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Orient the unique path from $s$ to $t$ in a uniform spanning tree $T$, and put $j_{xy}^T=1$ when it uses the edge from $x$ to $y$, $-1$ when it uses the reverse direction, and zero otherwise. Then the proposed current is $i_{xy}=\mathbb{E}{j}_{xy}^T$; set it to zero on nonedges. It is antisymmetric.

For each tree path, every intermediate graph vertex has one incoming and one outgoing path edge, while the source has one outgoing edge and the sink one incoming edge. HenceTaking expectations givesthe Kirchhoff node law for a unit flow.

To establish the Kirchhoff cycle law, it is not enough to check it separately for each tree path: those path currents need not satisfy the voltage law on the original graph. Instead, let $\mathcal{T}$ be all spanning trees and let ${\mathcal{F}}_{st}$ be all spanning forests with two components, one containing $s$ and the other $t$. For $F\in {\mathcal{F}}_{st}$ denote the source component by $C_s(F)$.

For an existing edge $xy$, deleting it from a tree whose $s$-$t$ path uses $x\to y$ produces precisely a forest $F\in {\mathcal{F}}_{st}$ with $x\in C_s(F)$ and $y\notin C_s(F)$. Conversely, adding $xy$ to any such forest produces exactly that tree and path orientation. This is a bijection. Applying it to both orientations givesDefine $h(x)=|\mathcal{T}{|}^{-1}{\sum }_F{1}_{x\in C_s(F)}$. Then $i_{xy}=h(x)-h(y)$ on every edge. Since the resistances are one, this is Ohm's law, and the voltage drops telescope around every oriented cycle:Together with the node law, this proves that the mean tree-path current is the electrical unit current. This argument is the mean spanning-tree path current identity.

For clarity, these laws determine the current uniquely. The difference of two solutions is a potential gradient $k(x)-k(y)$ with zero divergence. Multiplying its divergence by $k(x)$ and summing over graph vertices gives ${\sum }_{\{x,y\}}(k(x)-k(y){)}^2=0$. Thus every difference current is zero.

Fix an edge $e=\{u,v\}$ and orient it from $u$ to $v$. For each spanning tree $T$, let ${\gamma }_T$ be its unique graph path from $u$ to $v$. Assign a signed unit graph path flow $j_T$ to oriented edges: value $+1$ in the direction used by ${\gamma }_T$, value $-1$ on the reversed orientation, and zero otherwise. Define its meanEach $j_T$ has net divergence one at $u$, minus one at $v$ and zero elsewhere, so $J$ is a unit flow. The mean spanning-tree path current also satisfies the Kirchhoff cycle law. Although its verification is permitted to be omitted, a short counting argument makes it explicit. Let ${\mathcal{F}}_{uv}$ be the two-component spanning forests separating $u$ and $v$, write $A_F$ for the component containing $u$, and let $N_T$ be the number of spanning trees. Deleting an edge used by the oriented tree graph path gives one such forest. Conversely adjoining an edge across its two components gives a unique tree with that edge on its $u$-to-$v$ graph path. ConsequentlyThe gradient form proves the Kirchhoff cycle law directly and is Ohm's law for unit edge resistance. The Kirchhoff node law then gives $Lh={\delta }_u-{\delta }_v$, and the electrical unit flow is unique: the difference between two such voltages is a harmonic function on a graph and hence constant on the connected finite graph.

If $e\in T$, its single-edge route is the unique tree graph path from $u$ to $v$, so $j_T(u,v)=1$. If $e\notin T$, that tree graph path does not use $e$, so $j_T(u,v)=0$. ThereforeBy Ohm's law, $J(u,v)=h(u)-h(v)$; for a unit terminal current, this voltage difference is exactly the effective resistance between the terminals. We have proved the requested edge-inclusion formula for a uniform spanning tree:Now use linearity of expectation and the fact that every spanning tree has exactly $n-1$ edges. This gives Foster's theorem:The sums here use unoriented edges exactly once. For unit conductance this is the displayed identity; with conductances $c_e$ the inclusion probability and summand become $c_e{R}_{\mathrm{eff}}(e)$.

A two-component spanning forest of a finite connected graph is an acyclic spanning subgraph with exactly two connected components of a graph. It has $|V|-2$ edges. For distinct terminals $a,z$, a separating two-component forest places them in different components. Deleting an edge of the $a$-to-$z$ graph path of a spanning tree produces such a forest; adjoining any original edge across its cut reverses this operation. This is the counting correspondence behind the mean spanning-tree path current.