User blog comment:Vel!/Formal logic challenge/@comment-32876686-20171023182635/@comment-1605058-20171023190056

Nice job! Some of the things could be phrased better:

When talking about a function from some A to some B, we usually say that it assigns some element of B to some element of A, so in this case, it assigns a computable function to every natural number. [by the way, in logic, set theory and computability theory, 0 is usually included in the set of natural numbers. However, this is just a convention and is not too relevant.]

Since the function is surjective, you could've just said that \(Q\) is the set of all computable functions. You could've also used the notation used in the theorem - \(\textbb{P}^1\). There is no need to introduce new symbols if you are given some.

\(D\) should be the problem of whether the output of the function \(P\) is in \(A\) given some input.