Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Complete nonedge pairs force purity on induced E graphs

Statement

Let G be a finite simple co-Bird-free graph. Let x,y,u be distinct vertices outside the indicated induced subgraph, with xyE(G), uxE(G) and uyE(G), and with x,y complete to that subgraph. Let S={p1,p2,p3,p4,p5,q} induce E, with edges exactly p1p2,p2p3,p3p4,p4p5,p3q. Then u is pure to S. In particular, in an (,w)-comb with an outside vertex v complete to all blocks and anticomplete to all teeth, each uBk, ki, is pure to every induced E in Bi.

Facts & Assumptions

[F1]

A mixed vertex on E is pure on both terminal edges supplies the following statement: Let G be a finite simple co-Bird-free graph. Let x,y,u be distinct vertices outside the indicated induced subgraph, with xyE(G), uxE(G) and uyE(G), and with x,y complete to that subgraph. Let S={p1,p2,p3,p4,p5,q} induce E, with edges exactly p1p2,p2p3,p3p4,p4p5,p3q. If u is mixed on S, it is nevertheless pure to both terminal edges {p1,p2} and {p4,p5}.

[F2]

The edge-plus-isolate co-Bird obstruction supplies the following statement: Let G be a finite simple co-Bird-free graph. Let x,y,u be distinct vertices outside the indicated induced subgraph, with xyE(G), uxE(G) and uyE(G), and with x,y complete to that subgraph. If H={a,b,c} induces just the edge ab, then u cannot be mixed on {a,b} and nonadjacent to c.

[F3]

The path-plus-isolate co-Bird obstruction supplies the following statement: Let G be a finite simple co-Bird-free graph. Let x,y,u be distinct vertices outside the indicated induced subgraph, with xyE(G), uxE(G) and uyE(G), and with x,y complete to that subgraph. If H={a,b,c,d} induces the path abc and isolated vertex d, then u cannot be adjacent to d and two consecutive vertices of the path and nonadjacent to the remaining endpoint.

[F4]

The E-graph and co-E supplies the following definition: The E-graph is the graph on vertices {p1,p2,p3,p4,p5,q} with edge set {p1p2,p2p3,p3p4,p4p5,p3q}. Thus p1p2p3p4p5 is a five-vertex path and q is a leaf attached to its middle vertex p3. The co-E graph is the complement of this graph.

[F5]

Combs in a graph supplies the following definition: Let N with 1, and let w>0. An (,w)-comb in a graph G is a sequence of pairs ((ai,Bi):i[]) satisfying the conditions below. Here a vertex a is complete to (respectively, anticomplete to) a set B when the pair ({a},B) is complete (respectively, anticomplete) in the sense of def-edges-between-sets-and-pure-mixed-pairs. 1. (B1,,B) is an (,w)-blockade; 2. the vertices a1,,a are distinct; 3. the set {a1,,a} is disjoint from every block Bi; and 4. for every i[], the vertex ai is complete to Bi; and 5. for all distinct i,j[], the vertex ai is anticomplete to Bj. The vertices ai are the teeth of the comb.

Proof

Given: The graph, vertices, sets and hypotheses in the statement.

1.1

Write P={p1,,p5} using the five-edge definition. If u is complete to P but misses q, the path qp3p4 and isolate p1 violate the second obstruction. If u is anticomplete to P but sees q, the edge p3q and isolate p1 violate the first. Thus purity on P implies purity on S.

F2F3F4
1.2

It remains to exclude mixing on P. By terminal-edge purity, the adjacency values on p1,p2 agree and those on p4,p5 agree.

F1
1.3

If both terminal pairs are complete to u, mixing on P forces up3 absent. The path p3p2p1 and isolate p5 violate the second obstruction. If both are anticomplete, mixing forces up3 present, and the edge p2p3 with isolate p5 violates the first.

F2F3
1.4

In the remaining case reflect the path so u sees p1,p2 and misses p4,p5. The edge p2p3 with isolate p5 forces up3 present. The path qp3p4 with isolate p1 forces uq absent. But then the edge p3q with isolate p5 violates the first obstruction. This exhausts the possibilities and proves purity on S.

F2F3F4
2.1

For the comb assertion substitute (x,y)=(v,ai). These are distinct nonadjacent vertices outside Bi, both complete to it; uBk is outside Bi and both teeth and is adjacent to v but not to ai. All hypotheses of the proved assertion hold.

F5given

Depends on

Used by

Dependency tree · two levels

9 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