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Which of the following regular expression identities is/are TRUE?

1. $r^{(^\ast)} =r^\ast$
2. $(r^\ast s^\ast)=(r+s)^\ast$
3. $(r+s)^\ast = r^\ast + s^\ast$
4. $r^\ast s^\ast = r^\ast+s^\ast$

Anser is given as B only . I think A is also True . Correct me if wrong
A's RHS can accept null strings, LHS can't.
if r=ab then r*=abababa.....
r(*)=ab(*)=ab

why option a is not correct ?

r(*) = r*

and

r(*)=r*

pls tell the difference between above two

u have problem in r(*) right?

take an example a(a*) where the strings are {a,aa,aaa....}

Similarly r(*) where * is separate from r, it cannot accept epsilon

but r* can accept epsilon too

if r=a(a*)

a(a*)(*) in this what is the meaning of  concatenation of  * with a(a*) .what does it represent?

a(a*) and a(a*)(*) same
but similarly (a+b)* and (a*b*)* are both similar

got it.if

r(*) = r* this is the case then both are same

confirm this too pls

does a(*) means a.(fi*)..ie - a ??
@PEKKA how b is correct..?

R1= (r*s*)*  !=   R2=(r+s)*

In R1 "sr" can never be generated but in R2 " sr " can be generated
how can we get "sr " using lhs of option B?

(r*s*)* = (r*s*)(r*s*) = (s)(r)
thanks a lot got it

How $\left ( r^{*}s^{*} \right )^{*}$ = $\left ( r + s \right )^{*}$ ? Any proof in simple words !!

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2. RHS generates $\Sigma^*$ while LHS can't generate strings where $r$ comes after $s$ like $sr, srr$ etc. LHS $\subset$ RHS.
3. LHS generates $\Sigma^*$ while RHS can't generate strings where $r$ comes after an $s$. RHS $\subset$ LHS.
4. 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.
by

Can you please give any source which contain  r(*)=r*.Previous year question book gave answer b

and their b is (r*s*)*=(r+s)*

search question no (XVII)

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

@Anu No. I don't have a proof. In fact I choose i only because all other options are false. In case (b) is with a *, I would choose (b) as there is ambuiguity over (a). I doubt if they meant (empty string*) with (*) in which case (a) is false.

**** Disclaimer : I am not sure that its right or wrong . So if its wrong please dont 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

^But why you took {}* and not $\epsilon^*$ ? That seems more appropriate. $\epsilon^* = \epsilon$, so answer would still be same.
edited
yes answer might be same but i think ( for safety purpose i say " i think " what if i got wrong :D ) its not right thing to do .

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
Option B has to be right option because,

(r+s)* = ϵ + (r+s)*

Thus it can produce Σ*

And obviously (r*s*)* can also produce Σ*

================

For those saying strings starting with s can't be produced:

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

== s(r+s)*
can anyone explain the difference between the LHS and RHS of option A… I did not get the difference

B is only correct

why not A?

A's RHS can accept null strings, LHS can't.
if $r=ab$ then $r^*=abababa$.....
$r(^*)=ab(^*)=ab$

r(*) =ab(*) =ab

Is this the only string possible in LHS?

Or abababa...  Also possible?

What does it exactly mean (*)...?

@Amitabh Tiwari 1

@Anirudh
What does it exactly mean by (*)...?
A is not correct got it but how  B is correct ??

(P* q*)* = (P* + q*)* Then how B option is correct ??
someone please clarify what does (*) mean?

Wrong answer.   LHS can't generate strings where r comes after s like sr. where RHS can generate any combination of r and s.  LHS $\subset$ RHS . B,C,D are FALSE for sure.

a write ans bro

## Only option B is Correct.

There is no *  outside LHS of option B.

ANS: A.

B) LHS: (r*s*)*={ε ,r,s,rs....}

RHS:(r+s)*={ε ,r,s,rs,sr...} ,sr can't be generated by LHS...so wrong.

C)(r+s)* - generates all string over {r,s}

(r*+s*) can't generate anything as combination of sr, so wrong.

D) r*s*= {ε ,r,s,rs..} but r*+s* cannot generate rs , so wrong.

by

### 1 comment

Option B is only true .

how???
Sir,

A is not true as

r (*) cant accpt null string, the precedence is for () first, means it will be on an empty string rather than r

Whereas r* accepts null string

But i am nit getting how C is not true ?

I mean (r+s)* accepts all strings as r*+s* . Doesnt it ?

(r+s)*  accept rs but

r*+s* cant

(r+s)* means any string of a,b

r* + s* means any string of any length of either 'r' or 's'

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 .

### 1 comment

can u please tell me what's the difference between LHS & RHS in option A ???

can we do like that. I mean here * is considered as an operator(kleen closure), not a constant. I'm not sure about it, but may be.....

(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.