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I have searched quite a few posts here so that I can prove that the language $$L=\{a^kb^l: 1\leq k\leq l\}$$ is not regular (using Myhill-Nerode's theorem). I know that I must find an infinite number of pairs $(c,d)$ such that if $R$ is the equivalence relation defined by Myhill-Nerode's theorem, there is a string $z \in \{a,b\}^*$ s.t. if $cRd \Rightarrow \neg (cz R dz)$

I tried $c=a^{n+3}b^n$ and $d=a^{n+1}b^n$, using $z=b^2$.

What troubles me is: I can't assume that $c R d$ since it might not be holding like in this answer. However if I chose $c, d\in L$ I can't find a string $z$. What could I do?

And if I find $c,d$ how do I get that I have infinite number of classes? Can't all $c$ belong at the same class for example?

John L.
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Νικολέτα Σεβαστού
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    It looks you did not understand Myhill-Nerode's theorem correctly. To prove $L$ is nonregular, you should find an infinite number of strings such that for any two of them, say $c$ and $d$, there exists a string $z$ such that exactly one of $cz$ and $dz$ is in $L$. In other words, $c$ and $d$ are distinguished by $z$. – John L. Jun 15 '23 at 19:57

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The idea is to find infinitely many strings that are distinguishable from each other with respect to the Myhill-Nerode equivalence relation defined by $L$.

Consider the strings $a, a^2, a^3, \cdots$. Let us check they are distinguishable from each other.

Let $a^{k_1}$ and $a^{k_2}$ be two of them. WLOG, assume $k_1<k_2$.
Then $a^{k_1}b^{k_1}\in L$ but $a^{k_2}b^{k_1}\notin L$.
That is, $a^{k_1}$ and $a^{k_2}$ can be distinguished by appending $b^{k_1}$ to them. That is, these two strings are distinguishable. That is, these two strings are in different Myhill-Nerode equivalence classes.

John L.
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