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If a finitely generated group embeds in a finitely presented group, its null words in any fixed finite generating set form a recursively enumerable set. Choose words for the generator images in the larger group. Enumerate finite products of conjugates of its defining relators and their inverses, freely reduce, and compare with substituted words. Dovetailing over input words enumerates exactly the subgroup's identities. Choosing the finite image words is nonuniform, but their existence suffices for existence of an enumeration algorithm.

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.

The halting problem asks whether program $e$ halts on input $x$, given effective codes for both. Suppose a total decision algorithm $H(e,x)$ returned one exactly when that computation halts. Construct a program $D$ which, on input $e$, computes $H(e,e)$ and then loops forever if the answer is one, but halts if it is zero. Let $d$ be the code of $D$. If $H(d,d)=1$, the specification of $D$ makes it diverge, while if $H(d,d)=0$ the same specification makes it halt. Both alternatives contradict correctness. Therefore the halting problem is $\boxed{\text{undecidable}}$. This is a diagonal argument, not merely the observation that individual simulations can run forever.

Let $K=\{e:\text{program }e\text{ halts on input }e\}$. Simulating the specified program semidecides membership, or dovetailing all simulations enumerates $K$, so $K$ is a recursively enumerable set. The same diagonal program rules out a total decision algorithm for $K$. Thus $\boxed{K\text{ is recursively enumerable but not recursive}}$. Encoding a pair $(e,x)$ also gives an enumerable nonrecursive set of all halting program-input pairs.

A recursive function $f:{\mathbb{N}}^k\to \mathbb{N}$ is a total computable function: an algorithm halts on every input and returns the specified value. Equivalently it is an everywhere-defined member of the partial recursive functions, obtained from the initial zero, successor and projection functions by composition, primitive recursion and unbounded minimization. In minimization $\mu y[g(\mathbf{x},y)=0]$, the sequential search must find a zero after all earlier tests have been defined; otherwise the result is undefined. The Church–Turing thesis allows the algorithmic formulation here.

A recursively enumerable set $E\subseteq {\mathbb{N}}^k$ is the domain of a partial computable function. Thus an algorithm halts and accepts exactly on members of $E$, and may run forever on nonmembers. Equivalently an algorithm lists exactly its elements, allowing repetition and allowing an empty output for the empty set. To pass from a listing to a membership semidecision, wait for the input to appear. To pass back, use dovetailing on all inputs and list each whose recognizing computation halts. This differs from a recursive set, whose zero-one characteristic function is total recursive and decides both membership and nonmembership.

Take a recursively enumerable set $S\subseteq {\mathbb{N}}_{>0}$ that is not recursive, for example the diagonal halting set from Question 2 shifted by one. To obtain a finitely generated group with recursive presentation but undecidable word problem for a group, first define the countable Right-angled Artin groupTranslation of the indices gives an automorphism $a_i\mapsto a_{i+1}$. The semidirect product with its infinite cyclic shift is generated by $a=a_0,t$ and has presentationTo verify this presentation, the semidirect product satisfies its relators, while the reverse map sends $a_i$ to $t^{i}a{t}^{-i}$; all translated commutation relations follow by conjugation. These maps are inverse on their generators. Thus $B_S$ embeds in $G_S$.

For $n\in S$, the word $[a_0,a_n]$ is trivial. For $n\notin S$, map $a_0,a_n$ to the free generators $x,y$ of $F(x,y)$ and every other $a_i$ to one. Every defining commutator of $B_S$ maps to one, since the only possible relation involving both surviving generators would have separation $n$, which is absent. The resulting group homomorphism sends $[a_0,a_n]$ to the nontrivial reduced word $[x,y]$. HenceThis proves the shift-invariant graph groups encode arbitrary sets construction rather than supposing that a missing relator automatically means a nontrivial word. A word algorithm for $G_S$ would decide $S$, so none exists. Since $S$ is enumerable, the displayed two-generator presentation is a recursive presentation of a group. By the Higman embedding theorem, $G_S$ embeds in a finitely presented group $G_0$. A word algorithm for $G_0$, composed with substitution of the two fixed generator-image words, would solve the word problem for $G_S$. Therefore $\boxed{G_0\text{ is finitely presented with insoluble word problem}}$.

False. We construct a two-generated countable group that cannot embed in any finitely presented group. Choose a subset $S\subseteq {\mathbb{N}}_{>0}$ which is not a recursively enumerable set, for example a positive-integer coding of the complement of the halting set. Such sets exist because there are only countably many finite programs but uncountably many subsets of the positive integers. On generators $b_i$ indexed by $i\in \mathbb{Z}$, form the possibly infinite-graph Right-angled Artin groupThe shift $\sigma (b_i)=b_{i+1}$ preserves the defining relations and has inverse shift $i\mapsto i-1$, so it is an automorphism. Form the semidirect productIts base embeds by the definition of a semidirect product. Since $b_i=t^i{b}_0{t}^{-i}$, the group is generated by just $b_0,t$ and is countable.

