@@ -1307,7 +1307,13 @@ with $n$ occurences of $f$ and $m$ occurences of $g$.

\item$F(\alpha)=1_{(1,1)}$.

\end{itemize}

This last equation makes sense since $(1,1)=(0,1)+(1,0)=(1,0)+(0,1)$. For any

$1$\nbd{}cell $w$ of $P$ (encoded as a finite words on the alphabet $\{f,g\}$) such that $f$ appears $n$ times and $g$ appears $m$ times, we have $F(w)=(n,m)$. Let us now prove that $F$ is a Thomason equivalence using a dual of \cite[Corollaire 5.26]{ara2020theoreme} (see Remark 5.20 of op.\ cit.). If we write $\star$ for the only object of $B^1(\mathbb{N}\times\mathbb{N})$, what we need to show is that the canonical $2$\nbd{}functor from $P/{\star}$ (\ref{paragr:comma}) to the terminal $2$\nbd{}category

$1$\nbd{}cell $w$ of $P$ (encoded as a finite words on the alphabet $\{f,g\}$)

such that $f$ appears $n$ times and $g$ appears $m$ times, we have

$F(w)=(n,m)$. Let us now prove that $F$ is a Thomason equivalence using a dual

of \cite[Corollaire 5.26]{ara2020theoreme} (see Remark 5.20 of op.\ cit.). If

we write $\star$ for the only object of $B^1(\mathbb{N}\times\mathbb{N})$,

what we need to show is that the canonical $2$\nbd{}functor from $P/{\star}$

(see \ref{paragr:comma}) to the terminal $2$\nbd{}category

\[

P/{\star}\to\sD_0

\]

...

...

@@ -1315,9 +1321,48 @@ with $n$ occurences of $f$ and $m$ occurences of $g$.

\begin{itemize}[label=-]

\item A $0$\nbd{}cell is a $1$\nbd{}cell of $B^1(\mathbb{N}\times\mathbb{N})$.

\item For $(n,m)$ and $(n',m')$ two $0$\nbd{}cells of $P/{\ast}$, a

$1$\nbd{}cell

\item

\end{itemize}

$1$\nbd{}cell from $(n,m)$ to $(n',m')$ is a $1$\nbd{}cell $w$ of $P$ such