Let L = L1∩L2, where L1 and L2 are languages as defined below:
L1 = {am bm can bn | m, n >= 0}
L2 = {ai bj ck | i, j, k >= 0}
Then L is

Question

Let L = L1∩L2, where L1 and L2 are languages as defined below: L1 = {am bm can bn | m, n >= 0} L2 = {ai bj ck | i, j, k >= 0} Then L is

Accepted Answer

Correct answer: C — L1 is context-free but not regular, and L2 is regular. The intersection of a context-free language with a regular language is context-free. Since L1 ∩ L2 results in strings of the form am bm c an bn, which requires matching two pairs of counts (m and n), this language is not regular. However, it can be generated by a context-free grammar (e.g., S → A C B where A generates am bm, C is 'c', and B generates an bn), so it is context-free. Therefore, L is context-free but not regular.

Answer

A. Not recursive

Answer

B. Regular

Answer

C. Context free but not regular

Answer

D. Recursively enumerable but not context free.

Let L = L1∩L2, where L1 and L2 are languages as defined below: L1 = {am bm can…