LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01

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A maximal acyclic spanning subgraph of a connected graph is a spanning tree

Statement

Let $G$ be a finite connected graph. If $F$ is an acyclic spanning subgraph of $G$ maximal under edge inclusion among such subgraphs, then $F$ is a spanning tree of $G$ .

Facts & Assumptions

Given: A connected graph $G$ and an edge-maximal acyclic spanning subgraph $F$ .

[F1]

The connected components of $F$ partition $V(G)$ (The connected components of a graph partition its vertex set and are its maximal connected subgraphs).

[F2]

A spanning tree is a connected acyclic spanning subgraph (Spanning trees of a graph, Trees, forests, leaves and isolated vertices).

[F3]

Connectedness supplies a path between vertices (Connected graphs and connected components defined by the existence of vertex paths).

Proof

technique · contradiction

1.1

Suppose $F$ is disconnected, and choose vertices $u,v$ in distinct components of $F$ .

assume-contraF1

1.2

A $u$ - $v$ path in $G$ has a first edge $xy$ whose endpoints lie in distinct components of $F$ .

F3F1

2.1

Adding $xy$ to $F$ cannot create a cycle, because a cycle through $xy$ would contain an $x$ - $y$ path already in $F$ , contrary to the choice of distinct components.

step 1.2

3.1

Thus $F+xy$ is a strictly larger acyclic spanning subgraph, contradicting maximality.

step 2.1

4.1

Hence $F$ is connected, and therefore it is a spanning tree.

F2discharge-contradiction∎

Depends on

Used by

A finite graph is connected if and only if it has a spanning tree Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 59 results over 21 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

Reinhard Diestel, Graph Theory, Preview Chapter 1 (standard reference, not scraped)