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Given a language $L$ and a set of strings $\Sigma^* = \{0, 1\}^*$, how can I find a regular set that expresses

$L = \{ w \in \Sigma^* \mid w$ ends with $00$ and does not contain $11\}$?

Well, the part that states that w must end with 00 is easy and I (think I) managed to find the regular set for it. But I can't modify it so that w won't contain 11.

I didn't find many articles about this subject on the internet, and the ones I did didn't help me much in this "does not contain" kind of problem. So it'd help a lot if you guys could mention some articles on regular sets to me.

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  • $\begingroup$ Can you show what you have so far? If you want "does not contain 11", that means you can have zero or more of "0" or "01". $\endgroup$ – C8H10N4O2 Mar 10 at 4:00
  • $\begingroup$ isn't the $L$ you got already a regular language? Whats wrong with how its defined? $\endgroup$ – nir shahar Mar 10 at 6:40
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If a string doesn't contain $11$, then any $1$ is followed by $0$, unless it is the final $1$ in the string. Therefore you can break the string up into pieces of the forms $0,10,1$, the last one appearing only at the very end. This leads to the following regular expression for the set of binary strings excluding $11$: $$ (0+10)^*(\epsilon+1). $$ If we furthermore want the string to end with $0$, then we have no terminal $1$. On the other hand, we also want to disallow the empty string. Altogether, we get $$ (0+10)^+. $$ A similar case analysis leads to a regular expression for your language. Details left to you.

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