Let \(L_{1}\) and \(L_{2}\) be two languages over a finite alphabet, such that \(L_{1} \cap L_{2}\) and \(L_{2}\) are regular languages.
Which of the following statements is/are always true?
GATE CS - 2026,
Answer: C
The statement always true is \(L_{2}\) is context-free.
\(L_{2}\) is given to be regular, and every regular language is context-free.
Nothing useful is known about \(L_{1}\). Take \(L_{2} = \emptyset\), which is regular; then \(L_{1} \cap L_{2} = \emptyset\) is regular whatever \(L_{1}\) is. With \(L_{1} = \{a^{n}b^{n}\}\), \(L_{1}\) is not regular and \(L_{1} \cup L_{2} = L_{1}\) is not regular either. With \(L_{1} = \{a^{n}b^{n}c^{n}\}\), \(L_{1}\) is not even context-free. So the three statements about \(L_{1}\) can all fail.
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