Proof. We build a computable copy of $D\mathord {\upharpoonright }\mathcal {L}$ via a finite injury argument. First, some background assumptions:

• • We see the structure D as being given by a computable process, which outputs elements and c.e. indices for the atomic $\mathcal {L}'$ -type of the elements output until now, and sometimes erases previously constructed elements. We will be executing this construction in parallel in the background of ours.

• • An element of D which is added in this manner and never removed will be referred to as a real element of D.

• • We will build the $\mathcal {L}$ -structure $M\cong D\mathord {\upharpoonright }\mathcal {L}$ via a computable process that outputs a sequence of elements, and with each element, the quantifier-free $\mathcal {L}$ -type of the elements output until now.

• • We well-order the elements of M and the elements of D by time of addition. This includes “fake” elements of D. When we refer to “the first element of M (or D) that satisfies such-and-such property,” we mean the one that was added earliest.

• • We will imagine an assortment of agents, referred to as “agent i” for $i=0,1,2,\dots $ , each of which will be responsible for one step in a back-and-forth construction of an isomorphism between D and the structure M that we are constructing. Roughly speaking, agent i will be responsible for satisfying the requirement: The i-th real element of D is matched to an element of M, and the i-th element of M is matched to an element of D, and the matchings created by agents $0$ to i form an isomorphism of finite $\mathcal {L}$ -structures. We describe requirement i in more detail below, by describing how it may be injured, after we have introduced some further notation about the construction.

• • We assume that only one agent is active at each time. When agent i is done with its task, it will activate agent $i+1$ . When an injury occurs, we divert execution back to a prior agent, say agent j, and delete the state of all agents past agent j.

• • The internal state of this construction, that is, the information we will be keeping track of, consists of the following data (plus whatever auxiliary information is necessary to keep track of the agent mechanism):

  • – The current status of the structures M and D.

  • – One-to-one matchings between some elements of M and some elements of D, the latter of which may or may not be real.

  • – To some elements m of M, we match a triple of formulas $(q(\vec {x},y,\vec {z}),\varphi (\vec {x}), \tau (\vec {x},y))$ that have been output by the process from Remark 2.4, together with a decomposition of the elements of M at the time that this matching was created into tuples $\vec {x}$ , $y=m$ , and $\vec {z}$ .

  • – To each matching of the previous two kinds, we keep track of which agent created it and when.

• • The internal state of this construction may be modified via the following operations:

  • – We may add a new element to M, in which case we immediately match it to an element of D.

  • – We may match a triple of formulas $(q,\varphi ,\tau )$ to an unmatched element $y\in M$ , together with the decomposition $M_{\mathrm {current}}=(\vec {x},y,\vec {z})$ , where $\vec {x}$ denotes the elements of M that have a match in D at this moment.

  • – We may create a matching between an element of M and an element of D, if the element of M has a triple $(q,\varphi ,\tau )$ assigned.

  • – In the event of injury, we may remove matches of the three kinds above.

• • We say that requirement i is injured if any of the following three things happen:

  • – An element of D which was matched by agent i to an element of M is deleted.

  • – There is an element of M, to whom agent i matched a triple of formulas $(q,\varphi ,\tau )$ , for which the triple of formulas has been deleted from the enumeration of the process of Remark 2.4.

  • – There is an element of M, to whom agent i matched a triple of formulas $(q,\varphi ,\tau )$ , and a matched element $d\in D$ , such that our guess about whether $\tau $ holds of this element has changed to ‘no’.

  In any of these cases, all matches created after the “faulty match” will be deleted.

Now we list some conditions to be maintained, barring injury. Together with each assumption, we point out the operations that we need to be careful in exercising—the remaining operations will always preserve the given assumption. Moreover, the assumptions will not be broken with the passage of time unless injury happens, in which case the injury-handling part of the algorithm will rectify any issues.

• Assumption (A) If $M=(\vec {x},\vec {y})$ , where $\vec {x}$ denotes the elements of M that currently admit matches in D, $\vec {b}$ denotes these matches, and $q(\vec {x},\vec {y})$ is the type of the elements of M, we demand that $D\vDash \chi _{q}(\vec {b})$ .

  We need to be careful to ensure that this is true when we add new elements to M, and when we create matches between elements of M and elements of D. We would also need to be careful in the event that matches are deleted, but this is where Remark 2.5 becomes relevant, as it ensures that this is not a concern.

• Assumption (B) If $y\in M$ is matched with a triple of formulas $(q,\varphi ,\tau )$ and decomposition $(\vec {x},y,\vec {z})\subseteq M$ , we demand that $M\vDash q(\vec {x},y,\vec {z})$ and that all elements of $\vec {x}$ admit matches $\vec {b}$ in D, which satisfy $D\vDash \varphi (\vec {b})$ .

  We need to be careful to ensure that this is true when we assign a triple to an element. We don’t need to be careful when removing matches or elements of D, because our injury process will, in that case, unmatch the formulas assigned to y.

• Assumption (C) At most one unmatched element m of M will have an assigned triple at any given time, and in this event, the variables $\vec {x}$ correspond to the elements of M currently matched to elements of D, and the next action to be taken will be to match m to an element of D.

• Assumption (D) If y is an element of M with assigned triple $(q(\vec {x},y,\vec {z}),\varphi (\vec {x}), \tau (\vec {x},y))$ and y admits a match d in D, if $\vec {b}$ is the tuple of elements matched with $\vec {x}$ , it must be the case that $D\vDash \tau (\vec {b},d)$ at the current approximation (in the sense of Remark 2.2).

