\overline{\varphi}_n(x_0,\ldots,x_{n-1}+t)\,.$$ Suppose that $\delta(x_0\cdots x_{n-1})=-1$. Then it is decreasing, and for any $m$ $$\begin{aligned}
\overline{\varphi}_{n+m}(x_0,\ldots,x_{n+m-1}) &=
\overline{\varphi}_n(x_0,\ldots,x_{n-1}+
\overline{\varphi}_m(x_n,\ldots,x_{n+m-1}))\\
&\leq \overline{\varphi}_n(x_0,\ldots,x_{n-1})\,.\end{aligned}$$ In particular the subsequence $\{\overline{\varphi}_n(\ud{x})\}_n$ of all $n$ such that $\delta(x_0\cdots x_{n-1})=-1$ is decreasing with limit[^5] $y_\downarrow(\ud{x})$. When there is no $n$ such that $\delta(x_0\cdots x_{n-1})=-1$, we set $y_\downarrow(\ud{x}):=a_{x_0+1}$. Similarly, the subsequence $\{\overline{\varphi}_n(\ud{x})\}_n$ of all $n$ such that $\delta(x_0\cdots x_{n-1})=1$ is increasing with limit $y_\uparrow(\ud{x})\leq y_\downarrow(\ud{x})$. When there is no $n$ such that $\delta(x_0\cdots x_{n-1})=1$, we set $y_\uparrow(\ud{x}):=a_{x_0}$. Since any $\overline{\varphi}_n(\ud{x})$ appears in one of these sequences, there are at most two cluster points for $\{\overline{\varphi}_n(\ud{x})\}_n$.

Let $x\in X\backslash S$; $x=\varphi(\varphi^{-1}(x))$ and by $$\begin{aligned}
\label{identity}
x&=\varphi(\i_0(x)+Tx)=
\varphi(\i_0(x)+\varphi(\varphi^{-1}(Tx)))=
\varphi(\i_0(x)+\varphi(\i_1(x)+T^2x))=\cdots \nonumber\\
&=\varphi\big(\i_0(x)+\varphi(\i_1(x)+\ldots
+\varphi(\i_{n-1}(x)+T^{n}x))\big)\,.\end{aligned}$$ By monotonicity $$\label{id1}
\big(\text{$x\in X\backslash S$ and
$\delta(\i_0(x)\cdots\i_{n-1}(x))=-1$}\big)\implies
\overline{\varphi}_n(\i_0(x),\cdots,\i_{n-1}(x))\geq x\,,$$ and $$\label{id2}
\big(\text{$x\in X\backslash S$ and
$\delta(\i_0(x)\cdots\i_{n-1}(x))=1$}\big)\implies
\overline{\varphi}_n(\i_0(x),\cdots,\i_{n-1}(x))\leq x\,.$$ The inequalities of Lemma \[lem2.2\] follow from , and $\overline{\varphi}(\i_0(x)+t)\in [a_{x_0},
a_{x_0+1}]$.

Suppose that the $\varphi$-expansion is valid and that $\overline{\varphi}_\infty(\ud{x})=y\in X\backslash S$. We prove that $\ud{x}=\i(y)$. By hypothesis $y\in I_{x_0}$; using and the fact that $I_{x_0}$ is open, we can write $$y=\overline{\varphi}\big(x_0+\overline{\varphi}(x_1
+\overline{\varphi}(x_{2}+\ldots))\big)
=\varphi\big(x_0+\overline{\varphi}(x_1
+\overline{\varphi}(x_{2}+\ldots))\big)\,.$$ This implies that $$\varphi^{-1}(y)=\i_0(y)+Ty= x_0+ \overline{\varphi}(x_1
+\overline{\varphi}(x_{2}+\ldots))\,.$$ Since $Ty \in X\backslash S$, we can iterate this argument.

\[lem2.3\] Let $\ud{x},\ud{x}^\prime\in\tA^{\Z_+}$ and $\ud{x}\preceq\ud{x}^\prime$. Then any 