= Solution
Let $K\subseteq\mathbb N$ be the undecidable set denoted $\mathbb K$ in the question. The language satisfies the <pumping lemma for regular languages> with pumping length one. For any nonempty $s\in L$, write $s=xyz$ with $x=\varepsilon$, $y$ its first symbol, and $z$ the remaining suffix.
If $s\in b^*$, then $xy^iz\in b^*$ for every $i\geq0$. Otherwise $s=wab^n$ for some $n\in K$. If the chosen first symbol lies in $w$, pumping it preserves the same final suffix $ab^n$. If $w=\varepsilon$, then $s=ab^n$: pumping down gives $b^n\in b^*$, while every positive pump gives $a^ib^n$ and all but its last $a$ can be absorbed into $w$. Thus every sufficiently long word has a valid pumping decomposition.
Nevertheless, if $L$ were <regular language>, closure under intersection would make
$$
L\cap ab^*=\{ab^n:n\in K\}
$$
regular. A finite automaton deciding this unary language would decide whether $n\in K$ by running it on $ab^n$, contradicting the <halting problem>. Therefore
$$
\boxed{L\text{ satisfies the pumping lemma but is not regular}.}
$$
This is a <nonregular language satisfying the pumping lemma>.
Back to article page
