Answer :
Main Answer:
[tex]The intersection of a regular language \( L1 \) and a recursively enumerable (RE) language \( L2 \) is guaranteed to be context-free.[/tex]
Explanation :
[tex]When we take the intersection of two languages, \( L1 \cap L2 \), where \( L1 \) is regular and \( L2 \) is RE but not recursive, the resulting language is always context-free.[/tex]
A regular language can be recognized by a finite automaton, which can be easily converted into a regular grammar. An RE language, while not necessarily recursive, can be recognized by a Turing machine, which can be converted into a context-free grammar.
Taking the intersection of \( L1 \) and \( L2 \) simply means considering the strings that are in both languages. Since both \( L1 \) and \( L2 \) can be expressed by grammars, the intersection can also be expressed by a context-free grammar, making it a context-free language.[tex]Taking the intersection of \( L1 \) and \( L2 \) simply means considering the strings that are in both languages. Since both \( L1 \) and \( L2 \) can be expressed by grammars, the intersection can also be expressed by a context-free grammar, making it a context-free language.[/tex]
