High School

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Let [tex]L1[/tex] be a regular language and [tex]L2[/tex] be a recursively enumerable language but not recursive. Which of the following is a context-free language?

A. [tex]L1 \cup L2[/tex]
B. [tex]L1 \cap L2[/tex]
C. [tex]L1 \ast L2[/tex]
D. [tex]L1(L2)[/tex]

Answer :

Final answer:

Among the options given for the combination of a regular language (L1) and a recursively enumerable language but not recursive (L2), L1 ∪ L2 is the most likely to result in a context-free language. Therefore the correct option is a. L1 ∪ L2.

Explanation:

The question asks which of the following is a context-free language if L1 is a regular language and L2 is a recursively enumerable language but not recursive: a. L1 ∪ L2, b. L1 ∩ L2, c. L1 * L2, d. L1(L2). Among these options, the operation that most likely preserves the property of being a context-free language (CFL) when applied to a regular language and a recursively enumerable (RE) language, which is not recursive, would be a union of a regular language with another language. Therefore, option a. L1 ∪ L2 is the most likely to be a context-free language. It is important to note that the intersection, concatenation and Kleene star with a non-recursive RE language do not guarantee the result will be context-free.

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