Alphabeta Math
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18 results · all verified · 0 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 18 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Matchings, Covers, Menger and Network Flows

1 · Prerequisites

2 · Summary

Finite graphs provide the common language for matchings, covers, paths, and cuts. The development uses the finite simple-graph, bipartite-graph, path, deletion, connectivity, directed-walk, and finite-cardinality conventions supplied by the declared prerequisites. In particular, finite sets make maximum and minimum cardinalities available, while labelled arc sets permit parallel and antiparallel directed arcs to remain distinguishable.

The page defines matching and cover invariants, develops augmenting paths into Berge's theorem, and proves finite Hall, the SDR criterion, the deficiency formula, König's equality, and Gallai's edge-cover identity. It then builds finite integral flow theory from labelled residual arcs through max-flow min-cut and flow decomposition. Unit-capacity reductions give the directed and undirected edge forms of Menger, vertex splitting gives the nonadjacent-terminal forms, and the final corollary characterises finite kk-connected graphs by internally disjoint paths.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Matchings, saturated vertices, maximal and maximum matchings, perfect matchings and ν(G)\nu(G)

Definition

Let G=(V,E)G=(V,E) be a finite simple graph. A matching is a set MEM\subseteq E such that no two distinct edges of MM have a common endpoint. A vertex is MM-saturated when it is an endpoint of an edge of MM, and MM-exposed otherwise. The matching is perfect when every vertex is MM-saturated.

A matching is maximal when no strictly larger matching contains it, and is maximum when it has greatest cardinality among all matchings of GG. The matching number is ν(G):=max{M:M is a matching of G}.\nu(G):=\max\{\lvert M\rvert:M\text{ is a matching of }G\}. The maximum exists because EE is finite and the empty set is a matching.

Remarks

  • A maximal matching need not be maximum; the companion counterexample makes the distinction concrete.
RemarkRemark: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Maximal means inclusion-maximal; maximum means largest cardinality

Remark

Every maximum matching is maximal: a strict matching extension has strictly larger cardinality. The converse is not a definition and is false in general; maximal only rules out adding edges while preserving the ones already chosen.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Vertex covers, edge covers and the covering numbers τ(G)\tau(G) and ρ(G)\rho(G)

Definition

For a finite simple graph G=(V,E)G=(V,E), a vertex cover is a set CVC\subseteq V such that every edge has an endpoint in CC; its least possible size is τ(G):=min{C:C is a vertex cover of G}.\tau(G):=\min\{\lvert C\rvert:C\text{ is a vertex cover of }G\}. An edge cover is a set FEF\subseteq E such that every vertex is incident with an edge of FF; if GG has no isolated vertices, its least possible size is ρ(G):=min{F:F is an edge cover of G}.\rho(G):=\min\{\lvert F\rvert:F\text{ is an edge cover of }G\}. Both extrema exist whenever their displayed families are nonempty, since VV and EE are finite. An isolated vertex prevents every edge cover.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Alternating and augmenting paths relative to a matching

Definition

Let MM be a matching of GG. A path is MM-alternating if consecutive edges alternately lie outside and inside MM. It is MM-augmenting if it is MM-alternating, its first and last edges lie outside MM, and its two endpoints are MM-exposed. Thus an augmenting path has odd length, including the possible one-edge path between two exposed vertices.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Switching along an augmenting path increases a matching by one edge

Statement

If PP is an MM-augmenting path in a finite graph, then M:=ME(P)M':=M\mathbin{\triangle}E(P) is a matching and M=M+1|M'|=|M|+1.

Facts & Assumptions

Given: A matching MM and an MM-augmenting path PP.

[F1]

An augmenting path alternates between edges outside and inside MM, begins and ends outside MM, and has exposed endpoints (Alternating and augmenting paths relative to a matching).

Proof

technique · constructive
1.1

Along PP there is one more edge outside MM than inside MM, and every internal vertex of PP is incident with precisely one edge of ME(P)M\cap E(P).

F1construct
1.2

Replacing the MM-edges of PP by its non-MM edges leaves every internal vertex incident with one new matching edge, and its exposed endpoints incident with one; all edges away from PP are unchanged, so MM' is a matching.

F1
2.1

The replacement deletes ME(P)|M\cap E(P)| edges and adds ME(P)+1|M\cap E(P)|+1 edges, hence M=M+1|M'|=|M|+1.

step 1.1discharge-construct

Remarks

  • The use of symmetric difference is literal: it deletes the old matching edges on PP and inserts the new ones, with no choice of any further edges.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The symmetric difference of two matchings is a disjoint union of alternating paths and even cycles

Statement

For matchings M,NM,N of a finite graph GG, every nontrivial connected component of the spanning subgraph with edge set MNM\triangle N is an alternating path or an even alternating cycle. The components are edge-disjoint.

Facts & Assumptions

Given: Matchings M,NM,N in a finite graph GG.

[F1]

Proof

technique · direct
1.1

Each vertex is incident with at most one edge of MM and at most one edge of NN, hence has degree at most two in the spanning subgraph on MNM\triangle N.

