Pumping Lemma on Language with subtracted length

Question

My study group and I have had some back and forth on one exercise and I haven't found any matching solution online. The task looks as follows: Prove that $L$ is not regular given

$$ L = \{ a^k b a^{m-1} \mid k,m \in \mathbb{N} \} $$

Questions:

1. How does $m-1$ affect my choice of the word to be pumped? There must be a way to rule out $m=0$ because otherwise my length would be negative.
2. Would it be wrong to choose $w = a^{n-1}ba^n$ with the intent of pumping down $y = b$?

Every answer is highly appreciated.

Answer 1

You could argue with a reasonably straight face that L is not well-defined, since it supposedly contains the word $a^0ba^{-1}$, so L wouldn't be a language at all, therefore not a regular language and not an irregular language either.

If you change it slightly to $$ L = \{ w: w = a^k b a^{m-1} \mid k,m \in \mathbb{N} \} $$ then it is definitely regular; there's a very simple FSM with just two states for it.

Answer 2

Your language is regular. You won't be able to prove that it's not regular using the pumping lemma.

Regarding your first question, you're right that the language is present in a sloppy manner. We should add the condition $m \geq 1$. We can then notice another way of writing the language: $$ L = \{ a^k b a^m \mid k,m \in \mathbb{N} \}. $$ This is just the language of the regular expression $a^*ba^*$.

The reason that you can't "pump down" the word $a^{n-1}ba^n$ is that you don't get to choose the decomposition of the pumping lemma. This word can be decomposed as $a^{n-2} . a . ba^n$ (we can force $n \geq 2$ by choosing the pumping constant), and then the middle part can be pumped down or up while keeping the word in $L$.