Commit 961a179e authored by Leonard Guetta's avatar Leonard Guetta
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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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