F1
1.2

A nontrivial connected finite graph whose degrees are at most two is a path or a cycle; here its consecutive edges alternate between MNM\setminus N and NMN\setminus M.

F1
2.1

A cyclic component has even length, because alternation returns to the initial edge-type only after an even number of edges.

step 1.2
3.1

The nontrivial components partition the edge set MNM\triangle N, while degree-zero vertices contribute no edge, which proves the stated disjoint union.

step 1.2step 2.1
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

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

Statement

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

Facts & Assumptions

Given: A matching MM of a finite graph GG.

[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 MM-augmenting path exists, [L1] gives a matching larger than MM, so MM is not maximum.

L1
1.2

Conversely, suppose MM is not maximum and choose a matching NN with N>M|N|>|M|; decompose MNM\triangle N using [L2].

L2
1.3

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

L2
2.1

Such a path begins and ends with NN-edges, so both endpoints are MM-exposed and it is an MM-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.
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Bipartite neighbourhoods, Hall's condition and systems of distinct representatives

Definition

Let GG be a finite bipartite graph with specified parts XX and YY. For SXS\subseteq X, its neighbourhood in YY is N(S):={yY:xyE(G) for some xS}.N(S):=\{y\in Y:xy\in E(G)\text{ for some }x\in S\}. The pair (X,Y)(X,Y) satisfies Hall's condition (on XX) if N(S)S|N(S)|\ge |S| for every SXS\subseteq X.

For any indexed family (Ax)xX(A_x)_{x\in X}, write U:=xXAxU:=\bigcup_{x\in X}A_x. A system of distinct representatives (SDR) is an injection r:XUr:X\to U such that r(x)Axr(x)\in A_x for every xXx\in X.

When XX and UU are finite, the incidence graph of the family is the finite bipartite graph with the disjoint tagged parts XL:={(x,L):xX},UR:={(u,R):uU}.X_{\mathrm L}:=\{(x,\mathrm L):x\in X\},\qquad U_{\mathrm R}:=\{(u,\mathrm R):u\in U\}. and an edge (x,L)(u,R)(x,\mathrm L)(u,\mathrm R) exactly when uAxu\in A_x. For SXS\subseteq X, its left tagged copy has neighbourhood N(SL)={(u,R):uxSAx}.N(S_{\mathrm L})=\{(u,\mathrm R):u\in\bigcup_{x\in S}A_x\}.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Hall's condition passes to the strict and tight induction subinstances

Statement

Let (X,Y)(X,Y) satisfy Hall's condition. If X2|X|\ge2, then exactly one of the following usable reductions holds.

  1. Strict case: if N(S)>S|N(S)|>|S| for every nonempty proper SXS\subset X, then for every edge xyxy the graph obtained by deleting xx and yy satisfies Hall's condition on X{x}X\setminus\{x\}.
  2. Tight case: if some nonempty proper SXS\subset X has N(S)=S|N(S)|=|S|, then both the subgraph on SN(S)S\cup N(S) and the subgraph on (XS)(YN(S))(X\setminus S)\cup(Y\setminus N(S)) satisfy Hall's condition on their respective left parts.

Facts & Assumptions

Given: A finite bipartite graph with parts (X,Y)(X,Y) satisfying Hall's condition.

[F1]

Hall's condition says N(T)T|N(T)|\ge |T| for every left subset TXT\subseteq X (Bipartite neighbourhoods, Hall's condition and systems of distinct representatives).

Proof

technique · direct
1.1

In the strict case, let xyxy be an edge and TX{x}T\subseteq X\setminus\{x\}; if TT is nonempty then TT is proper in XX, so N(T)>T|N(T)|>|T| and deleting yy leaves at least T|T| neighbours.

F1
1.2

Thus the graph with x,yx,y deleted satisfies Hall's condition on X{x}X\setminus\{x\}, including T=T=\varnothing.

F1
2.1

In the tight case, TST\subseteq S has N(T)N(S)=N(T)T|N(T)\cap N(S)|=|N(T)|\ge|T|, while TXST\subseteq X\setminus S with fewer than T|T| neighbours outside N(S)N(S) would make N(TS)<TS|N(T\cup S)|<|T\cup S|; both induced subinstances therefore satisfy Hall.

step 1.2
3.1

Steps 1.1--2.1 establish the strict and tight reductions.

step 1.2step 2.1

Remarks

  • The two alternatives are exhaustive by whether a nonempty proper left subset is tight; the one-vertex case is kept in Hall's theorem rather than forced into this reduction.
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

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.
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

A finite family has an SDR if and only if every subfamily has a union at least as large as its index set

Statement

Let (Ax)xX(A_x)_{x\in X} have finite index set XX and finite union U:=xXAxU:=\bigcup_{x\in X}A_x. It has an SDR if and only if xSAxS(SX).\left|\bigcup_{x\in S}A_x\right|\ge |S|\qquad(S\subseteq X).

