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Let L1 be a regular language, L2 be a deterministic context-free language and L3 a recursively enumerable but not recursive language.
Which one of the following statements is false?
- a)L1 ∩ L2 is a deterministic CFL
- b)L3 ∩ L1 is recursive
- c)L1. L2 is context free
- d)L1∩ L2∩ L3 is recursively enumerable
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Let L1 be a regular language, L2 be a deterministic context-free lang...
A) L1∩L2 is a deterministic CFL, because Regular ∩ DCFL is
always DCFL
B) L3∩L1 is need not be recursive, But always REL
C) L1.L2 is context free, because Regular. DCFL is always DCFL
D) L1∩L2∩L3 is recursively enumerable, because all are REL
Most Upvoted Answer
Let L1 be a regular language, L2 be a deterministic context-free lang...
Explanation:
To determine which of the given statements is false, let's analyze each option one by one.
Option 'a': L1 ∩ L2 is a deterministic CFL
The intersection of a regular language (L1) and a deterministic context-free language (L2) can be either regular or context-free. Since both regular languages and deterministic context-free languages are closed under intersection, the intersection of L1 and L2 will also be either regular or deterministic context-free.
Therefore, option 'a' is true.
Option 'b': L3 ∩ L1 is recursive
The intersection of a recursively enumerable language (L3) and a regular language (L1) can be either recursively enumerable or recursive. However, the option states that L3 ∩ L1 is recursive, which may not always be true. It is possible for the intersection to be recursively enumerable but not recursive.
Therefore, option 'b' is false.
Option 'c': L1 . L2 is context free
The concatenation of a regular language (L1) and a deterministic context-free language (L2) will always result in a context-free language. Regular languages are a subset of context-free languages, and the concatenation operation preserves the context-free property.
Therefore, option 'c' is true.
Option 'd': L1 ∩ L2 ∩ L3 is recursively enumerable
The intersection of three languages, L1, L2, and L3, can be recursively enumerable. Since recursively enumerable languages are closed under intersection, the intersection of L1, L2, and L3 will also be recursively enumerable.
Therefore, option 'd' is true.
In summary, the false statement is option 'b': L3 ∩ L1 is recursive.
To determine which of the given statements is false, let's analyze each option one by one.
Option 'a': L1 ∩ L2 is a deterministic CFL
The intersection of a regular language (L1) and a deterministic context-free language (L2) can be either regular or context-free. Since both regular languages and deterministic context-free languages are closed under intersection, the intersection of L1 and L2 will also be either regular or deterministic context-free.
Therefore, option 'a' is true.
Option 'b': L3 ∩ L1 is recursive
The intersection of a recursively enumerable language (L3) and a regular language (L1) can be either recursively enumerable or recursive. However, the option states that L3 ∩ L1 is recursive, which may not always be true. It is possible for the intersection to be recursively enumerable but not recursive.
Therefore, option 'b' is false.
Option 'c': L1 . L2 is context free
The concatenation of a regular language (L1) and a deterministic context-free language (L2) will always result in a context-free language. Regular languages are a subset of context-free languages, and the concatenation operation preserves the context-free property.
Therefore, option 'c' is true.
Option 'd': L1 ∩ L2 ∩ L3 is recursively enumerable
The intersection of three languages, L1, L2, and L3, can be recursively enumerable. Since recursively enumerable languages are closed under intersection, the intersection of L1, L2, and L3 will also be recursively enumerable.
Therefore, option 'd' is true.
In summary, the false statement is option 'b': L3 ∩ L1 is recursive.
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Let L1 be a regular language, L2 be a deterministic context-free language and L3 a recursively enumerable but not recursive language.Which one of the following statements is false?a)L1 ∩ L2 is a deterministic CFLb)L3 ∩ L1 is recursivec)L1. L2 is context freed)L1∩ L2∩ L3 is recursively enumerableCorrect answer is option 'B'. Can you explain this answer?
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