Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + 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.

Deleting the high-degree vertices of a γ-self-regular set of density d leaves more than (1γ) of it, and that remainder is ((d+γ)/(1γ))-sparse

Statement

Let 0<γ<1, let WV(G) be nonempty, and suppose (W,W) is a γ-regular pair of density d. Then there is a subset WW with W>(1γ)W such that W is ((d+γ)/(1γ))-sparse.

Facts & Assumptions

Given: A finite simple graph G, a real 0<γ<1, a nonempty set WV(G), and a density d=dG(W,W) such that (W,W) is γ-regular.

[L1]

If (X,Y) is a γ-regular pair of density d and a fixed set YY has YγY, then fewer than γX vertices of X have more than (d+γ)Y neighbours in Y (In a regular pair, fewer than ϵX vertices have too small a degree into a large subset, and fewer than ϵX have too large a degree, ϵ-regular pairs and self-regular vertex sets).

[L2]

A set is c-sparse exactly when every vertex of the induced graph on it has degree at most c times its size (A set is c-sparse exactly when the maximum degree of the graph it induces is at most c times its size, c-sparse, c-dense and c-restricted vertex sets).

Proof

technique · direct
1.1

Apply [L1] to the pair (W,W) with Y=W. Since WγW, fewer than γW vertices of W have more than (d+γ)W neighbours in W.

L1
2.1

Let W be the remaining vertices. Then W>(1γ)W.

step 1.1choose
3.1

For every xW, one has degG[W](x)degG[W](x)(d+γ)W<((d+γ)/(1γ))W. Therefore [L2] makes W ((d+γ)/(1γ))-sparse.

step 2.1L2algebra

Depends on

Used by

Dependency tree · two levels

11 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