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 $