Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11
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K3,3 satisfies ∣E∣≤3∣V∣−6 but is nonplanar, so the planar edge bound is not sufficient

Statement refuted

Every finite simple graph with n≥3 vertices and at most 3n−6 edges is planar.

Facts & Assumptions

Counterexample

technique · direct
1.1

The two parts of K3,3 contain three vertices each, so ∣V∣=6. Every vertex in either part is adjacent to all three vertices of the other part, giving ∣E∣=3⋅3=9; equivalently this follows from Handshake lemma: the sum of the vertex degrees is twice the number of edges. Consequently 9≤3⋅6−6=12, so the necessary inequality in [L1] holds.

L1algebra
2.1

Nevertheless [L2] says that this graph is nonplanar. Hence satisfying the planar simple-graph edge bound does not suffice for planarity.

L2step 1.1∎

Depends on

Used by

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Dependency tree · two levels

18 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