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

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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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An edge of a finite graph is a bridge if and only if it lies on no cycle

Statement

For every edge $e$ of a finite graph $G$ , $e$ is a bridge (A bridge as an edge whose deletion increases the number of connected components) if and only if $e$ lies on no cycle (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges). Connectivity and deletion have the meanings of Connected graphs and connected components defined by the existence of vertex paths and Vertex and edge deletion, edge contraction, graph minors, subdivisions and topological minors.

Facts & Assumptions

Given: A finite graph $G$ and an edge $e=uv$ .

[F1]

A bridge is an edge whose deletion increases the number of connected components (A bridge as an edge whose deletion increases the number of connected components).

[F2]

A cycle is a closed walk with no repeated vertices apart from its first and last vertex (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

Proof

technique · direct

1.1

If $e$ lies on a cycle, the remaining edges of that cycle form a $u$ - $v$ path in $G-e$ . Every path in $G$ that used $e$ can replace that occurrence by this path, so deleting $e$ separates no formerly connected pair. Thus $e$ is not a bridge.

F1F2

1.2

Conversely, if $e$ is not a bridge, $u$ and $v$ remain in the same component of $G-e$ and hence are joined there by a path. Adding $e$ to that path gives a cycle containing $e$ .

F1F2

2.1

Step 1.1 says an edge on a cycle is not a bridge, and step 1.2 says an edge not a bridge lies on a cycle. Taking the contrapositive of either implication and combining them proves the biconditional componentwise, including when $G$ is disconnected.

step 1.1step 1.2∎

Depends on

Used by

Face frontiers are unions of whole edges; a cycle edge borders two faces and a bridge borders one Lemma

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 12 results over 9 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. Diestel, Graph Theory, 6th ed., Chapter 4, Section 4.2 (standard reference, not scraped)