# computability – How to prove that \$ A = {x in mathbb {N} | W_ {x} = [0..x]} \$ is a productive set by functional reduction?

You probably need to exploit the second recursion theorem, which states that for any total recursive function $$f$$ there are some $$n$$ such as $$varphi_n = varphi_ {f (n)}$$.

Given $$k$$, using s-m-n, you can first build $$f_k$$ as follows, where $$# ( ldots)$$ means "an index of".
$$f_k (x) = # left (y mapsto begin {cases} 0 & mbox {if} y leq x \ varphi_k (k) & mbox {else} end {cases} right)$$

This is a recursive total, as it always returns an index (the fact that it is for a partial function is unimportant). So, by the second recursion theorem, it must exist $$n_k$$ For who $$varphi_ {n_k} = varphi_ {f (n_k)}$$.

Examining the proof of the second recursion theorem, we can also note that $$n_k$$ is actually calculable from $$k$$, that is, there is a total recursive function $$g$$ such as $$varphi_ {g (k)} = varphi_ {f (g (k))}$$.

Let's check that $$g$$ m-reduced $$bar K$$ to your set $$A$$.

assume $$k in bar K$$and prove $$g (k) in A$$. The latter means $$W_ {g (k)} = [0..g(k)]$$, which is true because
$$varphi_ {g (k)} (y) = varphi_ {f (g (k))} (y) = begin {cases} 0 & mbox {si} y leq g (k) \ varphi_k (k) & mbox {else} end {cases} = begin {cases} 0 & mbox {si} y leq g (k) \ uparrow & mbox {else} end {cases}$$

Now, let's suppose $$k notin K bar$$and prove $$g (k) notin A$$. The latter means $$W_ {g (k)} neq [0..g(k)]$$, which is true because

$$varphi_ {g (k)} (y) = varphi_ {f (g (k))} (y) = begin {cases} 0 & mbox {si} y leq g (k) \ varphi_k (k) & mbox {else} end {cases}$$
Since $$varphi_k (k) downarrow$$we have that $$W_ {g (k)} = mathbb N neq [0..g(k)]$$