I think to prove that $\{a^ib^i | i\geq 0\}$ is not regular, we just have to consider the string $a^nb^n$ (which is in the language) and apply the pumping lemma. But I'm not sure how to proceed using the pumping lemma (even though I know it applies with our choice of string, since the string is at least $n$ long).


To apply the pumping lemma for regular languages:

  1. You assume some pumping length $p > 0$: Suppose a pumping length $p$.
  2. You pick a string $s$ such that $|s| \geq n$: Indeed, $a^pb^p$ is a good one since $|a^pb^p| = 2p \geq p$.
  3. Now, you have to consider all the partitions of $s$ as $xyz$, such that
    1. $|y|>0$ and
    2. $|xy| \leq p$.

In this case, the 2nd inequality restricts $y$ to contain only $a$'s. Therefore, all the possible partitions are of the form $y = a^k$, $x = a^{p-k-r}$ and $z = a^rb^p$ for $k$ and $r$ such that $k > 0$ and $p \geq k + r$.

  1. For every possible partition of the previous step, you have to show that there is some $i\in\mathbb{N}$ such that $xy^iz$ is not in the language.

Here, consider $i = 0$. Then $xy^0z = xz = a^{p-k-r}a^rb^p = a^{p-k}b^p$. It is now a matter of showing that $p-k\neq p$, or, equivalently, that $k > 0$ which is already true.

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  • $\begingroup$ Apologies for the late response, I just wanted to say thank you! Your response helped my understanding a lot! $\endgroup$ – James Ronald Mar 2 at 15:11

In this particular case, you can use the pumping lemma with any word of length at least $p$ (the pumping length). Indeed, consider any word $w = a^nb^n$, and suppose that we are given some decomposition $w = xyz$ where $y$ is non-empty and $xy^tz \in \{ a^ib^i \mid i \geq 0 \}$ for all $t$ (we don't even need the bound on $|xy|$). If $y$ consists only of $a$'s, then $xy^2z$ will have more $a$'s than $b$'s; if it consists only of $b$'s, then $xy^2z$ will have more $b$'s than $a$'s; and if it consists of both, then $xy^2z \notin a^*b^*$.

If instead of your language we consider the language of all words with equally many $a$'s and $b$'s (but not necessarily in the format $a^*b^*$) then the above argument would fail, and you would have tobe more careful when choosing your word; however, $a^pb^p$ would still work.

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