LemmaStatement: 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.

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 $s$ - $t$ path exists for a feasible flow $f$ , let $S$ be the vertices reachable from $s$ in its residual network. Then $S$ is an $s$ - $t$ cut and $c (S) = ∣ f ∣$ .

Facts & Assumptions

Given: A feasible flow $f$ whose labelled residual network has no $s$ - $t$ 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 $∣ f ∣$ 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 $t$ is not reachable, $S$ contains $s$ and excludes $t$ , so it is an $s$ - $t$ cut; an original arc leaving $S$ has no forward residual copy, hence is saturated.

1.2

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

1.3

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

2.1

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

step 1.1step 1.2step 1.3∎

Depends on

Used by

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

Dependency tree · two levels

6 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 flows and cuts (standard reference, not scraped)