CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck pass

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.

Two dense lobes meeting at one cut vertex give $κ (G) = 1 < λ (G) = 2 < δ (G) = 3$

Statement refuted

The false statement FALSE: vertex connectivity, edge connectivity and minimum degree are always equal claims that vertex connectivity, edge connectivity and minimum degree always agree.

Facts & Assumptions

Given: For $i = 1, 2$ , take vertices $a_{i}, b_{i}, c_{i}, d_{i}$ spanning $K_{4}$ with the edge $a_{i} b_{i}$ deleted. Add one vertex $v$ , add the edges $v a_{i}, v b_{i}$ for both $i$ , and add no edge between the two four-vertex lobes.

[F1]

$κ$ and $λ$ are the least sizes of vertex and edge cuts (Vertex cuts, edge cuts, vertex connectivity $κ (G)$ and edge connectivity $λ (G)$ , with conventions for complete and one-vertex graphs).

[F2]

$δ (G)$ is the least vertex degree (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

Counterexample

technique · direct

1.1givenF1

The graph is connected, and deleting $v$ separates the two lobes. No deletion of zero vertices disconnects a connected graph, so $κ (G) = 1$ .

1.2givenF2

In each lobe, $a_{i}$ and $b_{i}$ have two neighbours inside the lobe and the neighbour $v$ , so degree $3$ ; $c_{i}, d_{i}$ have degree $3$ inside the lobe; and $v$ has degree $4$ . Hence $δ (G) = 3$ .

1.3givenF1

Deleting $v a_{1}$ and $v b_{1}$ separates the first lobe from the rest, so $λ (G) \leq 2$ .

1.4givenF1

Every edge lies on a cycle. The edges $v a_{i}$ and $v b_{i}$ lie on the $4$ -cycle $v, a_{i}, c_{i}, b_{i}, v$ . Each internal edge incident with $a_{i}$ lies on the triangle $a_{i}, c_{i}, d_{i}, a_{i}$ , and each internal edge incident with $b_{i}$ lies on $b_{i}, c_{i}, d_{i}, b_{i}$ . Thus deleting one edge leaves an alternate path between its endpoints and cannot disconnect the graph, so $λ (G) \geq 2$ .

2.1step 1.1step 1.2step 1.3step 1.4∎

Steps 1.3 and 1.4 give $λ (G) = 2$ . Together with steps 1.1 and 1.2, this proves $κ (G) = 1 < λ (G) = 2 < δ (G) = 3$ and refutes equality in both Whitney inequalities.

Remarks

Depends on

Used by

Nothing in the library uses this result yet.

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

Wolfram MathWorld, Vertex Connectivity (standard reference, not scraped)

Two dense lobes meeting at one cut vertex give κ(G)=1<λ(G)=2<δ(G)=3κ(G)=1<λ(G)=2<δ(G)=3