Nobody knows yet if $P=NP$. Consider the language $L$ defined as follows.$$L = \begin{cases} (0+1)^* & \text{ if } P = NP \\ \phi & otherwise \end{cases} $$Which of the following statements is true?

- $L$ is recursive
- $L$ is recursively enumerable but not recursive
- $L$ is not recursively enumerable
- Whether $L$ is recursively enumerable or not will be known after we find out if $P=NP$

### 1 comment

## 4 Answers

### 9 Comments

if p=np( which is till now not solve ie. we dont know either it is yes or no) then L= (a+b)*

else L = phi

here condition dont matter in problem B/ c either it is true or false language is always regular.

Since (a+b)* and Phi both are regular , and since regular are subset of recusive which is decidable.

if P=NP problem becomes decidable then L will becomes Regular.

if we don't know whether P=NP or not , we can't comment about regularity but at least we can say L is REC at least.

(0+1)^{* }and ∅ both are regular , means there won't be any possibility of TM hang (looping) and in that we we can design algo for L. so with this argument L=REC.

tell me if someone find anything wrong in my arguments.

we have two problems one is P and another is NP

now we give these 2 problems to TM then acc to condition given if both P and NP are equal then it will give (0+1)∗ otherwise ϕ , not others will be given

so we make total TM for this becoz it say "YES" or "NOT", never fall into loop

and if we make Total TM for language then that language is recursive

hence **Ans is A **