  We need to be careful to ensure that this is true when we create a match between an element of M with an assigned triple $(q,\varphi ,\tau )$ and an element of D. We also need to be careful if the current approximation (in the sense of Remark 2.2) ever changes, which is why ‘the approximation of the value of $\tau (\vec {b},d)$ changes’ is a cause of injury.

• Assumption (E) If m is an element with assigned formulas $(q(\vec {x},y,\vec {z}),\varphi (\vec {x}), \tau (\vec {x},y))$ , with $\vec {b}$ denoting the matches of the $\vec {x}$ , and at this stage M is decomposed as $M=(\vec {x},y,\vec {z},\vec {w})$ with quantifier-free $\mathcal {L}$ -type $Q(\vec {x},y,\vec {z},\vec {w})$ , it must be the case that, if the $\vec {b}$ are real elements of D, we have $D\vDash \varepsilon \tau _{q\varphi Q}(\vec {b})$ .

  We need to be careful to ensure that this is true when new elements are added to M, and when we assign triples $(q,\varphi ,\tau )$ to elements of M.

The process starts with the initialization of agent $0$ .

Now we describe the algorithm that will be followed by agent i. When agent i is initialized:

1. Step I. Find the first element of M, if any, that does not have a match in D. If there is none, skip to Step V. If there is one, call it m.

2. Step II. If m has an assigned triple of formulas $(q,\varphi ,\tau )$ , skip to Step IV.Footnote ⁶

3. Step III. If m does not have an assigned triple: Let $(\vec {x},y,\vec {z})$ be the current elements of M, with $\vec {x}$ corresponding to the elements that currently have a match in D—call their matches $\vec {b}$ —, y corresponding to m, and $\vec {z}$ corresponding to the remaining elements. Look through the triples $(q,\varphi ,\tau )$ enumerated as per Remark 2.4 until you find such a triple that satisfies $M\vDash q(\vec {x},y,\vec {z})$ and $D\vDash \varphi (\vec {b})$ . If all elements of $\vec {b}$ are real, such a triple will be found in finite time by Assumption (A) in conjunction with Item (c) of the QETP. If not all are real, injury will happen. Once such a triple is found, we assign to m the triple $(q,\varphi ,\tau )$ with the decomposition $(\vec {x},y,\vec {z})$ above.

   We need to make sure that Assumption (B) and Assumption (E) are preserved.

   • • Assumption (B)—By design.

   • • Assumption (E)—We need to guarantee that, if the $\vec {b}$ are all real, we have $D\vDash \varepsilon \tau _{q\varphi q}(\vec {b})$ . This is a consequence of the note at the end of Item (d) of the QETP.

4. Step IV. Let $(q,\varphi ,\tau )$ be the formula assigned to m. The formula $q(\vec {x},y,\vec {z})$ will not, in general, be the quantifier-free $\mathcal {L}$ -type of the current elements of M, but it is definitely the quantifier-free $\mathcal {L}$ -type of some subset of elements of M. By Assumption (C), the elements corresponding to $\vec {x}$ are exactly the ones that currently admit matches in D. Let $\vec {b}$ denote their matches.

   We let $Q(\vec {x},y,\vec {z},\vec {w})$ denote the quantifier-free type of the current elements of M, where $\vec {x}$ , y, and $\vec {z}$ are the same as above.

   Look through the elements of D until you find an element d satisfying $D\vDash \tau (\vec {b},d)$ to current approximation (in the sense of Remark 2.2).

   Technical note: This search must be done in some systematic way so that, if all $\vec {b}$ are real, we eventually find a real element d satisfying this formula in such a way that the approximation stays fixed. An example of such a systematic way: Let’s say we give the n-th element added to D, say $d_{n}$ , a penalty of n stages in the time it takes to compute whether $D\vDash \tau (\vec {b},d_{n})$ , and select the d such that $D\vDash \tau (\vec {b},d)$ is verified (and not disproven until now) in the least amount of stages. This will be important when proving finite injury.

   We claim that, unless an injury occurs, this search will eventually terminate. By this we mean: Assume that all $\vec {b}$ are real, and that the triple $(q,\varphi ,\tau )$ will never be removed from the enumeration from Remark 2.4. Then, by Assumption (A) it must be the case that $D\vDash \chi _{Q}(\vec {b})$ and by Assumption (E) it must be the case that $D\vDash \varepsilon \tau _{q\varphi Q}(\vec {b})$ . Thus, by Item (d) of the QETP, it must be the case that $D\vDash (\exists y)\tau _{q}(\vec {b},y)$ , and any witness to this statement will serve as the d above.

   Once such a d is found, we match m to d.

   We need to make sure that Assumption (A) and Assumption (D) hold.

   • • Assumption (A)—If d is real, this is direct by Item (e) of the QETP.

   • • Assumption (D)—By construction.

5. Step V. Next, find the first element of D—call it d—that does not have a match in M. If there is no such element as of right now, proceed to Step VI.

   Let $q(\vec {x},\vec {y})$ be the quantifier-free $\mathcal {L}$ -type of the current elements of M, with $\vec {x}$ denoting the elements that currently have matches in D, and let $\vec {b}$ denote their matches. Let $r(\vec {x},z)$ be the quantifier-free $\mathcal {L}$ -type of the tuple $(\vec {b},d)$ .Footnote ⁷ In parallel, look through the quantifier-free $\mathcal {L}$ existential types $Q(\vec {x},z,\vec {y})$ until you find such a Q extending $q(\vec {x},\vec {y})\land r(\vec {x},z)$ such that:

   • • $D\vDash \chi _{Q(\vec {x},z,\cdot )}(\vec {b},d)$ , and

   • • for every m with an assigned triple $(q_{0}(\vec {x}_{0},y_{0},\vec {z}_{0}),\varphi (\vec {x}_{0}),\tau (\vec {x}_{0},y_{0}))$ , if $\vec {b}_{0}$ are the elements of M corresponding to $\vec {x}_{0}$ , $D\vDash \varepsilon \tau _{q_{0}\varphi Q}(\vec {b}_{0})$ .

