I've been struggling with this problem for quite a while now and every explanation I have managed to find doesn't seem to correctly solve it.
Question
Proving $L = \{a^nb^m \mid n, m≥0, n \neq m\}$ is not regular by use of Pumping Lemma. I know the correct way of prove with factorial or using closure. However today someone explain a way to me and I couldn't find what is wrong with it. I think it is wrong but if it is true, is it possible to explain why?
Proof (Maybe correct):
we suppose that that $L$ is regular and there is a machine with $u$ state that accept it. If it is a NFA, we can convert any NFA to DFA with for example $k$ state.
We have a group of strings like $a^kb^{k + t}$ for every $1 \le t \le k$ which accepted with machine.
There is only one way for read $a^k$ because our machine is DFA so we can write $a^k = a^sa^ta^p$ that $|a^sa^t| \le k$ and $|a^t| \ge 1$ and $a^t$ is cycle, so with pumping lemma we can say that $a^{k + t}b^{k + t}$ (with one more time of cycle) belongs to $L$ and it is contradiction. (one of our strings in our first group will fail) so our language is not regular.
My Problem
Because we we get a group of strings (instead just one strings which is normal in proves) and use same $t$ when break $a^k$ to $a^sa^ta^p$, is it true anymore? If no or yes, please explain.
My explanation
I think we can construct a NFA that accept each of $a^kb^{k + t}$ (for each $t$) with different size of cycle (cycle that is used in pumping lemma) however when convert it to a DFA, I can't understand how to prove it.