# Pumping Lemma on Language with subtracted length

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 $$y = b$$?

• Your language is regular! – Yuval Filmus Feb 11 '19 at 3:17

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.
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 $$L$$.