Problem Detail: I’ve converted an NFA to a DFA. But even after checking over it a few times, it still doesn’t feel right. I’m sure this is trivial, but I’d like someone to give me an idea where I went totally wrong on this. NFA: 
DFA: 
DFA:
Asked By : stackuser
Answered By : J.-E. Pin
Well, of course you have to merge the two empty states and there should be two transitions $({q_5}, a, emptyset)$ and $({q_5}, b, emptyset)$ if you want a complete automaton, but otherwise, it looks right and I agree with Subhayan: both automata accept $ab^+ cup ab^+a$.
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Question Source : http://cs.stackexchange.com/questions/14458
