[proofplan] Turing equivalence means mutual Turing reducibility. We unpack $A \equiv_T B$ into the two reductions $A \le_T B$ and $B \le_T A$, apply [monotonicity of the Turing jump](/theorems/5418) to each reduction, and then repack the two resulting reductions as $A' \equiv_T B'$. [/proofplan] [step:Unpack Turing equivalence into two reductions] Let $A$ and $B$ be sets of natural numbers. For sets $X,Y \subseteq \mathbb{N}$, write $X \le_T Y$ to mean that $X$ is [Turing reducible](/page/Turing%20Reducibility) to $Y$, namely that there exists an oracle Turing machine which computes the characteristic function of $X$ using $Y$ as an oracle. Write $X \equiv_T Y$ to mean [Turing equivalence](/page/Turing%20Equivalence), namely the conjunction $X \le_T Y$ and $Y \le_T X$. The [Turing jump](/page/Turing%20Jump) of a set $X \subseteq \mathbb{N}$ is denoted by $X'$. Assume $A \equiv_T B$. By the definition just recalled, this means \begin{align*} A \le_T B \quad\text{and}\quad B \le_T A. \end{align*} [guided] We first fix the meanings of the reducibility notation. For sets $X,Y \subseteq \mathbb{N}$, the notation $X \le_T Y$ means that $X$ is [Turing reducible](/page/Turing%20Reducibility) to $Y$: an oracle Turing machine can compute the characteristic function of $X$ when it is allowed to query membership in $Y$. The notation $X \equiv_T Y$ means [Turing equivalence](/page/Turing%20Equivalence), which is mutual Turing reducibility. The notation $X'$ denotes the [Turing jump](/page/Turing%20Jump) of $X$. The hypothesis $A \equiv_T B$ therefore says exactly that $A$ and $B$ compute each other. Unpacking the definition gives the two reductions \begin{align*} A \le_T B \quad\text{and}\quad B \le_T A. \end{align*} These two reductions are the inputs needed for monotonicity of the Turing jump. [/guided] [/step] [step:Apply monotonicity of the Turing jump in both directions] We use the monotonicity of the [Turing jump](/page/Turing%20Jump): for all sets $X,Y \subseteq \mathbb{N}$, if $X \le_T Y$, then $X' \le_T Y'$. Applying this result first with $X=A$ and $Y=B$ gives \begin{align*} A' \le_T B'. \end{align*} Applying the same result with $X=B$ and $Y=A$ gives \begin{align*} B' \le_T A'. \end{align*} [guided] The monotonicity principle for the [Turing jump](/page/Turing%20Jump) says that a Turing reduction lifts to the jumps: for all sets $X,Y \subseteq \mathbb{N}$, if $X \le_T Y$, then $X' \le_T Y'$. The hypotheses needed to apply it are precisely reductions between sets of natural numbers, and the previous step supplied those reductions. Apply monotonicity with $X=A$ and $Y=B$. Since $A \le_T B$, the conclusion is \begin{align*} A' \le_T B'. \end{align*} Apply monotonicity again with $X=B$ and $Y=A$. Since $B \le_T A$, the conclusion is \begin{align*} B' \le_T A'. \end{align*} Thus the jump reductions have been obtained in both directions. [/guided] [/step] [step:Repack the two jump reductions as Turing equivalence] By the definition of [Turing equivalence](/page/Turing%20Equivalence), the two reductions \begin{align*} A' \le_T B' \quad\text{and}\quad B' \le_T A' \end{align*} are precisely the assertion \begin{align*} A' \equiv_T B'. \end{align*} This proves that the Turing jump depends only on the Turing degree of the set. [guided] The definition of [Turing equivalence](/page/Turing%20Equivalence) is mutual Turing reducibility. We have just proved both required reductions between the jumps: \begin{align*} A' \le_T B' \quad\text{and}\quad B' \le_T A'. \end{align*} Therefore, by that definition, \begin{align*} A' \equiv_T B'. \end{align*} This is exactly the theorem's conclusion. [/guided] [/step]