TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31

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Every nonempty finite simple graph contains a path with at least $δ (G)$ edges

Statement

Every nonempty finite simple graph $G$ contains a path having at least $δ (G)$ edges.

Facts & Assumptions

Given: A nonempty finite simple graph $G = (V, E)$ .

[F1]

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

[F2]

A path has no repeated vertex, and appending a neighbour not already on it produces a longer path (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

[F3]

A finite graph has only finitely many paths, so a path of greatest length exists (The cardinality $∣ A ∣$ of a finite set).

Proof

technique · constructive

1.1

Choose a path $P = (v_{0}, v_{1}, \dots, v_{ℓ})$ of greatest length in $G$ .

givenF3chooseconstruct

2.1

Every neighbour of $v_{0}$ lies among $v_{1}, \dots, v_{ℓ}$ : a neighbour outside this list could be prepended to $P$ , giving a longer path and contradicting step 1.1.

step 1.1F2

3.1

The $\deg_{G} (v_{0})$ distinct neighbours of $v_{0}$ therefore occupy distinct positions among the $ℓ$ vertices $v_{1}, \dots, v_{ℓ}$ , so $ℓ \geq \deg_{G} (v_{0}) \geq δ (G)$ by [F1]. Thus $P$ has at least $δ (G)$ edges.

step 2.1F1discharge-construct∎

Depends on

Used by

Every finite simple graph of minimum degree at least two contains a cycle Corollary

Dependency tree · two levels

11 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

Cornell Mathematics, Graph Theory notes (standard reference, not scraped)

Every nonempty finite simple graph contains a path with at least δ(G)δ(G) edges