Correct Option: $C$
Given,
- $L_1-$ Regular Language (RL)
- $L_2-$ Context-Free Language (CFL)
$(A)$ $L_1 \cap L_2 \to$ CFL
Because intersection operation with regular languages is closed under CFLs.
Hence, True.
$(B)$ $L_1.L_2\to$ CFL
Every regular language is a CFL and CFLs are closed under concatenation.
Hence, True.
$(C)$ $L_1-L_2$ $\equiv$ $L_1 \cap L_2 ^{c}$.
Suppose, let’s consider $L_1 = \Sigma^*$ and $L_2$ as any CFL and we get $L_1 \cap \overline{L_2} = \overline{L_2}.$
Since CFLs aren’t closed under complementation, this means $L_1 – L_2$ NEED NOT be a CFL!
Hence, False.
$(D)$ $L_1\cup L_2 \to$ CFL
Since CFLs are closed under union operation and a regular language is also a CFL.
Hence, True.
Ref: Closure Property of Language Families