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 $ barand 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)]$