Is the language {⟨A⟩∣A is an NFA and L(A)={0,1}∗} decidable? decidable?

Question

How would one go about proving/disproving the language {⟨A⟩∣A is an NFA and L(A)={0,1}∗} is/isn't decidable?

I assumed at first since it was an NFA involved it would be decidable, but since there is no input string to simulate does this change things? If so, how? I can't think of a turing machine that would decide this. Since {0, 1}* is theoretically infinite does mean a turing machine may never halt thus the language is undecidable? If so how do I go about proving this?

Answer 1

Speaking informally, you can show this by constructing a Turing Machine constructs a DFA D_A equivalent to NFA A. It then constructs the DFA D_0 that accepts the language {0,1}*, we can then simulate the decider for EQ_DFA on .

Formally speaking, construct TM S: S = "On input :

1. Construct DFA D_A equivalent to A
2. Construct DFA D_0 that accepts {0,1}*
3. Simulate decider F for EQ_DFA on where EQ_{DFA} = { | A and B are DFAs and L(A)=L(B)} (we know that EQ_{DFA} is a decidable language).
4. Accept if F accepts; reject if F rejects."

Answer 2

Less formally:

1. we can algorithmically determine check that the input represents an NFA according to our format
2. we can algorithmically construct a DFA equivalent to the NFA using the subset construction
3. we can algorithmically minimize the DFA using any of several known algorithms
4. we can algorithmically compare the resulting DFA to the one-state DFA for {0, 1}*
5. if equal, output yes; otherwise, output no.

Because we can describe an algorithm to do this, and because we don't suppose we have more computational power than Turing machines (at least the above computations do not require such), the problem must be decidable.