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Comb Structure in co-Bird-Free Graphs

1 · Prerequisites

2 · Summary

Two small induced configurations force a vertex selected by a complete nonadjacent pair to be pure on every induced E. Explicit adjacency proofs isolate the terminal-edge cases and provide the local obstruction needed for the structural argument.

Inside each comb block, overlapping E copies form anticonnected classes. Repeated mixed-block quotients preserve external purity and end in a pure blockade with an E-free pattern. The construction includes the empty-overlap case by selecting a singleton; it assumes the additional vertex complete to the comb blocks and anticomplete to the teeth.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The edge-plus-isolate co-Bird obstruction

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.

Facts & Assumptions

[F1]

The Bird graph and co-Bird supplies the following definition: The Bird graph is the graph on vertices {x1,x2,x3,y,z,w} with edge set {x1x2,x2x3,x1x3,x1y,x2z,yw}. So {x1,x2,x3,y,z} spans the bull, and w is a new leaf attached to the horn vertex y. The co-Bird graph is the complement of the Bird graph.

[F2]

Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs supplies the following definition: Let G be a finite simple graph and let A,BV(G) be disjoint. An edge between A and B is an edge ab with aA and bB. The pair (A,B) is: - complete when every aA is adjacent to every bB; - anticomplete when no aA is adjacent to any bB; - pure when it is complete or anticomplete; and - mixed when it is neither complete nor anticomplete. Adjacency is the symmetric edge relation of G (def-finite-simple-graph, def-graph-adjacency-incidence-neighbourhood-and-degree). If A= or B=, the pair is both complete and anticomplete, hence pure and not mixed.

[F3]

H-free and F-free graphs under the induced-subgraph convention supplies the following definition: For finite graphs H and G, the graph G is H-free when G has no induced copy of H (def-induced-embedding-and-induced-copy). Equivalently, indH(G)=0 (def-induced-copy-number). For a family F of finite graphs, a finite graph G is F-free when it is H-free for every HF. Throughout this page, “free” always refers to induced subgraphs. It does not merely prohibit ordinary subgraph copies.

Proof

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

1.1

If the forbidden pattern holds, exchange a,b if needed so that ua is an edge and ub,uc are nonedges. This exhausts the two meanings of mixed on the edge.

givenF2
1.2

On the six distinct vertices (x,y,u,a,b,c) the edges are exactly xa,xb,xc,ya,yb,yc,xu,ua,ab. The other six pairs are xy,yu,ub,uc,ac,bc.

given
2.1

The bijection (x,y,u,a,b,c)(w,y,x1,z,x3,x2) sends those six nonedges to wy,yx1,x1x3,x1x2,zx2,x3x2, exactly the Bird edges. It therefore sends edges to co-Bird edges as well. This induced co-Bird contradicts the forbidden-induced-copy hypothesis.

F1F3
LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The path-plus-isolate co-Bird obstruction

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.

Facts & Assumptions

[F1]

The Bird graph and co-Bird supplies the following definition: The Bird graph is the graph on vertices {x1,x2,x3,y,z,w} with edge set {x1x2,x2x3,x1x3,x1y,x2z,yw}. So {x1,x2,x3,y,z} spans the bull, and w is a new leaf attached to the horn vertex y. The co-Bird graph is the complement of the Bird graph.

[F2]

Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs supplies the following definition: Let G be a finite simple graph and let A,BV(G) be disjoint. An edge between A and B is an edge ab with aA and bB. The pair (A,B) is: - complete when every aA is adjacent to every bB; - anticomplete when no aA is adjacent to any bB; - pure when it is complete or anticomplete; and - mixed when it is neither complete nor anticomplete. Adjacency is the symmetric edge relation of G (def-finite-simple-graph, def-graph-adjacency-incidence-neighbourhood-and-degree). If A= or B=, the pair is both complete and anticomplete, hence pure and not mixed.

