### modif

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 ... ... @@ -1359,7 +1359,7 @@ In practice, we will use the following criterion to detect discrete Conduché $n \item Similarly, if$C$is free and if$F_k : C_k \to D_k$is surjective for every$k \in \mathbb{N}$, we know that$D$is also free. If we write$\Sigma^C_n$for the$n$\nbd{}basis of$C$, then it follows once again from Proposition \ref{prop:uniquebasis} and Lemma \ref{lemma:conducheindecomposable} that the$n$\nbd{}basis of$D$is given by $\Sigma^D_n :=\left\{ y \in D_n \vert \exists x \in \Sigma_n^C \text{ with } F(x)=y \right\}. \Sigma^D_n :=\left\{ F(x) \in D_n \vert x \in \Sigma_n^C \right\}.$ \end{enumerate} \end{paragr} ... ... @@ -1491,7 +1491,8 @@ The converse of the previous lemma is obviously not true. However, Lemma \ref{le \] Hence$P_w(l_a)\leq 0$, which is impossible since$w$is well formed and since${l_a < l-1}$. \end{proof} \begin{lemma}\label{lemmapartialconverse} \begin{lemma}\label{lemmapartialconverse} Let$w$be a well parenthesized word. If$w\$ is a subword of a well formed word, then it is also well formed. \end{lemma} \begin{proof} ... ...
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