   We argue that either such a Q is found in finite time or otherwise an injury will occur. Indeed, suppose that both d and the elements of $\vec {b}$ are all real, and that every matched triple $(q_{0},\varphi ,\tau )$ will never be injured. Now, since we assume $D\vDash \chi _{q}(\vec {b})$ , there must be a tuple of real elements $\vec {c}$ in D such that $D\vDash q(\vec {b},\vec {c})$ . Now, we claim that the type Q of the tuple $(\vec {b},d,\vec {c})$ satisfies the desired requisites:

   • • Since $D\vDash Q(\vec {b},d,\vec {c})$ , we must have $D\vDash (\exists \vec {y})Q(\vec {b},d,\vec {y})$ and hence $D\vDash \chi _{Q}(\vec {b},d)$ .

   • • If m has the triple $(q_{0},\varphi ,\tau )$ assigned, with decomposition $(\vec {x}_{0},y_{0},\vec {z}_{0})\subseteq M$ , with $\vec {b}_{0}$ being the elements of D corresponding to $\vec {x}_{0}$ and $d_{0}$ being the element corresponding to m, we have $D\vDash \chi _{Q}(\vec {b}_{0},d_{0})$ as a consequence of $D\vDash Q(\vec {b},d,\vec {c})$ , and moreover $D\vDash \tau (\vec {b}_{0},d_{0})$ by Assumption (D), hence $D\vDash (\exists y)\left [\chi _{Q}(\vec {b}_{0},y)\land \tau _{q\varphi }(\vec {b}_{0},y)\right ]$ , and thus $D\vDash \varepsilon \tau _{q_{0}\varphi Q}(\vec {b}_{0})$ by Item (d) of the QETP.

   Once such a $Q(\vec {x},z,\vec {y})$ is found, one of two things occurs. Either it includes $z=y_{j}$ for some value of j, in which case we match d to the element corresponding to $y_{j}$ , or it includes the information that all its variables represent distinct elements, in which case we add a new element to M, whose relations to the previous elements are those dictated by Q, and match it to d.

   We need to make sure that Assumption (A) and Assumption (E) hold.

   • • Assumption (A)—By construction, since we ensured $D\vDash \chi _{Q}(\vec {b},d)$ .

   • • Assumption (E)—By construction.

6. Step VI. Once all the above steps are done, agent i will pause its execution, and initialize agent $i+1$ .

We now describe the process of injury. There are three means of injury:

• • Suppose that an element of D is seen to be removed. If this element does not have a match in M, no action is taken. On the other hand, suppose that this element d is matched to an element m in M. In that event, suppose that agent i is the one that made this match, and halt the execution of all agents with index greater than i, erasing matches and assigned formulas created past the moment that m and d were matched. If the match between d and m was created by Step I, restart the execution of agent i. If the match between d and m was created by Step V, initialize agent $i+1$ (and do not restart agent i).

• • Suppose that an element $m\in M$ has an assigned triple $(q,\varphi ,\tau )$ , which is seen to be deleted from the enumeration described in Remark 2.4 at a given stage. Suppose agent i is the one who has assigned this formula. Then, we halt the execution of all agents with index greater than i, and erase all matches and assigned formulas created past the moment that this formula was assigned to m. Then, execute agent i starting in Step III.

• • Suppose that an element $m\in M$ has an assigned triple $(q(\vec {x},y,\vec {z}),\varphi (\vec {x}),\tau (\vec {x},y))$ and a matched element d. Suppose agent i is the one who has created these matches. Suppose that the elements $\vec {x}$ are matched to the tuple $\vec {b}$ , and suppose that the current approximation of $D\vDash \tau (\vec {b},d)$ (in the sense of Remark 2.2) has changed to ‘no’. Then, we halt the execution of all agents with index greater than i, erase all matches and assigned formulas created past this point (including the match between m and d, but we retain the assigned triple $(q,\varphi ,\tau )$ ), and execute agent i starting in Step IV.

This concludes the construction. It remains to show that the resulting structure M is isomorphic to $D\mathord {\upharpoonright }\mathcal {L}$ , and this is done via a finite injury argument. We show that each requirement will be injured finitely many times. This will show that every element in D is eventually matched to an element of M (the i-th element of D (counting both real and fake elements) will be matched by, at worst, the i-th agent past the moment when this element is added to D), and that every element of M is eventually matched to an element of D (given an orphan element of M, inductively consider the first agent activated after all previous elements of M have been matched), and since these matches all preserve the predicates in $\mathcal {L}$ , the resulting map will be an isomorphism.

Suppose, for the sake of induction, that all requirements prior to i are injured finitely many times. We show that requirement i itself also suffers finite injury. We do this individually for each of the matches potentially created by agent i:

• • First, we show that the matching created by the first orphan element of m at this moment, if there is any, is injured finitely many times. Indeed, per the notation of Remark 2.4 and using Item (c) of the QETP, eventually a triple $(q,\varphi ,\tau )$ will be enumerated and never removed, such that the relevant elements of D satisfy $\chi _{q}\land \varphi $ . Past this point, after all previously-added false triples have been removed, this triple (or another) will be permanently assigned to m. Afterward, we know (see the technical note in Step IV) that after enough steps we match m with a real element d of D that satisfies $\tau (\vec {b},d)$ at the current approximation, which will then stay fixed. This match will never cause injury.