Facts & Assumptions

Given: A family (Ax)xX(A_x)_{x\in X} with finite XX and finite union UU, and its tagged incidence graph with parts XL,URX_{\mathrm L},U_{\mathrm R}.

[F1]

The tagged incidence graph is finite and bipartite, its left-neighbourhood of SLS_{\mathrm L} is the tagged copy of xSAx\bigcup_{x\in S}A_x, and an SDR is an injection choosing one adjacent right tag for each left tag (Bipartite neighbourhoods, Hall's condition and systems of distinct representatives).

[L1]

A finite bipartite graph has a matching saturating its left part exactly when Hall's condition holds (Hall's marriage theorem for a finite bipartite graph).

Proof

technique · direct
1.1

The displayed union inequality is exactly Hall's condition for the finite tagged incidence graph.

F1
1.2

By [F1] the tagged incidence graph is finite, so [L1] makes that condition equivalent to a matching that saturates XLX_{\mathrm L}.

F1L1
1.3

Such a matching assigns each xx the underlying element of its unique matched right tag, and conversely an SDR gives those pairwise disjoint tagged matching edges.

F1
2.1

Combining steps 1.1--1.3 proves the stated equivalence.

step 1.1step 1.2step 1.3
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The maximum matching from XX has size XmaxSX(SN(S))|X|-\max_{S\subseteq X}(|S|-|N(S)|)

Statement

Let (X,Y)(X,Y) be a finite bipartite graph and put d:=maxSX(SN(S)).d:=\max_{S\subseteq X}\bigl(|S|-|N(S)|\bigr). Then the greatest number of vertices of XX saturable by a matching is Xd|X|-d.

Facts & Assumptions

Given: A finite bipartite graph with parts (X,Y)(X,Y) and the displayed dd.

[L1]

Hall's theorem supplies a matching saturating a finite left part exactly under Hall's inequalities (Hall's marriage theorem for a finite bipartite graph).

Proof

technique · direct
1.1

Any matching saturating rr vertices of XX leaves at least SN(S)|S|-|N(S)| vertices of every SXS\subseteq X unmatched, so rXdr\le|X|-d.

given
1.2

Adjoin dd new right vertices, adjacent to every xXx\in X; then every SXS\subseteq X has at least N(S)+dS|N(S)|+d\ge|S| neighbours, so [L1] gives a matching saturating XX in the enlarged graph.

L1
2.1

At most dd of its matching edges use new vertices, so deleting those edges leaves a matching of the original graph saturating at least Xd|X|-d vertices of XX.

step 1.2
3.1

The upper bound in step 1.1 and lower bound in step 2.1 establish the formula.

step 1.1step 2.1
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Alternating reachability from unmatched left vertices produces a vertex cover of the same size as a maximum matching

Statement

Let GG have bipartition (X,Y)(X,Y) and let MM be a maximum matching. From every MM-exposed vertex of XX, follow alternating paths that begin with a nonmatching edge. Let ZXXZ_X\subseteq X and ZYYZ_Y\subseteq Y be the reachable vertices in the two parts. Then C:=(XZX)ZYC:=(X\setminus Z_X)\cup Z_Y is a vertex cover and C=M|C|=|M|.

Facts & Assumptions

Given: A finite bipartite graph (X,Y)(X,Y) and a maximum matching MM.

[F1]

A vertex cover meets every edge, and its size is its cardinality (Vertex covers, edge covers and the covering numbers τ(G)\tau(G) and ρ(G)\rho(G)).

Proof

technique · direct
1.1

No vertex of ZYZ_Y is MM-exposed, for an alternating path from an exposed vertex of XX to it would be MM-augmenting, contrary to [L1].

L1
1.2

Every edge xyxy is met by CC: if xZXx\notin Z_X it is met by xx, while if xZXx\in Z_X, a nonmatching edge reaches yy and a matching edge reaches yy by the preceding alternating step; hence yZYy\in Z_Y.

L1F1
2.1

The matching edges pair ZYZ_Y with ZXZ_X and pair XZXX\setminus Z_X with YZYY\setminus Z_Y; exposed left vertices lie in ZXZ_X, so this gives M=ZY+XZX=C|M|=|Z_Y|+|X\setminus Z_X|=|C|.

step 1.1
3.1

Step 1.2 makes CC a vertex cover and step 2.1 gives its size.

step 1.2step 2.1

Remarks

  • The direction of alternation is essential: left-to-right steps are nonmatching edges and right-to-left steps are matching edges.
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

König's theorem: ν(G)=τ(G)\nu(G)=\tau(G) for every finite bipartite graph

Statement

For every finite bipartite graph GG, its matching number and vertex-cover number agree: ν(G)=τ(G)\nu(G)=\tau(G).

Facts & Assumptions

Given: A finite bipartite graph GG and a maximum matching MM.

[L1]

Alternating reachability from a maximum matching constructs a vertex cover with exactly M|M| vertices (Alternating reachability from unmatched left vertices produces a vertex cover of the same size as a maximum matching).