[F3]

H-free and F-free graphs under the induced-subgraph convention supplies the following definition: For finite graphs H and G, the graph G is H-free when G has no induced copy of H (def-induced-embedding-and-induced-copy). Equivalently, indH(G)=0 (def-induced-copy-number). For a family F of finite graphs, a finite graph G is F-free when it is H-free for every HF. Throughout this page, “free” always refers to induced subgraphs. It does not merely prohibit ordinary subgraph copies.

Proof

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

1.1

Reverse the path if necessary to write the prohibited neighbourhood as N(u)H={b,c,d}. Completeness of y and the prescribed path give exactly the edges ya,yb,yc,yd,ub,uc,ud,ab,bc on (y,u,a,b,c,d).

givenF2
1.2

The six remaining pairs are yu,ua,ac,ad,bd,cd. Under (y,u,a,b,c,d)(w,y,x1,z,x3,x2) they map to wy,yx1,x1x3,x1x2,zx2,x3x2.

given
2.1

These are exactly the six Bird edges; the other nine pairs are therefore exactly co-Bird edges. The induced copy is forbidden.

F1F3
LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A mixed vertex on E is pure on both terminal edges

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}.

Facts & Assumptions

[F1]

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.

[F2]

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.

[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

Use the exact five-edge description of S. Suppose u mixes on p1p2. Each of q,p4,p5 is isolated from this edge; the edge-plus-isolate obstruction forces all three to be neighbours of u.

givenF1F3
1.2

If up1 is present and up2 absent, then up3 must be present: otherwise the path p3p4p5 with isolate p1 has precisely the prohibited neighbourhood. Now the path p2p3q with isolate p5 has that same prohibited pattern, a contradiction.

F2
1.3

If up2 is present and up1 absent, the edge p3q with isolate p1 forces up3 to be present. The path p1p2p3 with isolate p5 then violates the path-plus-isolate obstruction.

F1F2
2.1

The two possibilities exhaust mixing on p1p2. Reflection pjp6j fixing q preserves all five edges and the external hypotheses, and proves the same conclusion for p4p5.

F3
LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

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
DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

E overlap chains inside one comb block

Definition

Fix a block Bi of a finite graph comb (Combs in a graph). Let Ei consist of all six-vertex subsets of Bi inducing the graph in The E-graph and co-E. 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.

Throughout this definition, “the graph in The E-graph and co-E” means the E-graph defined there, not the co-E graph.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

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
LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Purity propagates through E overlap chains

Statement

Fix a comb block Bi and an E overlap class A. If uA is pure to every induced E contained in A, then u is pure to A. In particular, any uBi pure to every induced E in Bi is pure to every overlap class.

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]

Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs supplies the following definition: Let G be a finite simple graph and let A,BV(G) be disjoint. An edge between A and B is an edge ab with aA and bB. The pair (A,B) is: - complete when every aA is adjacent to every bB; - anticomplete when no aA is adjacent to any bB; - pure when it is complete or anticomplete; and - mixed when it is neither complete nor anticomplete. Adjacency is the symmetric edge relation of G (def-finite-simple-graph, def-graph-adjacency-incidence-neighbourhood-and-degree). If A= or B=, the pair is both complete and anticomplete, hence pure and not mixed.

Proof

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

1.1

Every induced E meeting A lies in A, because any two of its vertices have a length-one overlap chain. On each such six-vertex copy, purity means that all six adjacency values to u are equal. If two copies share a vertex, their values coincide at that vertex and therefore agree everywhere.

givenF1F2
2.1

For any two vertices of a class choose a defining finite chain. Each consecutive pair has equal adjacency to u because it lies in a common copy. Equality propagates along the chain, including a zero-length chain. Thus every vertex in the class has the same adjacency value. This is exactly purity. If there are no classes the assertion is vacuous.

F1F2
DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The E overlap blockade and mixed quotient sequence

Definition

Fix a comb block Bi with nonempty E overlap support Xi. By E overlap classes form an anticonnected partition, its overlap classes are nonempty anticonnected sets partitioning Xi. Fix an enumeration of the finite set Bi, and order the classes by their least enumerated vertex to obtain L1. Define recursively Ls+1=Ls/M for s1, using The quotient blockade obtained from mixed-block reachability and its least-member ordering. Thus one replaces each mixed-reachability class of blocks by its union. This construction is used only when Xi.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

E overlap quotients terminate at a pure blockade

Statement

For nonempty overlap support, put n=L1. Every stage Ls partitions the same support into nonempty anticonnected blocks and coarsens L1. There is a least q1 for which Lq is pure, with qn. At most n1 strict transitions occur, and all stages from q onward are identical.

