ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11

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.

$R (3, 3) = 6$ in both directions: the six-vertex argument and the red $5$ -cycle whose blue complement is another $5$ -cycle

Example

The equality in The Ramsey number $R (3, 3) = 6$ can be read directly on labelled complete graphs. Complete graphs are those of Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices, and the blue graph in the lower witness is the complement in the sense of Graph isomorphisms, automorphisms and graph complements.

Facts & Assumptions

Given: Vertices $0, \dots, 5$ for the upper witness and $0, \dots, 4$ for the lower witness; finite pigeonhole is If $∣ A ∣ > k ∣ B ∣$ then every $f : A \to B$ has a fibre with more than $k$ elements, and for nonempty $B$ some fibre has at least $⌈ ∣ A ∣ / ∣ B ∣ ⌉$ elements.

[L1]

The Ramsey number satisfies $R (3, 3) = 6$ (The Ramsey number $R (3, 3) = 6$ ).

Verification

technique · direct

1.1

At vertex $0$ of a red-blue $K_{6}$ , three incident edges share a colour. If they are $01, 02, 03$ and red, then a red edge among $12, 13, 23$ closes a red triangle, while the absence of such an edge makes $123$ a blue triangle. Exchanging colours covers the other case.

2.1

On $K_{5}$ , colour $01, 12, 23, 34, 40$ red and the other edges blue. The red graph is the cycle $0, 1, 2, 3, 4, 0$ ; the blue graph is the cycle $0, 2, 4, 1, 3, 0$ . Neither cycle has a triangle. This gives a five-vertex avoidance colouring and, together with step 1.1, verifies both sides of [L1].

step 1.1L1construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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

Douglas West, Combinatorial Game Theory, Ramsey example (standard reference, not scraped)

R(3,3)=6R(3,3)=6 in both directions: the six-vertex argument and the red 55-cycle whose blue complement is another 55-cycle