• • Now, assuming that we’ve reached a time where the matching discussed in the previous bullet point or created by previous agents will never be removed, suppose that there is an unmatched element in D for agent i to match with an element of M. If this element is real, there is no injury. If the element is not real, then it is eventually erased. Then execution will move to agent $i+1$ , and requirement i is never injured.

This concludes the proof.

Proof. Omitted. See [Reference Maia14].

Proof. Omitted. See [Reference Maia14].

Proof. Omitted. See [Reference Maia14].

Proof. Omitted. See [Reference Maia14].

Proof. Given a c.e. presentation of a group G, we first build a $0'$ -computable c.e.-typed copy of G over an appropriate signature $\mathcal {L}'$ , and we will then show that any nontrivial torsion-free abelian group has the QETP (uniformly) over $\mathcal {L}\subseteq \mathcal {L}'$ , where $\mathcal {L}$ is the signature containing only the ternary predicate $S(x,y,z)\equiv {\text{"}{x+y=z}\text{"}}$ (and its negation, and equality and its negation).

The signature $\mathcal {L}'$ contains $\mathcal {L}$ plus:

• • The countably many predicates $Q_{n,\vec {c}}(\vec {x})$ , parametrized over $n\in \mathbb {N}$ and finite tuples $c_{1},\dots ,c_{k}\in \mathbb {Z}$ :

  $$\begin{align*}Q_{n,\vec{c}}(\vec{x})\equiv\text{"}{n\mid c_{1}x_{1}+\dots+c_{k}x_{k}}.\text{"} \end{align*}$$

  Note that we allow n to equal zero, in which case (since $0\mid y$ iff $y=0$ ) $Q_{0,\vec {c}}(\vec {x})$ corresponds to the predicate $\vec {c}\cdot \vec {x}=0$ .

• • The countably many predicates $\neg Q_{0,\vec {c}}(\vec {x})$ , parametrized over finite tuples $c_{1},\dots ,c_{k}\in \mathbb {Z}$ :

  $$\begin{align*}\neg Q_{0,\vec{c}}(\vec{x})\equiv{\text{"}{c_{1}x_{1}+\dots+c_{k}x_{k}\neq0}.\text{"}} \end{align*}$$

• • Countably many predicates $\{n\Delta (x)\}_{n\in \mathbb {Z}}$ , whose meaning is as follows: We will carefully pick a single designated nonzero element of G, which we refer to as $\delta $ , and $n\Delta (x)$ should be interpreted as “ $x=n\delta $ .”

We need some linear algebra of $\mathbb {Z}$ -modules to show that, from a c.e. presentation of G, we can uniformly obtain a $0'$ -computable c.e.-typed copy of G. Then, we will explain how we choose the element of G that will be tagged with $\Delta $ . To obtain the $0'$ -computable c.e.-typed copy of G, here is the basic idea: Consider the computable process that outputs all possible linear combinations of generators. We want to, with each new linear combination of generators, say d, first check whether it is equal to any of the combinations previously produced (this is an easy check with $0'$ ), and if it is not, we want to output a c.e. index for its type relative to the elements previously output. This is easy for the positive predicates—to check if, for example, $5\mid 4d+6c$ (where c is a previously output combination of generators), it suffices to check if there is a further combination of generators u that satisfies the relation $5u-4d-6c=0$ . This is a c.e. check. The difficulty lies in the negative predicates, of which $\neg Q_{0,\vec {c}}(\vec {x})$ is the general case. In other words, we need to encode into a single c.e. index the full information about what linear combinations of elements-output-so-far equal zero.

Here is a fact from commutative algebra, whose proof can be found in any standard book that covers modules over principal ideal domains (e.g., [Reference Dummit and Foote5]):

1. Fact A: Any submodule M of $\mathbb {Z}^{k}$ admits a finite basis.

Here is the effective version of this fact that we will use:

1. Fact B: Given a c.e. index for a submodule M of $\mathbb {Z}^{k}$ , we may $0'$ -effectively find a basis of M.

Proof idea: Consider the k-width and possibly-infinite-height matrix whose rows are the elements of M. Use $0'$ to find the GCD of the elements of the first column, call it d. If $d=0$ (i.e., all elements of the first column are zero), skip to the next step. If $d\neq 0$ , find a combination of rows of M that has d in its first coordinate, and add this combination to the basis. Then, if $k\geq 1$ , recursively apply this algorithm to the $(k-1)$ -width and possibly-infinite-height matrix whose elements consist of the rows of the previous matrix that have a $0$ as a first entry. Upon finding a combination of rows that has the new GCD as its first coordinates, add the corresponding combination of the rows of the original matrix to the basis. After k steps, we have a basis of M.

