 Solution

Consider a finite directed flow network with source $s$, sink $t$, and nonnegative finite flow network edge capacities $c_e$. A feasible flow assigns $0\le f_e\le c_e$ to each arc and satisfies flow conservation at every vertex except $s,t$. Its value $|f|$ is net outflow from $s$, equivalently net inflow to $t$. A maximal flow here means a flow of greatest value, also called a maximum flow. A source–sink cut of a flow network is a partition $(S,S^c)$ with $s\in S$, $t\notin S$; its cut capacity is $c(S,S^c)={\sum }_{u\in S,v\notin S}{c}_{uv}$, summing directed arcs from $S$ to its complement.

The max-flow min-cut theorem statesFor any feasible flow and cut, summing flow conservation over vertices in $S$ cancels internal arcs and gives $|f|={\sum }_{S\to S^c}{f}_e-{\sum }_{S^c\to S}{f}_e\le c(S,S^c)$. This proves the upper bound by every cut.

For the reverse bound, a maximum exists: the feasible set is a nonempty closed subset of the bounded product ${\prod }_e[0,c_e]$, hence compact, and $|f|$ is continuous. Choose a maximizing flow $f$. Its residual network has a forward arc for unused capacity $c_e-f_e$ and a reverse arc for cancellable flow $f_e$, retaining original-arc identities if necessary. If a residual path joined $s$ to $t$, the minimum of its positive residual capacities would be positive. Increasing forward flows and cancelling reverse flows by this amount would preserve all constraints while increasing $|f|$, a contradiction.

Let $S$ consist of vertices reachable from $s$ through positive residual arcs. Then $t\notin S$. An original arc from $S$ to $S^c$ must be saturated, since otherwise its forward residual arc would make its endpoint reachable. An original arc from $S^c$ to $S$ must carry zero flow, since a positive reverse residual arc would again extend reachability. ThusThe upper bound and this equality prove the theorem and exhibit a residual reachability certificate for maximum flow. The argument works for real capacities and does not assume that an arbitrary augmenting-path algorithm terminates in finitely many steps.