[proofplan]
We prove the non-reducibility by the usual diagonal argument. Assuming that $A'$ is decidable by an $A$-oracle machine, we build an $A$-oracle machine whose behavior on input $e$ is the opposite of the alleged answer to whether $e \in A'$. Applying this machine to its own index gives the contradiction $d \in A' \iff d \notin A'$. We then show separately that $A \leq_T A'$ by uniformly coding the computation that queries whether a given number belongs to $A$.
[/proofplan]
[step:Assume an $A$-oracle decision procedure for $A'$]
Suppose, toward a contradiction, that $A' \leq_T A$. Then there exists an $A$-oracle computable total function
\begin{align*}
H^A: \mathbb{N} &\to \{0,1\}
\end{align*}
such that, for every $e \in \mathbb{N}$,
\begin{align*}
H^A(e) = 1 \iff e \in A',
\qquad
H^A(e) = 0 \iff e \notin A'.
\end{align*}
By the definition of $A'$,
\begin{align*}
e \in A' \iff \Phi_e^A(e)\downarrow.
\end{align*}
Thus $H^A$ decides, using oracle $A$, whether the $e$-th $A$-oracle program halts on its own index.
[/step]
[step:Build the diagonal $A$-oracle program]
Define an $A$-oracle partial computable function
\begin{align*}
D^A: \mathbb{N} &\rightharpoonup \mathbb{N}
\end{align*}
as follows. On input $e \in \mathbb{N}$, the program first computes $H^A(e)$. If $H^A(e)=0$, it halts and outputs $0$. If $H^A(e)=1$, it enters an infinite loop.
Since $H^A$ is total, this construction gives, for every $e \in \mathbb{N}$,
\begin{align*}
D^A(e)\downarrow
\iff H^A(e)=0
\iff e \notin A'.
\end{align*}
Because $(\Phi_e^A)_{e \in \mathbb{N}}$ is an acceptable enumeration of all partial $A$-oracle computable functions, there exists an index $d \in \mathbb{N}$ such that
\begin{align*}
\Phi_d^A = D^A.
\end{align*}
[guided]
The purpose of $D^A$ is to reverse the alleged decision procedure $H^A$. The function $H^A$ answers the question “does $\Phi_e^A(e)$ halt?” We define a new oracle program $D^A$ that halts exactly when the answer is no.
Formally, define
\begin{align*}
D^A: \mathbb{N} &\rightharpoonup \mathbb{N}
\end{align*}
by the following $A$-oracle algorithm. Given $e \in \mathbb{N}$, compute $H^A(e)$. This computation always halts because $H^A$ is total. If the result is $0$, output $0$ and halt. If the result is $1$, run forever.
Therefore, for each $e \in \mathbb{N}$,
\begin{align*}
D^A(e)\downarrow
&\iff H^A(e)=0 \\
&\iff e \notin A'.
\end{align*}
The first equivalence is the definition of $D^A$. The second equivalence is the assumption that $H^A$ decides $A'$.
The fixed acceptable enumeration $(\Phi_e^A)_{e \in \mathbb{N}}$ lists every partial $A$-oracle computable function. Since $D^A$ is partial $A$-oracle computable, there is some index $d \in \mathbb{N}$ with
\begin{align*}
\Phi_d^A = D^A.
\end{align*}
This index $d$ is the self-reference point of the diagonal argument.
[/guided]
[/step]
[step:Apply the diagonal program to its own index]
Using the definition of $A'$ and the equality $\Phi_d^A = D^A$, we obtain
\begin{align*}
d \in A'
&\iff \Phi_d^A(d)\downarrow \\
&\iff D^A(d)\downarrow \\
&\iff d \notin A'.
\end{align*}
This contradiction shows that the assumption $A' \leq_T A$ is false. Hence
\begin{align*}
A' \nleq_T A.
\end{align*}
[/step]
[step:Compute $A$ from its jump]
It remains to prove the stated strict inequality. For each $n \in \mathbb{N}$, let $P_n^A$ be the $A$-oracle program which, on input $x \in \mathbb{N}$, queries whether $n \in A$ and halts with output $0$ exactly if the oracle answer is yes; if the oracle answer is no, it runs forever. Thus
\begin{align*}
P_n^A(x)\downarrow \iff n \in A
\end{align*}
for every $x \in \mathbb{N}$.
By acceptability of the enumeration, there is a total computable function
\begin{align*}
s: \mathbb{N} &\to \mathbb{N}
\end{align*}
such that, for every $n \in \mathbb{N}$,
\begin{align*}
\Phi_{s(n)}^A = P_n^A.
\end{align*}
In particular,
\begin{align*}
s(n) \in A'
&\iff \Phi_{s(n)}^A(s(n))\downarrow \\
&\iff P_n^A(s(n))\downarrow \\
&\iff n \in A.
\end{align*}
Therefore an $A'$-oracle machine decides membership in $A$ as follows: on input $n$, compute $s(n)$, query the $A'$-oracle whether $s(n) \in A'$, and output the same answer. Hence
\begin{align*}
A \leq_T A'.
\end{align*}
Together with $A' \nleq_T A$, this proves
\begin{align*}
A <_T A'.
\end{align*}
[/step]
