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Recursive enumerable languages are not closed under
  • a)
    concatenation
  • b)
    union
  • c)
    intersection
  • d)
    set-difference
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Recursive enumerable languages are not closed undera)concatenationb)un...
Recursive Enumerable Languages
Recursive enumerable languages, also known as recursively enumerable or simply RE languages, are a class of languages in the field of theoretical computer science. These languages are recognized by Turing machines that may not halt on some inputs but will eventually accept if the input string belongs to the language. In other words, RE languages are those for which there exists a Turing machine that will accept all valid strings, but may either reject or run forever on invalid strings.

Closure Properties
Closure properties of languages refer to the properties that are preserved under certain operations. For example, if a class of languages is closed under a particular operation, it means that applying that operation to languages within the class will always result in a language that also belongs to the same class.

Set-Difference Operation
The set-difference operation, also known as relative complement, is an operation between two sets. Given two sets A and B, the set-difference operation A - B results in a set that contains all the elements of A that are not in B.

Explanation
The given question asks whether recursive enumerable languages are closed under set-difference. In other words, if we take two recursive enumerable languages A and B, is it always true that their set-difference A - B will also be a recursive enumerable language?

To answer this question, we need to consider the properties of recursive enumerable languages and the set-difference operation.

Concatenation, Union, and Intersection
Before we discuss set-difference, let's briefly consider the closure properties of recursive enumerable languages under other operations.

- Concatenation: Recursive enumerable languages are closed under concatenation. If we take two recursive enumerable languages A and B, their concatenation AB will also be a recursive enumerable language.
- Union: Recursive enumerable languages are closed under union. If we take two recursive enumerable languages A and B, their union A ∪ B will also be a recursive enumerable language.
- Intersection: Recursive enumerable languages are closed under intersection. If we take two recursive enumerable languages A and B, their intersection A ∩ B will also be a recursive enumerable language.

Set-Difference and Non-Closure
Now let's consider the set-difference operation. Suppose we have two recursive enumerable languages A and B, and we want to find their set-difference A - B.

If A and B are both recursive enumerable languages, it is possible to construct a Turing machine that will accept all strings in A and reject all strings in B. However, when we take the set-difference A - B, it is not guaranteed that the resulting language will still be recursive enumerable.

The reason behind this is that the set-difference operation can introduce strings that are not recognized by any Turing machine. In other words, the set-difference can result in a language that is not recursively enumerable.

Therefore, the correct answer to the given question is option 'D' - recursive enumerable languages are not closed under set-difference.
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Community Answer
Recursive enumerable languages are not closed undera)concatenationb)un...
Recursively enumerable languages are also known as Type 0 grammars. They are not closed under set difference.
Here, T - P (where T and P are two sets) may or may not be recursively enumerable.
If T is recursively enumerable, then the complement of T is recursively enumerable if and only if T is also recursive.
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Recursive enumerable languages are not closed undera)concatenationb)unionc)intersectiond)set-differenceCorrect answer is option 'D'. Can you explain this answer?
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