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Consider the language $$L=\{x y^{(n)} z y^{(n)} w: x,z,w \in \Sigma^*, y \in \Sigma, z\text{ does not contain }y, n \geq 0 \}.$$ To show that the language is not regular using the Myhill-Nerode Theorem, I have to show that there are infinite number of equivalence classes, right? But how can I find these equivalence classes?

user642796
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evinda
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    Are you sure this representation of the language correct? It seems to me that $L = \Sigma^*$. (For any string $u$ either some symbol $\mathtt{x}$ doesn't appear at least twice in which case you can take $u = x \mathtt{x}^0 z \mathtt{x}^0 w$ with no $\mathtt{x}$s in $z$, or if some symbol $\mathtt{x}$ appears twice you can express $u$ as $x \mathtt{x}^1 z \mathtt{x}^1 w$ where these represent consecutive occurrences of $\mathtt{x}$ in $u$.) Perhaps some condition on $L$ is missing? (No occurrences of $y$ in any of $x,z,w$?) – user642796 Dec 15 '13 at 12:00
  • This language is given from the exercise I am looking at.Do you think it is a misprint ? So,can we not apply the Myhill-Nerode Theorem? – evinda Dec 15 '13 at 12:07
  • My outline above seems to show that $L = \Sigma^$, which is clearly regular. (Of course, you could use Myhill-Nerode to show that $\Sigma^$ is regular, though that seems pretty pointless.) – user642796 Dec 15 '13 at 12:12
  • Could you tell me some of the equivalence classes of the language that is given??Are some of them: $xzw,xyzyw,xyyzyyw$ ?Or am I wrong? – evinda Dec 15 '13 at 12:26
  • If my outline is correct and $L = \Sigma^*$, then there is only one equivalence class: that corresponding to the empty string $\varepsilon$. (Note that $xyzyw$ and $xyyzyyw$ would be equivalent since for any $u$ we will have $xyzywu$ and $xyyzyywu$ in $L$.) – user642796 Dec 15 '13 at 12:29
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    yes, take $z=w=\epsilon$ and $n=0$ and you just have $x\in\Sigma^*$. I agree with @ArthurFischer. – Xoff Dec 15 '13 at 12:34
  • I tend to over-complicate things. Your outline is *much better*, @Xoff. – user642796 Dec 15 '13 at 12:38
  • @ArthurFischer,@Xoff could you explain me further why the only equivalence class is the empty string?I haven't understood it... – evinda Dec 15 '13 at 15:38
  • @evinda: Since every string is in $L$, then given any two strings $u , v \in \Sigma^$ and any string $w \in \Sigma^$ the biconditional $uw \in L \iff vw \in L$ is true, which implies that $u$ and $v$ are equivalent with respect to the "there is no distinguishing extension" relation. That is, all strings are equivalent with respect to this relation, or, there is only one equivalence class (represented by any string, in particular, by $\varepsilon$). – user642796 Dec 15 '13 at 16:07
  • @evinda: My guess is that it was intended to require that none of $x,z$, and $w$ contain $y$. Then you really do get one equivalence class for each $n\ge 0$. – Brian M. Scott Dec 16 '13 at 11:37

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