A partial function is partial recursive if and only if it is Turing computable

Assume first that $f$ is partial recursive. By [L3], choose $e$ with $f(x)=U(\mu s{\tau }_T(e,x,s))$ whenever $f(x)$ is defined. Every primitive-recursive function is Turing computable: the initial functions have immediate machines, machine composition realizes composition, and a finite loop realizes primitive recursion. Hence machines compute the primitive-recursive functions ${\tau }_T$ and $U$ from [L5]. A machine can therefore search successively through $s=0,1,2,\dots$, evaluate ${\tau }_T(e,x,s)$, and at the first zero compute $U(s)$ and halt with its binary numeral. If there is no successful $s$, the search continues forever. By [L1], this machine computes $f$.

Conversely, assume that $f$ is Turing computable. Then by [L1] there is a machine code $e$ whose halting computations on input $x$ produce exactly the values of $f(x)$. By [L5], the characteristic function of the relation $T(e,x,s)$ is primitive recursive, so the total search function $g_e(x,s)$ that returns $0$ exactly when $T(e,x,s)$ holds and $1$ otherwise is primitive recursive. By [L2], the partial search $m_e(x):=\mu s{g}_e(x,s)$ is therefore partial recursive, and composing with the primitive recursive output map $U$ shows that $x\mapsto U(m_e(x))$ is partial recursive as well. Its domain and values agree with $f$ because $m_e(x)$ exists exactly when the machine coded by $e$ has a halting history on $x$, and then [L5] says that $U(m_e(x))$ is exactly the machine's output. Hence $f$ is partial recursive.

Steps 1.1 and 1.2 prove both implications, so $f$ is partial recursive if and only if it is Turing computable.