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

As the title indicates, how can you prove that $$A$$ is productive? With $$W_ {x}$$ I mean the set of points in which the turing machine with index $$x$$ s & # 39; stop.

The standard approach that I am is the functional reduction: I first build a total recursive function $$psi (x, y)$$. Then, through the s-m-n theorem, I construct a partial recursive function $$phi_ {g (x)} (y) = psi (x, y)$$ with $$g$$ recursive total. Then I go on to show that $$overline {K}$$ recude to the set that I'm studying through $$g$$. The problem is that, in this set, the condition includes the index of the function ($$x$$) so I do not know how to build a $$psi$$ a function. As you can see in the appendix, I can not "talk" about $$g$$ in $$psi$$so I can not build $$g$$ behave like a member of $$A$$ put I think.

How can I overcome this problem?

APPENDIX: Evidence Example with Functional Reduction

Let's show that $$B = {x | W_x = emptyset }$$ is productive.

We do this by proving that $$overline {K}$$ reduced to $$B$$, so we can show that $$K$$ reduced to $$overline {B} = {x | W_x neq emptyset }$$.
Consider the function:
$$psi (x, y) = left { begin {array} {lr} 1, & text {for} x in K \ uparrow, & text {else} \ end {array} right }$$

We can build $$phi_ {g (x)} (y) = psi (x, y)$$ with $$g$$ recursive total (s-m-n theorem).

Yes $$x in K$$, $$phi_ {g (x)}$$ stop everywhere then $$W_x neq emptyset$$: $$g (x) in overline {B}$$.

Yes $$x notin K$$, $$phi_ {g (x)}$$ never stop, then $$W_x = emptyset$$: $$g (x) notin overline {B}$$.

So $$K$$ reduced to $$overline {B}$$ basically $$g$$: $$B$$ is productive. $$square$$