4, we build a monoid ${\cal S}^\ast$ (which in [@DP03], Section on ${\cal L}_c$ *and* ${\cal L}_\omega$, is called ${\cal L}_c^\ast$, while $\cal S$ is called ${\cal
L}_c$). On the arrows of $\cal S$, we define a total binary operation $\ast$ based on composition in the following manner: for $f\!:m\str n$ and $g\!:k\str l$, $$g\ast f=_{df}\left\{
\begin{array}{ll}
g\cirk F^{k-n}f & {\mbox{\rm if }}n\leq k
\\
F^{n-k}g\cirk f & {\mbox{\rm if }}k\leq n.
\end{array}
\right.$$

Next, let $f\equiv g$ if and only if for some $k$ and $l$ we have $F^kf=F^lg$ in $\cal S$. It is easy to check that $\equiv$ is an equivalence relation on the arrows of $\cal S$, which satisfies moreover $${\mbox{\rm if }}f_1\equiv f_2\; {\mbox{\rm and }} g_1\equiv g_2,
\; {\mbox{\rm then }}g_1\ast f_1\equiv g_2\ast f_2.$$

For every arrow $f$ of $\cal S$, let $[f]$ be $\{g\mid f\equiv
g\}$, and let ${\cal S}^\ast$ be $\{[f]\mid f$ is an arrow of $\cal S\}$. With $$\begin{array}{l}
\mj=_{df}[\mj_0],
\\[.1cm]
[g][f]=_{df}[g\ast f],
\end{array}$$ we can check that ${\cal S}^\ast$ is indeed a monoid, which in [@DP03] is shown to be isomorphic to the monoid ${\cal
L}_\omega$. The monoid ${\cal S}^\ast$ is built of syntactical material, coming from the category $\cal S$, 