then $$j+f_j\big(y_\uparrow(\ud{x})\big)=
\varphi^{-1}\big(\varphi(j+\lim_k\overline{\varphi}_{n_k-1}(\sigma\ud{x}))\big)=
j+y_\downarrow(\sigma\ud{x})\,.$$ Similar proofs for the other cases.

\[lem2.4quatro\] Let $\ud{x}\in\tA^{\Z_+}$.\
1) If $\{\overline{\varphi}_n(\ud{x})\}$ has two cluster points, and if $y\in \big(y_\uparrow(\ud{x}),
y_\downarrow(\ud{x})\big)$, then $y\in X\backslash S$, $\i(y)=\ud{x}$ and $y$ has no $\varphi$-expansion.\
Let $x\in X\backslash S$ and set $\ud{x}:=\i(x)$.\
2) If $\lim_n\overline{\varphi}_n(\ud{x})=y_\uparrow(\ud{x})$ and if $y\in (y_\uparrow(\ud{x}),
x)$, then $y\in X\backslash S$, $\i(y)=\ud{x}$ and $y$ has no $\varphi$-expansion.\
3) If $\lim_n\overline{\varphi}_n(\ud{x})=y_\downarrow(\ud{x})$ and if $y\in(x,y_\downarrow(\ud{x}))$, then $y\in X\backslash S$, $\i(y)=\ud{x}$ and $y$ has no $\varphi$-expansion.

Suppose that $y_\uparrow(\ud{x})<y< y_\downarrow(\ud{x})$. Then $y\in I_{x_0}$ and $\i_0(y)=x_0$. From Lemma \[lem2.4ter\] $$y_\uparrow(\sigma\ud{x})<Ty<y_\downarrow(\sigma\ud{x})\quad
\text{if $\delta(x_0)=1$}\,,$$ and $$y_\downarrow(\sigma\ud{x})>Ty>y_\uparrow(\sigma\ud{x})\quad
\text{if $\delta(x_0)=-1$}\,.$$ Iterating this argument we prove that $T^ny\in I_{x_n}$ and $\i_n(y)=x_n$ for all $n\geq 1$. Suppose that $y$ has a $\varphi$-expansion, $y=\overline{\varphi}_\infty(\ud{x}^\prime)$. If $\ud{x}^\p\prec\ud{x}$, then by Lemma \[lem2.3\] $\ophi(\ud{x}^\p)\leq y_\uparrow(\ud{x})$ and if $\ud{x}\prec
\ud{x}^\p$, then by Lemma \[lem2.3\] $y_\downarrow(\ud{x})\leq
\ophi(\ud{x}^\p)$, which leads to a contradiction. Similar proofs in cases 2 and 3.

\[lem2.4quinto\] Let $\ud{x}^\prime\in\tA^{\Z_+}$ and $x\in X\backslash S$. Then $$\text{$y_\downarrow(\ud{x}^\prime)<x$ $\implies$
$\ud{x}^\prime\preceq \i(x)$}
\quad\text{and}\quad\text{$x<y_\uparrow(\ud{x}^\prime)$ $\implies$
$\i(x)\preceq \ud{x}^\prime$.}$$

Suppose that $y_\downarrow(\ud{x}^\prime)<x$ and $y_\downarrow(\ud{x}^\prime)$ is a cluster point. Either $x^\prime_0<\i_0(x)$ or $x^\prime_0=\i_0(x)$ and by Lemma \[lem2.4ter\] $$y_\downarrow(\sigma\ud{x}^\prime) <Tx\quad\text{if
$\delta(x_0^\prime)=1$,}$$ or $$y_\uparrow(\sigma\ud{x}^\prime)
>Tx\quad\text{if $\delta(x_0^\prime)=-1$.}$$ Since $y_\downarrow(\sigma\ud{x}^\prime)$ or $y_\uparrow(\sigma\ud{x}^\prime)$ is a cluster point we can repeat the argument and 