Proof

technique · direct
1.1

Every vertex cover has at least M|M| vertices, because distinct edges of the matching MM require distinct cover vertices.

given
1.2

By [L1], some vertex cover has exactly M|M| vertices.

L1
2.1

The lower bound and exhibited cover show τ(G)=M=ν(G)\tau(G)=|M|=\nu(G).

step 1.1step 1.2
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Gallai's identity: ν(G)+ρ(G)=V(G)\nu(G)+\rho(G)=|V(G)| when GG has no isolated vertex

Statement

If a finite graph G=(V,E)G=(V,E) has no isolated vertex, then its matching and edge-cover numbers satisfy ν(G)+ρ(G)=V\nu(G)+\rho(G)=|V|.

Facts & Assumptions

Given: A finite graph G=(V,E)G=(V,E) without isolated vertices and a maximum matching MM.

[F1]

A matching has pairwise endpoint-disjoint edges, and M=ν(G)|M|=\nu(G) when MM is maximum (Matchings, saturated vertices, maximal and maximum matchings, perfect matchings and ν(G)\nu(G)).

[F2]

An edge cover meets every vertex and ρ(G)\rho(G) is the least cardinality of one (Vertex covers, edge covers and the covering numbers τ(G)\tau(G) and ρ(G)\rho(G)).

Proof

technique · constructive
1.1

The MM-exposed vertices are independent, or an edge between two would enlarge MM; choose one incident edge for each exposed vertex and adjoin them to MM, obtaining an edge cover of M+(V2M)=VM|M|+(|V|-2|M|)=|V|-|M| edges.

F1F2construct
1.2

Thus ρ(G)Vν(G)\rho(G)\le|V|-\nu(G).

F2
1.3

Let FF be an inclusion-minimal edge cover. Its graph has no cycle, since a cycle edge could be deleted, and no path of length three, since its middle edge could be deleted; hence every component is a star and choosing one edge from each component is a matching of VF|V|-|F| edges.

F1
2.1

Therefore ν(G)Vρ(G)\nu(G)\ge|V|-\rho(G) by step 1.3, while step 1.2 gives the reverse inequality, proving ν(G)+ρ(G)=V\nu(G)+\rho(G)=|V|.

step 1.2step 1.3discharge-construct

Remarks

  • The no-isolated-vertex hypothesis is load-bearing: an isolated vertex has no incident edge, so no edge cover exists.
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

Finite integral networks, feasible flows, values, cuts and residual networks

Definition

A finite integral ss-tt network is a tuple N=(V,A,tail,head,s,t,c)\mathcal N=(V,A,\operatorname{tail},\operatorname{head},s,t,c) where VV and the labelled arc set AA are finite, s,tVs,t\in V are distinct, tail,head:AV,\operatorname{tail},\operatorname{head}:A\longrightarrow V, every arc aAa\in A has tail(a)head(a)\operatorname{tail}(a)\ne\operatorname{head}(a), c:ANc:A\to\mathbb N, no arc enters ss, and no arc leaves tt. Parallel and antiparallel arcs are allowed because distinct elements of AA remain distinct.

A feasible integral flow is f:ANf:A\to\mathbb N with 0f(a)c(a)0\le f(a)\le c(a) and tail(a)=vf(a)=head(a)=vf(a)(vV{s,t}).\sum_{\operatorname{tail}(a)=v}f(a)=\sum_{\operatorname{head}(a)=v}f(a)\quad(v\in V\setminus\{s,t\}). Its value is f:=tail(a)=sf(a)|f|:=\sum_{\operatorname{tail}(a)=s}f(a). An ss-tt cut is a set SVS\subseteq V with sSs\in S, tSt\notin S; its capacity is c(S):=tail(a)S,head(a)Sc(a)c(S):=\sum_{\operatorname{tail}(a)\in S,\operatorname{head}(a)\notin S}c(a).

The residual network of ff has the labelled copies Rf={(a,+):f(a)<c(a)}{(a,):f(a)>0}.R_f=\{(a,+):f(a)<c(a)\}\cup\{(a,-):f(a)>0\}. On these copies, put tailf(a,+)=tail(a),headf(a,+)=head(a),rf(a,+)=c(a)f(a),tailf(a,)=head(a),headf(a,)=tail(a),rf(a,)=f(a).\begin{aligned}\operatorname{tail}_f(a,+)&=\operatorname{tail}(a),&\operatorname{head}_f(a,+)&=\operatorname{head}(a),&r_f(a,+)&=c(a)-f(a),\\\operatorname{tail}_f(a,-)&=\operatorname{head}(a),&\operatorname{head}_f(a,-)&=\operatorname{tail}(a),&r_f(a,-)&=f(a).\end{aligned} For x,yVx,y\in V, an xx-yy residual path is a sequence ρ1,,ρk\rho_1,\ldots,\rho_k of members of RfR_f, with k1k\ge1, for which tailf(ρ1)=x,headf(ρk)=y,headf(ρi)=tailf(ρi+1)(1i<k),\operatorname{tail}_f(\rho_1)=x,\qquad \operatorname{head}_f(\rho_k)=y,\qquad \operatorname{head}_f(\rho_i)=\operatorname{tail}_f(\rho_{i+1})\quad(1\le i<k), and the vertex list x,headf(ρ1),,headf(ρk)=yx,\operatorname{head}_f(\rho_1),\ldots,\operatorname{head}_f(\rho_k)=y has no repetitions. Thus the path records the selected labelled copies, so (a,)(a,-) never coincides with a forward copy of an antiparallel original arc.

Remarks

  • Capacities and flows are natural numbers here. No rational, real, or infinite-capacity termination assertion is part of this definition.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

A flow's value is its net flow across every source-sink cut and never exceeds the cut capacity

Statement

For a feasible flow ff and every ss-tt cut SS, f+tail(a)S,head(a)Sf(a)=tail(a)S,head(a)Sf(a)c(S).|f|+\sum_{\operatorname{tail}(a)\notin S,\operatorname{head}(a)\in S}f(a)=\sum_{\operatorname{tail}(a)\in S,\operatorname{head}(a)\notin S}f(a)\le c(S). In particular fc(S)|f|\le c(S).

Facts & Assumptions

Given: A finite integral network, a feasible flow ff, and an ss-tt cut SS.

[F1]

Flow is conserved at every vertex except s,ts,t, its value is the outgoing flow at ss, and a cut capacity sums outgoing capacities (Finite integral networks, feasible flows, values, cuts and residual networks).

Proof

technique · direct
1.1

Sum flow conservation over S{s}S\setminus\{s\} and cancel arcs whose two endpoints lie in SS; because no original arc enters ss, the resulting equality is the displayed equality of outgoing flow with f|f| plus incoming flow.

F1
1.2

Each outgoing flow term is at most its capacity, so the outgoing flow sum is at most c(S)c(S).

F1
2.1

The incoming sum is nonnegative, hence the equality and inequality of steps 1.1--1.2 give fc(S)|f|\le c(S).

step 1.1step 1.2
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Augmenting by the residual bottleneck preserves feasibility and increases the flow value

Statement

If PP is an ss-tt residual path for a feasible integral flow ff, and δ\delta is the least residual capacity of a copy on PP, then adding δ\delta on each original arc used as (a,+)(a,+) and subtracting δ\delta on each original arc used as (a,)(a,-) gives a feasible integral flow ff' with f=f+δ|f'|=|f|+\delta.

Facts & Assumptions

Given: A feasible integral flow ff, a labelled residual ss-tt path PP, and its bottleneck δ\delta.

[F1]

A forward copy (a,+)(a,+) has residual capacity c(a)f(a)c(a)-f(a), a reverse copy (a,)(a,-) has residual capacity f(a)f(a), and the source has no entering original arc (Finite integral networks, feasible flows, values, cuts and residual networks).

Proof

technique · constructive
1.1

For every forward copy on PP, δc(a)f(a)\delta\le c(a)-f(a), and for every reverse copy, δf(a)\delta\le f(a); the stated additions and subtractions therefore keep every new arc value in [0,c(a)]N[0,c(a)]\cap\mathbb N.

F1construct
1.2

At each internal vertex of the residual path exactly one δ\delta-change enters and one leaves, so the altered flow still satisfies conservation there.

F1
1.3

The first residual copy leaving ss is forward, since no original arc enters ss; it raises the outgoing source flow by δ\delta, while all other source incidences are unchanged.

F1
2.1

Steps 1.1--1.3 prove that ff' is feasible and that f=f+δ|f'|=|f|+\delta.

step 1.1step 1.2step 1.3discharge-construct

Remarks

  • The sign is attached to an arc label, not merely to its endpoints. This is what keeps a reverse copy separate from an antiparallel original arc.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

If the residual network has no source-sink path, its reachable set gives a cut whose capacity equals the flow value

Statement

If no residual ss-tt path exists for a feasible flow ff, let SS be the vertices reachable from ss in its residual network. Then SS is an ss-tt cut and c(S)=fc(S)=|f|.

Facts & Assumptions

Given: A feasible flow ff whose labelled residual network has no ss-tt path.

[F1]

A forward residual copy exists exactly on unused original capacity, and a reverse residual copy exists exactly on positive original flow (Finite integral networks, feasible flows, values, cuts and residual networks).

[L1]

For every cut, outgoing flow is f|f| plus incoming flow and is at most the cut capacity (A flow's value is its net flow across every source-sink cut and never exceeds the cut capacity).

Proof

technique · direct
1.1

Since tt is not reachable, SS contains ss and excludes tt, so it is an ss-tt cut; an original arc leaving SS has no forward residual copy, hence is saturated.

F1
1.2

An original arc entering SS from outside has no reverse residual copy starting in SS, hence carries zero flow.

F1
1.3

Thus outgoing flow across SS equals c(S)c(S) and incoming flow is zero; [L1] gives f=c(S)|f|=c(S).

