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If $L$ and $\bar L$ are recursively enumerable then $L$ is

1. regular
2. context-free
3. context-sensitive
4. recursive
retagged | 3k views

(D) recursive

$L$ is recursively enumerable means a $TM$ accepts all strings in $L$. $\bar L$ is recursively enumerable means a $TM$ accepts all strings in $\bar L$. So, we can always decide if a string is in $L$ or not, making $L$ recursive.

http://goo.gl/RtV8MO

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Let say if L is regular then L(complement) is also regular then both are RE.???

also if L is CFG then L(complement) is recursive(as CFG are not closed under complementation) then also both will be RE.???

same with CSG and recursive

Then why not all option is correct?

Plz explain
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@Hitesh
L = {$a^n b^n c^n$} is a CSL, hence it is RE language too as every CSL is RE as per the Chomsky hierarchy.

Complement of L, L' is also CSL as CSL is closed under complement operation. Hence L' is also RE.
Now, see the case, both L and L' are RE, but L is neither regular nor CFL.
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Ref:

+1 vote

Recursively enumerable language is NOT Closed under complementation where as Recursive lang is Closed Under complementation .

Ans D) Recursive , is correct option.

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why regular is not the answer all regular languages arealso closed in complement.
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@phelps L = {$a^n b^n c^n$} is a CSL, hence it is RE language too as every CSL is RE as per the Chomsky hierarchy.

Complement of L, L' is also CSL as CSL is closed under complement operation. Hence L' is also RE.
Now, see the case, both L and L' are RE, but L is not regular.
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Manu Thakur The logic is simple, If L is recursively enumerable, then the complement of L is recursively enumerable if and only if L is also recursive.For proof just see the best ans.Actually, this is one of the best methods to differentiate (or to check whether the lang is rec or rel) B/w Recursive language and Recursively Enumerable language.