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Let $X$ be a non-empty set. There exists sets $X'$ and $\varepsilon$ such that $\varepsilon \notin X \cup X'$, $X' \cap X = \varnothing$ and $|X|=|X'|$, ie, $X$ and $X'$ have the same cardinality. Hence, there exists a bijection $f:X \to X'$. Let $A = X \cup X' \cup \{ \varepsilon \} $.

For all $x \in A$, we define $$x' = \left\{ \begin{array}{c} \varepsilon , \text{ if } \ x= \varepsilon; \\ f(x) , \text{ if } \ x \in X ; \\ f^{-1} (x) , \text{ if } \ x \in X' . \end{array} \right.$$

The usual set of sequences of elements of $A$ is $A^{\omega}$, whose members are functions $\omega \to A$. It is usual to write $u_n = u(n)$, $\forall n \in \omega$, $\forall u \in A^{\omega}$.

The set $$W = A^{\omega}_{\varepsilon} = \{ u \in A^{\omega} : (\exists k \in \omega)(\forall n \in \omega)(n \geq k \to u_n = \varepsilon) \}$$ is the set of all the sequences "quasi-$\varepsilon$", which we will call "words".

For all $u \in W$, the set $\{ k \in \omega : (\forall n \in \omega)(n \geq k \to u_n = \varepsilon) \}$ is non-empty by definition and has a minimum element $\ell(u)$, which we call the "length" of the word $u$.

A word $u \in W$ is called "reduced" if, and only if, $(\forall n\in \omega)[n < \ell(u) \to u_n \neq \varepsilon ] $ and $(\forall n \in \omega)[n < \ell(u) \to u_{n+1} \neq u_n' ]$.

I need an algorithm that, for all $u \in W$, reduces $u$ till it can't be no more. Then I need to prove that every word can be reduced by this algorithm until I reach an unique reduced word.

Thanks in advance.

Gustavo
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  • It is usual to write $u = (u_0 , u_1 , u_2 , ...)$, for all $u \in A^{\omega}$. By definition, we have $$e = \omega \times { \varepsilon } = (\varepsilon , \varepsilon , \varepsilon , ...) \in A^{\omega}_{\varepsilon}$$ Then, $\forall u \in W$, $\ell(u) = 0$ if, and only if, $u=e$.

    For all $u \in W$, if $u \neq e$, then $\ell(u) \neq 0$. Hence $\nu = \ell(u) - 1 \in \omega$ and $u = (u_0 , u_1 , u_2 , ... , u_{\nu}, \varepsilon , \varepsilon , \varepsilon , ...)$, where $u_{\nu} \neq \varepsilon$. It is usual to write $u = u_0 u_1 u_2 ... u_{\nu}$.

     – Gustavo Feb 28 '14 at 17:24
  • In advance we can define $e'=e$ and $u' = u_{\nu}' u'_{\nu - 1} ... u'_2 u'_1 u'_0$, for all $u \in W$. – Gustavo Feb 28 '14 at 17:31
  • I made some progress since the last time I thought about it.

    Rather than setting $ W $ as $ A^{\omega}_{\varepsilon} $, I started setting

    $ W = { w \in A^{\omega}_{\varepsilon} : (\forall n \in \omega) (n < \ell_w \to u_n \neq \varepsilon) } $

    Where $\ell_w = \ell(w)$. So, words don't have blank spaces.

     – Gustavo Sep 01 '14 at 22:22
  • Then I invented an algorithm and I could demonstrate what is being asked. That is, I created a function $ R: W \to W $ such that every reduced word is in the image of $ R $, but I could not prove that every word in the image of $ R $ is reduced. In fact, I proved that $ R (w) = w $, for every reduced word $ w $. – Gustavo Sep 01 '14 at 22:25
  • News: I found the $R$ function I was looking for. It is some kind of string projection, as presented on this Wikipedia link

    https://en.wikipedia.org/wiki/String_operations#String_projection

    But I need to define this projection using some kind of recursion. How can I do that?

     – Gustavo Mar 02 '17 at 06:53

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