LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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

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 $t$ yields a unit $s$ - $t$ path flow after subtracting one on its arcs.

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 $c(a)\le1$ , 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∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 32 results over 17 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

MIT 6.854, Flow notes (standard reference, not scraped)