where the last inequality is from (R)$_{n,c}$.

To prove (\[inessential\]), we let $\beta_1$ be the smallest free return time $\leq \mu-1$ such that $$\label{deep-add2}
\sum_{\stackrel{\beta_1 +1\leq j\leq\mu}{\text{free return}}}2\log
d_C(c_j) > \log d_C(c_{\beta_1}).$$ We claim that no deep free return occurs during the period $[\beta_1+1,\mu]$. This is because if $i' \in [\beta_1 + 1,
\mu]$ is a deep return, then we must have $$\sum_{\stackrel{\beta_1 +1\leq j\leq\mu}{\text{free return}}}2\log
d_C(c_j) \leq \sum_{\stackrel{\beta_1 +1\leq j\leq
i'}{\text{free return}}}2\log d_C(c_j) \leq \log
d_C(c_{\beta_1}),$$ contradicting (\[deep-add2\]). If $\beta_1$ is a deep return, we are done. Otherwise we find a $\beta_2 < \beta_1$ so that $$\label{deep-add1}
\sum_{\stackrel{\beta_2+1 \leq j\leq \beta_1}{\text{free
return}}}2\log d_C(c_j) > \log d_C(c_{\beta_2}),$$ and so on. This process will end at a deep free return time, which we denote as $\beta_q : = i(\mu)$. (\[inessential\]) follows from adding (\[deep-add2\]), (\[deep-add1\]) and so on up to the time for $\beta_q = i(\mu)$.

We first prove (G1)$_{n,c}$. Let $0\leq i < j \leq
n+1$. Observe that the bound periods for all returns to $C_{\delta}$ for the orbit of $c_0$ up to time $n$ is $\leq \frac{2
\alpha}{\lambda}n \ll n$. This follows from (\[prop-s3.1\]) and Lemma \[reclem1\](a). Hence, it is possible to introduce the bound-free structure starting from $c_i$ 