October 6, 2023
October 6, 2023
October 6, 2023
###### Theory-of-Computation
October 6, 2023
 Question 12
Which of the following statements is/are TRUE?
 A Every subset of a recursively enumerable language is recursive. B If a language L and its complement L’ are both recursively enumerable, then L must be recursive. C Complement of a context-free language must be recursive. D If L1 and L2 are regular, then L1 ∩ L2 must be deterministic context-free
Question 12 Explanation:
Every subset of recursively enumerable need not be recursive. For example (a+b)* is a universal language and regular. Every language over alphabet {a,b} is subset of (a+b)*. This language is regular so it is recursively enumerable also and so many languages are known to be non-recursive which are definitely subset of (a+b)*. Thus every subset of recursively enumerable need not be recursive.
If L and L(complement) both are recursively enumerable then L must be recursive. It is a theorem.
Every CFL is CSL and CSL is closed under complement so complement of CFL must be CSL and every CSL is recursive. Thus the complement of CFL must be CSL, hence it must be recursive also.
If L1 and L2 are regular then their intersection must be regular, as regular languages are closed under intersection, so L1 intersection L2 must be regular, hence it must be DCFL, since every regular is also a DCFL.
Question 12 Explanation:
Every subset of recursively enumerable need not be recursive. For example (a+b)* is a universal language and regular. Every language over alphabet {a,b} is subset of (a+b)*. This language is regular so it is recursively enumerable also and so many languages are known to be non-recursive which are definitely subset of (a+b)*. Thus every subset of recursively enumerable need not be recursive.
If L and L(complement) both are recursively enumerable then L must be recursive. It is a theorem.
Every CFL is CSL and CSL is closed under complement so complement of CFL must be CSL and every CSL is recursive. Thus the complement of CFL must be CSL, hence it must be recursive also.
If L1 and L2 are regular then their intersection must be regular, as regular languages are closed under intersection, so L1 intersection L2 must be regular, hence it must be DCFL, since every regular is also a DCFL.