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To simplify, let $D$ be the domain of $T$ and let $R = \{\epsilon\} \cup (\Sigma^* \setminus \Sigma^*D\Sigma^*)$. Then by definition $$ N(T) = Id_R \quad \text{and} \quad R^{obl}(T) = N(T)(TN(T))^*. $$ Here is a formal way to justify your idea. Let $(u,v) \in \Sigma^* \times \Sigma^*$. By definition, $(u,v) \in R^{obl}(T)$ if and only if $(u,v)$ can be ...


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The conjecture does not hold: Let $L$ be the set of prefixes of $(c^*ac^*b)^*$. Then $L'=(c^*ac^*b)^\omega+ (c^*ac^*b)^*c^\omega+(c^*ac^*b)^*c^*ac^\omega$. Take the word $\eta=(c^1ac^1b)(c^2ac^2b)(c^3ac^3b)\dots \in L'$. For all $k\in\mathbb N$, we can show that $PF_k(\eta)\not\subseteq L'$, as witnessed by $$u_k=(c^1ac^1b)(c^2ac^2b)\dots (c^kac^kb)c^kb c^\...


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