Does pumping lemma ever fail?

Question

Consider the language F = {$a^i$$b^j$$c^k$ | i, j, k ≥ 0 and if i = 1 then j = k}.

1. Show that F is not regular.
2. Show that F acts like a regular language in the pumping lemma. In other words, give a pumping length p and demonstrate that F satisfies the three conditions of the pumping lemma for this value of p.
3. Explain why parts (a) and (b) do not contradict the pumping lemma

I tried to choose x=empty string,y=a and z=$b^p$$c^p$ to prove that a$b^p$$c^p$ is regular and it worked while the language is non-regular and it was supposed to fail with pumping lemma.What is the reason? Do I have to cut off the first non-repeating parts like a?

Answer 1

I just found out the solution and sharing it for others who have trouble with the same problem: The pumping lemma is not violated because it states only that regular languages satisfy the three conditions, and it doesn’t state that nonregular languages fail to satisfy the three conditions.