@ akash.dinkar12
yes, i understand ,
a*(ba*)* =(a+b)* possible
now a*(ba*)* = (a*b)* a * .....(i) so put a=0 b=1 and we get (0*1)*0*
given 0*(10*)* = (0*1)*0* = (0* 1) (0* 1) 0* = (0 1 ) (ϵ 1) ϵ = 011 [ by putting 0* = ϵ ] it generates all strings over Σ . Where Σ =(0,1)
now option A, (1*0)*1* = (1* 0) (1* 0) 1* = ( 1 0) (ϵ 0) ϵ = 100 [ by putting 1* = ϵ ] this also generates all strings over Σ . where Σ = (0,1)
so thats why 0* (1 0*)* = (1* 0)* 1* because from both we can get all strings of 0 and 1 generates over Sigma .
PS: There is no way we can directly show 0*(10*)* = (1*0)*1* ... we need to generate strings and compare are both same or not !
Hope you are getting my point .