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Is $E_{DFA}$ in the class of regular languages?

$\qquad E_{DFA} = \{ \langle D \rangle \mid D \text{ is a DFA }, L(D) = \emptyset\}$

My argument is that it is because all of the DFAs in $E_{DFA}$ can be reduced to a DFA that only accepts on no input. Is this right?

Raphael
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user3370201
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    Regular languages have a very limited structure. There's no reason for $E_{DFA}$ to be regular. It should be easy to show by multiple methods, e.g., using pumping lemma. – Ran G. Apr 27 '16 at 17:48
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    For other possible methods, refer to How to prove that a language is not regular? – Ran G. Apr 27 '16 at 18:04
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    What kind of encoding is $<.>$ ? – Klaus Draeger Apr 27 '16 at 19:07
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    What a weird exercise problem. Does it maybe say "minimal DFA" in the original? – Raphael Apr 27 '16 at 20:24
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    We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. – D.W. Apr 27 '16 at 20:43

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