RESEARCH / Focused target
Sahbi's Conjecture 4.2: bipartite two-independence and upper star formation
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/Combinatorics/Graph/StarFormingDissociationRefutation.resultAbsent from this graph
Problem
The question
Rafik Sahbi, Upper k-Star-Forming Sets, k-Independence, and Upper Domination, arXiv:2610.03785v2, Conjecture 4.2:
For every bipartite graph G, β₂(G) = SF₂(G).
For a finite simple graph, a set I is k-independent when every vertex of G[I] has degree less than k; βₖ(G) is the maximum such cardinality. Definition 2.1 calls S k-star-forming if, for each v outside S, G[S ∪ {v}] contains a copy of K₁,ₖ containing v. The copy need not be induced. A star-forming set is minimal if no proper subset is star-forming; SFₖ(G) is the maximum cardinality among these minimal sets.
The Lean claim quantifies over all n and all decidable simple graphs on
Fin n, assumes Colorable 2, and asserts beta G 2 = SF G 2.
The theorem result : ¬ claim refutes this statement. Fin n represents
every finite vertex type up to bijection, and all the defining adjacency,
subset and cardinality conditions are invariant under such a relabelling.
Motivation
Our foothold
The paper proves equality for complete bipartite graphs and chain graphs. Conjecture 4.2 proposes extending equality to all bipartite graphs. One bipartite graph with unequal parameters settles that extension negatively.
Gap
Missing bridges
The supplied statement source identifies Conjecture 4.2 as open and cites issue #14706 as its preregistration. No independent literature or priority check is available in this offline implementation. The result concerns the verbatim conjecture, without asserting a first settlement in the literature.
Route
Proposed approach
Take vertices aᵢ = i and bⱼ = 5+j for 0 ≤ i,j < 5. The cross-edge matrix is
1 1 0 0 1
1 1 0 1 0
0 0 0 1 1
0 1 1 1 1
1 0 1 1 1
There are 16 edges and no edges within either side. Colour vertices 0 through 4 by zero and vertices 5 through 9 by one.
For k = 2, the star definition is equivalent to this local criterion: each v outside S either has at least two neighbours in S, or has a neighbour u in S with at least one neighbour in S. Indeed, v is either the centre or a leaf of the star. Conversely either alternative gives a centre and two distinct leaves. When v is a leaf, its fellow leaf lies in S and therefore differs from v.
Set S = {0,1,2,5,6,7}. The local criterion holds for S and fails for all 63 proper subsets, so S is minimal two-star-forming. Thus SF₂ ≥ 6. All 210 six-element subsets fail two-independence. Any two-independent set of size at least six would contain a six-element subset that remains two-independent, because deleting vertices cannot increase the selected degree. Hence β₂ ≤ 5 < 6 ≤ SF₂.
Falsifier
What would falsify this route
The counterexample must have the specified edges, admit the stated
colouring, and use Definition 2.1 rather than an induced-star variant.
Both directions of the local equivalence are needed, as are failure on
every proper subset of S and the hereditary argument for larger
independent sets. Each obligation is part of the Lean proof of result.
Evidence
Evidence to collect
D5/S3/Combinatorics/Graph/StarFormingDissociationRefutation.result
has the closed type ¬ claim. Its general local equivalence explicitly
constructs and extracts two-leaf stars. The finite certificates use
kernel reduction of decidable propositions. No native_decide, sorry
or new axiom is used. The axiom closure and measured compilation costs
are recorded in the implementation report.
Triage
Scope assessment
- [proved: D5/S3/Combinatorics/Graph/StarFormingDissociationRefutation.result] Conjecture 4.2 is false: the displayed bipartite graph satisfies β₂ ≤ 5 and SF₂ ≥ 6.
- [proved: D5/S3/Combinatorics/Graph/StarFormingDissociationRefutation.starForming_two_iff] Lemma 2.2's two-star local criterion is equivalent to Definition 2.1 for arbitrary vertex types with decidable equality and adjacency.
- [computed: Python, enumerate all 1024 vertex masks with the displayed adjacency matrix, count selected neighbours, and test every proper submask for minimality] The exact values on this graph are β₂ = 5 and SF₂ = 6. These equalities are supplemental computations; the Lean theorem uses only the two separating bounds.
- [computed: read rows zero and one of the displayed matrix] This is not a chain graph. N(a₀) = {b₀,b₁,b₄} and N(a₁) = {b₀,b₁,b₃} are incomparable by inclusion, so neighbourhoods on this side are not nested. The example therefore does not contradict the paper's chain-graph theorem.
- [computed: supplied exhaustive C enumeration of every cross-edge set] Equality β₂ = SF₂ holds for every bipartite graph with part sizes (1,8), (2,7), (3,6), (4,5), (2,8), (3,7), and (4,6). Smaller splits embed into the nine-vertex splits by adding isolated vertices. Every isolated vertex belongs to every minimal star-forming set and can be added to a maximum two-independent set, increasing both parameters by one. Thus all bipartite counterexamples on at most nine vertices are excluded by the supplied computation and this reduction. The C program and its execution were not independently checked here.
- [derived: star plus isolated vertices, using the two-star local criterion] A graph with one part of size one is a star with d leaves plus t isolated vertices. If d ≤ 1, there is no two-leaf star; every vertex is mandatory for star formation and both parameters equal t+d+1. If d ≥ 2, the largest two-independent set in the star has size max(d,2). Its minimal two-star-forming sets are exactly all d leaves, or the centre together with one leaf. Indeed, omitting the centre forces every leaf to be selected; including it forces at least one leaf, and one leaf already suffices. Including the mandatory isolates gives β₂ = SF₂ = t+max(d,2). This includes the unenumerated (1,9) split. Together with the computed splits, every ten-vertex bipartite counterexample must have a 5+5 bipartition. The argument is mathematical prose and is not an additional Lean theorem.
- [computed: the same exhaustive C enumeration on the 5+5 split] Among the 2²⁵ edge sets between two labelled parts of size five, exactly 3600 give β₂ ≠ SF₂. Together with the exclusions above, this computation makes ten the smallest order of a bipartite counterexample.
- [computed: supplied exhaustive C subset computation on grids Pₘ□Pₙ] Equality β₂ = SF₂ holds for 1×1 through 1×10, 2×2 through 2×8, 3×3 through 3×7, 4×4, and 4×5. These are finite computations, not a proof for all grids; their execution was not repeated here.
- [open] Conjecture 8.4 on grids remains open. The complete-bipartite and chain-graph theorems are not formally verified here.
ASSUMED-UNVERIFIED
Unverified assumptions
The paper title, version, verbatim definitions and conjecture, and the stated complete-bipartite and chain-graph results are supplied source attributions. The paper and issue #14706 have not been retrieved in this offline implementation. No exhaustive literature or priority claim follows.