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Firstly, I've tried assuming $L$ is regular and find a contradiction with help of the pumping lemma's 3 conditions, I was not able to get to a contradiction.

I've tried thinking of a word $z\in \{0,1\}^*$ to attach to $wv^i$ and $wv^j$ for some $i\neq j$, and to combine that with the pumping lemma's $xy^2z\in L, k\geq 1$ yet I got stuck on the way.

I'm looking for a direction of thought, a colleague suggested looking at the example word $10000000110000000$ , yet I cant see the how this example promotes what I'm trying to prove.

Aishgadol
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2 Answers2

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This language is regular. Denote $L = \{wvw\mid v,w\in\{0,1\}^*\}$.

I claim $L = \{0,1\}^*$. Indeed, the inclusion $\subseteq$ is clear. Conversely, let $u\in \{0,1\}^*$. Then $u = \varepsilon u\varepsilon \in L$.

Nathaniel
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  • Apologizes, I've forgot to mention the fact that $\epsilon \notin L$ I've also saw how it is regular when $\epsilon \in L$. – Aishgadol Mar 25 '23 at 10:07
  • What you say makes no sense: the language ${wvw\mid v,w\in{0,1}^*}$ is uniquely defined and it contains $\varepsilon$. – Nathaniel Mar 25 '23 at 10:11
  • Even if you consider $L = {wvw\mid v,w\in{0,1}^*}\setminus{\varepsilon}$, it is still regular, because regular languages are closed under set difference. – Nathaniel Mar 25 '23 at 10:13
  • double apologizes, it was my misunderstanding of the instructions I was given.

    after further clarifications from my instructor, instead of $\epsilon \notin L$, what I've meant is that the $w$ in $w,v\in {0,1}^*$ satisfies $w\neq \epsilon $

     – Aishgadol Mar 25 '23 at 10:29
  • Then just write $L = {wvw\mid v\in {0,1}^*, w\in {0,1}^+}$ so that there is no ambiguity. – Nathaniel Mar 25 '23 at 10:52
  • I see, I can also understand this notation. Thank you for this clarification. – Aishgadol Mar 25 '23 at 10:56
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Of course it is regular, because it contains the complete set {0,1}*.

Note: This was the correct answer to the question before it was edited.

gnasher729
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