> 0$ is an occurrence of $w$ in $\uu$. This means that $\uu_{[0,i)}w$ is a prefix of $\uu$ and hence, by Lemma \[lem:unique\_preimage\_under\_fi\_b\] (note that $w$ begins with $0$), there must be $p'$ a prefix of $\uu'$ such that $\varphi_b(p') = \uu_{[0,i)}$ and $p'w'$ is a prefix of $\uu'$. Since $|p'|$ is an occurrence of $w'$ in $\uu'$, there is $j \in \N$ such that $p' = r'_{s_0}r'_{s_1}\cdots r'_{s_j}$. It follows that $$\uu_{[0,i)} = \varphi_b(r'_{s_0}r'_{s_1}\cdots r'_{s_j}) = r_{s_0}r_{s_1}\cdots r_{s_j}$$ and $i = |r_{s_0}r_{s_1}\cdots r_{s_j}|$.

So, again as in the previous proof, we have shown that the numbers $0$ and $|r_{s_0}r_{s_1}\cdots r_{s_j}|, j = 0,1,\ldots$, are all occurrences of $w$ in $\uu$. It follows that $r_0$ and $r_1$ are the two return words to $w$ in $\uu$ and $$d_{\uu}(w) = s_0s_1s_2 \cdots =  d_{\uu'}(w'). \qedhere$$

\[thm:preimages\_fi\_a\] Let $\uu$ and $\uu'$ be Sturmian words such that $\uu$ starts with the letter $1$ and $\uu = \varphi_{a}(\uu')$. Then $\uu'$ starts with $1$ and the sets of their derivated words coincide, i.e., $$\Der(\uu) = \Der(\uu')\,.$$ In particular, for any prefix $w$ of $\uu$ there exists a prefix $w'$ of $\uu'$ such that $|w'| \leq |w|$ and $d_{\uu}(w) = d_{\uu'}(w')$ (up to a permutation of 