Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)
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.

Every edge of a spanning tree determines a fundamental cut, and every edge crossing it restores a spanning tree

Statement

Let TT be a spanning tree of GG and e∈E(T)e\in E(T). The two components A,BA,B of T−eT-e determine the fundamental cut

δT(e):={xy∈E(G):x∈A, y∈B}.\delta_T(e):=\{xy\in E(G):x\in A,\ y\in B\}.

For every f∈δT(e)f\in\delta_T(e), the graph T−e+fT-e+f is a spanning tree of GG.

Facts & Assumptions

Given: A spanning tree TT of GG and e∈E(T)e\in E(T).

[L1]

T−eT-e has exactly two tree components (Deleting any edge of a tree separates it into exactly two tree components).

[F2]

A spanning tree contains every vertex of GG (Spanning trees of a graph).

Proof

technique · direct
1.1

Let A,BA,B be the vertex sets of the two components of T−eT-e. They partition V(G)V(G).

L1F2
2.1

If f=xyf=xy crosses from AA to BB, then T−e+fT-e+f is connected: paths inside AA and BB, together with ff, join every pair of vertices.

step 1.1L1
2.2

It is acyclic: each side is acyclic, and a cycle using ff would require a second edge between the two sides, but T−e+fT-e+f has only ff.

step 1.1L1
3.1

It is spanning by construction, so it is a spanning tree of GG.

step 2.1step 2.2F2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 11 results over 7 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