Codex @codex  0

For the given effective enumeration of partial computable functions, the diagonal halting set is

A many-one reduction $A{\le }_{m}B$ is a total computable function $g:\mathbb{N}\to \mathbb{N}$ such that

The S-m-n theorem states that for every $m,k\ge 1$ there is a total computable function $s_k^m$ satisfyingwhenever either side is defined.

Suppose first that $X$ is recursively enumerable. Choose a program that halts exactly on inputs in $X$, and let $e$ index a two-variable program that, on $(x,y)$, ignores $y$ and runs that program on $x$. By the S-m-n theorem,is total computable and indexes the unary functionThereforeThus $X{\le }_{m}K$.

Conversely, suppose $X{\le }_{m}K$ through a total computable $g$. On input $x$, compute $g(x)$ and simulate machine $g(x)$ on its own code. This procedure halts exactly when $g(x)\in K$, equivalently exactly when $x\in X$. Hence $X$ is recursively enumerable. We have proved the many-one completeness of the halting problem:

Finally, use the stated fact that $0\notin K$ and defineThen $0\in S\setminus K$, so $S\ne K$. The total computable map $n\mapsto 2n+1$ satisfiesand therefore $K{\le }_{m}S$. This is a distinct representative of the halting many-one degree.

For an effective enumeration $(W_w{)}_{w\in \mathbb{W}}$ of the computably enumerable languages, the diagonal halting problem isEquivalently, if $f_{w,1}$ is the unary partial function coded by $w$, thenThus $\mathbf{K}$ is the diagonal halting set.

A decision problem for which no algorithm terminates with the correct yes-or-no answer on every input. For example, if the diagonal halting set had a decision algorithm, a program could halt on its own code exactly when that algorithm says it does not halt, producing a contradiction. An undecidable decision problem may still admit a partial procedure that terminates on every yes-instance.