If L1 and L2 are context free languages and R a regular set, one of the…

Let's evaluate the algebraic closure properties for each operation listed in the options:

• Concatenation (L1 . L2): CFLs are closed under concatenation. If you have two context-free languages, combining them sequentially always results in another context-free language.

• Union (L1 U L2): CFLs are closed under union. If you combine all strings from two different context-free languages, the resulting language remains context-free.

• Intersection with a Regular Language (L1 ∩ R): CFLs are closed under intersection with a regular language. This is a special property: if you pass a context-free language through a regular filter, the result is guaranteed to stay context-free.

• Intersection of two CFLs (L1 ∩ L2): Context-Free Languages are NOT closed under intersection.

For example, consider these two context-free languages:

• L1 = { (an)(bn)(cm) | n, m >= 0 } (Context-free because it only matches a's and b's)

• L2 = { (am)(bn)(cn) | n, m >= 0 } (Context-free because it only matches b's and c's)

If we take their intersection (L1 n L2), all three symbols must have the exact same count:

• L1 ∩ L2 = { (an)(bn)(cn) | n >= 0 }

This resulting language is a well-known Context-Sensitive Language and cannot be parsed by a standard Pushdown Automaton. Therefore, the intersection of two CFLs is not necessarily a context-free language.

Correct Answer: B