Part OneApplying this new pumping lemma can require the same sort of creative decisions as the one for regular languages. Previously we showed this language was not regular:L={ap∈{a}∗|p is a prime number }𝐿={𝑎𝑝∈{𝑎}∗|𝑝 is a prime number }Can you prove that it is not context-free?Here is the start of the proof.Assume L𝐿 is context-free, then the context-free pumping lemma applies, with a pumping length n𝑛.Then consider ap𝑎𝑝 where p≥n𝑝≥𝑛 is a prime number.Then by the lemma, ap=uvxyz𝑎𝑝=𝑢𝑣𝑥𝑦𝑧 with the usual conditions.Let s=|uxz|𝑠=|𝑢𝑥𝑧|, and r=|vy|𝑟=|𝑣𝑦|. Notice that p=s+r𝑝=𝑠+𝑟.Then |uvixyiz|=s+ir|𝑢𝑣𝑖𝑥𝑦𝑖𝑧|=𝑠+𝑖𝑟, and by the context-free pumping lemma, as+ir∈L𝑎𝑠+𝑖𝑟∈𝐿, i.e. s+ir𝑠+𝑖𝑟 is prime.Now see if you can pick a value for i𝑖 that leads to a contradiction. You might want to use the proof of non-regularity for inspiration.