60.2k views
4 votes
Suppose language L₁ is regular and language L₂ is context free but not regular. Suppose both are languages over a common alphabet . From the list below, select all statements that are necessarily true.

O Both L₁ and L₂ are context free.
O L₁ and L₂ are both subsets of Σ*.
O L₁ ∪ L₂ is regular.
O L₁ • L₂ is context free.
O L₁ ⊆ L₂

User Pgregory
by
7.7k points

1 Answer

5 votes

Final answer:

The true statements are that both L₁ and L₂ are context-free, L₁ and L₂ are both subsets of Σ*, and L₁ ⋅ L₂ is context-free, while it is not necessarily true that L₁ ∪ L₂ is regular, and there is no evidence to support that L₁ is a subset of L₂.

Step-by-step explanation:

The question relates to formal language theory, a topic within the field of computer science and linguistics. When asked about two languages, L₁ and L₂, in this context, we are referring to sets of strings defined by certain rules, not spoken or written human languages. Language L₁ is regular, meaning it can be described by a regular expression or a finite automaton, while L₂ is context-free but not regular, meaning it requires a context-free grammar or a pushdown automaton for its description.

  • Both L₁ and L₂ are context free: True. Since regular languages are a subset of context-free languages, L₁ is also context-free by definition.
  • L₁ and L₂ are both subsets of Σ*: True. All languages over a given alphabet are subsets of the set of all strings over that alphabet, known as Σ*.
  • L₁ ∪ L₂ is regular: False. The union of a regular language and a context-free language is not necessarily regular. It is context-free because the class of context-free languages is closed under union, but it may not be regular if L₂ is not regular.
  • L₁ ⋅ L₂ is context-free: True. The concatenation of a regular language and a context-free language results in a context-free language because the class of context-free languages is closed under concatenation.
  • L₁ ⊆ L₂: This is False by necessity. There is no information provided that L₁ is a subset of L₂.

User Joe Beuckman
by
8.2k points