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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.

✓ 2 results · all verified · 2 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs; all 2 also cleared it.

Leaf Reducibility and Wonderful Families — Examples

1 · Prerequisites

2 · Summary

These examples spend the two finite checks that the A-page proofs actually use. The first isolates the one-subdivided-star model behind the E-graph route and the pendant-leaf deletion back to P5. The second records an explicit six-vertex witness graph for Bird, checks the homogeneous pair that gives it the Erdős-Hajnal property by substitution, and verifies the induced co-Bird models inside its H+ and H− extensions.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The E-graph sits inside a one-subdivided star and deletes to the five-vertex path

Example

The E-graph occurs as an induced subgraph of the 1-subdivision of K1,3, and deleting the remaining pendant leaf recovers the five-vertex path.

Facts & Assumptions

Given: The 1-subdivision of K1,3 with center c, subdivision vertices s1,s2,s3, and leaves t1,t2,t3.

[L1]

The E-graph has edge set {p1p2,p2p3,p3p4,p4p5,p3q} (The E-graph and co-E).

Verification

technique · direct finite check
1.1L1givenconstruct

On the six-vertex subset {t1,s1,c,s2,t2,s3}, the induced edges are t1s1, s1c, cs2, s2t2, and cs3. Relabelling p1=t1, p2=s1, p3=c, p4=s2, p5=t2, and q=s3 turns this induced subgraph into the edge set from [L1]. Hence the selected subset is an induced copy of the E-graph.

2.1step 1.1L2∎

Deleting the vertex q=s3 from that induced copy leaves the five vertices p1,p2,p3,p4,p5 with exactly the four path edges p1p2,p2p3,p3p4,p4p5. By [L2], this is P5.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

A six-vertex witness graph makes the Bird criterion explicit

Example

The six-vertex graph

H: E(H)={v1v2,v1v3,v1v5,v1v6,v2v3,v2v4,v2v5,v3v5,v4v5,v4v6}

has a homogeneous clique {v2,v5}, so H has the Erdős-Hajnal property; moreover H+−v3 and H−−v6 are both co-Bird.

Facts & Assumptions

Given: The graph H above, with distinguished vertices v1 and v2.

[L1]

Every graph on at most five vertices has the Erdős-Hajnal property (Every graph on at most five vertices has the Erdős-Hajnal property).

[L3]

The graph H+ adds a new vertex adjacent to the two distinguished vertices, while H− does the same after deleting the distinguished edge if it is present; co-Bird is the complement of Bird (The graphs H+ and H− for two distinguished vertices, The Bird graph and co-Bird).

Verification

technique · direct finite check
1.1givenconstruct

Outside the pair {v2,v5}, both vertices are adjacent exactly to v1,v3,v4 and are nonadjacent to v6. Hence {v2,v5} is a homogeneous clique. Let Q be the graph on {x,v1,v3,v4,v6} with edges {xv1,xv3,xv4,v1v3,v1v6,v4v6}. Replacing x by the clique {v2,v5} recovers all edges of H, so H=Q[x→K2].

1.2L3givenalgebra

In H−, delete v6. The remaining six vertices {v,v1,v2,v3,v4,v5} have edge set {vv1,vv2,v1v3,v1v5,v2v3,v2v4,v2v5,v3v5,v4v5}. Relabel them by x1=v4, x2=v, x3=v3, y=v1, z=v5, and w=v2. Again the only missing edges are x1x2,x1x3,x2x3,x1y,x2z,yw, so H−−v6 is also co-Bird.

2.1step 1.1L1L2

The graphs Q and K2 both have at most five vertices, so [L1] gives the Erdős-Hajnal property for each. By [L2], H also has the Erdős-Hajnal property.

3.1step 2.1L3algebra

In H+, delete v3. The remaining six vertices {v,v1,v2,v4,v5,v6} have edge set {vv1,vv2,v1v2,v1v5,v1v6,v2v4,v2v5,v4v5,v4v6}. Relabel them by x1=v, x2=v6, x3=v5, y=v4, z=v2, and w=v1. Then the only missing edges are x1x2,x1x3,x2x3,x1y,x2z,yw, which are exactly the Bird edges. Therefore H+−v3 is co-Bird.

4.1step 2.1step 3.1step 1.2∎

Steps 2.1-3.1 together with step 1.2 verify the finite witness data used in the Bird route: H has the Erdős-Hajnal property, and both H+ and H− contain induced co-Bird subgraphs.

Sources