Consider the following sets:
$$
\begin{aligned}
&A=\left\{x: \phi_{x}^{1}(0) \text { is defined }\right\} \\
&B=\left\{x: \phi_{x}^{1} \text { is a total function }\right\}
\end{aligned}
$$
(a) Show that the complement of $A$ is not recursively enumerable.
(b) Show that there exists a primitive recursive function $f \in \mathcal{F}_{1}$ such that for all $i, i \in A$ if and only if $\alpha(i) \in B$. Show that the complement of $B$ is not recursively enumerable.
(c) Let $F$ be the following partial function:
$$
F(x, y)= \begin{cases}1 & \text { if for all } z<y, \neg B^{1}(e, z, x) \\ \text { undefined } & \text { otherwise }\end{cases}
$$
where $B^{1}$ is the predicate defined in the proof of the enumeration theorem and $e$ is the index of a partial function whose domain is $A$.
Show that the partial function $\lambda y \cdot F(x, y)$ is total if and only if $x \notin A$. Conclude from this that $B$ is not recursively enumerable.
(d) By generalizing the results from (b) and (c), prove the following:
Proposition Let $f$ be a partial recursive function of one variable whose domain is infinite; then neither the set $\left\{x: \phi_{x}^{1}=f\right\}$ nor its complement is recursively enumerable.