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CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)
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A connected graph with pairwise distinct edge weights has a unique minimum spanning tree

Statement

If a connected graph has pairwise distinct real edge weights, then it has exactly one minimum spanning tree.

Facts & Assumptions

Given: A connected weighted graph (G,w)(G,w) whose edge weights are pairwise distinct.

[L2]

Exchanging an outside edge for an edge on its fundamental cycle preserves the spanning-tree property (The spanning-tree exchange lemma).

[L3]

Adding an edge outside a spanning tree creates its unique fundamental cycle (Every edge outside a spanning tree determines a unique fundamental cycle).

[F2]

Every finite nonempty set of edge weights has a least member, by order duality from Every nonempty finite set of reals has a maximum and a minimum.

Proof

technique · contradiction
1.1

Suppose TT and TT' are distinct MSTs. Choose the least-weight edge ee in the symmetric difference E(T)E(T)E(T)\mathbin\triangle E(T'), and assume after swapping names that eE(T)E(T)e\in E(T)\setminus E(T').

assume-contraF2
1.2

By [L3], T+eT'+e has a fundamental cycle. Apart from ee, that cycle cannot consist entirely of edges of TT, since then it would be a cycle in TT. Hence it contains an edge fE(T)E(T)f\in E(T')\setminus E(T).

chooseL3
2.1

The choice of ee and distinct weights gives w(e)<w(f)w(e)<w(f).

step 1.1step 1.2
3.1

The exchange Tf+eT'-f+e is a spanning tree of weight w(T)w(f)+w(e)<w(T)w(T')-w(f)+w(e)<w(T'), contradicting the minimality of TT'.

step 1.2step 2.1L2F1
4.1

Hence the MST is unique.

step 3.1F1discharge-contradiction

Depends on

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Sources