CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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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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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 $(G, w)$ whose edge weights are pairwise distinct.

[F1]

At least one MST exists (Real edge-weighted graphs, total tree weight and minimum spanning trees).

[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 $T$ and $T^{'}$ are distinct MSTs. Choose the least-weight edge $e$ in the symmetric difference $E (T) △ E (T^{'})$ , and assume after swapping names that $e \in E (T) ∖ E (T^{'})$ .

assume-contraF2

1.2

By [L3], $T^{'} + e$ has a fundamental cycle. Apart from $e$ , that cycle cannot consist entirely of edges of $T$ , since then it would be a cycle in $T$ . Hence it contains an edge $f \in E (T^{'}) ∖ E (T)$ .

chooseL3

2.1

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

step 1.1step 1.2

3.1

The exchange $T^{'} - f + e$ is a spanning tree of weight $w (T^{'}) - w (f) + w (e) < w (T^{'})$ , contradicting the minimality of $T^{'}$ .

step 1.2step 2.1L2F1

4.1

Hence the MST is unique.

step 3.1F1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

ISI Bangalore discrete mathematics notes, Minimal spanning trees (standard reference, not scraped)