Last Updated : 12 Nov, 2018
Let L = L1∩L2, where L1 and L2 are languages as defined below:
L1 = {\"a^{m}b^{m}ca^{n}b^{n}\" | m, n >= 0 }
L2 = {\"a^{i}b^{j}c^{k}\" | i, j, k >= 0 }
Then L is (A) Not recursive (B) Regular (C) Context free but not regular (D) Recursively enumerable but not context free.

Answer: (C)

Explanation: The language L1 accept strings {c, abc, abcab, aabbcab, aabbcaabb, …} and L2 accept strings {a, b, c, ab, abc, aabc, aabbc, … }. Intersection of these two languages is [TEX]L1 \\cap L2 = \\{a^{k}b^{k}c | k >= 0\\}[\\TEX] which is context free, but not regular.

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