letters). Moreover, if $w$ is right special, $w'$ is right special as well.

The morphisms $\varphi_a$ and $\varphi_b$ are conjugate, that is, $0\varphi_a(x)=\varphi_b(x)0$ for each word $x$. This means that for any prefix $u'_0u'_1u'_2 \cdots u'_k$ of $\uu'$ we have $0\varphi_a(u'_0u'_1u'_2 \cdots u'_k) = \varphi_b(u'_0u'_1u'_2 \cdots u'_k)0$. As this holds true for each $k$, we obtain $0\uu = 0\varphi_a(\uu') =\varphi_b(\uu')$.

Denote ${\bf v} = v_0v_1v_2\cdots  = 0u_0u_1u_2 \cdots$. We have $v_i = u_{i-1}$ for each $i\geq 1$. Let $w$ be a nonempty prefix of $\uu$ and $(i_n)$ be the increasing sequence of its occurrences in $\uu$. Note that $w$ starts with the letter $1$. This letter is in $\uu$ surrounded by $0$’s. Thus the sequence $(i_n) $ is also the sequence of occurrences of $0w$ in ${\bf v}$ and thus $d_{\bf v}({0w}) = d_{\bf u}({w})$. It follows that $$\Der(\uu)=\{d_{\bf v}(v) \colon v \text{ is a prefix of $\bf v$ and } |v|>1\}\,.$$ We finish the proof by applying Theorem \[thm:preimages\_fi\_b\] and Proposition \[prop:der\_of\_images\_fi\_b\] to the word ${\bf v} =\varphi_b(\uu')$.

The only case which is not treated by Theorems \[thm:preimages\_fi\_b\] and \[thm:preimages\_fi\_a\], namely the case when $\uu = \varphi_a(\uu')$ and $\uu$ begins with $0$, can be translated into one of the 