott and Seymour, Induced subgraph density I, 5.1 and paragraph on constant ell before it.

Depends on

Used by

Dependency tree · two levels

43 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