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.

E overlap classes form an anticonnected partition

Statement

The relation Ri on the overlap support Xi is an equivalence relation. Its classes partition Xi; every induced E in Bi lies within one class. Each class is nonempty and anticonnected. If Xi=, there are no classes.

Facts & Assumptions

[F1]

E overlap chains inside one comb block supplies the following definition: Fix a block Bi of a finite graph comb (def-comb-in-a-graph). Let Ei consist of all six-vertex subsets of Bi inducing the graph in def-e-graph-and-co-e-graph. Put Xi=SEiS and Yi=BiXi. For d,eXi, define dRie if there exist m0 and vertices d=d0,d1,,dm=e in Xi such that each consecutive pair is contained in some SEi. A zero-length chain is allowed. If Ei is empty then Xi and the relation are empty. We call this the E overlap chain relation.

[F2]

Anticonnected graphs and anticonnected components supplies the following definition: A graph G is anticonnected, or co-connected, when its complement G is connected (def-connected-graph-and-connected-component, def-graph-isomorphism-and-complement). An anticonnected component, or anticomponent, of G is a vertex set AV(G) that is the vertex set of a connected component of G. Equivalently, G[A] is anticonnected and A is inclusion-maximal with that property (def-subgraph-induced-subgraph-and-spanning-subgraph). Under the library convention, the null graph is not anticonnected, while a one-vertex graph is anticonnected.

[F3]

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.

Proof

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

1.1

Length zero gives reflexivity, reversal gives symmetry, and concatenating two finite chains gives transitivity. These assertions also hold on the empty support. Classes cover Xi because each vertex relates to itself; classes meeting at a vertex are equal by symmetry and transitivity.

F1
1.2

Any two vertices of one induced E have a length-one chain, so that copy lies in one class. In the complement of E, p1 is adjacent to p3,p4,p5,q and reaches p2 through p4. Thus the complement is connected.

F1F3
2.1

For two vertices of a class, take a defining chain. Each consecutive pair can be joined in the complement of its witnessing E; all vertices of that copy belong to the class. Concatenation gives a complement walk within the class, from which deleting closed portions gives a path. For an identical pair the length-zero path suffices. The class is nonempty and hence anticonnected.

F1F2

Depends on

Used by

Cited to discharge well-definedness by E overlap chains inside one comb block.

Dependency tree · two levels

7 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