Fact 2.7. If x and $\bar {y}$ are the variables occurring freely in $\varphi $ and if theory T implies $(\forall \bar y) (\exists x)\varphi (x,\bar y)$ then T implies that $(\forall x)(\varphi \to \psi )$ is equivalent to an h-formula.

Proof. In [Reference Palmgren56, Lemma 2], it was proved that every atomless reduced product has the simple cover property, and in [Reference Palmgren56, Lemma 3] it was proved that the simple cover property implies $\varphi $ is equivalent to a Boolean combination of h-formulas. Although the latter asserts that the theory of the reduced product implies $\varphi $ is equivalent to a Boolean combination of h-formulas, the proof gives that the simple cover property implies this, and that the Boolean combination depends only on $\varphi $ .

Proof. For simplicity, we denote the ideal $ {\mathrm {Fin}}({\mathbb I})$ by $ {\mathrm {Fin}}$ when the index set ${\mathbb I}$ is clear from the context.

We first assume that (2.1) holds for $\mathcal K(\mathfrak C)$ and prove (1) and (2).

Note that if $\mathcal K(\mathfrak C)$ satisfies (2.1), then so does $\mathcal K(\mathfrak C)^m$ (identified with the reduced product $\prod _{(i,j,T)\in m\times {\mathbb N}\times \mathbb T(\mathfrak C)} \mathcal N_{j,T}/ {\mathrm {Fin}}$ ) for every $m\geq 1$ , because $\mathcal K(\mathfrak C)\cong \mathcal K(\mathfrak C)^m$ by an isomorphism that lifts to an isomorphism between $\prod _{\xi \in T({\mathbb I})} N_\xi $ and $\prod _{\xi \in {\mathbb I}(\mathfrak C)\times m} N_\xi $ . To be precise, this isomorphism corresponds to a bijection $\alpha \colon T({\mathbb I})\to T({\mathbb I})\times m$ such that

$$\begin{align*}\alpha[\{(i,T): i\in {\mathbb N}\}]=\{ ((i,T),j): i\in {\mathbb N}, j<m\} \end{align*}$$

for every $T\in T(\mathfrak C)$ .

For simplicity of notation, we work with formulas in a single variable. Throughout, we let T be the common theory of all atomless reduced products of structures from $\mathfrak {C}$ and all implications are modulo this theory. We first prove Statement (1). Proposition 2.9 implies that there are h-formulas $A_1, \dots , A_n$ such that $ {\mathrm {Th}}(\mathcal K(\mathfrak C))$ implies that $\varphi $ is equivalent to a Boolean combination B of $A_1, \dots , A_n$ . We will show that $\varphi $ is equivalent to a subconjunction of $A_1,\dots , A_n$ . Let $l_1<\dots <l_m \leq n$ be such that $B \rightarrow A_{l_1} \wedge \cdots \wedge A_{l_m} $ and assume that $\{l_1,\dots ,l_m\}$ is a maximal collection of indices (under inclusion) with this property, in the sense that for all $l_{m+1}\leq n$ with $l_{m+1}\neq l_j$ , $j\leq m$ ,

$$\begin{align*}B \not \rightarrow \bigwedge_{j\leq m+1} A_{l_j}.\end{align*}$$

To simplify notation, by re-indexing we may assume $l_j=j$ for all $j\leq m$ . Therefore, for all $j>m$ , the formula $B \wedge \neg A_j$ is consistent. We will show that B is equivalent to $A_1 \wedge \cdots \wedge A_m$ . For each $m< j\leq n$ , let $\mathcal K_j$ be an isomorphic copy of $\mathcal K(\mathfrak C)$ ; denote by ${\mathbb I}_j$ its index set, and let $a_j$ be an element of $\mathcal K_j$ satisfying $B \wedge \neg A_j$ . Let $\mathcal K_{m}$ be another isomorphic copy of $\mathcal K(\mathfrak C)$ , with index set ${\mathbb I}_m$ , and let a be an element of $\mathcal K_m$ satisfying $A_1 \wedge \cdots \wedge A_m$ . As B is consistent, there is an element b in $\mathcal K_{m}$ satisfying B (and thus it also satisfies $A_1 \wedge \cdots \wedge A_m$ ). The structure