Facts & Assumptions

[F1]

The E overlap blockade and mixed quotient sequence supplies the following definition: Fix a comb block Bi with nonempty E overlap support Xi. By lem-e-overlap-classes-form-an-anticonnected-partition, its overlap classes are nonempty anticonnected sets partitioning Xi. Fix an enumeration of the finite set Bi, and order the classes by their least enumerated vertex to obtain L1. Define recursively Ls+1=Ls/M for s1, using def-quotient-blockade-by-mixed-block-reachability and its least-member ordering. Thus one replaces each mixed-reachability class of blocks by its union. This construction is used only when Xi.

[F2]

A quotient block of connected or anticonnected blocks is again connected or anticonnected supplies the following statement: Let L be a blockade and let D be a block of the quotient blockade L/M. 1. If every block of L contained in D induces a connected subgraph, then G[D] is connected. 2. If every block of L contained in D induces an anticonnected subgraph, then G[D] is anticonnected.

[F3]

The well-ordering principle supplies the following statement: Every nonempty subset SN has a least element: there is S with s for all sS.

Proof

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

1.1

The initial classes have the asserted properties by the construction. Every quotient is a partition of the same union into unions of old blocks, so coarsening and nonemptiness persist at every stage. The anticonnected clause of the quotient preservation lemma, applied successively, preserves anticonnectedness.

F1F2
1.2

If a stage has a mixed pair, those two blocks belong to the same reachability class, so the next stage has strictly fewer blocks. If it has no mixed pair, every reachability class is a singleton, so the next ordered blockade is identical. Conversely an identical stage cannot have a mixed pair.

F1
2.1

The positive integer block count begins at n; after n1 strict decreases it is at most one, when no mixed pair exists. Thus a pure stage exists among 1,,n. The well-ordering principle gives a least such q, and the preceding fixed-stage argument makes all later stages equal. This also covers n=1,q=1 and zero strict transitions.

F3F1
LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A separated anticonnected block pair forbids mixing in one direction

Statement

Let G be finite simple and co-Bird-free. Let D1,D2 be disjoint nonempty vertex sets, with D2 anticonnected. Suppose distinct x,y,zD1D2 satisfy xyE(G), both x,y are complete to D1D2, zxE(G), zyE(G), and z is complete to D1 and anticomplete to D2. Then no vertex of D1 is mixed on D2.

Facts & Assumptions

[F1]

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.

[F2]

A vertex mixed on an anticonnected set yields opposite adjacency on a nonedge supplies the following statement: Let G be a finite graph, let AV(G) be anticonnected, and let vV(G)A be mixed on A. Then there exist distinct vertices b,bA such that bbE(G),vbE(G),vbE(G).

[F3]

Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs supplies the following definition: Let G be a finite simple graph and let A,BV(G) be disjoint. An edge between A and B is an edge ab with aA and bB. The pair (A,B) is: - complete when every aA is adjacent to every bB; - anticomplete when no aA is adjacent to any bB; - pure when it is complete or anticomplete; and - mixed when it is neither complete nor anticomplete. Adjacency is the symmetric edge relation of G (def-finite-simple-graph, def-graph-adjacency-incidence-neighbourhood-and-degree). If A= or B=, the pair is both complete and anticomplete, hence pure and not mixed.

Proof

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

1.1

If b1D1 mixes on D2, the anticonnected witness lemma gives distinct b2,b2D2 with b2b2 absent, b1b2 present and b1b2 absent. Thus these three vertices induce exactly an edge and an isolate.

F2given
2.1

The vertices x,y are outside this triple and complete to it. The outside vertex z sees the edge endpoint b1 and misses the other endpoint b2 and isolate b2. This violates the edge-plus-isolate obstruction. Hence such a b1 cannot exist.

