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Please refer this Question - here

How in that Question - $(0+1)^*(0+1)(0+1)^* = (0+1)(0+1)^* = (0+1)^*(0+1) = (0+1)^+$

Answer 1

Two ways to handle:-

First see the language accepted:- At least 1 bit,so you can design DFA for tha and write as :-(0+1)(0+1)* or (0+1)*(0+1) as both are representing one bit, And i can say this as (0+1)+.

Other way ,use properties. :- a*a=aa*,a*a*=a*,  aa*=a+

 (0+1)*(0+1)(0+1)* 

Using above property i can exchange first two expressions:- (0+1)(0+1)* (0+1)*  

Then i can combine last two terms :(0+1)(0+1)* 

Then i can combine them in one :-(0+1)+ 

Comments

Please tell  in the below text - How can you say at least 1 bit?

First see the language accepted:- At least 1 bit,so you can design DFA for tha and write as :-(0+1)(0+1)* or (0+1)*(0+1) as both are representing one bit, And i can say this as (0+1)+.


Also where does this property comes - a*a=aa*,a*a*=a*,  aa*=a+ which topic??