= Solution
On a <tree>, every finite connected <induced subgraph> is itself a <tree>, and its only <spanning tree> contains all its edges. Thus the <free uniform spanning forest> is deterministic:
$$
\mathrm{FSF}=\delta_E.
$$
Suppose for a contradiction that $G$ is a <transient graph>. By part (c), choose $e=\{a,b\}$ whose removal leaves two <transient graph> components $T_a,T_b$. Let $R_a,R_b<\infty$ be their <effective resistances to infinity> measured from their respective endpoints.
In a wired finite exhaustion, the edge $e$ of resistance one is in parallel with the route from $a$ through $T_a$ to the wired boundary and back through $T_b$ to $b$. The latter route has resistance $R_{a,k}+R_{b,k}$, converging to $R_a+R_b$. Consequently the limiting wired <effective resistance> is
$$
R_{\mathrm{eff}}^{\mathrm w}(a,b)=\frac{R_a+R_b}{1+R_a+R_b}<1.
$$
The <edge-inclusion formula for a uniform spanning tree>, the one-edge case of the <transfer-current theorem>, states that an edge of conductance $c_e$ is present with probability $c_eR_{\mathrm{eff}}(a,b)$. Passing to the wired limit and using $c_e=1$ gives
$$
\boxed{\mathbb P_{\mathrm{WSF}}(e\notin F)=\frac1{1+R_a+R_b}>0.}
$$
But under the <free uniform spanning forest>, $e$ is present with probability one. This contradicts $\mathrm{WSF}=\mathrm{FSF}$. Therefore \b[the tree is recurrent]. Combined with part (b), this characterizes equality of the free and wired <uniform spanning forests> on locally finite unweighted <trees>.
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