[proofplan] We first verify that the relation on equivalence classes does not depend on the chosen representatives. This uses only reflexivity and transitivity of Turing reducibility on subsets of $\mathbb{N}$. Once well-definedness is established, reflexivity and transitivity descend directly from the preorder $\le_T$, while antisymmetry follows because mutual reducibility of representatives is exactly membership in the same Turing degree. [/proofplan] [step:Check that the order relation is independent of representatives] Let $A,A',B,B' \subseteq \mathbb{N}$ satisfy $A \equiv_T A'$ and $B \equiv_T B'$. Suppose $A \le_T B$. By the definition of $\equiv_T$, the assumptions give \begin{align*} A' \le_T A, \qquad A \le_T A', \qquad B' \le_T B, \qquad B \le_T B'. \end{align*} Using transitivity of $\le_T$ on subsets of $\mathbb{N}$, we obtain \begin{align*} A' \le_T A \le_T B \le_T B', \end{align*} and hence $A' \le_T B'$. Therefore the truth of $[A]_T \le [B]_T$ is independent of the chosen representatives $A$ and $B$, so $\le$ is well-defined on $\mathcal{D}_T$. [guided] We must show that the proposed definition \begin{align*} [A]_T \le [B]_T \iff A \le_T B \end{align*} does not change if we replace $A$ or $B$ by another set of the same Turing degree. Let $A,A',B,B' \subseteq \mathbb{N}$ be sets with $A \equiv_T A'$ and $B \equiv_T B'$, and assume $A \le_T B$. The equivalence $A \equiv_T A'$ means both reductions hold: \begin{align*} A \le_T A' \quad \text{and} \quad A' \le_T A. \end{align*} Similarly, $B \equiv_T B'$ means \begin{align*} B \le_T B' \quad \text{and} \quad B' \le_T B. \end{align*} To prove that the new representatives also satisfy the desired comparison, we need $A' \le_T B'$. The available reductions compose in the order \begin{align*} A' \le_T A, \qquad A \le_T B, \qquad B \le_T B'. \end{align*} By transitivity of Turing reducibility on subsets of $\mathbb{N}$, these three reductions imply \begin{align*} A' \le_T B'. \end{align*} Thus replacing $A$ by an equivalent representative $A'$ and $B$ by an equivalent representative $B'$ preserves the truth of the comparison. Therefore $\le$ is a well-defined relation on the quotient set $\mathcal{D}_T$. [/guided] [/step] [step:Prove reflexivity by choosing a representative] Let $a \in \mathcal{D}_T$. Since $\mathcal{D}_T = \mathcal{P}(\mathbb{N})/{\equiv_T}$, there exists $A \subseteq \mathbb{N}$ such that $a = [A]_T$. Reflexivity of Turing reducibility gives $A \le_T A$. Hence, by the definition of $\le$ on $\mathcal{D}_T$, \begin{align*} a = [A]_T \le [A]_T = a. \end{align*} Thus $\le$ is reflexive. [/step] [step:Prove transitivity by composing representative reductions] Let $a,b,c \in \mathcal{D}_T$ satisfy $a \le b$ and $b \le c$. Choose representatives $A,B,C \subseteq \mathbb{N}$ such that \begin{align*} a = [A]_T, \qquad b = [B]_T, \qquad c = [C]_T. \end{align*} By the definition of $\le$ on $\mathcal{D}_T$, the assumptions imply \begin{align*} A \le_T B \quad \text{and} \quad B \le_T C. \end{align*} Transitivity of $\le_T$ gives $A \le_T C$. Therefore \begin{align*} a = [A]_T \le [C]_T = c. \end{align*} So $\le$ is transitive. [/step] [step:Prove antisymmetry by identifying mutually reducible representatives] Let $a,b \in \mathcal{D}_T$ satisfy $a \le b$ and $b \le a$. Choose representatives $A,B \subseteq \mathbb{N}$ such that \begin{align*} a = [A]_T, \qquad b = [B]_T. \end{align*} By the definition of $\le$ on $\mathcal{D}_T$, the assumptions give \begin{align*} A \le_T B \quad \text{and} \quad B \le_T A. \end{align*} By the definition of Turing equivalence, this is exactly $A \equiv_T B$. Hence $A$ and $B$ belong to the same $\equiv_T$-equivalence class, so \begin{align*} a = [A]_T = [B]_T = b. \end{align*} Thus $\le$ is antisymmetric. [/step] [step:Conclude that the Turing degrees are partially ordered] The relation $\le$ on $\mathcal{D}_T$ is well-defined, reflexive, transitive, and antisymmetric by the preceding steps. Therefore $\le$ is a partial order on the set of Turing degrees $\mathcal{D}_T$. [/step]