If L1 ∩L2 and L2 are regular then which of the following is always true?

The correct option is: ¯ L₂ is CFL

Let us analyze the given conditions and verify why this option is always true step-by-step.

Step 1: Analyze the properties of L₂
We are given that L₂ is a regular language.

Step 2: Apply the closure properties of regular languages
One of the fundamental closure properties of regular languages is that they are closed under complementation. This means that if a language is regular, its complement is also guaranteed to be regular.
Since L₂ is regular, its complement:
$\overline{L_2}$
is also a regular language.

Step 3: Relationship between Regular Languages and Context-Free Languages (CFLs)
According to the Chomsky hierarchy, the class of regular languages is a strict subset of the class of context-free languages (CFLs). Therefore, every regular language is also a context-free language.
Since the complement of L₂ is regular, it must also be a context-free language (CFL).

Step 4: Counterexamples for the other options
To show why the other options are not always true, consider a case where L₂ is the empty language (which is regular, i.e., L₂ = ∅) and L₁ is any non-context-free language (for example, a highly complex non-recursive language).
In this case:
1. The intersection L₁ ∩ L₂ = L₁ ∩ ∅ = ∅, which is regular. This satisfies the problem constraints.
2. However, L₁ is not regular, so "L₁ is Regular" is false.
3. The complement of L₁ (¯ L₁) is not a CFL, so "¯ L₁ is CFL" is false.
4. The union L₁ ∪ L₂ = L₁ ∪ ∅ = L₁, which is not regular, so "L₁ ∪ L₂ is regular" is false.

Thus, the only statement that is guaranteed to always hold true is that ¯ L₂ is a CFL.