Direct proof that $K \leq_\mathrm{T} Rec$

Question

Soare's Recursively Enumerable Sets and Degrees (1987) shows that $Rec = \left\{ e : W_e \text{ is recursive} \right\}$ is $\Sigma^0_3$-complete via its relationship to other index sets, namely $Cof$ and $Cpl$, as discussed in this previous question on MSE. I assume there is a simple direct argument exhibiting a recursive function $f$ which reduces $K = \left\{ e : \varphi_e(e)\downarrow \right\}$ to $Rec$, but I'm having trouble coming up with it.

One thought I had was that each finite initial segment of $K$ is a recursive set and hence has a pair of indices in $Rec$, and that one could then some how use $Rec$ to take the union of these approximations, but I can't quite see how to make it work. Any help people could offer would be appreciated; I'm sure I'm just missing something obvious.

Answer 1

Consider the recursive function $\psi$ such that : $$\psi(e,x)=\mbox{Compute $\phi_e(e)$ on $x$ steps. If it halts in less than x steps then 1 else }\phi_x(x)$$

By smn, there is a recursive function $f$ such that $$\phi_{f(e)}(x)=\psi(e,x)$$

Then $K\le Rec$ by $f$ :

1. If $e\in K$ then $W_{f(e)}$ is cofinite and $f(e)\in Rec$
2. If $e\notin K$ then $W_{f(e)}=K$ and $f(e)\notin Rec$

Note that $f$ is obtained by smn, and thus can be made injective and increasing. Hence $K\le_1 Rec$