L1
2.1

The reachable set therefore has the asserted tight-cut property.

step 1.1step 1.2step 1.3
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Ford-Fulkerson terminates for finite integer capacities and proves max-flow min-cut with an integral maximum flow

Statement

In a finite integral ss-tt network, begin with the zero flow and repeatedly augment along any residual ss-tt path by its integral bottleneck. This process terminates. Its final integral flow ff^* is maximum, and f=min{c(S):S is an s-t cut}.|f^*|=\min\{c(S):S\text{ is an }s\text{-}t\text{ cut}\}.

Facts & Assumptions

Given: A finite integral ss-tt network and the augmenting-path process from its zero flow.

[L1]

Residual bottleneck augmentation preserves feasibility and raises the value by its positive integral bottleneck (Augmenting by the residual bottleneck preserves feasibility and increases the flow value).

[L2]

A flow with no residual ss-tt path has a cut of capacity equal to its value (If the residual network has no source-sink path, its reachable set gives a cut whose capacity equals the flow value).

[L3]

Proof

technique · constructive
1.1

The zero flow is feasible and integral, and every feasible flow has value at most the finite integer sum of capacities of arcs leaving ss.

givenconstruct
1.2

Each augmentation increases the integral value by at least one by [L1], so no more than that finite source-capacity bound many augmentations occur.

L1
1.3

At termination the final flow has no residual path, so [L2] supplies a cut SS with c(S)=fc(S)=|f^*|.

L2
1.4

By [L3], every feasible flow has value at most c(S)=fc(S)=|f^*|, and every cut has capacity at least f|f^*|.

L3
2.1

Steps 1.1--1.4 prove termination, integrality, maximality of ff^*, and equality with the minimum cut capacity.

step 1.1step 1.2step 1.3step 1.4discharge-construct

Remarks

  • Integer bottlenecks are the termination mechanism. This theorem makes no assertion about arbitrary real capacities.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

An integral flow decomposes into source-sink path flows and directed cycle flows; unit capacities make the paths arc-disjoint

Statement

Every feasible integral flow is a finite sum of unit ss-tt path flows and unit directed cycle flows. Here a labelled directed cycle flow also permits a two-arc antiparallel circuit, so that it records a circulation even when the underlying simple-digraph convention excludes two-cycles. If every capacity is at most one, the source-sink paths in such a decomposition are arc-disjoint.

Facts & Assumptions

Given: A feasible integral flow ff in a finite integral network.

[F1]

Positive flow is conserved at nonterminals, the source has no incoming original arc, and the sink has no outgoing original arc (Finite integral networks, feasible flows, values, cuts and residual networks).

Proof

technique · induction on the finite total flow $\sum_{a\in A}f(a)$
1.1

The zero flow is the empty sum of path and cycle flows.

base
1.2

If ff is nonzero, follow positive-flow arcs from a positive outgoing source arc; conservation continues the walk at every nonterminal until it reaches tt or repeats a vertex.

F1
1.3

A repeated segment is a positive labelled circulation; subtract one on it and record a unit directed cycle flow, while an unrepeated walk to tt yields a unit ss-tt path flow after subtracting one on its arcs.

F1
2.1

The remainder is feasible, integral, and has strictly smaller total flow, so the induction hypothesis decomposes it and adding the recorded term gives the desired decomposition.

step 1.3ih
3.1

If c(a)1c(a)\le1, no arc can occur in two recorded path terms; together with the base case and step 2.1 this proves the claim for every finite total flow.

step 1.1step 2.1discharge-induction
PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The unit network for a bipartite graph identifies integral flows with matchings and finite cuts with vertex covers

Statement

For a finite bipartite graph with parts (X,Y)(X,Y), give the arcs sxs\to x, each graph edge xyx\to y, and yty\to t unit capacity. Integral flow paths correspond to matching edges xyxy; a vertex cover CXCYC_X\cup C_Y gives the cut {s}(XCX)CY\{s\}\cup(X\setminus C_X)\cup C_Y of the same size. Consequently the maximum-flow/minimum-cut equality agrees with the matching/cover equality, and a flow of value X|X| agrees with Hall's saturating matching criterion.

Facts & Assumptions

Given: The displayed unit network of a finite bipartite graph (X,Y)(X,Y).

[L1]

Integral Ford--Fulkerson gives equal maximum flow and minimum cut values (Ford-Fulkerson terminates for finite integer capacities and proves max-flow min-cut with an integral maximum flow).

[L2]
[L3]

In a finite bipartite graph, matching and minimum vertex-cover sizes agree (König's theorem: ν(G)=τ(G)\nu(G)=\tau(G) for every finite bipartite graph).

[L4]

Hall's condition is equivalent to a matching saturating the left part (Hall's marriage theorem for a finite bipartite graph).

Proof

technique · direct
1.1

By [L2], each unit source-sink path has the form s,x,y,ts,x,y,t, and arc-disjointness makes its middle edges a matching; conversely a matching gives these unit paths.

