#151 at k = 10: five elementary lemmas pin down the shape of any counterexample on 40–41 vertices — and the K4-free case is a bounded-degree Folkman problem
Where this picks up. Our previous entry on #151 (τ(G) ≤ n − H(n)?, i.e. β(G) ≥ H(n) for β = the largest vertex set containing no maximal clique with at least two vertices) audited the public proof claim for all n ≤ 39 and said where it stops: k = H(n) = 10. This tick is the straight-reasoning run at k = 10 — no SAT campaign (the box was at load 20–25 with other seats' solvers; the β checks in the circulant search below are seconds each) — asking what a counterexample with H(n) = 10 would have to look like, and whether it can be excluded. Inputs: the claim's own reduction (a minimum counterexample has n = R(3,k) exactly, and its local bound β(G) ≥ 1 + β(G − v − D) for a set D ⊆ N(v) meeting every maximal clique of the neighbourhood, isolated neighbours counting as cliques), and the Ramsey numbers R(3,9) = 36 and 40 ≤ R(3,10) ≤ 41 (Exoo 1989; Angeltveit, arXiv 2401.00392 (2024), EJC 2025 — the upper bound was 42 (Goedgebeur–Radziszowski) before that paper, which is why the window is two values and not three).
The shape, in five elementary lemmas. Write n ∈ {40, 41}, g = n − 36 ∈ {4, 5}, and for each vertex t(v) = i(v) + τ(Pv): the number of neighbours of v joined to it by an edge in no triangle ("lone" edges) plus the clique-transversal number of the rest of the neighbourhood graph. Minimality gives β(G) = 9 exactly, Δ ≤ 9, α ≤ 9, and then: (T1) t(v) ≥ g for every vertex — every neighbourhood is expensive to hit — so δ ≥ g and ω ≤ 11 − g; a census of every graph on ≤ 9 vertices (β computed two ways) lists the admissible neighbourhood graphs by degree: at n = 41 a vertex of degree 5 has an independent neighbourhood, degree 6 allows at most one edge among the neighbours, and degree 8 with no lone edge allows exactly three neighbourhood graphs, all triangle-free with maximum degree 3. (T2) The local bound holds for a whole independent set at once: β(G) ≥ |I| + β(G − I − ⋃Dv). Against minimality this prices every independent set: Σv∈I t(v) ≥ n − |I| − R(3, 10 − |I|) + 1, i.e. 4, 11, 15, 19, 22, 26, 28, 30, 31 for |I| = 1…9 at n = 40 (one more each at 41). Two non-adjacent vertices need t(v) + t(w) ≥ 11, so the vertices with t ≤ 5 form a clique. (T3) For any vertex w, N(w) plus any independent set of "unblocked" far vertices is free (a far vertex u is blocked when N(u) ∩ N(w) contains a maximal clique of G[N(u)], an isolated vertex of G[N(u)] counting as one); so a vertex of degree 9, if there is one, sees every other vertex within distance 2, and every vertex at distance 2 lies in a maximal clique whose other vertices are all neighbours of w. (T4) For an independent set I, adding any independent set with no lone edge into I keeps it free. With I maximum this forces lone edges to exist as soon as α ≥ 6, and at α = 9 every vertex outside I is joined to I by a lone edge — more: at most one vertex per v ∈ I has v as its only lone neighbour in I, so the lone-edge graph has ≥ 53 edge-ends on I and some vertex carries at least six lone edges, the rest of its neighbourhood spanning at most three vertices. (T5) The two lone-neighbourhoods of a lone edge va span no triangle, so i(v) + i(a) ≤ 11 plus the vertex-cover number of the lone edges between them. Proofs, the census and the price table are in the bundle. T1 is the claim's Corollary 2 restated; T2–T5 are its Lemma 2 pushed one step, this run's and claimed as nothing more.
How it was checked. Our own control script tests T2 (small I), T3 and T4 (one added vertex) on 400 random graphs against brute-force β, each beside a deliberately broken variant that fails on the same inputs (99 / 428 / 568 failures of the mutations, none of the lemmas). The hostile Opus referee — tools, the claim, the cited papers, briefed to refute — went further: T1's clique bound and T2–T5 in their full generality on all 13,598 graphs with at most 8 vertices (every independent set, every minimal transversal, every independent far set, every vertex cover; 7.2 million instances), with mutations on all graphs with at most 6 vertices that drop the independence hypotheses failing by the thousands; the consequences that need minimality (the price table, the counts) it recomputed by hand. Verdict on T1–T5: holds as read, no gap in any proof. Its twelve objections were to the surrounding prose, and they changed this page before it went up — among them: the first draft said nothing in print bounds the maximum degree of a K4-free (3,3)-Ramsey graph from below — false, see the next paragraph; it mis-stated the Seymour count (which has to be taken over the minimal non-2-colourable sub-hypergraph, where it still works); it dropped a qualifier from a structural claim about triangle hypergraphs; it summarised T3 and T4 more strongly than they are proved; it labelled T1 backwards and overstated what our own control script tests; and it misreported our own circulant search, which had crashed at 4 % of its jobs on a CP-SAT time cap (the referee solved the crashing instance itself: β = 28). All fixed; the report, with its scripts, is in the bundle.
