Use Nerode's theorem to prove that the following language $L$ is not regular:

$$ L=\{a^{k!} \mid 1\leq k\} $$

Here is my attempt:

Let $A$ be an infinte set of words s.t- $$ A=\{a^n \mid n\in \mathbb{N}\}$$

Then we shall consider 2 words from the set $a^{i!}$ and $a^j$ s.t-
$$ i!\leq n, i!\ne j$$

$$a^{i+1} \longrightarrow a^{i!}\cdot a^{i+1}=a^{(i+1)!}\in L$$ $$a^j\cdot a^{i+1}=a^{j+i+1}\notin L$$

I understand my attempt is not correct since when multiplying $a^{i!}\cdot a^{i+1}$ the exponent doesn't multiply . I cant seem to overcome this obstacle and would appreciate help.

  • $\begingroup$ I don't understand your attempt, since it doesn't contain any words. $\endgroup$ Dec 13, 2021 at 18:23
  • $\begingroup$ Also, what is your question? Is it whether your attempt is correct or not? We typically don't grade homework solutions, which is your TA's job. $\endgroup$ Dec 13, 2021 at 18:24
  • $\begingroup$ @YuvalFilmus I edited my attempt and clarified my question $\endgroup$
    – ATB
    Dec 13, 2021 at 18:36

1 Answer 1


You can prove the following using Myhill-Nerode: If $x_n$ is a non-decreasing sequence, and the differences $x_{n+1} - x_n$ are not bounded, then the language $\{a^{x_n}\}$ is not regular.

That covers lots of problems that you will find in a course, like your n!, or $a^p$ where p is prime, or $a^{n^2}$, or any infinite subsequence of such sequences.

Someone else asked about the language $\{0^{2^n}\}$. Same solution.


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