Returning to our original problem: We have a collection of combinations of generators, say $d_{0}$ , $d_{1}$ , $\dots $ , $d_{k}$ , and we wish to encode all possible linear combinations of these elements that are equal to zero. We do this by applying Fact B to the c.e. submodule of $\mathbb {Z}^{k+1}$ given by

$$\begin{align*}M=\{\,\vec{c}\in\mathbb{Z}^{k+1}\mid\vec{c}\cdot\vec{d}=0\,\}. \end{align*}$$

With access to a basis of M, say $b_{1},\dots ,b_{\ell }$ (encoded into the columns of a matrix B), we can tell whether any set of coefficients $\vec {c}$ is in M (and hence, whether we should say yes or no to $Q_{0,\vec {c}}(\vec {d})$ ) by using standard algorithms to solve the linear equation $B\vec {x}=\vec {c}$ over $\mathbb {Q}$ , and conclude $\vec {c}\in M$ if and only if a solution exists and its coefficients are all integers. Coding this basis into a c.e. index, we obtain a c.e. index for the atomic $\mathcal {L}'$ -type of a tuple $\vec {d}$ of linear combinations of generators.

Now, let us explain how to choose the element $\delta $ , which we will use to decide the predicates $n\Delta $ . Of course, this has to have been done in advance, so that we can hardcode the combination of generators corresponding to $\delta $ into the c.e. indices of the types of the elements of G in advance.

There is an easy way and a hard way to choose $\delta $ . The easy way is simply to pick an arbitrary nonzero element of G. The issue is that to make this choice requires the usage of an oracle for $0'$ , which will lead to a non-uniform presentation. Here is a uniform way to choose the element. Let $\{C_{i}\}_{i\in \mathbb {N}}$ enumerate all possible combinations of generators. To start, we will guess $\delta =C_{0}$ , and look through the relations until (if ever) we see enough relations to conclude $C_{0}=0$ . When this happens, we will guess $\delta =N_{1}C_{1}$ , where $N_{1}$ is a nonzero number divisible by enough integers. This choice is made so that, for every number k for which we’d been able to conclude $k\mid C_{0}$ before we found that $C_{0}=0$ , we have $k\mid N_{1}$ and so $k\mid N_{1}C_{1}$ . We proceed in this manner, enumerating relations until we are able to conclude that $N_{1}C_{1}=0$ , or equivalently that $C_{1}=0$ . If this happens, we change our mind to guess $\delta =N_{2}C_{2}$ , where $N_{2}$ has enough divisibilities that, again, every divisibility we’d already concluded about $N_{1}C_{1}$ also holds for $N_{2}$ and hence $N_{2}C_{2}$ . Since, by assumption, G is a nontrivial group, eventually this procedure will run into a nonzero combination of generators, and $0'$ is able to figure out which element is $\delta $ in advance. The important fact about this way of choosing $\delta $ is that we can uniformly find a c.e. index for the type of $\delta $ .

Now that we have a $0'$ -computable c.e.-typed copy of our group over the signature $\mathcal {L}'$ , we will show that every nontrivial torsion-free abelian group with a designated nonzero element with c.e. atomic type satisfies the QETP uniformly.

• • First, we construct $\chi _{q}$ . This is an adaptation of the standard proof that the theory of torsion-free abelian groups with divisibility-by-integer predicates admits quantifier elimination, which can be found e.g. in [Reference Hodges9, Appendix A.2]. Very briefly, we have the following lemma.

  Lemma 3.8. Any existential statement is equivalent to a system of the type

  (1) $$ \begin{align} (\exists\vec{y})q(\vec{x},\vec{y})\Leftrightarrow\left\{ \begin{array}{@{}l} M_{0}\vec{x}=0,\\ n_{1}\mid\vec{c}_{1}\cdot\vec{x},\\ \vdots\\ n_{\ell}\mid\vec{c}_{\ell}\cdot\vec{x},\\ K_{0}\vec{x}\neq0, \end{array}\right. \end{align} $$

  with the system found effectively. We define $\chi _{q}(\vec {x})$ to be the right-hand side of (1).

  The effective transformation goes as follows:

  • – Express the type $q(\vec {x},\vec {y})$ as a system of integer coefficient equations and inequations

    $$\begin{align*}q(\vec{x},\vec{y})\equiv\left\{ \begin{array}{@{}l} M\vec{x}=N\vec{y},\\ K\vec{x}\neq L\vec{y}. \end{array}\right. \end{align*}$$

    Important: The notation $K\vec {x}\neq L\vec {y}$ means “every entry of the vector $K\vec {x}$ is distinct from the corresponding entry in $L\vec {y}$ .”

  • – Now, write $N=ADB$ , where D is a diagonal matrix padded with zeros (i.e. of the form $\left [\begin {smallmatrix}D_{0} & 0\\ 0 & 0 \end {smallmatrix}\right ]$ with $D_{0}$ diagonal) and A and B are invertible, all with integer entries [Reference Bapat1, Theorem 2.12]. Then, $q(\vec {x},\vec {y})$ is logically equivalent to a new system of equations and inequations

    $$\begin{align*}q(\vec{x},\vec{y})\Leftrightarrow\left\{ \begin{array}{@{}l} M'\vec{x}=DB\vec{y},\\ K'\vec{x}\neq LB^{-1}B\vec{y}. \end{array}\right. \end{align*}$$

  • – The existence of a solution to this system is, since B is invertible, equivalent to the existence of a solution to the following other system:

    $$\begin{align*}(\exists\vec{y})q(\vec{x},\vec{y})\Leftrightarrow(\exists\vec{z})\left\{ \begin{array}{@{}l} M'\vec{x}=D\vec{z},\\ K'\vec{x}\neq L'\vec{z}. \end{array}\right. \end{align*}$$

