[proofplan] We prove the two non-reducibilities separately. If $B \leq_T A$, then some oracle Turing functional computes the characteristic function $\mathbb{1}_B$ using oracle $A$; since $(\Phi_e)_{e \in \mathbb{N}}$ enumerates all oracle Turing functionals, that functional is some $\Phi_e$, contradicting $\mathcal{R}_e$. The proof that $A \nleq_T B$ is the same argument with $A,B$ and $\Phi,\Psi$ interchanged, using $\mathcal{S}_e$. [/proofplan] [step:Use the requirements $\mathcal{R}_e$ to rule out $B \leq_T A$] Assume, toward a contradiction, that $B \leq_T A$. By the definition of Turing reducibility, there exists an oracle Turing functional $\Gamma$ such that $\Gamma^A$ is total and \begin{align*} \Gamma^A = \mathbb{1}_B \end{align*} as functions $\mathbb{N} \to \{0,1\}$. Since $(\Phi_e)_{e \in \mathbb{N}}$ is an effective enumeration of all oracle Turing functionals, there exists $e \in \mathbb{N}$ such that $\Gamma = \Phi_e$. Hence \begin{align*} \Phi_e^A = \Gamma^A = \mathbb{1}_B, \end{align*} which contradicts the requirement $\mathcal{R}_e$. Therefore $B \nleq_T A$. [guided] We want to show that no computation of $B$ from oracle $A$ exists. Suppose the opposite: assume $B \leq_T A$. The meaning of this reducibility statement is that there is an oracle Turing functional $\Gamma$ which, when run with oracle $A$, computes the characteristic function of $B$. Formally, $\Gamma^A$ is a total map $\mathbb{N} \to \{0,1\}$ and \begin{align*} \Gamma^A(n) = \mathbb{1}_B(n) \end{align*} for every $n \in \mathbb{N}$. Now we use the enumeration hypothesis. The sequence $(\Phi_e)_{e \in \mathbb{N}}$ lists all oracle Turing functionals, so the particular functional $\Gamma$ must occur in the list. Thus there is some index $e \in \mathbb{N}$ with $\Gamma = \Phi_e$. Substituting this equality of functionals into the computation above gives \begin{align*} \Phi_e^A = \Gamma^A = \mathbb{1}_B. \end{align*} But this is exactly what the requirement $\mathcal{R}_e$ forbids: $\mathcal{R}_e$ states that $\Phi_e^A \neq \mathbb{1}_B$. The assumption $B \leq_T A$ therefore produces a contradiction. Hence $B \nleq_T A$. [/guided] [/step] [step:Use the requirements $\mathcal{S}_e$ to rule out $A \leq_T B$] Assume, toward a contradiction, that $A \leq_T B$. By the definition of Turing reducibility, there exists an oracle Turing functional $\Delta$ such that $\Delta^B$ is total and \begin{align*} \Delta^B = \mathbb{1}_A \end{align*} as functions $\mathbb{N} \to \{0,1\}$. Since $(\Psi_e)_{e \in \mathbb{N}}$ is an effective enumeration of all oracle Turing functionals, there exists $e \in \mathbb{N}$ such that $\Delta = \Psi_e$. Hence \begin{align*} \Psi_e^B = \Delta^B = \mathbb{1}_A, \end{align*} which contradicts the requirement $\mathcal{S}_e$. Therefore $A \nleq_T B$. [/step] [step:Conclude Turing incomparability] The previous two steps prove \begin{align*} B \nleq_T A \quad\text{and}\quad A \nleq_T B. \end{align*} By definition, this says exactly that $A$ and $B$ are Turing incomparable. [/step]