F1F3
LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

External purity survives every E overlap quotient

Statement

Let ((ak,Bk):k[]) be an (,w)-comb in a finite simple co-Bird-free graph G, and let v be outside all teeth and blocks, complete to every Bk and anticomplete to every tooth. Fix i with nonempty E overlap support. For every s1, every block L of Ls and every vertex ukiBk, the vertex u is pure to L.

Facts & Assumptions

[F1]

Complete nonedge pairs force purity on induced E graphs 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. 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.

[F2]

Purity propagates through E overlap chains supplies the following statement: Fix a comb block Bi and an E overlap class A. If uA is pure to every induced E contained in A, then u is pure to A. In particular, any uBi pure to every induced E in Bi is pure to every overlap class.

[F3]

The E overlap blockade and mixed quotient sequence supplies the following definition: Fix a comb block Bi with nonempty E overlap support Xi. By lem-e-overlap-classes-form-an-anticonnected-partition, its overlap classes are nonempty anticonnected sets partitioning Xi. Fix an enumeration of the finite set Bi, and order the classes by their least enumerated vertex to obtain L1. Define recursively Ls+1=Ls/M for s1, using def-quotient-blockade-by-mixed-block-reachability and its least-member ordering. Thus one replaces each mixed-reachability class of blocks by its union. This construction is used only when Xi.

[F4]

E overlap quotients terminate at a pure blockade supplies the following statement: For nonempty overlap support, put n=L1. Every stage Ls partitions the same support into nonempty anticonnected blocks and coarsens L1. There is a least q1 for which Lq is pure, with qn. At most n1 strict transitions occur, and all stages from q onward are identical.

[F5]

A separated anticonnected block pair forbids mixing in one direction supplies the following statement: Let G be finite simple and co-Bird-free. Let D1,D2 be disjoint nonempty vertex sets, with D2 anticonnected. Suppose distinct x,y,zD1D2 satisfy xyE(G), both x,y are complete to D1D2, zxE(G), zyE(G), and z is complete to D1 and anticomplete to D2. Then no vertex of D1 is mixed on D2.

[F6]

A vertex mixed on a quotient block but pure on each member block yields two mixed member blocks with opposite adjacency supplies the following statement: Let D be a block of the quotient blockade L/M, and let uD be a vertex. Suppose that u is mixed on D but is pure to every original block of L contained in D. Then there are two original blocks A1,A2 of L, both contained in D, such that 1. A1 and A2 are mixed; and 2. u is complete to A1 and anticomplete to A2.

[F7]

A quotient-level mixed-block witness descends to two mixed member blocks supplies the following statement: Let L be a blockade in a graph G, and suppose that every block of L is connected or every block is anticonnected. Let D1,D2 be distinct mixed blocks of the quotient blockade L/M. Assume there are vertices x,y,uD1D2 such that: 1. x and y are nonadjacent and both are complete to D1D2; 2. uN(x)N(y), with u complete to D1 and anticomplete to D2; and 3. no vertex of D1 is mixed on D2. Then there are mixed original blocks A1,A2 of L, both contained in D1, and vertices x,y,uA1A2 such that: 1. x and y are nonadjacent and both are complete to A1A2; and 2. uN(x)N(y), with u complete to A1 and anticomplete to A2.

[F8]

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

All stages consist of nonempty anticonnected subsets of Bi and are successive mixed quotients. Fix u in another comb block. The comb and special vertex give uvE(G), uaiE(G), vaiE(G), with v,ai complete to Bi.

F3F4F8
1.2

At stage one, the induced-E purity lemma makes u pure to every E in Bi, and overlap propagation makes it pure to every class. If there is no vertex in another block the whole assertion is vacuous.

F1F2base
1.3

Assume purity through stage s1, where s2, and suppose u mixes on a block L of stage s. It is pure to all member blocks by the induction assumption. The opposite-member-block witness gives mixed blocks D1,D2 of stage s1 inside L, with u complete to D1 and anticomplete to D2. Together with (x,y,z)=(v,ai,u) these form a separated witness: distinct outside vertices, xy absent, x,y complete to both blocks, zx present and zy absent, and opposite z adjacency to the blocks.

