Pumping Lemma for irregular language, how to choose and divide the string
w = abn-1abn.

We have to consider all the division xyz = w with |y| >= 1 and |xy| &lt;= n. So

x = apbr

y = aqbs

z = btabn

where:


p + q = 1
r > 0 implies p = 1
q + s > 0
r + s + t = n - 1


Let's now consider w' = xy0z=apbr+tabn. If x is the empty string, then w' is missing the initial 'a' (and maybe some b), so w' is not in L. Else if x is not empty, we know that y contains at least one occurrence of 'b', so w' can't be in the form abnabn+1 and w' is not in L.