The shift-invariant graph groups encode arbitrary sets construction has the exact testThe forward direction for $n\in S$ is a defining edge relation. If $n\notin S$, kill all vertex generators except $b_0,b_n$. Every defining commutator then either contains a killed generator or is trivial; no relation is imposed between these two retained generators. Thus there is a retraction of $B_S$ onto the rank-two free group on $b_0,b_n$. Their commutator is a nonempty reduced word there, so it is nonidentity in $B_S$ and remains so in $G_S$.

Now suppose $G_S$ embedded in a finitely presented group $P$. Choose fixed words $u,v$ in its finite generators representing the images of $b_0,t$. The null words of $P$ are recursively enumerable: enumerate all finite products of conjugates of its relators and their inverses, then freely reduce. Substituting $u,v$ into $[b_0,t^n{b}_0{t}^{-n}]$ and dovetailing these enumerations for all $n>0$ would enumerate exactly $S$, contradicting its choice. These finitely many image words may be hard-coded; the argument does not require an algorithm finding an embedding. This proves the enumerable identities of finitely generated subgroups obstruction and supplies a concrete construction of the required countable counterexample.

A recursive set $E\subseteq {\mathbb{N}}_0$ has an indicator function computed by a register machine that halts on every input. A recursively enumerable set is the exact halting set of some register machine; equivalently, it is the domain of a partial computable function.

If $E$ is recursive, its decider immediately gives machines that halt exactly on $E$ and on ${\mathbb{N}}_0\setminus E$, so both sets are recursively enumerable. Conversely, suppose machines $M_E$ and $M_{E^c}$ recognize the two sets. On input $x$, use dovetailing to run one step of $M_E(x)$ and one step of $M_{E^c}(x)$ alternately. Exactly one must eventually halt. Return one if $M_E$ halts first and zero if $M_{E^c}$ halts first. This is a total decider for $E$, so $E$ is recursive.

A subset $E\subseteq \mathbb{N}$ is recursive when its indicator function is a total computable function. It is recursively enumerable when it is the domain of a partial computable function, equivalently when an algorithm can enumerate its members.

If $E$ is recursive, its indicator decides both $E$ and its complement, so both sets are recursively enumerable. Conversely, suppose programs enumerate $E$ and $\mathbb{N}\setminus E$. On input $n$, run the two enumerations in parallel by dovetailing. Exactly one must eventually print $n$; accept in the first case and reject in the second. This algorithm always halts and decides $E$. 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.

By assumption the set of combinators is recursively enumerable. Finite beta reduction sequences, and hence finite certificates of beta equivalence, are also recursively enumerable.

For a closed term $Y$, choose a fresh variable $f$. The term $Y$ is a fixed-point combinator exactly whenIndeed, substitution then gives the required equivalence for every $F$, and the forward direction follows by taking $F=f$. Dovetail the enumeration of closed terms with all finite beta-equivalence certificates. Whenever a certificate of the displayed equivalence is found, output $Y$. This enumerates exactly the fixed-point combinators, proving recursively enumerable fixed-point combinators.

Fix an effective enumeration of register-machine programs. The $n$th register machine is $P_n$, and its domainis the $n$th recursively enumerable set. A many-one reduction $A{\le }_{m}B$ is a total computable function $h$ satisfying $x\in A⟺h(x)\in B$. Rice theorem says that every nontrivial property depending only on the computed partial function, or equivalently on an r.e. set in its extensional form, has an undecidable index set.

There is no total equality algorithm: it would decide whether $W_n$ is empty by comparing it with a fixed index for the empty set, contradicting Rice's theorem.

There is no partial algorithm that halts exactly when $W_m=W_n$ either. Given $e$, effectively construct an index $h(e)$ whose machine enumerates nothing unless $P_e(e)$ halts, after which it enumerates $0$. ThenA semialgorithm for equality with a fixed empty index would enumerate the complement of the halting problem $K$. Since $K$ is r.e., both it and its complement would then be r.e., making $K$ recursive, a contradiction.

The set of fixed-point combinators is recursively enumerable. Enumerate closed lambda terms and finite proofs of beta equivalence, and output $Y$ whenever a proof of $Yf{\equiv }_{\beta }f(Yf)$ appears for a fresh variable $f$.

A recursive presentation has an effectively enumerable alphabet of generators and a recursively enumerable set of relators. Usually the alphabet is finite; when it is countably infinite this is specified explicitly. Enumerating finite products of conjugates of relators and their inverses enumerates all null words. Conversely the enumerable null words can themselves serve as the relators. Recursive presentability does not imply a soluble word problem for a group.