words $u$ and $v$ and $\ell \in \N$ such that $q = uv, p = (uv)^\ell u$ and $F(q) = vu$. This implies that $vu = F(u)F(v)$ and we can apply the induction hypothesis as $|uv| < |pz|$. Therefore, there are $x$ and $s,r \in \N$ such that $v = x\bigl(F(x)x\bigr)^s$ and $u = \bigl(F(x)x\bigr)^tF(x)$. Putting this altogether we obtain $$\begin{aligned}
        q & = uv =\bigl(F(x)x\bigr)^tF(x)x\bigl(F(x)x\bigr)^s = \bigl(F(x)x\bigr)^{t+s+1}, \\
        p & = (uv)^\ell u = \bigl(F(x)x\bigr)^jF(x), \quad \text{with} \ j={\ell(t+s+1) + t} \\
        z & = F(p)q =x \bigl(F(x)x\bigr)^i,  \quad \text{with} \ i=\ell(t+s+1) + 2t + s + 1.
    \end{aligned}$$

To prove Item $(ii)$, we apply on $zp = pF(z)$. We have $z = uv, F(z) = vu$ and $p = (uv)^\ell u$ for some words $u$ and $v$ and $\ell \in \N$. It follows that $vu = F(u)F(v)$ and so, by Item $(i)$, there is $x$ such that $v = x\bigl(F(x)x\bigr)^i$ and $u = \bigl(F(x)x\bigr)^jF(x)$ for some $i,j \in \N$. Using all these equations we finish the proof by stating that $$z = uv = \bigl(F(x)x\bigr)^jF(x)x\bigl(F(x)x\bigr)^i = \bigl(F(x)x\bigr)^{j+i+1} \quad \text{and} \quad p = (uv)^\ell u = \bigl(F(x)x\bigr)^{\ell(j+i+1) + j}F(x). \qedhere$$

Derivated words of Sturmian preimages {#sec:der_wo_St_preim}
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In this 