TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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.

Berge's theorem: a matching is maximum if and only if it has no augmenting path

Statement

A matching $M$ in a finite graph is maximum if and only if no $M$ -augmenting path exists.

Facts & Assumptions

Given: A matching $M$ of a finite graph $G$ .

[L1]

Switching along an augmenting path produces a matching with exactly one more edge (Switching along an augmenting path increases a matching by one edge).

[L2]

The nontrivial components of the symmetric difference of two matchings are alternating paths and even cycles (The symmetric difference of two matchings is a disjoint union of alternating paths and even cycles).

Proof

technique · direct

1.1

If an $M$ -augmenting path exists, [L1] gives a matching larger than $M$ , so $M$ is not maximum.

1.2

Conversely, suppose $M$ is not maximum and choose a matching $N$ with $∣ N ∣ > ∣ M ∣$ ; decompose $M △ N$ using [L2].

1.3

Every alternating cycle and every alternating path with equally many $M$ - and $N$ -edges has zero contribution to $∣ N ∣ - ∣ M ∣$ , so some path component has one more $N$ -edge than $M$ -edge.

2.1

Such a path begins and ends with $N$ -edges, so both endpoints are $M$ -exposed and it is an $M$ -augmenting path.

step 1.3

3.1

Steps 1.1 and 1.2--2.1 prove both directions of the equivalence.

step 1.1step 2.1∎

Remarks

Finiteness is used only to ensure that a nonmaximum matching has a larger matching and to make the component count finite.

Depends on

Used by

Dependency tree · two levels

7 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

M. Goemans, Lecture notes on bipartite matching (standard reference, not scraped)