$$\begin{align*}\mathcal K=\prod_{m\leq j\leq n} \mathcal K_j \end{align*}$$

is isomorphic to $\mathcal K(\mathfrak C)^{n-m+1}$ and (as mentioned at the beginning of this proof) also to $\mathcal K(\mathfrak C)$ , via an isomorphism that respects the reduced product structure. Specifically, with ${\mathbb J}= (n-m+1)\times {\mathbb N}\times \mathbb T(\mathfrak C) = {\mathbb I}_m \sqcup \cdots \sqcup {\mathbb I}_n $ we have that $\prod _{(i,j,T)\in {\mathbb I}}\mathcal N_{j,T}/ {\mathrm {Fin}}$ is naturally isomorphic to $\mathcal K$ . For convenience, we write ${\mathbb I}=(n-m+1)\times {\mathbb N}\times \mathbb T(\mathfrak C)$ and $\mathcal M_l$ for $\mathcal N_{(j,T)}$ if $l=(i,j,T)$ , so that

$$\begin{align*}\mathcal K=\prod_{l\in {\mathbb I}} \mathcal M_l/{\mathrm{Fin}}. \end{align*}$$

Since the isomorphism respects the reduced product structure, (2.1) holds for this representation of $\mathcal K$ .

In $\mathcal K$ the element $b^\frown a_{m+1}^\frown \cdots ^\frown a_n$ satisfies B (by condition (2.1)). Moreover, since the $A_i$ ’s are h-formulas, multiple applications of Theorem 2.8 imply that both $a^\frown a_{m+1}^\frown \cdots ^\frown a_n$ and $b^\frown a_{m+1}^\frown \cdots ^\frown a_n$ satisfy

$$\begin{align*}A_1\wedge \cdots \wedge A_m \wedge \neg A_{m+1} \wedge \dots \wedge \neg A_n.\\[-24pt]\nonumber\end{align*}$$

Therefore, as B is a Boolean combination of the $A_i$ ’s, $a^\frown a_{m+1}^\frown \cdots ^\frown a_n$ also satisfies B. By condition (2.1), if we let $(c_i)_i\in \prod _{i\in {\mathbb I}} \mathcal M_i$ be a representative of $a^\frown a_{m+1}^\frown \cdots ^\frown a_n$ , then

$$\begin{align*}\{i\in \mathbb{J}\mid \mathcal{M}_i\models \neg \psi(c_i) \} \in {\mathrm{Fin}}({\mathbb J}). \end{align*}$$

It follows that

$$\begin{align*}\{i\in \mathbb{I}_m \mid \mathcal{M}_i\models \neg \psi(c_i) \}\in {\mathrm{Fin}}({\mathbb I}_m). \end{align*}$$

By Condition (2.1) again, $\mathcal K=\prod _{\mathcal {I}}\mathcal {M}_i $ satisfies $\varphi (a).$ As the element a was chosen arbitrarily to satisfy $A_1 \wedge \cdots \wedge A_m$ , this shows that

$$\begin{align*}A_1 \wedge \cdots \wedge A_m \rightarrow B,\end{align*}$$

and thus we conclude that $\varphi $ is equivalent to an h-formula (namely, the conjunction of the formulas $A_1,\dots ,A_m$ ). We let $\Phi $ be this h-formula. We now prove Statement (2). By our hypothesis and Theorem 2.8, we have for any $(a_i)_i$ ,

(*) $$ \begin{align} \{i \in \mathbb{I} \mid \mathcal{M}_i \models \neg \psi(a_i)\}\in \mathcal{I} \ \Leftrightarrow \ \{i \in \mathbb{I} \mid \mathcal{M}_i \models \neg \Phi(a_i)\} \in \mathcal{I}. \end{align} $$

Assume towards a contradiction that there is an $\mathcal {I}$ -positive set $J \subseteq \mathbb {I}$ such that, for any $i\in J$ , we have,

$$\begin{align*}\mathcal{M}_i \models (\exists x ) \neg (\psi(x) \leftrightarrow \Phi(x)). \end{align*}$$

For $i\in J$ , let $ a_i$ be an element in $\mathcal {M}_i$ such that $\neg (\psi (a_i) \leftrightarrow \Phi (a_i))$ holds. At the cost of restricting to a smaller positive set and swapping the role of $\psi $ and $\Phi $ , we may assume that for any $i\in J$ , $\neg \psi (a_i) \wedge \Phi (a_i)$ holds in $\mathcal {M}_i$ . For $i \in \mathbb {I}\setminus J$ , we choose $a_i$ in $\mathcal {M}_i$ such that $\Phi (a_i)$ holds (if it exists), otherwise take $a_i$ to be anything. Note that since $\Phi $ is a satisfiable h-formula, the set of indices in which $\Phi $ are not satisfiable belongs to $\mathcal {I}$ . Then we conclude that

$$\begin{align*}\{i \in \mathbb{I} \mid \mathcal{M}_i \models \neg \psi(a_i)\}\notin \mathcal{I}, \end{align*}$$

and

$$\begin{align*}\{i \in \mathbb{I} \mid \mathcal{M}_i \models \neg \Phi(a_i)\}\in \mathcal{I}, \end{align*}$$

contradicting $(*)$ .

