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Assume first that deleting an edge leaves two transient graph components. One component supports a finite-energy unit flow from its endpoint to infinity by the finite-energy flow criterion for transience. Extend that flow by zero across the deleted edge and throughout the other component. It remains a finite-energy unit flow on the whole tree, so the tree is a transient graph.

Conversely, root a transient tree at $o$ and take its finite-energy unit electrical flow to infinity. On a tree this flow can be chosen nonnegative in every direction away from the root: it is obtained as the limit of the currents to wired boundaries, each of which sends nonnegative current into descendant subtrees. At each nonroot vertex its incoming current equals the sum of its outgoing currents, by flow conservation.

If every vertex reached by positive current had exactly one positive-current child, the unit current would run along a single infinite ray. Every edge of that ray would carry current one, so its energy of a flow would be ${\sum }_{j\ge 1}1=\infty$, a contradiction. Thus there is a vertex $v$ with two positive-current child edges, say $(v,x)$ and $(v,y)$.

The restricted flow below $x$, rescaled by its incoming current, is a finite-energy unit flow from $x$ to infinity; hence the component below $x$ is a transient graph. After removing $(v,x)$, the other component contains $v$, the edge $(v,y)$, and the descendant subtree below $y$. Send unit current along $(v,y)$ and then use the rescaled descendant flow below $y$, with zero current elsewhere. Its energy of a flow is finite, so this component is a transient graph too. ThereforeUnit edge resistances are essential here: an infinite ray with sufficiently rapidly decreasing resistances can be transient without any such splitting edge.

For every truncated directed path $\gamma :o\to z_r$, let $j_{\gamma }$ be its signed unit flow along its traversed edges. Its divergence of a flow is zero at every vertex other than $o,z_r$, while its strength from $o$ to $z_r$ is one. The displayed expression in the question isIt is a nonnegative linear combination of such path flows. Hence it obeys flow conservation at every interior vertex, is antisymmetric on oppositely directed edges, and is supported on open edges. It is therefore a flow from $o$ to $z_r$ for every configuration $\omega$.

For a feasible flow $f$, each original directed edge $u\to v$ supplies a forward residual network edge of capacity $c_{uv}-f_{uv}$ and a backward one $v\to u$ of capacity $f_{uv}$; zero-capacity residual edges may be omitted. The forward edge permits an increase, while the backward edge permits cancellation of existing flow. In particular, backward residual capacity is the amount already sent, rather than unused original capacity. If both original orientations exist, keep their residual contributions separate, or add capacities for contributions with the same direction. An augmenting path uses these permitted changes while preserving flow conservation and the capacity constraints.

To see why cancellation matters, give all five directed edges $s\to a$, $s\to b$, $a\to b$, $a\to t$, $b\to t$ unit flow network edge capacities. Starting from zero flow, an augmenting path $s\to a\to b\to t$ carrying $1$ yieldsThe remaining unused original edges are $s\to b$ and $a\to t$, which do not form a source-to-sink graph path. However, the backward residual edge $b\to a$ has capacity $f_{ab}=1$. Consequently the residual network contains the augmenting path $s\to b\to a\to t$, with residual capacities $1,1,1$.

Augment by $1$: increase $f_{sb}$ and $f_{at}$ from $0$ to $1$, and decrease $f_{ab}$ from $1$ to $0$. The result isNow flow conservation is $1=0+1$ at $a$ and $1+0=1$ at $b$. The final flow sends one unit along each of $s\to a\to t$ and $s\to b\to t$. The cut of a flow network $S=\{s\}$ has cut capacity $1+1=2$, proving this is a maximum flow. Traversing $b\to a$ in the residual network revised the earlier routing decision; it did not send actual traffic along a nonexistent original edge.

In a finite flow network, a maximum flow exists because the feasible flow set contains the zero flow and is a compact set: the capacity constraints bound every coordinate, and they and flow conservation define a closed set. The strength of a flow is a continuous function, so the extreme value theorem gives an attained maximum. An augmenting path in the residual network would permit positive augmentation, contradicting maximality. The source-reachable vertices in the residual network therefore define a cut of a flow network whose outgoing original edges are saturated and incoming original edges carry zero flow. Flow conservation makes the strength of a flow equal to this cut capacity, proving the max-flow min-cut theorem.

Existence and strong duality for finite real flow network edge capacities do not imply finite termination for every choice of augmenting paths in the Ford-Fulkerson algorithm. With irrational capacities, some choices can continue forever. With integer capacities, Integrality of the Ford-Fulkerson algorithm instead forces finite termination; rational capacities reduce to that case by a common denominator. Choosing shortest augmenting paths as in the Edmonds–Karp algorithm avoids dependence on capacity magnitudes.^[ref]

A unit flow from $a$ to $b$ is an antisymmetric flow with net outward strength of a flow one at $a$, net inward strength one at $b$, and zero divergence of a flow elsewhere. For a unit flow to infinity, only the finite source has nonzero divergence.

An internal vertex capacity bounds the total flow through that vertex, equivalently its total inflow or outflow by flow conservation. Replace the vertex by an input and output vertex, direct incoming edges to the input and outgoing edges from the output, and connect input to output with an edge of the vertex capacity. This converts the constraint into an ordinary flow network edge capacity.