L2
1.2

A vertex cover CXCYC_X\cup C_Y yields the stated cut: no xyx\to y arc crosses it, and its only cut arcs are the CX|C_X| source arcs and CY|C_Y| sink arcs.

given
1.3

Thus [L1] makes maximum flow equal minimum cut, and [L3] identifies their common value with ν(G)=τ(G)\nu(G)=\tau(G).

L1L3
2.1

A flow of value X|X| uses every unit arc out of ss, so step 1.1 makes a matching saturating XX; [L4] gives the equivalent Hall condition.

L4
3.1

The path, cut, matching, cover, and Hall correspondences in steps 1.1--2.1 prove the proposition.

step 1.1step 1.2step 1.3step 2.1
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-02Open item page →

Arc-disjoint, edge-disjoint and internally vertex-disjoint path families and their local separators

Definition

For the following definitions, a directed path occurrence in a finite labelled directed graph (V,A,tail,head)(V,A,\operatorname{tail},\operatorname{head}), where V,AV,A are finite and tail,head:AV\operatorname{tail},\operatorname{head}:A\to V, is an alternating list v0,a1,v1,,a,vv_0,a_1,v_1,\ldots,a_\ell,v_\ell with aiAa_i\in A, tail(ai)=vi1\operatorname{tail}(a_i)=v_{i-1}, and head(ai)=vi\operatorname{head}(a_i)=v_i for every ii, whose displayed vertices are distinct. An undirected path occurrence in a finite undirected multigraph (Multigraphs, loops and directed graphs as variants distinct from the default finite simple graph) is an alternating list v0,e1,v1,,e,vv_0,e_1,v_1,\ldots,e_\ell,v_\ell whose displayed vertices are distinct and whose displayed edge eie_i has endpoint set {vi1,vi}\{v_{i-1},v_i\} for every ii. Thus each path records its arc or edge labels, including when parallel labels have the same endpoints. An ordinary graph uses each edge eEe\in E as its own label with endpoint set ee, while an ordinary digraph uses each ordered-pair arc (u,v)(u,v) as its own label with tail(u,v)=u\operatorname{tail}(u,v)=u and head(u,v)=v\operatorname{head}(u,v)=v.

For distinct s,tVs,t\in V, a family of directed ss-tt path occurrences is arc-disjoint if no arc label occurs in two paths; an arc separator is a set of arc labels whose deletion leaves no directed ss-tt path occurrence. For an undirected multigraph, replace arc labels by edge labels to obtain edge-disjoint paths and edge separators. Deleting labels means removing exactly those labels while retaining the other incidence data.

An ss-tt path family is internally vertex-disjoint if no vertex other than s,ts,t occurs in two paths. A vertex separator is a set SV{s,t}S\subseteq V\setminus\{s,t\} for which deleting SS, together with every incident arc or edge label, leaves no ss-tt path occurrence. The local vertex comparison is used only when the terminals are nonadjacent (or, in the directed case, when no arc label has tail ss and head tt), so that such a separator can exist.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Menger's theorem: the finite directed and undirected arc, edge and nonadjacent-vertex forms

Statement

All graphs below are finite and the terminals s,ts,t are distinct.

  1. In a directed graph, the maximum number of arc-disjoint directed ss-tt paths equals the minimum size of an arc separator.
  2. In an undirected graph, the maximum number of edge-disjoint ss-tt paths equals the minimum size of an edge separator.
  3. In a directed graph with no arc sts\to t, the maximum number of internally vertex-disjoint directed ss-tt paths equals the minimum size of a vertex separator.
  4. In an undirected graph with nonadjacent s,ts,t, the maximum number of internally vertex-disjoint ss-tt paths equals the minimum size of a vertex separator.

Facts & Assumptions

Given: The path packings and separator notions of the statement.

[F1]

Arc, edge, and vertex separators are deleted sets that eliminate the corresponding ss-tt paths; the local vertex form excludes a direct terminal connection (Arc-disjoint, edge-disjoint and internally vertex-disjoint path families and their local separators).

[L1]

Finite integral max-flow min-cut applies to a network with no original arc entering ss or leaving tt, and supplies a maximum integral flow and a minimum cut of equal value (Ford-Fulkerson terminates for finite integer capacities and proves max-flow min-cut with an integral maximum flow).

[L2]

Every feasible integral flow decomposes into unit source-sink path flows and circulation flows; when every capacity is at most one, its source-sink paths are arc-disjoint (An integral flow decomposes into source-sink path flows and directed cycle flows; unit capacities make the paths arc-disjoint).

Proof

technique · constructive
1.1

For a directed graph, delete arcs entering ss and arcs leaving tt; no directed ss-tt path uses one, so this preserves both the path packing number and the minimum arc-separator size. Give every remaining labelled arc capacity one.

F1construct
1.2

In that unit network, cut arcs are arc separators; conversely, after deleting an arc separator, the vertices reachable from ss determine a cut whose outgoing arcs are contained in that separator. Thus minimum cut capacity equals minimum arc-separator size.

