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Most UNIX regular expressions have, besides the usual **,+,?* operators a backslash operator where \1,\2,... match whatever's in the last parentheses, so for example *L=(a*)b\1* matches the (non regular) language *a^n b a^n*.
On one hand, this seems to be pretty powerful since you can create (a*)b\1b\1 to match the language *a^n b a^n b a^n* which can't even be recognized by a stack automaton. On the other hand, I'm pretty sure *a^n b^n* cannot be expressed this way.
I have two questions:
*a^n b^n* cannot be expressed this way?
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In the theory of formal languages, the pumping lemma may refer to: Pumping lemma for regular languages, the fact that all sufficiently long strings in such a language have a substring that can be repeated arbitrarily many times, usually used to prove that certain languages are not regular.
Explanation: Pumping lemma is used to prove a language is regular or not. So, option (D) is correct.
Pumping Lemma is used as a proof for irregularity of a language. Thus, if a language is regular, it always satisfies pumping lemma. If there exists at least one string made from pumping which is not in L, then L is surely not regular. The opposite of this may not always be true.
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and of course follow their citations forward and backward to find more literature on this subject.
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