This proves that if $\mathcal K(\mathfrak C)$ satisfies (2.1), then $\varphi $ and $\psi $ are equivalent to the same h-formula $\Phi $ modulo $T(\mathfrak C)$ . By Palyutin’s theorem (Theorem 2.8), (1) and (2) hold in every reduced product of structures in $\mathfrak C$ . This concludes the proof.

Proof. The proposition is an immediate consequence of the following two claims.

Claim 2.15. With respect to the theory of any structure $\mathcal {M}$ with at least two elements, the formula $x \neq y$ is equivalent to

$$ \begin{align*}\psi_{\neq}(x,y):= (\forall z) \ (x=y \rightarrow z=y),\end{align*} $$

and is therefore equivalent to an h-formula.

Claim 2.16. With respect to the theory of any structure $\mathcal {M}$ with at least three elements, the formula $x=z \rightarrow y=w$ is equivalent to the following formula:

(2.2) $$ \begin{align} \begin{aligned}(\exists u)(\exists x')(\exists y') & \big(\psi_{\neq}(u,x)\wedge \psi_{\neq}(u,y)\big. \\ \wedge &(x=z \leftrightarrow x'=z)\\ \wedge &(\forall x") \left ( (x=z \leftrightarrow x"=z) \rightarrow (x"=u \rightarrow x'=u) \right)\\ \wedge & (y=w \leftrightarrow y'=w)\\ \wedge &(\forall y") \left ( (y=w \leftrightarrow y"=w) \rightarrow (y"=u \rightarrow y'=u)\right)\\ \big.\wedge &y'=u \rightarrow x'=u\big). \end{aligned} \end{align} $$

Proof. To facilitate the argument, we view the variables $x,y,z,w,\dots $ as functions from a set A to the structure $\mathcal {M}$ . The claim then follows if we take A to be a singleton and identify $\mathcal {M}$ with the set of functions from A to $\mathcal {M}$ . The first line of (2.2) introduces u, a function nowhere equal to x and y. The second line defines $x'$ as the function which coincides with x whenever x coincides with z, and coincides with u everywhere else. The third line similarly defines $y'$ as the function which coincides with y whenever y coincides with w, and coincides with u everywhere else. The fourth line compares where $x'$ and $y'$ coincide with u.

To see that (2.2) is equivalent to the original formula, notice that $x'$ coincides with u more often than $y'$ does if and only if y coincides with w more often than x coincides with z.

We proceed to prove Proposition 2.14. We may assume that $\mathcal M_i$ , for $i\in {\mathbb I}$ , are structures with at least three elements each and that $x=z\to y=z$ is equivalent to an h-formula modulo $ {\mathrm {Th}}_{\mathcal I}(\mathcal M_i, i\in {\mathbb I})$ . In order to prove that $x=z \rightarrow y=w$ is equivalent to an h-formula modulo $ {\mathrm {Th}}_{\mathcal {I}}(\mathcal {M}_i, i\in \mathbb {I})$ it suffices to prove that each of the conjuncts in the body of the formula (2.2) is equivalent to an h-formula. This is true for the first line by Claim 2.15.

Note that $x=z \leftrightarrow x'=z$ is an abbreviation for

$$\begin{align*}x=z \rightarrow x'=z \wedge x'=z \rightarrow x=z, \end{align*}$$

and is therefore equivalent to an h-formula. This takes care of the second, fourth and sixth conjuncts.

Since the formula

$$\begin{align*}(\exists x') ~(x=z \leftrightarrow x'=z )\end{align*}$$

trivially holds for all x and z, we deduce that the remaining two conjuncts in (2.2) are equivalent to h-formulas.