F1
1.3

For an undirected graph, bidirect every edge, delete the now-irrelevant arcs entering ss or leaving tt, and give the remaining labelled arcs unit capacity; a cut then contains exactly one forward copy of each crossing undirected edge.

F1
1.4

Cancel the smaller of the two opposed flow values on every original edge. This preserves feasibility and value, since the preprocessing leaves no opposed pair incident with ss or tt; [L1] and [L2] then give paths which are edge-disjoint after orientations are forgotten, proving clause 2.

L1L2
1.5

For the directed vertex form, first make the same source/sink preprocessing. With n:=Vn:=|V|, replace every internal vertex vv by a capacity-one arc vv+v^-\to v^+, and replace each original arc by a transport arc of capacity nn from the appropriate source copy to target copy.

F1
1.6

Deleting every internal vertex arc gives a cut of capacity n2<nn-2<n, because no sts\to t arc exists. Hence a minimum cut uses no transport arc; its vertex arcs are exactly a vertex separator, while every vertex separator gives such a cut.

F1
1.7

By [L1], the split network has an integral maximum flow equal to that separator number. By [L2] its unit source-sink path terms number exactly that flow value, while cycle terms have value zero; capacity one on every vv+v^-\to v^+ prevents two such terms from sharing an internal vertex. Projecting them proves clause 3.

L1L2
1.8

For the undirected vertex form, bidirect first and then apply steps 1.5--1.7; nonadjacency gives no sts\to t arc, and projection gives internally vertex-disjoint undirected paths, proving clause 4.

F1
2.1

By [L1] and [L2], a maximum unit flow decomposes into as many arc-disjoint source-sink paths as its value, so clause 1 follows from step 1.2.

L1L2
3.1

The unit arc construction proves clause 1, its bidirected cancellation proves clause 2, and the capacity-one vertex-arc construction proves clauses 3 and 4.

step 2.1step 1.4step 1.7step 1.8discharge-construct

Remarks

  • The undirected edge reduction cancels opposed flow before decomposition. This prevents two output paths from consuming the two orientations of one original undirected edge.
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02Open item page →

A finite graph on at least k+1k+1 vertices is kk-connected if and only if every two vertices have kk internally disjoint paths

Statement

Let k1k\ge1 and let GG be a finite graph on at least k+1k+1 vertices. Then GG is kk-connected, meaning it is connected and κ(G)k\kappa(G)\ge k, if and only if every two distinct vertices of GG are joined by at least kk internally vertex-disjoint paths.

Facts & Assumptions

Given: A finite graph GG, an integer k1k\ge1, and V(G)k+1|V(G)|\ge k+1.

[L1]

For nonadjacent terminals, finite vertex Menger equates the largest internally disjoint path family with the smallest vertex separator (Menger's theorem: the finite directed and undirected arc, edge and nonadjacent-vertex forms).

[F1]

For a connected graph on at least two vertices, vertex connectivity is the minimum size of a set whose deletion disconnects the graph or leaves at most one vertex (Vertex cuts, edge cuts, vertex connectivity κ(G)\kappa(G) and edge connectivity λ(G)\lambda(G), with conventions for complete and one-vertex graphs).

[F2]

The open neighbourhood NG(x)N_G(x) is the set of neighbours of xx, and its size is the degree of xx (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

Proof

technique · direct
1.1

Suppose GG is kk-connected and x,yx,y are nonadjacent. Every xx-yy separator has at least kk vertices by [F1], so [L1] gives kk internally disjoint xx-yy paths.

L1F1
1.2

Suppose instead that xyxy is an edge and that SS separates xx from yy in GxyG-xy with Sk2|S|\le k-2. If the xx-component of (Gxy)S(G-xy)-S contains a vertex other than xx, deleting S{x}S\cup\{x\} disconnects that vertex from yy, contrary to [F1].

F1
1.3

If that xx-component is the singleton {x}\{x\}, then NG(x)S{y}N_G(x)\subseteq S\cup\{y\}, so NG(x)k1|N_G(x)|\le k-1; deleting NG(x)N_G(x) leaves the isolated xx and another vertex because V(G)k+1|V(G)|\ge k+1, again contradicting [F1].

F1F2
1.4

Thus every xx-yy separator in GxyG-xy has size at least k1k-1. By [L1], GxyG-xy contains k1k-1 internally disjoint xx-yy paths, and adjoining xyxy gives kk such paths in GG.

L1
1.5

Conversely, suppose every pair has kk internally disjoint paths. If S<k|S|<k and x,ySx,y\notin S, at most S|S| of those paths have an internal vertex in SS, so one survives in GSG-S; as at least two vertices survive, no such SS is a vertex cut.

F1
2.1

Taking S=S=\varnothing in step 1.5 gives connectivity, and then [F1] makes the absence of cuts of size below kk exactly κ(G)k\kappa(G)\ge k; steps 1.1--1.4 prove the converse direction.

F1step 1.1step 1.4step 1.5

5 · Examples, counterexamples and false statements

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