# Why is this choice of $y$ not permitted in using pumping lemma?

Consider this snippet shown below from, An Introduction to Formal Languages and Automata 6th Edition by Peter Linz.

As per the text, choosing a value of $$y = a^k$$, where $$k$$ is odd is not permitted since this violates the condition of pumping lemma. This text simply says that the assumption on $$k$$ (number of $$a$$'s in $$y$$ being odd) is not permitted.

Now, as far as my understanding goes, the condition on choosing $$xy$$ is $$|xy| \le p$$, where $$p$$ is the pumping length, and $$|y|>1$$. And if we can show that one valid choice of $$y$$ satisfies the pumping lemma, we have been able to show that this language does not violate the pumping lemma, though the language can not be claimed to be regular. When I need to show that the language is not regular, I need to explore all possible $$y$$'s.

I fail to understand, why a $$y$$, with odd number of $$a$$'s, and $$|y| \le p$$, $$x=\epsilon$$, will not be permitted here.

• If you are using the pumping lemma to prove $L$ is not regular then you are not permitted to make the assumption that $y$ has odd length, or any other assumption of this kind. It is not clear what you want to achieve. – Michal Adamaszek Dec 5 '18 at 9:05
• @MichalAdamaszek Would you mind elaborating, "any other assumption of this kind", part? – Masroor Dec 5 '18 at 9:15
• I mean anything that is not in the conclusion of the pumping lemma. – Michal Adamaszek Dec 5 '18 at 9:24
• @MichalAdamaszek Do you want to put a reference in support of your claim? As far as my understanding goes, if I take $y$ to be of odd length, it does not violate the acceptable conditions for $y$. – Masroor Dec 5 '18 at 9:39
• What do you mean by "you take y"? You get y from the pumping lemma, and not choose it yourself. I am referring to how you use the pumping lemma to prove a language is not regular. Are you trying to do something else? – Michal Adamaszek Dec 5 '18 at 9:56