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In a finite unweighted connected graph, take distinct graph vertices $a\ne z$ and orient the unique graph path from $a$ to $z$ in a uniform spanning tree and assign signed unit current to its edges. Its mean over trees is the electrical unit flow from $a$ to $z$. The Kirchhoff node law follows by averaging graph path divergences. For the graph cycle law, let ${\mathcal{F}}_{az}$ be the two-component spanning forests separating the terminals and $A_F$ the component containing $a$. Deleting a tree-path edge and conversely adjoining an edge across a forest cut are inverse operations. Hence the mean signed current on an edge $u\to v$ isThis is the gradient of $h(u)=|\mathcal{T}(G){|}^{-1}{\sum }_F{1}_{A_F}(u)$, proving the graph cycle law. In particular $R_{\mathrm{eff}}(a,z)=|{\mathcal{F}}_{az}|/|\mathcal{T}(G)|$. For an oriented existing edge $e=(a,z)$, that edge is used by the tree graph path exactly when it belongs to the tree, and then has positive orientation. Thus its mean current is $\mathbb{P}(e\in T)$, giving the edge-inclusion formula for a uniform spanning tree by Ohm's law.

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.