  • – The rows of the equation $M'\vec {x}=D\vec {z}$ that have nonzero entries of D yield equations of the type $nz_{i}=\vec {a}\cdot \vec {x}$ . This allows us to remove (while preserving logical equivalence) every instance of $z_{i}$ from every other (in)equation: Any other (in)equation can be multiplied by n preserving logical equivalence (this uses $n\neq 0$ and the fact that we’re working with torsion-free groups), resulting in an instance of $\text {(coefficient)}{\times}nz_{i}$ , which can be substituted for $\text {(coefficient)}\times (\sum a_{i}x_{i})$ . Thus, we may eliminate all z variables corresponding to nonzero rows of D, and our system now becomes

    $$\begin{align*}(\exists\vec{y})q(\vec{x},\vec{y})\Leftrightarrow(\exists\vec{z})(\exists\vec{w})\left\{ \begin{array}{@{}l} M_{0}\vec{x}=0,\\ n_{1}z_{1}=\vec{c}_{1}\cdot\vec{x},\\ \vdots\\ n_{\ell}z_{\ell}=\vec{c}_{\ell}\cdot\vec{x},\\ K'\vec{x}\neq L'\vec{w}. \end{array}\right. \end{align*}$$

  • – The equations including the z variables become divisibility statements, i.e., $(\exists z_{i})nz_{i}=\vec {c}\cdot \vec {x}\Leftrightarrow n\mid \vec {c}\cdot \vec {x}\equiv Q_{n,\vec {c}}(\vec {x})$ , so we now need only deal with the inequations $K'\vec {x}\neq L'\vec {w}$ .

  • – Now, let us stratify the equations $K'\vec {x}\neq L'\vec {w}$ as follows: First, consider the equations that use no w-variable. Then, consider the ones that use $w_{1}$ and no other. Then, consider the ones that use only $w_{1}$ and $w_{2}$ , and so on. The equations that use no w-variable are important—keep them. These are a requirement on $\vec {x}$ that we cannot get rid of. Let us refer to these equations as $K_{0}\vec {x}\neq 0$ . On the other hand, at every new stage of the strata, the resulting equations will doubtlessly be satisfied by some $w_{i}$ . For example, if the first stratum has e equations, this is forbidding at most e possible values of $w_{1}$ . But the group is nontrivial and hence infinite, whereby there is at least one valid choice for $w_{1}$ . Then, in the next stratum, contingent on this choice of $w_{1}$ , there are once again at most (number of equations) forbidden values for $w_{2}$ . Simply choose $w_{2}$ to not be one of those. Continuing in this manner, we see that we can ensure that all equations that include a w variable can be automatically satisfied and are therefore adding no information. This concludes the effective transformation.

    Remark 3.9. If it is known that not every $x_{i}$ equals zero, one might in fact choose the $w_{i}$ variables to be linear combinations of the $x_{i}$ . This will be important later.

• • Next, we construct $\tau $ . Intuitively, given $q(\vec {x},y,\vec {z})$ , we consider all possible assignments of y of the form $y=\frac {1}{a}\vec {c}\cdot \vec {x}$ (if such an element exists). We will prove below, when establishing the QETP, that assuming not every element of $\vec {x}$ is null there is such a combination that provably satisfies $\chi _{q}(\vec {x},\frac {1}{a}\vec {c}\cdot \vec {x})$ . We will choose one such combination, and set $\tau _{q\varphi }(\vec {x},y)\equiv (ay=\vec {c}\cdot \vec {x}$ ). If every element of $\vec {x}$ is null, we set $\tau (\vec {x},y)\equiv N\Delta (y)$ , where N is large enough to encompass all divisibilities demanded by the tuple of elements $\vec {z}$ (e.g., if $q(y,z)\equiv (z+z=y)$ , we would set $y=2\delta $ ).

  Let’s discuss this process more formally. Given $q(\vec {x},y,\vec {z})$ :

  • – If q includes the data $y=0$ , set $\tau _{q\top }(\vec {x},y)\equiv (y=0)$ . Otherwise:

  • – Set $\varphi _{0}(\vec {x})\equiv \land _{i}(x_{i}=0)$ and set $\tau _{q\varphi _{0}}(\vec {x},y)\equiv \varphi _{0}(\vec {x})\land N\Delta (y)$ , where N is positive and large enough to ensure $(\forall d\neq 0)\chi _{q}(\vec {0},Nd)$ .

  • – For every nonzero integer a and tuple of integers $\vec {c}$ , set $\varphi _{a\vec {c}}(\vec {x})\equiv {\text{"} {\chi _{q}(\vec {x},\frac {1}{a}\vec {c}\cdot \vec {x})}\text{"}}$ and $\tau _{q\varphi _{a\vec {c}}}(\vec {x},y)\equiv (ay=\vec {c}\cdot \vec {x})$ . We now explain precisely what we mean by $\chi _{q}(\vec {x},\frac {1}{a}\vec {c}\cdot \vec {x})$ .

    Since q is assumed to be a type, we have by definition that $\chi _{q}(\vec {x},y)$ is a system of equations as in (1). By multiplying all equations by a, we obtain an equivalent system of equations where every instance of y is being multiplied by a. In other words, we can rewrite $\chi _{q}(\vec {x},y)\Leftrightarrow \psi (\vec {x},ay)$ , and as a consequence the formula “ $\chi _{q}(\vec {x},\frac {1}{a}\vec {c}\cdot \vec {x})$ ” may be taken to mean

    $$\begin{align*}{\text{"}{\chi_{q}(\vec{x},\frac{1}{a}\vec{c}\cdot\vec{x})}\text{"}}\equiv(a\mid\vec{c}\cdot\vec{x})\land\psi(\vec{x},\vec{c}\cdot\vec{x}). \end{align*}$$

