ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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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.
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A four-vertex integral network with an exhibited maximum flow and minimum cut of value five

Example

Let the vertices be $s,u,v,t$ , with labelled arcs and capacities $s\!\to u:3,\quad s\!\to v:2,\quad u\!\to v:1,\quad u\!\to t:2,\quad v\!\to t:3.$ The flow with values $3,2,1,2,3$ in that order has value five and is maximum.

Facts & Assumptions

Given: The displayed finite integral network and flow $f$ .

[L1]

Finite integral max-flow min-cut equates a maximum flow value and a minimum cut capacity (Ford-Fulkerson terminates for finite integer capacities and proves max-flow min-cut with an integral maximum flow).

Verification

Verification technique: direct.

1.1

At $u$ , incoming flow $3$ equals outgoing flow $1+2$ , and at $v$ , incoming flow $2+1$ equals outgoing flow $3$ ; every arc value is within its capacity.

given

1.2

The flow value is $3+2=5$ , while the cut $S=\{s\}$ has capacity $3+2=5$ .

1.3

By [L1], the exhibited feasible flow and cut of common value five are respectively maximum and minimum.

2.1

This network therefore has max-flow value and min-cut capacity both equal to five.

step 1.1step 1.2step 1.3∎

Depends on

Used by

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