orbits. The orbit of $0$ is coded by $\ud{u}^{1,\beta}=(1)^\infty$, that is $\ud{u}^{1,\beta}_j=1$ for all $j\geq 0$. Similarly, if $\gamma=0$ the orbit of $1$ is coded by $\ud{v}^{\alpha,\beta}=(k-2)^\infty$. [*We always assume that $\alpha\in[0,1]$, $\gamma\in[0,1]$ and $\beta\geq1 $*]{}.

\[lemelementary\] The equation $$y=\oph^\ab(x_k+t)\,, \quad y\in[0,1]$$ can be solved uniquely if $y\not\in S_0$, and its solution is $x_k=\i_0(y)$ and $t=T_\ab(y)\in(0,1)$.\
If $y<y^\prime$, then the solutions of the equations $$y=\oph^\ab(x_k+t)\quad\text{and}\quad
y^\prime=\oph^\ab(x^\prime_k+t^\p)$$ are such that either $x_k=x^\prime_k$ and $T_\ab(y^\prime)-T_\ab(y)=\beta(y^\prime-y)$, or $x_k<x^\prime_k$.

The proof is elementary. It suffices to notice that $$y\not\in S_0\implies y= \varphi^\ab(x_k+t)\,.$$ The second statement follows by monotonicity.

\[pro3.4\] Let $0\leq\alpha< 1$ and assume that the $\varphi$-expansion is valid. The following assertions are equivalent.\
1) There is a unique solution ($\ud{u}=\ud{u}^\ab$) of the equations $$\label{eqalpha}
\ophi^\ab(\ud{u})=0\quad\text{and}\quad\ophi^\ab(\sigma\ud{u})=\alpha\,.$$ 2) The orbit of $0$ is not periodic or $x=0$ is a fixed point of $T_\ab$.\
3) $\ud{u}^\ab$ is not periodic or $\ud{u}^\ab=\ud{0}$, where $\ud{0}$ is the string $\ud{x}$ with $x_j=0$ $\forall j\geq 0$.

\[pro3.4bis\] Let $0<\gamma\leq 1$ and assume that the $\varphi$-expansion is valid. The following assertions are equivalent.\
1) There is a unique solution ($\ud{v}=\ud{v}^\ab$) of the equations $$\label{eqbeta}
\ophi^\ab(\ud{v})=1\quad\text{and}\quad\ophi^\ab(\sigma\ud{v})=\gamma\,.$$ 2) The orbit of $1$ is not periodic or $x=1$ is a fixed point of $T_\ab$.\
