# Understanding how to decide whether a language for a DFA is decidable

So from what I understand, if a language is recognisable then using a TM it can be accepted and halted or rejected or halted, however a language that is decidable can be accepted and always halts on rejection?

Given the language $$A_{\mathrm{DFA}} = \{\langle A \rangle\mid A \text{ is a DFA and } L(A) = \{0, 1\}^∗\}$$ , I want to determine if it is decidable.

What I have come up with so far is that it is a recognisable language, I don't believe it can be rejected, only accepted and halted due to the fact once the read head comes across a blank it can just accept it. Is this the way I should be thinking? Am I correct in thinking it is not decidable or is this not a valid way to argue the point?

What should my process of thinking be when I am trying to prove it by contradiction if it is not decidable?

Thanks in advance for any tips!

• I can just about understand your question but the way you've written it suggests to me that you're rather confused about the basic concepts. For example, you say "I have been given a DFA = {< A >|A is a DFA and L(A) = {0, 1}∗}" but that isn't a DFA: it's a language (a set of strings). A DFA is a kind of simple computer, not a set of strings. And I can't understand your attempt at proving recognizability at all. I think your first step is to review the basics, such as what a DFA and a language is, and how DFAs work. The language is decidable. – David Richerby Apr 13 '17 at 11:03
• Apologies it is a A DFA, I don't know how to format my text in the way it should be, a capital A and then DFA below that. Does that mean a language in relation to the DFA A? – HCF3301 Apr 13 '17 at 11:09
• Ahh, gotcha. I'll edit it for you. Think about what it means if $L(A)\neq\{0,1\}^*$ and how you might test for that condition. – David Richerby Apr 13 '17 at 11:10
• That's exactly what you need to do, yes. – David Richerby Apr 13 '17 at 12:01

Is there an algorithm that, given a DFA, decides whether its language is $\{0,1\}^*$?