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Rice's theorem, in its formal language version, says: every nontrivial property of the computably enumerable language accepted by a Turing machine has an undecidable set of machine descriptions. Here nontrivial means that some computably enumerable language has the property and some does not. The property must be an extensional property of programs: any two Turing machines accepting the same formal language must agree about it. It need not be a property of all formal languages, including those no Turing machine accepts.

Let $\mathcal{P}$ be such a property and writeWe prove the theorem by a many-one reduction from the halting problem. First suppose the empty language is not in $\mathcal{P}$. Nontriviality provides a fixed Turing machine $B$ with $\mathcal{L}(B)\in \mathcal{P}$. Given a Turing machine $M$ and a word over an alphabet $w$, effectively construct a Turing machine $N_{M,w}$ which, on input $x$, first simulates $M$ on $w$. If that simulation ever halts, it simulates $B$ on $x$ and accepts precisely when $B$ accepts. If the first simulation never halts, it never accepts $x$.

The Turing machine encoding of $N_{M,w}$ is computable from those of $M,w$: the description merely embeds $M,w$ and the fixed simulator for $B$. Constructing it does not require predicting whether $M$ halts. Its accepted formal language isConsequentlyA decider for $I_{\mathcal{P}}$ would therefore decide the halting problem, a contradiction. Invalid descriptions can be mapped to a fixed description of a machine accepting the empty language, if the many-one reduction is formulated on all strings.

If the empty language does belong to $\mathcal{P}$, apply the preceding argument to the opposite property, within the computably enumerable languages. This opposite property is still nontrivial and excludes the empty language. A hypothetical decider for $I_{\mathcal{P}}$ would decide the opposite property's index set by first rejecting malformed Turing machine encodings and otherwise reversing its answer. The opposite property is undecidable by the first case, giving the required contradiction in this case too.

For example, takeThis is a decidable language: check that the input is a valid Turing machine encoding, count its declared states, and test the count's integer parity. There are Turing machines with even state counts and Turing machines with odd state counts, so the formal language contains some but not all valid descriptions.

There is no contradiction with Rice's theorem because this is a syntactic property of the description. Add one unreachable state to a Turing machine, assigning any permitted transitions to it. The accepted formal language is unchanged, while the integer parity of the state count changes. Thus the property is not an extensional property of programs, which is an essential hypothesis of Rice's theorem.

${\mathcal{L}}_{\ell 3}$ is undecidable. The condition depends only on the accepted formal language. It holds for a Turing machine accepting the empty language, and fails for a Turing machine accepting the singleton $\{aaaa\}$, where $a$ is an alphabet symbol. It is therefore a nontrivial extensional property of programs, and Rice's theorem applies. In particular, the condition does not assert that the machine halts on every input; nonacceptance of long words over an alphabet may result from rejection or from an infinite computation.

${\mathcal{L}}_{1\mathrm{T}}$ is a decidable language: it consists of every valid Turing machine encoding. Every multitape Turing machine has an equivalent single-tape Turing machine. To see this, represent the finitely used portions of its tapes on one tape, separated by delimiters, and mark each simulated head position. To simulate one step, scan these portions to read the marked symbols, retain this fixed number of symbols and the control state in finite control, and scan again to update the symbols and head marks. Extend a represented portion with a blank cell when necessary. Each simulated step takes finitely many real steps.

The single-tape Turing machine accepts exactly when the simulated multitape Turing machine accepts; infinite computations remain infinite. The simulation may increase time complexity, but the question only asks about accepted formal languages. Hence a decider simply checks the syntax of the Turing machine encoding and accepts every valid description. Rice's theorem does not make this property undecidable, because the property is true of every computably enumerable language and is therefore trivial in the theorem's sense.

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.

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.

No. The propertyis extensional: it depends only on the partial function computed, not on its program index. It is nontrivial, since some partial recursive functions have a nine-element domain and others do not. The Rice theorem therefore implies that its index setis undecidable. Hence

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.

A finite string describing a Turing machine's states, alphabets, transition rules and designated initial and halting states. Standard encodings have a decidable language of valid descriptions. A syntactic property of an encoding need not be an extensional property of programs; for example, adding an unreachable state changes state count without changing the accepted formal language.