Why doesn't $|uv|\le k$ break the pumping lemma?

Asked Mar 20 '21 at 19:47

Active Mar 20 '21 at 19:50

Viewed 35 times

Let $N = \{ab^x | x \in\mathbb{N}\}$.

Let the pumping length be $k$. So $ab^k$ belongs to $N$.

Let $u = a, v = b^k, w = \operatorname{empty}$.

Then $|uv|\le k$ does not hold. No other splitting I can find satisfies the requirements and I'm pretty sure this should be the one to work. Why doesn't it?

edited Mar 20 '21 at 19:50

vitamin d

5,783

asked Mar 20 '21 at 19:47

Bob

Try the splitting $u = a$, $v = b$ and $w = b^{k-1}$? – Anon Mar 20 '21 at 19:57
Thank you. Had not occurred to me I could do that. – Bob Mar 20 '21 at 20:08
FYI, the pumping length of your language is 2. – Anon Mar 20 '21 at 20:20

Why doesn't $|uv|\le k$ break the pumping lemma?

0 Answers0