Please answer all questions! We have seen in class that the sets of both regular

Question

Please answer all questions!

We have seen in class that the sets of both regular and context-free languages are closed under the union, concatenation, and star operations. We have also seen in A2 that the regular languages are closed under intersection and complement. In this question, you will investigate whether the latter also holds for context-free languages. Use the languages A = {a^m b^n c^n | m, n greaterthanorequalto 0} and B = {a^n b^n c^m | m, n greaterthanorequalto 0} to show that the class of context-free languages is not closed under intersection. You may use the fact that the language C = {a^n b^n c^n | n greaterthanorequalto 0} is not context-free. Using part (a) above, show now that the set of context-free languages is not closed under complement.

Explanation / Answer

The Given Expression is as follows: A=  { am bn cn| m, n 0,} and B=  { an bn cm| m, n 0,).

The expression is as follows: L(G) = {a, aa, ab, aaa, aab ,abb, …….}

Next the starting layout has to take at least one ‘a’ carry on by any number of ‘b’ and 'c'

.Finally, The Grammer expressions is:-

G: ({S, A, B,C}, {a, b,c}, S, {S aA, A aA | B,C, B , C--> | bB,cC })

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