Using pumping lemma to prove that $L=\{ab^nab^k \mid n>k>3\}$ is not regular

Question

I want to use pumping lemma to prove that $L=\{ab^nab^k \mid n>k>3\}$ is not regular.

Can anyone help me?

Answer 1

Here's a possible hint:

Suppose that $L$ is regular and let $p$ be the pumping length as in the pumping lemma. Let $\gamma = \max\{p, 4\}$ and the string $ab^{\gamma + 1}ab^{\gamma}$ is in $L$.

The pumping lemma implies that the pumpable portion of the string has to be a substring of the $ab^{\gamma + 1}$ portion. Try to derive a contradiction from here.

Answer 2

You have to use the general form of the pumping lemma.

Assume that the given language is regular. Let $p$ be a pumping length ( $p$ can be the number of states of an automaton that recognizes the language). Consider the word of the language $$a b^{q} a b^r$$ with $q >r > 3$ and $r\ge p$. Then there exists a subword of $b^r$ that is pumpable, of length $s$, $1\le s \le p$. Then all of the words $$a b^r a b^{r+ (k-1) s}$$ are in the language, for all $k\ge 1$. For $k$ large enough we have $r+ (k-1) s > q$, contradiction.