LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31

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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.

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Every walk between two vertices contains a path between the same endpoints

Statement

Let $W = (v_{0}, \dots, v_{ℓ})$ be a walk in a graph. By deleting zero or more closed segments from its vertex list, one obtains a path from $v_{0}$ to $v_{ℓ}$ of length at most $ℓ$ . If $W$ repeats a vertex, the resulting path can be chosen to have length strictly less than $ℓ$ .

Facts & Assumptions

Given: A walk $W = (v_{0}, \dots, v_{ℓ})$ in a finite simple graph.

[F1]

A path is a walk with no repeated vertex (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

[F2]

If $v_{i} = v_{j}$ with $i < j$ , deleting $v_{i + 1}, \dots, v_{j}$ leaves the list $(v_{0}, \dots, v_{i}, v_{j + 1}, \dots, v_{ℓ})$ , which is a walk with the same endpoints and length $ℓ - (j - i) < ℓ$ .

[F3]

A finite vertex list has only finitely many lists obtainable by deleting segments, so any nonempty family of their lengths has a least member (The cardinality $∣ A ∣$ of a finite set).

Proof

technique · constructive

1.1

Let $W$ be the nonempty finite family of walks obtained from $W$ by deleting zero or more closed segments, and choose $P \in W$ of least length.

givenF2F3chooseconstruct

2.1

The walk $P$ has no repeated vertex: if two positions carried the same vertex, [F2] would delete the intervening closed segment and produce a shorter member of $W$ , contrary to the choice of $P$ .

step 1.1F2

3.1

By [F1], $P$ is a path with the same endpoints as $W$ , and its length is at most $ℓ$ because $W \in W$ . If $W$ repeats a vertex, applying [F2] once before minimizing gives a member of $W$ shorter than $W$ , so $P$ is strictly shorter than $W$ .

step 1.1step 2.1F1F2discharge-construct∎

Depends on

Used by

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

University of Lethbridge, Combinatorics: A Guided Tour, Paths and Cycles (standard reference, not scraped)