and $E$-CNR, then it is also $(E,m)$-NS.

\(A) Fix two points $x,y\in\ball_L(u)$ with $|x-y|\ge L_{k-1}^{1+\rho}$. By assumption,

1.  either $\ball_{L_k}(u)$ does not contain any $(E,m)$-S ball of radius $L_{k-1}$,

2.  or there is a ball $\ball_{L_{k-1}}(w)\subset\ball_{L_k}(u)$ such that any ball $\ball_{L_{k-1}}(v)\subset\ball_{L_k}(u)$ with $d(v, w)> 2L_{k-1}$ is $(E,m)$-NS.

To treat both cases with one argument, in the simpler case (1) set, formally, $w=y$ (although no exclusion is actually necessary). Set \[eq:r.r.IID\] r’ &= (d(u, w) - L\_[k-1]{}-1, 0),\
r” &= ((y, w) - L\_[k-1]{}-1, 0). Assume first that $r', r'' \ge L_{k-1}$. By triangle inequality, \[eq:r.r\] r’ + r” |x-y| - 2L\_[k-1]{} L\_[k-1]{}\^[1+]{} - 2L\_[k-1]{}.

Consider the function $f: \ball\times\ball \to \DR_+$ defined by $
 f(x', x'') = |G_{\ball_{L_k}}(x', x''; E)|.
$ By construction, $f$ is $(L_{k-1}, q_k)$-subharmonic both in $x'\in\ball_{r'}(x)$ and in $x''\in\ball_{r''}(y)$, with $
q \le e^{-\gamma(m, L_{k-1})},
$ since $\ball_{L_k}(u)$ is $E$-NR, and all balls of radius $L_{k-1}$ both in $\ball_{r'}(x)$ and in $\ball_{r''}(y)$ are $(E,m)$-NS (being disjoint from $\ball_{L_{k-1}(w)}$). Collecting the assertion (B) of Lemma \[lem:cond.SubH.IID\], Lemma \[lem:BiSubH.IID\] and inequality , one can write, with the convention $-\ln 0 = +\infty$: \[eq:proof.lem:few.nloc.and.few.R.imply.nloc.IID\] -f(x,y) & -\
& m (1+ L\_[k-1]{}\^[-1/8]{} ) ( R -2L\_[k-1]{} ) - L\_[k-1]{}\^\
& 