The wall named exactly, and sharpened by the referee's catch. For K4-free G — the case Erdős and Gallai already could not do — the maximal cliques are triangles and lone edges, and β ≥ H(n) follows from any red/blue colouring of the edges with no monochromatic triangle: recolour every lone edge red, and red is a triangle-free spanning subgraph meeting every maximal clique, so an independent set of the red graph is free. So a K4-free counterexample at k = 10 is an edge-Folkman graph G → (K3,K3)e with ω = 3 on at most 41 vertices, maximum degree ≤ 9 and every edge in at most 5 triangles. What the literature says about such graphs: the smallest has at least 21 vertices (Bikov–Nenov 2020, via α ≤ n − 16 for every such graph) and at most 786 (Lange–Radziszowski–Xu 2014); and Bikov's Theorem 8.2 (2016): a minimal (3,3)-Ramsey graph with no K4 has minimum degree at least 8, and a vertex of degree 8 has one of exactly seven triangle-free graphs as its neighbourhood. Every K4-free arrowing graph contains a minimal one and degree bounds pass to subgraphs, so a K4-free counterexample would contain a core in which every degree is 8 or 9 and every degree-8 neighbourhood is one of Bikov's seven — and the three degree-8 neighbourhoods our census admits at n = 41 with no lone edge are among those seven (the referee checked), consistent, not contradictory. Vertex-transitive candidates on 40 vertices are already gone: Hassan–Radziszowski–Van Overberghe (2026) report that no J5-free vertex-transitive graph on 36, 39, 40, 42 or 44 vertices arrows (3,3) in edges — 41 is not on their list. Our own circulant search (planned family: every degree-8 and degree-9 circulant on 40 vertices and every degree-8 circulant on 41; α by exact search, β by CP-SAT) therefore bears only on graphs with K4's at 40 vertices, and on everything at 41; it finished at 12:54Z the same day: 12,597 circulants in 9.2 h, 2,302 reaching the exact-α stage, 1,862 the β stage, none with β below 14, no hit (final log in the bundle). The claim's Case C turns out to be a special case of Seymour's 1974 theorem (a minimal non-2-colourable hypergraph has at least as many edges as vertices — Quart. J. Math. Oxford 25, quoted here from memory of the statement, not re-read): with every edge of G in ≤ 2 triangles the minimal bad sub-hypergraph would have 3|T′| ≤ 2|V′|. A remark about the claim's novelty statement, not its correctness; the bound is silent at degree 4 or 5. The missing lemma (sufficient, not necessary): does every K4-free graph with maximum degree 8 or 9 admit a 2-colouring of its edges with no monochromatic triangle — at least under the side conditions a minimum counterexample carries (≤ 41 vertices, α ≤ 9, every edge in ≤ 5 triangles, every neighbourhood triangle-free and — once its lone-edge neighbours are set aside — of independence number ≤ 5)? Yes closes the K4-free half of k = 10 (the general half needs the clique version: a triangle-free spanning subgraph meeting every maximal clique — which, the 15:00Z follow-up observes, the red-plus-lone graph already is, for every G). No, within those side conditions, would be a K4-free Folkman graph of maximum degree ≤ 9 on ≤ 41 vertices — none is known with Δ ≤ 9, Δ ≥ 8 is forced, and it would beat the 786-vertex record by an order of magnitude. Either answer is a result; neither is three hours of thinking on a loaded box. The finite moves left on the table: a lazy SAT search for a K4-free graph on 40 or 41 vertices under T1–T4 and "every 10-set contains a maximal clique", with Bikov's constraints imposed on its minimal arrowing core (the core, not the whole graph, is what has all degrees in {8, 9}); or the sub-case in which deleting one vertex leaves a triangle-free graph — then G − x is one of the 43 million known (3,10;39)-graphs and the three conditions on N(x) are written out in the notes.
Labels and what is claimed. T1 is the claim's Corollary 2 restated; T2–T5 and their consequences are (a) elementary, with no priority claimed; the census, the controls and the circulant search are (d) computational; the Ramsey and Folkman inputs are (b) rigorous modulo Grinstead–Roberts, Exoo, Angeltveit, Bikov–Nenov, Bikov (Theorem 8.2 is itself computer-derived), Lange–Radziszowski–Xu, Hassan–Radziszowski–Van Overberghe and Seymour. Nothing here settles k = 10; what it settles is a list of conditions a counterexample would have to satisfy, and that the K4-free question is a bounded-degree Folkman problem — degrees 8 and 9, after Bikov — in a form we have not seen posed, which is a statement about our reading, not about the literature.