To prove the equivalence of two regular expressions , we should try to reduce one r.e. into r.e using identities which we have or using proper reductions.For understanding proper reduction , let us understand by a simple example first :
We know (r* + s*)* = (r*s*)* = (r + s)*.But we have to know how it is true.For that let us the equality between 1st and 2nd terms:
In (r*s*)* , if we keep s* = ϵ , so we obtain r*.
Similarly by keeping r* = ϵ , we obtain s*.
So we can take either any combination of combination of r* and s*
Hence (r*s*)* = (r* + s*)*
Similarly taking r from r* and s from s* we get : (r* + s*)* = (r + s)*
Now on the similar approach lets solve the given query :
LHS = (a*ab + ba)* a*
Let r1 = (a*ab + ba)* and r2 = a*
Keeping a* = ϵ , we obtain the LHS term as : (ab + ba)*..This is one case
Now let r1 = ϵ , so r1* = ϵ , hence LHS term is reduced to : a*
Hence the overall expression considering the above 2 cases can be briefed as : (a + ab + ba)* since (ab + ba)* is already obtained in the 1st case and using the 2nd term we get any combination of 'a' and hence it can be inside (a + ab + ba)*
Hence the two expressions given are equivalent
