Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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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.

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A set is c-sparse in G exactly when it is c-dense in G‾, so c-restrictedness is complement-invariant

Statement

Let G be a finite simple graph, let c≥0, and let X⊆V(G) be nonempty. Then X is c-sparse in G if and only if X is c-dense in G‾. Consequently X is c-restricted in G if and only if it is c-restricted in G‾.

Facts & Assumptions

Given: A finite simple graph G, a real c≥0, and a nonempty set X⊆V(G).

[L1]
[L2]

The definitions of c-sparse, c-dense, and c-restricted are those of c-sparse, c-dense and c-restricted vertex sets.

Proof

technique · direct
1.1L1

For x∈X, the set NG‾(x)∩X is exactly (X∖{x})∖NG(x) by [L1].

2.1step 1.1L2

Therefore the inequality defining c-sparsity in G‾ is exactly the inequality defining c-density in G, and vice versa, by [L2].

3.1step 2.1L2∎

Since c-restricted means the disjunction of the sparse and dense conditions, step 2.1 shows that restrictedness is unchanged by complementation.

Depends on

Used by

Dependency tree · two levels

7 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