Proof. Assume Condition (1). Fix $a,a',b$ and $b'$ in $\prod _{\mathcal {I}} \mathcal {M}_i$ with representatives $(a_i)_{i},(b_i)_{i},(a^{\prime }_i)_{i}$ and $(b^{\prime }_i)_{i}$ respectively. Then,

$$ \begin{align*} {\mathrm{supp}}_=(a,a') \subseteq {\mathrm{supp}}_=(b,b') & \Leftrightarrow \{i : a_i = a_i'\ \mathrm{and}\ b_i\neq b_i^{\prime}\} \in \mathcal{I}\\ & \Leftrightarrow \{i \in \mathbb{I} \mid \neg (a_i = a_i' \rightarrow b_i = b_i')\} \in \mathcal{I}\\ & \Leftrightarrow \{i \in \mathbb{I} \mid \mathcal{M}_i \models \neg \Phi(a_i,a_i',b_i,b_i')\} \in \mathcal{I} \\ & \Leftrightarrow \prod_{\mathcal{I}} \mathcal{M}_i \models \Phi(a,a',b,b'). \end{align*} $$

It follows that $\mathcal {P}(\mathbb {I})/\mathcal {I} $ is interpretable in $\prod _{\mathcal {I}} \mathcal {M}_i$ as a poset. Moreover, all Boolean operations are interpretable in this poset. For example, $A\cap B$ is the maximal (under $\leq $ ) C such that $C\leq A$ and $C\leq B$ , and $A\setminus B$ is the maximal C such that $C\leq A$ and $C\cap B=\emptyset $ . Therefore, $\mathcal {P}(\mathbb {I})/\mathcal {I} $ is interpretable as a Boolean algebra. The natural projection is the relative support function; therefore Condition (2) holds.

Proof. The implication $(1)\Rightarrow (2)$ is obvious and $(3)\Rightarrow (1)$ follows from Lemma 2.17. It remains to show that (2) implies (3). Assume that (2) holds and consider $\mathcal K(\mathfrak C)$ as in Definition 2.10. Let $\varphi (x,x',y,y')$ be the $\mathcal {L}$ -formula defining the relation $ {\mathrm {supp}}_=(a,a') \subseteq {\mathrm {supp}}_=(b,b')$ in $\prod _{\mathcal {I}} \mathcal {M}_i$ . It follows that for all $a,a',b,b' \in \prod _{\mathcal {I}} \mathcal {M}_i$ , with respective representatives $(a_i)_i,(a^{\prime }_i)_i,(b_i)_i,(b^{\prime }_i)_i$ :

$$\begin{align*}\prod_{\mathcal{I}} \mathcal{M}_i \models \varphi(a,a',b,b') \Leftrightarrow \{i \in \mathbb{I} \mid \neg (a_i = a_i' \rightarrow b_i = b_i')\} \in \mathcal{I}. \end{align*}$$

By Proposition 2.11 (1), in $ {\mathrm {Th}}(\mathcal K(\mathfrak C))$ the formula $\varphi (x,x',y,y')$ is equivalent to an h-formula $\Phi (x,x',y,y')$ , and by Theorem 2.8 we have

$$\begin{align*}\prod_{\mathcal{I}} \mathcal{M}_i \models \Phi(a,a',b,b') \Leftrightarrow \{i \in \mathbb{I} \mid \neg \Phi(a_i,a_i',b_i,b_i')\} \in \mathcal{I}. \end{align*}$$

By Proposition 2.11 (2), for all but finitely many $i\in \mathbb {I}$ we have that

$$\begin{align*}\mathcal{M}_i \models \forall x,x',y,y'~\left(\Phi(x,x',y,y') \leftrightarrow (x=x'\rightarrow y=y')\right). \end{align*}$$

Since $ {\mathrm {Th}}_{ {\mathrm {Fin}}({\mathbb I}(\mathfrak C))} (\mathcal N_{j,T}:(j,T)\in {\mathbb I}(\mathfrak C))= {\mathrm {Th}}(\mathfrak C)$ (Lemma 2.13), we have the equivalence in $ {\mathrm {Th}}(\mathfrak C)$ , as required, and this concludes the proof.

Proof. Let ${\mathcal U}$ be an ultrafilter on a set ${\mathbb J}$ and let ${\mathcal I}_j$ , for $j\in {\mathbb J}$ , be a family of ideals. We will produce an ideal ${\mathcal J}$ such that the ultraproduct $\prod _{\mathcal U} (\prod _{{\mathcal I}_j} \mathcal {M}_{ij})$ is isomorphic to the reduced product $\prod _{\mathcal J} \mathcal {M}_{ij}$ for every (appropriately indexed) family of structures $\mathcal {M}_{ij}$ . We will also show that if all ${\mathcal I}_j$ are atomless then so is ${\mathcal J}$ , thus proving both parts of Proposition 2.19.

