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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Hall's marriage theorem for a finite bipartite graph

Statement

For a finite bipartite graph with parts (X,Y)(X,Y), there is a matching saturating XX if and only if N(S)S|N(S)|\ge |S| for every SXS\subseteq X.

Facts & Assumptions

Given: A finite bipartite graph with specified parts (X,Y)(X,Y).

[L1]

The strict and tight Hall subinstances in the induction have Hall's condition (Hall's condition passes to the strict and tight induction subinstances).

[L2]

The induction principle proves a statement for every natural number from its base case and successor step (The principle of mathematical induction).

Proof

technique · induction on $|X|$
1.1

For X=0|X|=0 the empty matching saturates XX; for X=1|X|=1, Hall gives a neighbour and its incident edge saturates XX.

L2base
1.2

Any matching that saturates XX assigns distinct neighbours to each SXS\subseteq X, hence has N(S)S|N(S)|\ge|S|.

given
1.3

In the strict case with X2|X|\ge2, choose any edge xyxy; [L1] gives Hall after deleting x,yx,y, so induction supplies a matching there saturating X{x}X\setminus\{x\}, and adjoining xyxy saturates XX.

L1ih
1.4

In the tight case, [L1] gives Hall on the two smaller left parts SS and XSX\setminus S; induction gives saturating matchings in each, and their disjoint vertex sets let their union saturate XX.

L1ih
2.1

The base cases and the two alternatives prove Hall's sufficient direction by [L2], and step 1.2 proves its necessary direction.

L2step 1.2step 1.3step 1.4discharge-induction

Remarks

  • This is the finite theorem only. No infinite-family or choice-principle claim is being made here.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 25 results over 14 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