LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

If two distinct faces of a connected plane graph have the same boundary subgraph, then the graph is a cycle

Statement

If two distinct faces of a connected plane graph have the same boundary subgraph, then the whole graph is a cycle in the sense of Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges and Connected graphs and connected components defined by the existence of vertex paths.

Facts & Assumptions

Given: A connected plane graph $G$ and distinct faces $f_{1}, f_{2}$ with the same boundary subgraph $H$ .

[L1]

A cycle edge borders two faces and a bridge borders one face (Face frontiers are unions of whole edges; a cycle edge borders two faces and a bridge borders one).

[L2]

A polygon has exactly two regions, each with frontier the polygon (Polygonal Jordan curve theorem: a polygon has exactly two complementary regions and is the frontier of each).

Proof

technique · direct

1.1

No edge of the common boundary $H$ is a bridge, because [L1] gives a bridge only one incident face. Hence every boundary edge lies on a cycle, and $H$ contains a cycle $C$ .

2.1

By [L2], $C$ has exactly two complementary regions. Since both $f_{1}$ and $f_{2}$ have all of $H$ as boundary, they lie on opposite sides of $C$ . Any edge, vertex, chord or attached component of $H$ outside $C$ would lie on only one side of $C$ and could not lie in the frontier of the face on the other side. Thus $H = C$ .

step 1.1L2

3.1

If $G$ contained an edge or vertex outside $H = C$ , connectedness would attach it through one side of $C$ and alter only that face boundary, contradicting the assumed equality. Therefore $G = C$ .

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 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

R. Diestel, Graph Theory, 6th ed., Lemma 4.2.5 (standard reference, not scraped)