holds if (R$)_{n,c}$ holds for every $c\in C$.

\[derive0\] If (R$)_{n,c}$ holds, then (G1$)_{n,c}$, (G2$)_{n,c}$ hold.

The first step is to show $$\label{prop-s3.1}
\prod_{\stackrel{N_0 < i \leq n}{\text{\rm free return }}} \ d_C(c_i)\geq L^{- \alpha n}.$$ We call a free return [*shallow*]{} if it is not a deep free return. Let $\mu\in(0,n)$ be a shallow free return time, and $i(\mu)$ be the largest deep free return time $< \mu$. We claim that $$\label{inessential}
\sum_{\stackrel{i(\mu) +1\leq j\leq\mu}{\text{free return}}} \log
d_C(c_j) \geq \log d_C(c_{i(\mu)}).$$ We finish the proof of (\[prop-s3.1\]) assuming (\[inessential\]). Let $\mu_1$ be the largest free shallow return time in $(0, n]$, and $i_1$ be the largest deep free return time $< \mu_1$. We then let $\mu_2$ be the largest shallow free return time $< i_1$, and $i_2$ be the largest deep free return time $<\mu_2$, and so on. We obtain a sequence of deep free return times $i_1 > i_2 \cdots
> i_q$, and we have $$\label{f4-s3.1}
\sum_{\stackrel{0\leq j\leq n}{\text{shallow return}}} \log
d_C(c_{j}) \geq \sum_{j=1}^q\log d_C (c_{i_j})\geq
\sum_{\stackrel{0\leq i\leq k}{\text{deep return}}}\log
d_C(c_{i})$$ where the first inequality is from (\[inessential\]). We then have $$\sum_{0\leq j\leq n\colon \text{free return}} \log d_C(c_{j}) \geq
2\sum_{\stackrel{0\leq j\leq n}{\text{free return}}} \log
d_C(c_{j})
  \geq \alpha n,$$ 