Solution

The correct answer is L1 ∩ L2 is Recursively Enumerable

Concept:

To determine the properties of the intersection of two context free languages L1 and L2 , we need to analyze each option provided.

Context-Free Languages

1. L1 and L2 are Context-Free Languages (CFLs): 
    Context-free languages are closed under union and concatenation but not closed under intersection.

Explanation:

1. L1 ∩ L2 is context-free:  
    False. The intersection of two context-free languages is not guaranteed to be context free. For example, if \(L_1 = \{ a^n b^n | n \geq 0 \} and L_2 = \{ b^n c^n | n \geq 0 \} , then L_1 \cap L_2 = \{ b^n | n \geq 0 \}\) , which is not context-free in this context.

2. L1 ∩ L2 is regular:  
    False. The intersection of two context-free languages is not necessarily regular. Regular languages are a subset of context free languages, but the intersection of two CFLs can produce a language that is not regular.

3. L1 ∩ L2 is recursively enumerable:  
    True. Context-free languages are a subset of recursively enumerable languages. Since both L1 and L2 are context-free, their intersection L1 ∩ L2 is recursively enumerable.

4. L1 ∩ L2 is context-sensitive:  
    True, but it is a broader statement. All context-free languages are context-sensitive, and since the intersection can produce a language that fits within the context-sensitive category, this statement is also valid, but it doesn't specifically apply only to CFLs.

Conclusion: The correct answer is: 3) L1 ∩ L2 is Recursively Enumerable