Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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FALSE: vertex connectivity, edge connectivity and minimum degree are always equal

Statement

FALSE. Every nontrivial connected finite simple graph satisfies κ(G)=λ(G)=δ(G)\kappa(G)=\lambda(G)=\delta(G).

Facts & Assumptions

Given: The bowtie graph GG with two triangles xabxxabx and xcdxxcdx sharing only the vertex xx.

[L1]

Whitney's theorem guarantees only κ(G)λ(G)δ(G)\kappa(G)\le\lambda(G)\le\delta(G) (Whitney's inequalities: κ(G)λ(G)δ(G)\kappa(G)\le\lambda(G)\le\delta(G) for every nontrivial connected graph).

Refutation

technique · direct
1.1

Deleting xx separates the two edges abab and cdcd, while the connected graph has no vertex cut of size 00. Hence κ(G)=1\kappa(G)=1.

givenF1
1.2

Deleting the two edges xaxa and xbxb separates a,ba,b from the other triangle, so λ(G)2\lambda(G)\le2. No single edge disconnects GG, because every edge lies on one of the two triangles and the other two edges of that triangle give an alternate path between its endpoints. Hence λ(G)=2\lambda(G)=2.

givenF1
1.3

The four noncentral vertices have degree 22 and xx has degree 44, so δ(G)=2\delta(G)=2.

given
2.1

Thus κ(G)=1<2=λ(G)=δ(G)\kappa(G)=1<2=\lambda(G)=\delta(G), a strict instance of the first Whitney inequality and a counterexample to the asserted equality.

step 1.1step 1.2step 1.3L1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 23 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources