TITLE: Countable choice and term extraction
QUESTION [13 upvotes]: The constructive Axiom of Countable Choice (ACC) is widely accepted due to its computational content. It states that:
$$ \forall n\in \mathbb{N} . \exists x \in X . \varphi [n, x] \implies \exists f: \mathbb{N} \longrightarrow X . \forall n \in \mathbb{N} . \varphi [n, f(n)]$$
to mean: if we can (constructively) prove the antecedent than there exists a function that produces $x$'s satisfying the "certificate" $\varphi [n, x]$.
The famous realizability interpretation claims that in order to prove the antecedent, we already have to construct a term (a function) witnessing $\varphi [n, x]$. Therefore, ACC amounts to simple reformatting. However, realizability is a metamathematical concept in which sense it is fair to ask how the claimed functions come into play.
Question: What does a computable realization of ACC precisely mean and how to convert a proof involving ACC into a program?

REPLY [3 votes]: We are looking at the scheme 
$$(\forall n\in \mathbb{N})(\exists x \in X)\psi(n,x) \to (\exists f \in X^{\mathbb{N}})(\forall n) \psi(n,f(n)).$$
There are two viewpoints: from pure (classical) computability theory, and from a constructive viewpoint.
In computability theory
The classical computability of the choice scheme depends greatly on the set $X$ and the class of formulas that $\psi$ comes from. In general, $f$ need not be computable, even when $X$ is just $\mathbb{N}$. 

When $X = \mathbb{N}$ and $\psi$ is $\Sigma^0_1$, the function $f$ obtained is always computable. We can define $f(n) = (\mu m)\psi(n,m)$, and this will be computable because $\phi$ is $\Sigma^0_1$. 
If $X = \mathbb{N}$ and $\psi$ is $\Pi^0_2$, then the function $f$ is not computable in general. For example, if we let $\phi_n$ be a standard enumeration of the computable functions, we might let 
$$\psi_0(m,k) \equiv [(\phi_{n}(n) \text{ halts in $k$ steps}) \Leftrightarrow (\exists s) (\phi_{n}(n) \text{ halts in $s$ steps})].$$  Certainly we have $(\forall n)(\exists k)\psi_0(n,k)$, but there is no computable selector $f$. 

Quite a bit is known about the strength of the scheme for the cases $X = \mathbb{N}$, $X = \mathbb{N}^\mathbb{N}$, and $X = P(\mathbb{N})$, when $\psi$ comes from various classes in the arithmetical or analytical hierarchy.  Depending on these choices, the strength can vary enormously. In come cases, the choice function may require strong axioms in second-order arithmetic. In other cases, as we have seen, $f$ may be computable. 
In constructive mathematics
In constructive mathematics, the situation can be quite different. The reason for this difference is that a proof of $(\forall m)(\exists k)\psi(m,k)$ must be more uniform in constructive mathematics, in various ways, than in classical mathematics. The classical proof of $(\forall m)(\exists k)\psi_0(m,k)$ from above requires a nonconstructive choice between two cases (whether $\phi_n(n)$ halts or not). Such a choice is impossible in constructive systems. 
Bishop seems to have thought, essentially, that the right side of the choice scheme is a synonym for the left side, in his perspective on constructivism. This is evident in his comment about the choice scheme following from "the very meaning of existence". Similarly, the BHK interpretation validates the choice scheme, as do several kinds of realizability. 
In these settings, I would not view the choice scheme as a "mere existence claim". Instead, I would view as stating an equivalence that follows directly from the meanings that its components have been given, much like the classical equivalence $$\lnot (\exists x)\phi(x) \Leftrightarrow (\forall x)\lnot \phi(x)$$ follows directly from the meanings of its components.