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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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  • Have you tried $(\mathsf{ab})^p(\mathsf{ba})^p\mathsf{a}$, where $p$ is $L_3$'s pumping length? To me it looks promising. – Kai Sep 25 '23 at 05:16
  • @Kai Indeed I have. We have to consider both the cases when p is even and p is odd, as it determines where we insert additional the y. This works for even p, but for odd, the string has form $(ab)^{\lfloor p/2 \rfloor} ab (ab)^{\lfloor p/2 \rfloor} (ba)^p a$, where the place to insert y is after the "a" of the middle ab. Then, if $y = (ab)^{\lfloor p/2 \rfloor} a = a(ba)^{\lfloor p/2 \rfloor}$, then we have $(ab)^{\lfloor p/2 \rfloor} a a(ba)^{\lfloor p/2 \rfloor} \ldots$, which is in the language – Edward Shanahan Sep 25 '23 at 05:30
  • So your $w=xyz$ split has $x = \epsilon$ and $y=\mathsf{a(ba)}^{\lfloor{p/2}\rfloor}$. Is $xy^0z = z = \mathsf{(ba)}^{\lfloor{p/2}\rfloor+p}$ in $L_3$? – Kai Sep 25 '23 at 05:44
  • Hmm, I'm not sure what you mean that $z$ from the string $(ab)^{\lfloor p / 2 \rfloor} ab (ab)^{\lfloor p / 2 \rfloor} (ba)^p a$ is equal to $(ba)^{\lfloor p / 2\rfloor + p}$. Assuming you meant $xy^0z = z = (ba)^p a$, you are correct in that it is not in $L_3$. Moreover, the pumping I described in my comment works for all other values of $y$ except for $(ab)^{\lfloor p / 2 \rfloor} a$. Can we use both pumping and deflation for different values of $y$ when applying the pumping lemma? I'm learning about it at the moment, and that wasn't mentioned anywhere – Edward Shanahan Sep 25 '23 at 06:01
  • Yes, you can choose $i$ to prove $xy^iz\notin L_3$ after 1. you chose the word $w$ and 2. the "demon" chose the $xyz$-split of $w$. – Kai Sep 25 '23 at 06:09
  • Oh, wait a second. I think I still don't follow your second comment. Let $s = (ab)^p(ba)^p a$. Let $y = (ab)^{\lfloor p/2 \rfloor} a = a (ba)^{\lfloor p/2 \rfloor}$ and $x = \epsilon$. Then, if we deflate to $xz$, we get $s' = b (ab)^{\lfloor p/2 \rfloor} (ba)^p a$, which simplifies to $(ba)^{\lfloor p/2 \rfloor} bb (ab)^{p - 1} aa$, and this string actually is in $L_3$. – Edward Shanahan Sep 25 '23 at 07:31
  • Some suggestions how to handle this language are given here: Proving non-regularity of $uu^Rv$? – Hendrik Jan Sep 25 '23 at 18:02
  • You're right. I seem to have lost a $\mathsf{b}$ in my $z$. – Kai Sep 25 '23 at 20:38

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