It will be convenient to assume that all ${\mathcal I}_j$ are ideals on the same index set ${\mathbb I}$ . Towards this, let ${\mathbb I}=\bigsqcup _{j\in {\mathbb J}} {\mathbb I}_j$ (the disjoint union of ${\mathbb I}_j$ ) and replace ${\mathcal I}_j$ with the ideal on ${\mathbb I}$ generated by ${\mathcal I}_j$ and the set $\bigcup _{k\in {\mathbb J}, k\neq j} {\mathbb I}_k$ for every j. Clearly this does not affect $\prod _{{\mathcal I}_j} \mathcal {M}_{ij}$ (take $\mathcal {M}_{ij}$ to be arbitrary for $i\in {\mathbb I}\setminus {\mathbb I}_j$ ).

Let ${\mathcal U}_*$ be the ideal on ${\mathbb J}$ dual to ${\mathcal U}$ and note that for any indexed family of $\mathcal {L}$ -structures $(A_j)_{j \in \mathbb {J}}$ , the ultraproduct $\prod _{{\mathcal U}} A_j$ is literally the same as the reduced product $\prod _{{\mathcal U}_*} A_j$ . For $X\subseteq {\mathbb J}\times {\mathbb I}$ and $j\in {\mathbb J}$ , let $X_j=\{i\mid (j,i)\in X\}$ . On the set ${\mathbb J}\times {\mathbb I}$ define ${\mathcal J}$ to be the set of all $X\subseteq {\mathbb J}\times {\mathbb I}$ such that,

$$\begin{align*}\{j\mid X_j\notin {\mathcal I}_j\}\in {\mathcal U}_*. \end{align*}$$

The conclusion of the first part of Proposition 2.19 follows from the two claims.

Proof. We first observe that the sort,

$$\begin{align*}B:= (\mathcal{P}(\mathbb{I})/\mathcal{I} ;\cap ,\cup, 0, 1, \{c_\theta : \theta \text{ an } \mathcal{L}\text{- sentence} \} ), \end{align*}$$

is closed in the following sense: in the language, there are neither functions from B to the main sort $\mathcal {M}$ , nor predicates in $B^n\times \mathcal {M}^m$ for strictly positive integers $n,m$ . Any (parameter-free) $\mathcal {L}^{+}$ -formula is equivalent, in the theory of the reduced product, to a Boolean combination of formulas of the form:

• • $\psi _B( {\mathrm {supp}}_{\varphi }(\bar {x}))$

• • $\varphi (\bar {x})$

where $\varphi $ is an $\mathcal {L}$ -formula, and $\psi _B$ is a formula in the language $\mathcal {L}_{\textrm {Bool}} \cup \{c_\theta :~\theta \text { an }\mathcal {L}\text {-sentence} \}$ . By [Reference Chang and Keisler9, Theorem 6.3.2], a formula of the second kind can be expressed as a formula of the first kind. Therefore, we have eliminated all quantifiers in the main sort $\prod _{\mathcal {I}} \mathcal {M}_i$ . Since the sort is closed, relative quantifier elimination follows by [Reference Rideau-Kikuchi63, Remark A.8].

Proof. It is enough to show that every function of the form $ {\mathrm {supp}}_{\psi (\bar x)}$ , where $\psi (\bar x)$ is any $\mathcal {L}$ -formula, is equivalent to a quantifier-free $\mathcal {L}^+_\Phi $ -formula. Fix an $\mathcal {L}$ -formula $\psi (\bar x)$ . By assumption, $\psi (\bar x)$ is equivalent to a Boolean combination of formulas from $\Phi $ , say $\varphi _1(\bar x),\dots ,\varphi _k(\bar x)$ . By Proposition 2.26, $ {\mathrm {supp}}_{\psi (\bar x)}$ is the same Boolean combination of $ {\mathrm {supp}}_{\varphi _i(\bar x)}$ , $i\leq k$ . Therefore, it is expressed without quantifiers in the language $\mathcal {L}^+_\Phi $ , as desired.

