ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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.

$K_{n}$ and ${\overline{K}}_{n}$ both have homogeneous number $n$

Example

For every $n \in N$ , $hom (K_{n}) = hom ({\overline{K}}_{n}) = n .$

Facts & Assumptions

Given: A natural number $n$ .

[L1]

$hom (G) = \max {ω (G), α (G)}$ , with both numbers equal to $0$ for the null graph (Homogeneous vertex sets and the homogeneous number $hom (G) = \max {ω (G), α (G)}$ ).

[L2]

The graph $K_{n}$ has every possible edge and ${\overline{K}}_{n}$ has none (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

Verification

technique · direct

1.1

If $n = 0$ , both graphs are null and the assertion is [L1].

L1L2

1.2

If $n > 0$ , all $n$ vertices form a clique in $K_{n}$ and a stable set in ${\overline{K}}_{n}$ , while neither graph has a vertex set larger than its whole vertex set.

2.1

Thus the maximum in [L1] is $n$ for both graphs.

step 1.1step 1.2L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 13 results over 7 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

M. Chudnovsky, The Erdos-Hajnal Conjecture: A Survey, sec. 1 (standard reference, not scraped)

KnKn​ and K‾nKn​ both have homogeneous number nn