$$\ud{x}^+=(k-1,k-1,\ldots)\quad\text{and}\quad
\ud{x}^{-}=(0,0,\ldots)\,.$$

\[thm2.2\][[@P2]]{} A necessary and sufficient condition for a $\varphi$-expansion to be valid is that $S$ is dense in $[0,1]$. A sufficient condition is $\sup_t|\varphi^\prime(t)|<1$.

For each $j\in\tA$ we define (the limits are taken with $x\in
X\backslash S$) $$\label{2.12}
\ud{u}^j:=\lim_{x\downarrow a_{j}}\i(x)\quad\text{and}\quad
\ud{v}^j :=\lim_{x\uparrow a_{j+1}}\i(x)\,.$$ The strings $\ud{u}^j$ and $\ud{v}^j$ are called [virtual itineraries]{}. Notice that $\underline{v}^j
\prec\underline{u}^{j+1}$ since $v^j_0<u_0^{j+1}$. $$\label{2.12b}
\sigma^k\ud{u}^j=\sigma^k(\lim_{x\downarrow a_{j}}\i(x))=
\lim_{x\downarrow a_{j}}\sigma^k\i(x)=\lim_{x\downarrow
a_{j}}\i(T^kx)\qquad(x\in X\backslash S)\,.$$

\[prof2.2\] Suppose that $\ud{x}^\prime\in\tA^{\Z_+}$ verifies $\ud{u}^{x_n^\prime}\prec \sigma^n\ud{x}^\prime\prec
\ud{v}^{x_n^\prime}$ for all $n\geq 0$. Then there exists $x\in
X\backslash S$ such that $\i(x)=\ud{x}^\prime$.

Notice that we do not assume that the $\varphi$-expansion is valid or that the map $\overline{\varphi}_\infty$ is well-defined. For unimodal maps see e.g. Theorem II.3.8 in [@CoE]. Our proof is different.

If $y_\uparrow(\ud{x}^\prime)<y_\downarrow(\ud{x}^\prime)$ are two cluster points, then this follows from Lemma \[lem2.4quatro\]. Therefore, assume that $\lim_{n}\overline{\varphi}_n(\ud{x}^\prime)$ exists. Either there exists $m>1$ so that $y_\uparrow(\sigma^m\ud{x}^\prime)<y_\downarrow(\sigma^m\ud{x}^\prime)$ are two cluster points, or $\lim_{n}\overline{\varphi}_n(\sigma^m\ud{x}^\prime)$ exists for all $m\geq 1$.

In the first case, there exists $z_m\in X\backslash S$, $$y_\uparrow(\sigma^m\ud{x}^\prime)<z_m<y_\downarrow(\sigma^m\ud{x}^\prime)
\quad\text{and}\quad \i(z_m)=\sigma^m\ud{x}^\prime\,.$$ Let $$z_{m-1}:=\overline{\varphi}(x^\prime_{m-1}+z_m)\,.$$ We show that $a_{x^\prime_{m-1}}< z_{m-1}<a_{x^\prime_{m-1}+1}$. This implies that $z_m\in {\rm int(dom} \varphi)$ so that $$\varphi^{-1}(z_{m-1})=x^\prime_{m-1}+Tz_{m-1}=x^\prime_{m-1}+z_m\,.$$ Suppose that $\delta(x^\prime_{m-1})=1$ and $a_{x^\prime_{m-1}}=z_{m-1}$. Then for any $y\in X\backslash S$, $y>a_{x^\prime_{m-1}}$, we have $Ty>z_{m}$. Therefore, 