• • Finally, we construct $\varepsilon \tau $ . Given $q(\vec {x},y,\vec {z})$ and $\varphi (\vec {x})$ as one of the above cases, as well as $Q(\vec {x},y,\vec {z},\vec {w})$ extending q, we wish to find a quantifier-free expression that is equivalent to $(\exists y)\left [\chi _{Q}(\vec {x},y)\land \tau _{q\varphi }(\vec {x},y)\right ]$ . Intuitively, the idea is that in either case our formula $\tau $ determines y exactly (as a function of $\vec {x}$ ), so we can just “plug that y into $\chi _{Q}(\vec {x},y)$ .” To do this more precisely, we split into cases:

  • – If $\tau _{q\varphi }(\vec {x},y)\equiv (ay=\vec {c}\cdot \vec {x})$ , we apply the same “multiply every equation by a” trick to write a formula that corresponds to “ $\left (a\mid \vec {c}\cdot \vec {x}\right )\land \chi _{Q}(\vec {x},\frac {1}{a}\vec {c}\cdot \vec {x})$ .” The resulting formula is the desired $\varepsilon \tau $ .

  • – If $\tau _{q\varphi }(\vec {x},y)\equiv \varphi _{0}(\vec {x})\land N\Delta (y)$ , then, intuitively, we know that for a tuple to satisfy $\tau $ it must be that all $x_{i}$ are zero, and that $y=N\delta $ . Therefore, the existential quantifier may be eliminated, and we wish to determine whether $\chi _{Q}(\vec {0},N\delta )$ . This is where it is important that we have a c.e. index for the type of $\delta $ —and, if we desire a uniform result, that we have this index without recourse to any oracle. Knowing the c.e. index for the type of $\delta $ , we can just consult whether $\chi _{Q}(\vec {0},N\delta )$ holds—more precisely, there is a computable sequence $\{B_{n}\}_{n\in \mathbb {N}}$ such that

    $$\begin{align*}\left\{ \begin{aligned} & G\nvDash\chi_{Q}(\vec{0},N\delta) & \Rightarrow & B_{n}=\bot\text{ for all } {n},\\ & G\vDash\chi_{Q}(\vec{0},N\delta) & \Rightarrow & B_{n}=\top\text{ for all but finitely many } {n}. \end{aligned} \right. \end{align*}$$
    In this scenario, we set
    

Now we verify that the above definitions are witnesses to the QETP.

1. (a) ( $\mathcal {L}$ is finite and closed under negation) Obvious.

2. (b) ( $G\vDash \chi _{q}(\vec {x})\leftrightarrow (\exists \vec {y})q(\vec {x},\vec {y})$ ) By construction.

3. (c) There are two bullet points here. In reverse order:

   • • ( $G\vDash \chi _{q}(\vec {x})\land \varphi (\vec {x})\rightarrow (\exists y)\tau _{q\varphi }(\vec {x},y)$ ) Direct from construction.

   • • (Every tuple $\vec {b}$ of elements of G that satisfies $\chi _{q}(\vec {b})$ satisfies some $\varphi (\vec {b})$ with $(q,\varphi )$ in domain of $\tau $ )

     • – If $\vec {b}=\vec {0}$ this is obvious.

     • – If some $b_{i}\neq 0$ : We wish to show that there are $a\in \mathbb {Z}_{\neq 0}$ and $\vec {c}\subseteq \mathbb {Z}$ such that $a\mid \vec {c}\cdot \vec {b}$ and $\chi _{q}(\vec {b},\frac {1}{a}\vec {c}\cdot \vec {b})$ . To this effect, we use the following related lemma, which obviously implies the desired claim.

       Lemma 3.10. If $q(\vec {x},\vec {y})$ is a quantifier-free type in the smaller signature $\mathcal {L}$ , and $\vec {b}$ is a tuple in G such that $G\vDash (\exists \vec {y})q(\vec {b},\vec {y})$ , and some element of $\vec {b}$ is nonzero, then there exist integer vectors $\vec {c}_{1},\dots ,\vec {c}_{\ell }$ and nonzero integers $a_{1},\dots ,a_{\ell }$ such that, for every i, $a_{i}\mid \vec {c}_{i}\cdot \vec {b}$ , and $G\vDash q(\vec {b},\frac {1}{a_{1}}\vec {c}_{1}\cdot \vec {b},\dots ,\frac {1}{a_{\ell }}\vec {c}_{\ell }\cdot \vec {b})$ .

       The proof of this lemma uses a change of variables, similar to the one used in the construction of $\chi _{q}$ , followed by noticing Remark 3.9. The details are omitted.

4. (d) ( $G\vDash (\exists y)\left [\chi _{Q}(\vec {x},y)\land \tau _{q\varphi }(\vec {x},y)\right ]\leftrightarrow \varepsilon \tau _{q\varphi Q}(\vec {x})$ ) True by construction.

5. (e) ( $G\vDash \varepsilon \tau _{q\varphi Q}(\vec {x})\land \tau _{q\varphi }(\vec {x},y)\rightarrow \chi _{Q}(\vec {x},y)$ ) Essentially, this is true because $\tau _{q\varphi }(\vec {x},y)$ always determines y uniquely in terms of $\vec {x}$ . As such, if it is the case that there is some $y_{0}$ that satisfies $\chi _{Q}(\vec {x},y_{0})\land \tau _{q\varphi }(\vec {x},y_{0})$ (i.e., $\varepsilon \tau _{q\varphi Q}(\vec {x})$ holds), and it is the case that some specific y satisfies $\tau _{q\varphi }(\vec {x},y)$ , then it must be the case that $y=y_{0}$ and therefore $\chi _{Q}(\vec {x},y)$ .

