we have $$\begin{aligned}
J^{n_k+p_k-i}(c_i)d_C(c_i)d_S(c_i)
\geq K^{-1} J^{n_k+p_k-i}(\tilde c_{i-n_k-1})d_C(\tilde c_{i-n_k-1})d_S(\tilde c_{i-n_k-1}) \geq
\sigma^2L^{-5\alpha(i-n_k-1)}\end{aligned}$$ where the last inequality is obtained by using (G1), Lemma \[add-lem1-s2.3B\](a) and (G3) for $\tilde c$. Summing this estimate over all $i$ and combining the result with (\[f1-deep\]), $$\label{sublem2}
|(f^{n_k+p_k})'c_0|^{-1} \Theta_{n_k}\leq
 \sigma^{-1} (d_C(c_{n_k}))^{- \frac{16 \alpha}{\lambda}} + \sigma^{-2}L^{6\alpha
p_k} \leq (d_C(c_{n_k}))^{-\frac{18 \alpha}{\lambda}}.$$ Here for the last inequality we use $\sigma \gg \delta$ and $L^{6\alpha p_k}\leq|d_C(c_{n_k})|^{-\frac{12\alpha}{\lambda}}$ from Lemma \[reclem1\](a).

[*Step 2 (Estimate for free segments):*]{} By definition, $$\frac{\Theta_0}{J^{\nu}(c_0)} = \sum_{i  \in [0, \nu-1] \setminus
(\cup [n_k, n_k + p-1])}\frac{1}{J^{\nu - i}(c_i)d_C(c_i)
d_S(c_i)}.$$ Here we can not simply use (G3) for $d_S(c_i)$ in proving (\[quatro\]). We observe, instead, that either we have $v_i \not
\in S_{\sigma}$, for which $d_S(v_i)
> \sigma$; or $c_i \in S_{\sigma}$ for which we have

$$J^{\nu - i}(c_i) =J^{\nu-i+1}(c_{i+1})J(c_i)\geq  
K^{-1} L^{\frac{\lambda}{3} (\nu - i + 1)} (d_S(c_i))^{-1}.$$ It then follows, by using $d_C(c_i) > \delta$, that $$\label{deep}
\frac{\Theta_0}{J^{\nu}(c_0)} \leq   \sum_{i  \in [0, \nu-1]
\setminus (\cup [n_k, n_k + p-1])} K L^{-\frac{\lambda}{3}
(\nu-i)} (\sigma \delta)^{-1} \leq \frac{1}{\sigma \delta}.$$ This estimate is unfortunately not good enough for (\[quatro\]). To obtained (\[quatro\]), we need to use $C_{\delta^{\frac{1}{20}}}$ in the place of $C_{\delta}$ to define a new bound/free structure for each 