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There are incomparable Turing degrees strictly between the computable degree and the degree of the diagonal halting set. An oracle-assisted finite-extension construction alternately defeats each Turing functional in the two directions. Since the sets are computable in the halting oracle, their degrees are at most $0^{\prime }$; incomparability makes them nonzero and strictly below $0^{\prime }$. This theorem does not require the constructed sets to be computably enumerable. The Friedberg–Muchnik theorem supplies that stronger conclusion.

For subsets $A,B\subseteq \mathbb{N}$, a many-one reduction $A{\le }_{m}B$ is a total computable function $f:\mathbb{N}\to \mathbb{N}$ such thatA Turing reduction $A{\le }_{T}B$ is an oracle machine that, using membership queries to $B$, halts on every input and computes the indicator function of $A$. It may make several adaptive queries, and need not preserve the yes/no answer to a single query. A many-one reduction yields a Turing reduction by computing $f(x)$ and querying $B$ once. Two sets have the same Turing degree when each is Turing reducible to the other. Write $0^{\prime }$ for the degree of the diagonal halting set $K$.

We prove the Kleene–Post incomparability theorem by an oracle-assisted finite-extension construction. Fix an effective list $({\Phi }_e)$ of all Turing functionals. We shall build increasing finite binary strings ${\alpha }_s,{\beta }_s$ whose unions are the indicator functions of $A,B$. A finite oracle computation ${\Phi }_e^{\tau }(n)\downarrow =v$ means that the computation halts with output $v$ and every oracle query is strictly below $|\tau |$. Its answers are supplied by $\tau$. By the finite oracle use property, every infinite oracle extending $\tau$ preserves this computation.

Here is the decision available from $K$. Given $e,n$ and a finite string $\beta$, considerThis predicate is semidecidable: enumerate all finite extensions $\tau$ of $\beta$ and dovetail their finite oracle simulations, stopping at the first halting binary output. There is a computably produced program which, on its own index as well as on every other input, performs this search and halts exactly if it succeeds. Membership of its code in $K$ therefore decides $E$. In a yes case, running the search finds an actual witness $(\tau ,v)$. The decision oracle is essential in a no case; merely waiting would not produce a terminating construction.

Start with empty strings. At stage $s$ first ensureChoose $n=|{\alpha }_s|$, which is not yet assigned in $A$. Ask whether $E(s,n,{\beta }_s)$ holds. If yes, find a witness $(\tau ,v)$, extend the current $B$ string to $\tau$, and append the bit $1-v$ to the current $A$ string. If no, leave the current $B$ string unchanged and append $0$ to the current $A$ string. The yes case permanently forces disagreement at $n$, since later strings extend both commitments. In the no case no final oracle extending the current $B$ string can give a binary output at $n$: any such convergent computation would use finitely many bits and hence supply a forbidden finite witness. Thus it cannot be the indicator function of $A$ either, regardless of whether it diverges or returns a nonbinary value.

Next, using the strings just obtained, ensureby the same procedure with the roles reversed. Choose $m$ equal to the current length of the $B$ string, ask the analogous extension question about the current $A$ string, and, if it has witness output $v$, extend $A$ to that witness and append $1-v$ to $B$. Otherwise append $0$ to $B$. Finally pad both strings with zeros if necessary so that each length is at least $s+1$, and call the resulting strings ${\alpha }_{s+1},{\beta }_{s+1}$. Every operation is computable using the membership oracle $K$ and terminates. Earlier disagreements remain protected because no already fixed bit is ever changed.

Set $1_A={\bigcup }_s{\alpha }_s$ and $1_B={\bigcup }_s{\beta }_s$. The length condition makes these total binary functions. To compute either bit $n$ with oracle $K$, run the construction through stage $n$, when both strings have length at least $n+1$, and read that bit. Consequently $A,B{\le }_{T}K$. Every Turing functional has been defeated in both directions, soNeither set is computable, since a computable set is Turing reducible to every oracle. Neither can have degree $0^{\prime }$: if $K{\le }_{T}A$, then $B{\le }_{T}K{\le }_{T}A$, contrary to incomparability; the other case is symmetric. ThereforeThis proves the requested theorem. The argument does not claim $A$ or $B$ is computably enumerable; it constructs sets computable in $K$. Obtaining incomparable computably enumerable degrees is the stronger Friedberg–Muchnik theorem, which is not needed for the printed request.