559 views

### 3 Comments

Use Rice's Theorems here
edited
How?plz explain.

I get that it's not decidable but how is it semidecidable
Had this Question been set properly, answer would be $D.$

## 1 Answer

Best answer

If someone knows even only the basic definitions of Automata Theory, will realize How Poorly this question has been constructed.

Now coming to the Question, Why is it Poorly constructed?

$L$ is the set of all Regular languages over some alphabet, say $\{ 0,1\}.$ So, $L$ itself is NOT a language But a collection of languages. A Language is a set of Strings over some Alphabet. What is a string in $L$ ? What is the alphabet of $L$ ?

So, Now you know what is Wrong with this Question. Whoever set this Question has very poor knowledge of Automata Theory.

Now let me change this Question and make it a valid one.

Consider $<M>$ be the encoding of a Turing Machine as a string over alphabet $Σ = \{0, 1\}$

$L = \{ <M> | L(M) \,\, is \,\, Regular\}$

Now this language $L$ is Non-RE language.

We can prove it using Rice's Theorem.

Property $P(L) = L \,\, is\,\, Regular$ is Non-trivial and Non-monotonic Property of RE languages. And by Rice's Theorem we know that Any Non-trivial and Non-monotonic Property of RE languages is Undecidable and Unrecognizable.

Hence, $L$ is Not-RE.

### 1 comment

Nice explanation.thanks a lot

4 votes
2 answers
1
575 views
0 votes
1 answer
2
299 views
3 votes
2 answers
3
813 views
3 votes
2 answers
4
769 views