Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
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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.
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τ(Cn)=n\tau(C_n)=n for every n≥3n\ge3

Statement

For every n≥3n\ge3, the cycle graph CnC_n has exactly nn spanning trees.

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Facts & Assumptions

Given: n≥3n\ge3 and the cycle graph CnC_n.

[F1]

CnC_n has nn edges and deleting an arbitrary edge gives the path PnP_n (Empty and complete graphs, complete bipartite graphs, and the convention that PnP_n and CnC_n have nn vertices).

[L1]

A spanning tree on nn vertices has n−1n-1 edges (Equivalent characterisations of a nonempty tree by unique paths, edge count, minimal connectivity and maximal acyclicity).

[F2]

τ\tau counts spanning trees (The spanning-tree number τ(G)\tau(G), Spanning trees of a graph).

Verification

technique · direct
1.1

Deleting any one edge of CnC_n gives a connected acyclic spanning graph, hence a spanning tree. The nn choices yield distinct trees.

F1F2
1.2

Conversely, every spanning tree uses n−1n-1 of the nn cycle edges, so it is obtained by deleting exactly one edge.

L1F1
2.1

Therefore τ(Cn)=n\tau(C_n)=n.

step 1.1step 1.2F2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 16 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