F6ih
1.4

Consider such a separated witness on any level r2. Its second block is anticonnected, so the no-forward-mixing lemma says that no vertex of D1 mixes on D2. Apply the descending-witness lemma to Lr1: every original block is anticonnected, the two blocks are mixed blocks of its quotient, and all outside adjacency hypotheses hold. It gives mixed blocks of level r1 and new outside vertices satisfying exactly the same separated-witness conditions. The outside vertices are distinct: xy also follows from their completeness to a nonempty set and the relation zN(x)N(y); zx follows from adjacency, and zy from their opposite adjacency on the nonempty second block.

F4F5F7
1.5

Repeat this descent finitely until level one (or do nothing if s1=1). Write the resulting blocks as A1,A2 and vertices as x,y,z. Again no vertex of A1 mixes on A2. Each such vertex is therefore complete or anticomplete to A2. Since the pair is mixed, both types occur; otherwise the pair itself would be pure. Choose any u2A2. It sees every vertex of the first type and none of the second, so it mixes on A1.

F5given
1.6

Now y,z are nonadjacent outside vertices both complete to A1, while u2 sees y and misses z. For every induced E contained in A1, apply induced-E purity with (x,y,u)=(y,z,u2). Thus u2 is pure to every such copy. Every copy meeting the initial overlap class A1 lies wholly in it, and every defining chain between its vertices stays in it. The overlap propagation proof therefore applies within A1, and makes u2 pure to A1, a contradiction.

F1F2F3
2.1

The assumed mixing at stage s is impossible. Starting from stage one and repeating this implication proves the assertion for every positive integer stage, including the fixed terminal stages.

step 1.3F4discharge-induction
LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The terminal E overlap pattern is E-free

Statement

For a comb block with nonempty E overlap support, the pattern graph of its terminal pure quotient blockade Lq is E-free.

Facts & Assumptions

[F1]

E overlap quotients terminate at a pure blockade supplies the following statement: For nonempty overlap support, put n=L1. Every stage Ls partitions the same support into nonempty anticonnected blocks and coarsens L1. There is a least q1 for which Lq is pure, with qn. At most n1 strict transitions occur, and all stages from q onward are identical.

[F2]

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. Throughout this definition, “the graph in def-e-graph-and-co-e-graph” means the E-graph defined there, not the co-E graph.

[F3]

The pattern graph of a pure blockade supplies the following definition: Let B=(B1,,Bt) be a pure blockade in a graph G. Its pattern graph is the graph P(B) with vertex set [t] in which i and j are adjacent exactly when Bi is complete to Bj. Because the blockade is pure, every unordered pair of distinct blocks is either complete or anticomplete, so this graph is well defined. A pattern graph is called P4-free when it contains no induced four-vertex path.

[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]

The E overlap blockade and mixed quotient sequence supplies the following definition: Fix a comb block Bi with nonempty E overlap support Xi. By lem-e-overlap-classes-form-an-anticonnected-partition, its overlap classes are nonempty anticonnected sets partitioning Xi. Fix an enumeration of the finite set Bi, and order the classes by their least enumerated vertex to obtain L1. Define recursively Ls+1=Ls/M for s1, using def-quotient-blockade-by-mixed-block-reachability and its least-member ordering. Thus one replaces each mixed-reachability class of blocks by its union. This construction is used only when Xi.

Proof

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

1.1

All terminal blocks are nonempty, pairwise pure, and unions of initial overlap classes. Suppose six distinct pattern vertices induce E. Choose one vertex in each corresponding block. The six choices are possible because each block is nonempty, and the chosen vertices are distinct because the blocks are disjoint.

F1given
1.2

By the definition of the pattern, complete block pairs supply edges between the representatives; nonadjacent pattern pairs, being pure and not complete, are anticomplete. Thus all fifteen pairs of representatives have exactly the adjacency of E, including its ten nonedges.

