= Solution
Use $T_y=\inf\{n\ge0:X_n=y\}$ and the positive return time $T_x^+=\inf\{n\ge1:X_n=x\}$. The walk has transitions $p_{uv}=c_{uv}/d(u)$, $d(u)=\sum_vc_{uv}$; its transition matrix defines a <reversible Markov chain>, although the <transition probabilities> need not be symmetric.
The voltage $h(u)=P_u(T_x<T_y)$ equals one at $x$, zero at $y$, and is a <harmonic function on a graph> away from the terminals: $h(u)=\sum_vp_{uv}h(v)$. For this unit <electric potential difference> the <electric current> is
$$
I=\sum_vc_{xv}(1-h(v))=\frac1{R(x,y)}.
$$
The <first-step analysis> therefore proves the <return probability from effective resistance> identity:
$$
P_x(T_x^+<T_y)=\sum_vp_{xv}h(v)
=1-\frac{I}{d(x)}
=\boxed{1-\frac1{d(x)R(x,y)}}.
$$
Finite irreducibility ensures eventual hitting of one of the terminals, so no missing escape event appears in this finite-network calculation.
For an infinite locally finite <connected graph>, exhaust it by finite sets and wire their exteriors to one sink. The event of exiting before the first return decreases to the event of never returning. Thus
$$
P_o(T_o^+=\infty)=\frac1{d(o)R(o,\infty)},
$$
with $1/\infty=0$. In particular <recurrence> is equivalent to infinite <effective resistance to infinity>.
After deleting an <edge>, consider any infinite component and choose a root $o$ away from that <edge>'s endpoints. Its <vertex degree> $d(o)$ is unchanged. On every finite wired exhaustion the <Rayleigh monotonicity principle> makes the resistance no smaller; the <limit of a sequence> does too. A <recurrent> original network consequently leaves a <recurrent> component. Finite components give <recurrent> walks as well, with an isolated <graph vertex> viewed as an <absorbing state>. This proves that <edge deletion preserves recurrence>. The phrase “no less <recurrent>” refers to this <recurrence> comparison: at a deleted-edge endpoint the <vertex degree> factor also changes, so monotonicity of its first-return <probability> alone should not be inferred.
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