Regular Expression

Question

Can I write $a^* + b^* = (a + b)^*$ ????

Comments

NO  you cannot write

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To check equivalence of two regex create minimal DFA for both regex and if they are same then yes they are equivalent.

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sumit goyal 1 please explain the difference 

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a* + b *  wil give { a , aa ,aaa ,aaaa ,......  , b,bb,bbb,bbbb,bbbb  , epsilon }

(a+b)* will give all possible combinations

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so how to create a DFA for this? a* + b *

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It is drawn like this

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@ sumit goyal 1 You forgot dead state.$4$ states will be there moreover first state is final

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i cretaed a nfa , so i didnot take dead state ,

Answer 1

NO,

we can write (a+b)*=(a*+b*)*

                               =(a*+b)*

                               =(a+b*)*

                               =(a*b*)*

                               =a*(ba*)* 

                               =b*(ab*) *

 

or

make minimal dfa for both the RE and check LHS != RHS ,infact  language both are different 

 

Comments

$(a+b)^* \neq (ab^*)^*$

And last one should be b*(ab*)*

 

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yes..by mistake correct one is

(a+b)*=(a*b*)*

           =b*(ab*)*

           =a*(ba*)*

Answer 2

$1^{st}$ one is RHS.
$2^{nd}$ one is LHS.
LHS $\neq$ RHS

Answer 3

@MRINMOY_HALDER  check this

https://gateoverflow.in/169116/regular-expression

Answer 4

No ,because both are not same as u can see by the 1st REstring produce like ab,aabb.a,b,aa,bbetc but u can't get string like aba,aaabbabab,ababa etc .

whereas by the 2nt RE u will get all string produce by a and b.