Is the language of DFAs which do not accept themselves recognizable? I understand how a the language of turing machines which do not accept themselves is not recognizable but I'm not sure if the same proof could be used to describe a DFA... i.e a proof by contradiction in which a recognizer M recognizes a TM which does not accept itself on input ...creates a contradiction in which if M accepts then M does not accept itself.
DFAs always terminate, so if you want to check whether a given DFA accepts or rejects a given word (say its representation), just run it on the word. You can even do this in polynomial time.