by Lemma \[lem2.4\], $\i(Ty)\succeq \i(z_m)=\sigma^m\ud{x}^\prime$; $\i_0(y)=x^\prime_{m-1}$ when $y$ is close to $a_{x^\prime_{m-1}}$, so that $$\lim_{y\downarrow
a_{x^\prime_{m-1}}}\i(y)=\ud{u}^{x^\prime_{m-1}}\succeq
\sigma^{m-1}\ud{x}^\prime\,,$$ which is a contradiction. Similarly we exclude the cases $\delta(x^\prime_{m-1})=1$ and $a_{x^\prime_{m-1}+1}=z_{m-1}$, $\delta(x^\prime_{m-1})=-1$ and $a_{x^\prime_{m-1}}=z_{m-1}$, $\delta(x^\prime_{m-1})=-1$ and $a_{x^\prime_{m-1}+1}=z_{m-1}$. Iterating this argument we get the existence of $z_0\in
X\backslash S$ with $\i(z_0)=\ud{x}^\prime$.

In the second case, $\lim_{n}\overline{\varphi}_n(\sigma^m\ud{x}^\prime)$ exists for all $m\geq 1$. Let $x:=\lim_{n}\overline{\varphi}_n(\ud{x}^\prime)$. Suppose that $x^\prime_0=j$, so that $\ud{u}^{j}\prec\ud{x}^\prime\prec
\ud{v}^{j}$. By Lemma \[lem2.4\] and definition of $\ud{u}^{j}$ and $\ud{v}^{j}$ there exist $z_1,z_2\in I_j$ such that $$z_1<x<z_2\quad\text{and}\quad\ud{u}^{j}\preceq\i(z_1)\prec
\ud{x}^\prime\prec\i(z_2)\preceq \ud{v}^{j}\,.$$ Therefore $a_j< x<a_{j+1}$, $\i_0(x)=x_0^\prime$ and $Tx=
\overline{\varphi}_\infty(\sigma\ud{x}^\prime)$ (Lemma \[lem2.4ter\]). Iterating this argument we get $\ud{x}^\prime=\i(x)$.

\[thm2.5\] Suppose that the $\varphi$-expansion is valid. Then

1.  $\Sigma:=\{\i(x)\in\tA^{\Z_+}\colon x\in X\backslash
    S\}=\{\ud{x}\in\tA^{\Z_+}\colon \ud{u}^{x_n}\prec
    \sigma^n\ud{x}\prec \ud{v}^{x_n}\quad\forall\,n\geq 0\}$.

2.  The map $\i\colon X\backslash S\ra \Sigma$ is bijective, $\overline{\varphi}_\infty\cirk\i={\rm id}$ and $\i\cirk\overline{\varphi}_\infty={\rm id}$.\
    Both maps $\i$ and $\overline{\varphi}_\infty$ are order-preserving.

3.  $\sigma(\Sigma)=\Sigma$ and $\overline{\varphi}_\infty(\sigma\ud{x})=T\overline{\varphi}_\infty(\ud{x})$ if $\ud{x}\in\Sigma$.

4.  If $\ud{x}\in\tA^{\Z_+}\backslash\Sigma$, then there exist $m\in\Z_+$ and $j\in\tA$ such that $\overline{\varphi}_\infty(\sigma^m\ud{x})=a_j$.

5.  $\forall n\geq 0\,,\,\forall j\in\tA\colon
    \quad\ud{u}^{u^j_n}\preceq \sigma^n\ud{u}^j\prec \ud{v}^{u^j_n}$ if $\delta(\ud{u}^j_0\cdots\ud{u}^j_{n-1})=1$ and $\ud{u}^{u^j_n}\prec \sigma^n\ud{u}^j\preceq \ud{v}^{v^j_n}$ if $\delta(\ud{u}^j_0\cdots\ud{u}^j_{n-1})=-1$.

6.  $\forall n\geq 0\,,\,\forall j\in\tA\colon
    \quad\ud{u}^{u^j_n}\preceq \sigma^n\ud{v}^j\prec \ud{v}^{u^j_n}$ if $\delta(\ud{v}^j_0\cdots\ud{v}^j_{n-1})=-1$ and $\ud{u}^{u^j_n}\prec \sigma^n\ud{v}^j\preceq \ud{v}^{v^j_n}$ if $\delta(\ud{u}^j_0\cdots\ud{u}^j_{n-1})=1$.

Let $x\in X\backslash S$. Clearly, by monotonicity, $$\ud{u}^{\i_k(x)}\preceq \sigma^k\i(x)\preceq \ud{v}^{\i_k(x)}\quad
\forall\;k\in\Z_+\,.$$ Suppose 