TheoremStatement: Literature-sourcedProof: 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.

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

Statement

The Ramsey number of The off-diagonal Ramsey number $R (s, t)$ as the least $N$ with $N \to (s, t)^{2}$ , for positive $s, t$ satisfies $R (3, 3) = 6$ . Complete graphs and cycles use Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices and Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges.

Facts & Assumptions

Given: Red-blue colourings of the edges of $K_{6}$ and $K_{5}$ .

[L1]

If $A, B$ are finite, $k \in N$ , and $f : A \to B$ satisfies $∣ A ∣ > k ∣ B ∣$ , then there is $b \in B$ with $∣ f^{- 1} [{b}] ∣ > k$ (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).

Proof

technique · direct

1.1

At a fixed vertex $v$ of $K_{6}$ , at least three of its five incident edges have one colour by [L1]. Call their other endpoints $a, b, c$ and suppose that colour is red. If one of $a b, b c, c a$ is red it closes a red triangle with $v$ ; if none is red, then $a, b, c$ form a blue triangle. The same argument with the colour names exchanged proves $6 \to (3, 3)^{2}$ .

1.2

On five cyclically ordered vertices, colour the five cycle edges red and the remaining five edges blue. The red graph is a $5$ -cycle and has no triangle; the blue graph is also a $5$ -cycle, in the order obtained by stepping two places at a time, and has no triangle. Thus $5 \to̸ (3, 3)^{2}$ .

construct

2.1

Step 1.1 gives $R (3, 3) \leq 6$ and step 1.2 gives $R (3, 3) > 5$ . Since $R (3, 3)$ is a natural number, it equals $6$ .

step 1.1step 1.2∎

Depends on

Used by

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

Dependency tree · two levels

24 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)

The Ramsey number R(3,3)=6R(3,3)=6