F3F4
2.1

These six vertices form an induced E inside Bi, so every pair has an overlap chain of length one and they all belong to a single overlap class. That class is one block of L1, and coarsening places it in a single terminal block, contradicting the six distinct chosen blocks. If the pattern has fewer than six vertices the prohibited selection is already impossible.

F2F5F1
TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A special-vertex co-Bird-free comb admits an E-free structural partition

Statement

Let ((ak,Bk):k[]) be an (,w)-comb in a finite simple co-Bird-free graph G, and let v be outside all teeth and blocks, complete to every Bk and anticomplete to every tooth. For every i[] there exist disjoint sets Xi,Yi with Bi=XiYi such that G[Yi] is E-free, and Xi has a partition into a nonempty ordered sequence (A1i,,Atii) of nonempty sets satisfying: the sequence is a pure blockade, its pattern is E-free, and each individual vertex of every other comb block is pure to each Aji. The blocks Aji may additionally be chosen anticonnected.

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]

E overlap quotients terminate at a pure blockade supplies the following statement: For nonempty overlap support, put n=L1. Every stage Ls partitions the same support into nonempty anticonnected blocks and coarsens L1. There is a least q1 for which Lq is pure, with qn. At most n1 strict transitions occur, and all stages from q onward are identical.

[F3]

The terminal E overlap pattern is E-free supplies the following statement: For a comb block with nonempty E overlap support, the pattern graph of its terminal pure quotient blockade Lq is E-free.

[F4]

External purity survives every E overlap quotient supplies the following statement: Let ((ak,Bk):k[]) be an (,w)-comb in a finite simple co-Bird-free graph G, and let v be outside all teeth and blocks, complete to every Bk and anticomplete to every tooth. Fix i with nonempty E overlap support. For every s1, every block L of Ls and every vertex ukiBk, the vertex u is pure to L.

[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.

[F6]

H-free and F-free graphs under the induced-subgraph convention supplies the following definition: For finite graphs H and G, the graph G is H-free when G has no induced copy of H (def-induced-embedding-and-induced-copy). Equivalently, indH(G)=0 (def-induced-copy-number). For a family F of finite graphs, a finite graph G is F-free when it is H-free for every HF. Throughout this page, “free” always refers to induced subgraphs. It does not merely prohibit ordinary subgraph copies.

[F7]

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.

[F8]

The Bird graph and co-Bird supplies the following definition: The Bird graph is the graph on vertices {x1,x2,x3,y,z,w} with edge set {x1x2,x2x3,x1x3,x1y,x2z,yw}. So {x1,x2,x3,y,z} spans the bull, and w is a new leaf attached to the horn vertex y. The co-Bird graph is the complement of the Bird graph.

Proof

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

1.1

The forbidden pattern here is the complement of the six-vertex Bird, and freeness means absence of an induced copy. Fix i. The comb definition ensures Bi is nonempty. Use its overlap support Xi0 and complement Yi0.

F8F6F5F1
1.2

If Xi0 is nonempty, set Xi=Xi0 and Yi=Yi0. An induced E in Yi would put all its vertices in the overlap support, contradicting disjointness. Take the terminal quotient as the partition of Xi; it is a nonempty pure blockade of anticonnected sets.

F1F2
1.3

The terminal pattern is E-free. Every vertex of another comb block is pure to every terminal block by external quotient purity, with the same given comb and special vertex. This verifies all claims in the nonempty-support case.

F3F4
1.4

If Xi0 is empty, there is no induced E anywhere in Bi. Choose the first vertex b in a fixed finite enumeration of this nonempty block, and set Xi={b}, Yi=Bi{b}. Then Yi is E-free. The one-block sequence ({b}) is pure, anticonnected and has a one-vertex pattern, which cannot contain the six-vertex E. Any outside vertex is either adjacent or nonadjacent to b, so is pure to this block.

F1F5F7
2.1

The two support cases exhaust every i. Use a fixed enumeration of the finite ambient vertex set for all choices and block orderings. When =1 there are no other-block vertices, so that clause is vacuous; Yi is allowed to be empty. The constructions establish the assertion for all blocks.

given

5 · Examples, counterexamples and false statements

None yet.

Sources