Let L1 be a recursive language. Let L2 and L3 be languages that are re...
ANSWER B L1−L3=L1∩L3
Recursively enumerable languages are not closed under complement. So, L3 may or may not be recursively enumerable and hence we can't say anything if L1∩L3 is recursively enumerable or not.
View all questions of this test
Let L1 be a recursive language. Let L2 and L3 be languages that are re...
A) Always True (Recursively enumerable - Recursive) is Recursively enumerable
B) Not always true L1 - L3 = L1 intersection (Complement L3) L1 is recursive , L3 is recursively enumerable but not recursive Recursively enumerable languages are NOT closed under complement.
C) and D) Always true Recursively enumerable languages are closed under intersection and union.
Let L1 be a recursive language. Let L2 and L3 be languages that are re...
Explanation:
To understand why option B is not necessarily true, let's break down the properties of each type of language mentioned in the question.
Recursive Language (L1):
A recursive language is a language for which there exists a Turing machine that can decide whether a given input belongs to the language or not. In other words, there is an algorithm that can always halt and provide a definite answer (either "yes" or "no") for any input.
Recursively Enumerable Language (L2 and L3):
A recursively enumerable language is a language for which there exists a Turing machine that can recognize whether a given input belongs to the language or not. However, this Turing machine may not always halt for inputs that do not belong to the language. It may either halt and provide a definite answer (either "yes" or "no") or loop indefinitely for inputs that do belong to the language.
Now, let's analyze each option:
a) L2 ⊆ L1 is recursively enumerable:
This statement is necessarily true because if L2 is a subset of L1, then any Turing machine that recognizes L1 can also recognize L2. Thus, L2 is recursively enumerable.
b) L1 ⊆ L3 is recursively enumerable:
This statement is not necessarily true because L3 may not have a Turing machine that can recognize L1. L3 being recursively enumerable does not guarantee that it can recognize all recursive languages.
c) L2 ⊆ L1 is recursively enumerable:
This statement is necessarily true for the same reason mentioned in option a. If L2 is a subset of L1, then any Turing machine that recognizes L1 can also recognize L2. Thus, L2 is recursively enumerable.
d) L2 ⊆ L1 is recursively enumerable:
This statement is necessarily true for the same reason mentioned in option a. If L2 is a subset of L1, then any Turing machine that recognizes L1 can also recognize L2. Thus, L2 is recursively enumerable.
Therefore, the correct answer is option B because L1 being a subset of L3 does not guarantee that L3 can recognize L1.
To make sure you are not studying endlessly, EduRev has designed Computer Science Engineering (CSE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Computer Science Engineering (CSE).