0 votes 0 votes $(a+b)^* =a^*(ba^*)^*$ As this identity already proved. But $a^*(ba^*)^*$ couldn't generate "bab" . But $(a+b)^*$ could generate all strings over {a,b}. Then the above identity seen to be proved false. Please discuss how this is possible. Dhananjay15 asked Aug 19, 2018 Dhananjay15 471 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Shaik Masthan commented Aug 20, 2018 reply Follow Share @Dhananjay15, Brother if it is already proved means it should be correct... If we didn't get that string, it means I hope we did somewhere mistake @Vikas Verma, added it as answer, just check it. 0 votes 0 votes Swapnil Naik commented Aug 20, 2018 reply Follow Share Let's generate bab - (ba*)* = { ε , (ba*), (ba*)(ba*) ....... } a* = { ε , a, aa, aaa.....} I am expanding (ba*)* twice i.e. (ba*)* = (ba*)(ba*) a*(ba*)* = a*(ba*)(ba*) = ε(b a )(b ε ) = bab .......... [ ε .b = b or b . ε = b] 1 votes 1 votes Please log in or register to add a comment.
2 votes 2 votes a^* (ba^*)^* =( ba^*)^* = (ba)(ba^*) = bab Vikas Verma answered Aug 20, 2018 Vikas Verma comment Share Follow See all 0 reply Please log in or register to add a comment.