TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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Every planar graph has a proper vertex colouring with at most six colours

Statement

Every planar graph has a proper vertex colouring with at most six colours (Proper vertex colourings and chromatic number). Vertex deletion is from Vertex and edge deletion, edge contraction, graph minors, subdivisions and topological minors, and the proof is finite induction The principle of mathematical induction.

Facts & Assumptions

Given: A finite simple planar graph $G$ .

[L1]

Every nonnull simple planar graph has a vertex of degree at most five (Every nonnull simple planar graph has a vertex of degree at most five).

[F1]

A proper $k$ -vertex-colouring is a function $c:V\to k$ such that $c(u)\ne c(v)$ whenever $\{u,v\}\in E$ (Proper vertex colourings and chromatic number).

Proof

technique · induction

1.1

The null graph has the empty proper colouring.

baseF1

1.2

For a nonnull graph choose by [L1] a vertex $v$ of degree at most five. The planar graph $G-v$ has a proper six-colouring by the induction hypothesis.

ihL1

2.1

At most five colours appear on the neighbours of $v$ , so one of the six colours is absent there. Give $v$ that colour. Edges not incident with $v$ remain proper, and every edge incident with $v$ has differently coloured endpoints by construction.

step 1.2F1

3.1

This extends the induction colouring at every nonnull stage, so every planar graph is six-colourable.

step 1.1step 2.1discharge-induction∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 18 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

R. Grassl and O. Levin, Exploring Combinatorial Mathematics, Activity 306 (standard reference, not scraped)