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So a common method used to construct non-zero $\omega$-REA arithmetic degrees with various properties is to build an $\omega$-REA operator $J$ satisfying the constraints that (for all $X$)

$$\tag{1} J(X') \equiv_T J(X) \oplus X'$$ $$\tag{2} J(X) >_T X$$

Inductively, 1 implies that $J(X^n) \equiv_T J(\emptyset) \oplus X^n$. Thus, together, these constraints ensure that $J(\emptyset)$ isn't arithmetic (if $J(\emptyset) \leq_T 0^n$ then $J(0^n) \equiv_T 0^n$).

My question is whether this is fully general, i.e., if $A$ is $\omega$-REA is there some $\omega$-REA operator $J$ satisfying 1 such that $A$ and $J(\emptyset)$ have the same arithmetic degree? And can we also assume 2 holds if $A$ isn't arithmetic?

Basically, I'm hoping someone will let me know if I'm missing some obvious elementary result or known result before I spend any time trying a hard construction.

So a common method used to construct non-zero $\omega$-REA arithmetic degrees with various properties is to build an $\omega$-REA operator $J$ satisfying the constraints that (for all $X$)

$$\tag{1} J(X') \equiv_T J(X) \oplus X'$$ $$\tag{2} J(X) >_T X$$

Inductively, 1 implies that $J(X^n) \equiv_T J(\emptyset) \oplus X^n$. Thus, together, these constraints ensure that $J(\emptyset)$ isn't arithmetic (if $J(\emptyset) \leq_T 0^n$ then $J(0^n) \equiv_T 0^n$).

My question is whether this is fully general, i.e., if $A$ is $\omega$-REA but non-arithmetic is there some $\omega$-REA operator $J$ satisfying 1 and 2 such that $A$ and $J(\emptyset)$ have the same arithmetic degree?

Basically, I'm hoping someone will let me know if I'm missing some obvious elementary result or known result before I spend any time trying a hard construction.

So a common method used to construct non-zero $\omega$-REA arithmetic degrees with various properties is to build an $\omega$-REA operator $J$ satisfying the constraints that (for all $X$)

$$\tag{1} J(X') \equiv_T J(X) \oplus X'$$ $$\tag{2} J(X) >_T X$$

Inductively, 1 implies that $J(X^n) \equiv_T J(\emptyset) \oplus X^n$. Thus, together, these constraints ensure that $J(\emptyset)$ isn't arithmetic (if $J(\emptyset) \leq_T 0^n$ then $J(0^n) \equiv_T 0^n$).

My question is whether this is fully general, i.e., if $A$ is $\omega$-REA is there some $\omega$-REA operator $J$ satisfying 1 such that $A$ and $J(\emptyset)$ have the same arithmetic degree? And if we can also assume 2 holds if $A$ isn't arithmetic?

Basically, I'm hoping someone will let me know if I'm missing some obvious elementary result or known result before I spend any time trying a hard construction.

So a common method used to construct non-zero $\omega$-REA arithmetic degrees with various properties is to build an $\omega$-REA operator $J$ satisfying the constraints that (for all $X$)

$$\tag{1} J(X') \equiv_T J(X) \oplus X'$$ $$\tag{2} J(X) >_T X$$

Inductively, 1 implies that $J(X^n) \equiv_T J(\emptyset) \oplus X^n$. Thus, together, these constraints ensure that $J(\emptyset)$ isn't arithmetic (if $J(\emptyset) \leq_T 0^n$ then $J(0^n) \equiv_T 0^n$).

My question is whether this is fully general, i.e., if $A$ is $\omega$-REA is there some $\omega$-REA operator $J$ satisfying 1 such that $A$ and $J(\emptyset)$ have the same arithmetic degree? And can we also assume 2 holds if $A$ isn't arithmetic?

Basically, I'm hoping someone will let me know if I'm missing some obvious elementary result or known result before I spend any time trying a hard construction.