## 7 Answers

- Is the answer
- RHS generates $\Sigma^*$ while LHS can't generate strings where $r$ comes after $s$ like $sr, srr$ etc. LHS $\subset$ RHS.
- LHS generates $\Sigma^*$ while RHS can't generate strings where $r$ comes after an $s$. RHS $\subset$ LHS.
- LHS contains all strings where after an $s$, no $r$ comes. RHS contains all strings of either $r$ or $s$ but no combination of them. So, RHS $\subset$ LHS.

### 8 Comments

http://gateforum.com/wp-content/uploads/2013/01/CS-1992.pdf

search question no (XVII)

(*r***s****)*=(*r*+*s*)* but in case of (*r***s****) != (*r*+*s*)*

**** Disclaimer : I am not sure that its right or wrong . So if its wrong please don`t blame me :P

I think *r*(∗) != *r*∗ ,

let take a example A(B*) . In this case (B*) will be evaluated first because its with in the bracket . And now consider this R(*) then (*) will be evaluated first but it contain only ∅ and ∅*= ∈ . So R.(∈) = R . peace

suppose L= {} this language means two things either there is no final state or final state is unreachable from initial state .

And for L={∈} this language means without any input we can reach the final state ,So ∅ and ∈ is very much different . But in this case ∅*=∈*=∈ . peace

People who are confused what's going on in this question please refer this

Option A is wrong as right hand side can produce null string while left hand side cannot.

Now coming to answer B , there can be 2 situation with this

1. Mistake in presenting option B

Option given there is wrong as (r*s*) cannot generate rsr

2. Answer will be correct if we have (r*s*)* =(r+s)* you can refer Ullman section 3.4.6

3. If we think like this option B can be correct

(r*s*)=(r+s)*

Taking closure on both the sides we get

(r*s*)*=((r+s)*)*

We know thst (a*)*= a* check identities

Hence (r*s*)*= (r+s)*

Others please check if my third interpretation is valid and correct me if I am wrong .

(B) RHS generates Σ* while LHS can't generate strings where r comes after s like sr, srr, etc.

LHS ⊂ RHS.

(C) LHS generates Σ* while RHS can't generate strings where r comes after an s.

RHS ⊂ LHS.

(D) LHS contains all strings where after an s, no r comes. RHS contains all strings of either r or s but no combination of them.

So, RHS ⊂ LHS.