We are in condition to apply Theorem 2.6 to the $0'$ -computable c.e.-typed copy of G (plus designated element). Since this copy was obtained uniformly from the original presentation, the result is uniform.

Proof. By construction.

Proof. By construction.

Proof. We will apply Theorem 2.6 to D which, as an $X'$ -computable model over a finite signature, is also an $X'$ -computable X-c.e.-typed model. We need only establish that D has the QETP over $\mathcal {L}=(S^{i+1},\dots ,S^{n})$ (plus negations) and $\mathcal {L}'=(S^{i},\dots ,S^{n})$ (plus negations).

We construct $\chi $ , $\tau $ , and $\varepsilon \tau $ :

• • First, we construct $\chi _{q}$ for $q(\vec {x},\vec {y})$ a quantifier-free $\mathcal {L}$ -type with all variables distinct.

  We start by considering the case, where $q(\vec {x},y)$ only has one y-variable. In this case, the formula $(\exists y)q(\vec {x},y)$ is true if, and only if, there exists an element y satisfying a certain prescription of predicates $S^{i+1},\dots ,S^{n}$ . We show that the following three statements are equivalent:

  • – $P(\vec {x})\equiv (\exists y)q(\vec {x},y)$

  • – $Q(\vec {x})$ : For every instance of $\neg S^{j+1}(\vec {z}_{0},z_{1})$ in $q(\vec {x},y)$ :

    • * if $j>i$ , we require that $S^{j}(\vec {z}_{0})$ be in q, andFootnote ¹⁰

    • * if $j=i$ , $z_{1}\equiv y$ , and $\vec {z}_{0}$ is a subtuple of $\vec {x}$ , we require $S^{i}(\vec {z}_{0})$ .

    Moreover, we also demand $(q\mathord {\upharpoonright }\vec {x})(\vec {x})$ . In other words, whatever demand q makes of the tuple $\vec {x}$ , we also demand it.

  • – $R(\vec {x})\equiv $ There is a y such that $q(\vec {x},y)$ and the relations $S^{0}$ to $S^{i}$ hold for all tuples including elements from $(\vec {x},y)$ that include at least one instance of y.

  Indeed, $P(\vec {x})\rightarrow Q(\vec {x})$ follows from the fact that $(\forall \vec {z}_{0})((\exists z_{1})\neg S^{j+1}(\vec {z}_{0},z_{1}))\rightarrow S^{j}(\vec {z}_{0})$ (and the assumption that q is a type), $Q(\vec {x})\rightarrow R(\vec {x})$ follows by Lemma 4.3, and $R(\vec {x})$ obviously implies $P(\vec {x})$ .

  Now, still in the case where there is only one y-variable, we set $\chi _{q}(\vec {x})\equiv Q(\vec {x})$ as above. We will now handle the case where there are two y-variables, say $y_{1}$ and $y_{2}$ , and an obvious induction will handle the general case. We will use the formulas P, Q, and R constructed above for different formulas q. We will use a subscript to indicate which formula q is being used.

  Let $q(\vec {x},y_{1},y_{2})$ be a quantifier-free $\mathcal {L}$ -type. Then, let us look at $\chi _{q}(\vec {x},y_{1})$ . As above, it’s either $\bot $ (if the first bullet point of the definition of $Q_{q}(\vec {x},y_{1})$ fails), in which case the quantifier elimination is trivial, or otherwise it consists of $q\mathord {\upharpoonright }(\vec {x},y_{1})$ , together with a few positive requirements of type $S^{i}(\vec {z}_{0})$ with $\vec {z}_{0}$ a subtuple of $(\vec {x},y_{1})$ . Let $q'(\vec {x},y_{1})$ be $q\mathord {\upharpoonright }(\vec {x},y_{1})$ together with those requirements that include $y_{1}$ , and $q^{*}(\vec {x})$ be the remainder of these requirements. Note that

  $$\begin{align*}(\exists y_{1})(\exists y_{2})q(\vec{x},y_{1},y_{2})\leftrightarrow(\exists y_{1})\chi_{q}(\vec{x},y_{1})\leftrightarrow q^{*}(\vec{x})\land(\exists y_{1})q'(\vec{x},y_{1}), \end{align*}$$
  and crucially that, since all $S^{i}$ -requirements made by $q'$ are positive, $(\exists y_{1})q'(\vec {x},y_{1})$ is logically between $P_{q'}(\vec {x})$ and $R_{q'}(\vec {x})$ , and as such is equivalent to both, and as such is equivalent to $Q_{q'}(\vec {x})$ . Then, we conclude $(\exists \vec {y})q(\vec {x},\vec {y})\leftrightarrow q^{*}(\vec {x})\land Q_{q'}(\vec {x})$ , and so we define the latter as $\chi _{q}$ in this scenario. The induction continues in a similar fashion, in the event that there are more y-variables, with the essential point being that every $S^{i}$ -demand being made at any step is positive.

• • Now, we construct $\tau $ . Given $q(\vec {x},y,\vec {z})$ , we set $\tau _{q\top }(\vec {x},y)$ to be the demand that every $S^{i}$ -relation including y at least once holds, and that $\chi _{q}(\vec {x})$ holds.

• • Finally, $\varepsilon \tau _{q\top Q}(\vec {x})$ is simply $\chi _{Q}(\vec {x})$ .

The proof that these formulas are witnesses to the QETP is simple and follows the same P, Q, and R reasoning used to establish that $\chi _{q}(\vec {x})\leftrightarrow (\exists y)q(\vec {x},y)$ .