We have $\Sigma =\{0\}$ and $$L=\{0^{2^n} \mid n\ge 0\}$$ How to prove that $L$ is irregular by using Myhill–Nerode theorem?
At other languages with $\Sigma >1$ we can usually separate the word or something like this with combination and this how we can show that two words are not at the same equivalent class....
But what we can do at this case?
Here what I tried:
Assuming that $i\ne j$, then $0^{2^i+1}$ is not at the same equivalent class with $0^{2^j+1}$, why?
Let mark: $p$ - the amount of $0$'s that we need to add to $0^{2^i+1}$ to be $0^{2^{i+1}}$, $q$ - same thing but with $j$ instead of $i$.
Of course $p\ne q$, hence:
$0^p\in 0^{2^i+1}$ but $0^p\notin 0^{2^j+1}$.
I'd like to know if I'm right at my and I should continue or try a different way...
Thank you!