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Every recursively enumerable set many-one reduces to the diagonal halting set. If a program halts exactly on $X$, the S-m-n theorem produces from $x$ an index $g(x)$ for a unary program that ignores its input and performs that computation on $x$. ThenConversely, a computable preimage of a recursively enumerable set is recursively enumerable.

Rice theorem: every nontrivial extensional property of programs computing partial computable functions has an undecidable index set. More precisely, take an acceptable effective enumeration $({\phi }_e)$ of the unary partial computable functions. If $\mathcal{P}$ contains some but not all of these functions, then“Extensional” means that equality of partial functions, including their domains, preserves membership; no syntactic restriction on program codes is being considered.

We use two standard facts, stated explicitly. The diagonal halting set $K=\{e:{\phi }_e(e)\downarrow \}$ is undecidable. For the associated acceptable enumerations ${\phi }_e^{(r)}$ of $r$-ary partial computable functions, with ${\phi }_e={\phi }_e^{(1)}$, the S-m-n theorem says that for every $m,n\ge 1$ there is a total computable function $s_n^m$ such thatwhere $\simeq$ means equality of defined values and equality of undefinedness. Thus fixed parameters in a program can be compiled into its index effectively.

Let $\perp$ be the nowhere-defined partial computable function. Choose a partial computable function $h$ whose membership in $\mathcal{P}$ is opposite to that of $\perp$; this is possible by nontriviality. For each $e$, construct the programThis is a uniform partial computation. Take an index $d$ for the binary partial computable function $(e,y)\mapsto {\psi }_e(y)$. The S-m-n theorem gives the total computable function $q(e)=s_1^1(d,e)$ with ${\phi }_{q(e)}={\psi }_e$. Crucially,Even when $h(y)$ itself is undefined, the equality of partial functions remains correct. If $\perp \notin \mathcal{P}$, this gives a many-one reduction of $K$ to $I_{\mathcal{P}}$; if $\perp \in \mathcal{P}$, it gives a many-one reduction of $K$ to the complement of $I_{\mathcal{P}}$. A decision procedure for $I_{\mathcal{P}}$ would therefore decide $K$ in either case, a contradiction.

The equivalent language formulation says that every nontrivial property of computably enumerable languages depending only on the recognized language has an undecidable index set. Apply the function formulation to the property of a partial computable function that its domain has the specified language property: every computably enumerable language occurs as such a domain.

The S-m-n theorem says that parameters can be compiled effectively into program indices: for a standard effective numbering, there is a total computable function $s_m^n$ with ${\phi }_{s_m^n(e,\mathbf{a})}(\mathbf{x})={\phi }_e(\mathbf{a},\mathbf{x})$ for partial computable functions of $m+n$ arguments.

Write $K=\{e:{\phi }_e(e)\text{ halts}\}$, a recursively enumerable halting problem. If $X$ is recursively enumerable, make a program with parameter $x$ which ignores its input and runs a semidecision procedure for $x\in X$, halting when that procedure halts. By the S-m-n theorem its index $f(x)$ is total computable and $f(x)\in K⟺x\in X$. Conversely, $X{\le }_{m}K$ implies that $X$ is recursively enumerable by the previous construction. Thus $\boxed{X\text{ is r.e.}⟺X{\le }_{m}K}$.

Use uniform finite-stage enumerations $W_{n,s}$ and $K_s$. Define a uniformly recursively enumerable setThe S-m-n theorem supplies a total computable index function $f$ with $W_{f(n)}=V_n$. If $W_n$ is finite of size $h$, then $V_n\subseteq \{0,\dots ,h-1\}$, so $V_n$ is recursive. If $W_n$ is infinite, then for every $x\in K$ a sufficiently late stage has both $x\in K_s$ and $|W_{n,s}|>x$, while no $x\notin K$ is ever enumerated. Hence $V_n=K$, which is not recursive. Therefore $\boxed{n\in P⟺f(n)\in Q,P{\le }_{m}Q}$. As usual the effective numbering indexes all programs; if malformed codes are included, send them to a fixed index of $K$.

Fix an effective enumeration $f_{e,n}$ of the $n$-ary partial recursive functions.

The S-m-n theorem says that program inputs can be compiled into an index: for every $m,n\ge 1$ there is a total recursive $s_n^m$ such thatHere $\simeq$ means that either both sides are undefined or both are defined and equal.

The recursion theorem says that every total recursive transformation $h$ of program indices has a semantic fixed point: some $e$ satisfies

The Rice theorem says that every nontrivial extensional property of programs computed by partial recursive functions is undecidable.

By the S-m-n theorem, there is a total recursive function $h$ such thatfor all $x,y$. Apply the recursion theorem to $h$. It supplies an index $e$ for whichConsequently

Fix an acceptable effective enumeration $f_{e,n}$ of the $n$-ary partial computable functions. The S-m-n theorem states that for every $m,n\ge 1$ there is a total computable function $s_n^m$ such thatwhere $\simeq$ means that both sides have the same value when defined and are otherwise both undefined.

The recursion theorem states that every total computable transformation $h$ of program indices has a semantic fixed point: there is an $e$ such that

Rice theorem says that every nontrivial extensional property of programs computing partial functions is undecidable. Equivalently, a nontrivial index set cannot be a computable set.

Let $A$ be such an index set. Replacing $A$ by its complement if necessary, assume that the nowhere-defined function has no index in $A$. Since the property is nontrivial, choose an index $a\in A$. If membership in $A$ were decidable, we could decide the diagonal halting problem as follows. For each $x$, defineBy the S-m-n theorem, a total computable function $q$ produces an index $q(x)$ for ${\psi }_x$. If $f_{x,1}(x)$ diverges, then ${\psi }_x$ is nowhere defined and $q(x)\notin A$. If it halts, then ${\psi }_x\simeq f_{a,1}$ and extensionality gives $q(x)\in A$. A decider for $A$ would therefore decide whether $f_{x,1}(x)$ halts, a contradiction. Hence $A$ is undecidable.

Apply the S-m-n theorem to $g$. It gives a total computable function $h$ such thatThe recursion theorem supplies an index $e$ satisfying $f_{e,1}\simeq f_{h(e),1}$. Consequently

For each $x$, let $h(x)$ be an effectively obtained index of the partial functionThe existence of the total computable index transformation $h$ follows from the S-m-n theorem. By the recursion theorem, some $m$ satisfies $f_{m,1}\simeq f_{h(m),1}$. Therefore its domain isand hence

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.