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I have this language that I have to prove either regular or irregular. $$ L_3 = \{mm^rn | m^r \text{ is the reverse of } m,\ m,n \in \{a,b\}^+\} $$ It's trivial to prove that it is in fact irregular using Myhill-Nerode theorem. We let $$S = \{ab,abab,ababab,\ldots\}$$, and the suffix for two strings $s_1$ and $s_2$, $\text{"ba" }\cdot (\min(|s_1|, |s_2|)\div2) + \text{"a"}$, distinguishes between them. Ex: for "ab" and "abab", suffix "baa" will work for the former but not the latter.

However, is it possible to prove that this language is irregular using just the pumping theorem? I've tried a dozen string constructions with no avail.

Edward Shanahan
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No, because this is a non-regular language that satisfies the conditions of the pumping lemma.

The pumping lemma says that every regular language must satisfy certain conditions. But it's not a complete characterization—some irregular languages, like this one, also satisfy those conditions.

In fact, we can pump the first character of any string in the language $L_3$ to obtain another string in the language.

To see this, let $x=uu^Rv$ be any string in the language $L_3$; note that $u,v$ are non-empty.

  • The initial string is $x = uu^Rv$, which is in the language.
  • If we delete the first character of the string, we're left with $u_{2\ldots n}u_{n\ldots 2}u_1v$ which is in the language because $u_{2\ldots n}u_{n\ldots 2}$ is a palindrome and $u_1v$ can be the arbitrary string at the end.
  • If we duplicate the first character of the string any number of times $k\geq 2$, we're left with $u_1^k u_{2\ldots n}u_{n\ldots 2}u_1 v$ which is in the language because $u_1^k$ starts with $u_1u_1$, which is a simple palindrome, and then the rest of the string can be arbitrary.

Therefore, $L_3$ satisfies the condition of the pumping lemma with pumping length $p=1$ despite being irregular. You can't use the pumping lemma in this case to establish that it is irregular.

user326210
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