Question 2.28. For which indexed families of $\mathcal {L}$ -structures $(\mathcal {M}_i)_{i \in \mathbb {I}}$ and ideals $\mathcal {I}$ on $\mathbb {I}$ does the reduced product $\prod _{\mathcal {I}} \mathcal {M}_i$ interpret the Boolean algebra $\mathcal {P}(\mathbb {I})/\mathcal {I} $ and the support functions $ {\mathrm {supp}}_{\varphi (\bar {x})}$ in the language $\mathcal {L}$ ? In other words, in which reduced products is the language $\mathcal {L}^+$ a definable expansion of the language $\mathcal {L}$ ?

Proof. We may assume that $\bar x$ is a single variable, as the proof changes only notationally. Fix $b\in \prod _{\mathcal {I}} \mathcal {M}_i$ satisfying $\varphi (x)$ . For every element $a \in \prod _{\mathcal {I}} \mathcal {M}_i$ and $S\in \mathcal {P}(\mathbb {I})/\mathcal {I} $ , there is a unique element $a_S$ which coincides with a on S and with b on $S^\complement $ . Formally, $a_S$ is the unique element such that:

• • $ {\mathrm {supp}}_=(a_S,a) \supseteq S$ ,

• • $ {\mathrm {supp}}_=(a_S,b) \supseteq S^\complement $ .

Then, $\varphi (a_S)$ holds if and only if $\mathcal M_i\models \varphi (a_S(i))$ for all but ${\mathcal I}$ -many i’s, and this is equivalent to $S\setminus {\mathrm {supp}}_{\varphi (\bar {x})}(\bar {a}) \in {\mathcal I}$ . Therefore $ {\mathrm {supp}}_{\varphi (\bar {x})}(\bar {a})$ is the largest $S \in \mathcal {P}(\mathbb {I})/\mathcal {I} $ with respect to inclusion such that $\varphi (a_S)$ holds. Since S does not depend on the choice of b, we get that the $\varphi $ -support function $ {\mathrm {supp}}_{\varphi (x)}$ is interpretable without parameters.

Proof. We fix an interpretation of the Boolean algebra $\mathcal {P}(\mathbb {I})/\mathcal {I} $ and of the map $ {\mathrm {supp}}_=$ . By Lemma 2.29, for any h-formula $\varphi (x)$ the function $ {\mathrm {supp}}_{\varphi (\bar {x})}$ is definable in the language,

$$\begin{align*}\mathcal{L} \cup \mathcal{L}_{\textrm{Bool}}\cup \left\{{\mathrm{supp}}_{=}\right\}. \end{align*}$$

Relative quantifier elimination follows directly from Corollary 2.27.

Proof. We first recover the base set of $\mathcal {M} \restriction S$ as the quotient of $\mathcal {M}$ by the equivalence relation $\simeq $ given by: for all $a,b \in \mathcal {M}$ ,

$$\begin{align*}a\simeq b \Leftrightarrow {\mathrm{supp}}_=(a,b) \supseteq S.\end{align*}$$

It remains to interpret the $\mathcal {L}$ -structure of $\mathcal {M} \restriction S$ . Let R be a predicate in $\mathcal {L}$ . Since $\mathfrak C$ is full, R is a satisfiable h-formula. By Lemma 2.29, the support function $ {\mathrm {supp}}_{R(\bar {x})}$ is interpretable in $\mathcal {M}$ . Then, for any $\bar {a}\restriction S$ in $\left (\mathcal {M}\restriction S\right )^{|\bar {x}|}$ , we have that $\mathcal {M}\restriction S \models R( \bar {a}\restriction S )$ if and only if $ {\mathrm {supp}}_{R(\bar {x})} (\bar {a}) \supseteq S$ . The interpretations of function symbols f from $\mathcal {L}$ in $\mathcal {M}\restriction S$ are also straightforward: for any $\bar {a}\restriction {S}\in \left (\mathcal {M}\restriction S\right )^{|\bar {x}|}$ , set

$$ \begin{align*}f(\bar{a}\restriction{S})=f(\bar{a})\restriction{S}. \end{align*} $$

Proof. $(1) \to (2)$ . Suppose that $\varphi (x,x',y,y')$ is a definable set. Consider the equivalence relation given by $E := \varphi (x,x',y,y') \cap \varphi (y,y',x,x')$ . Then $\mathcal {M}^{2}/E$ is naturally isomorphic to $\mathcal {P}(\mathbb {I})/\mathcal {I} $ . Let $\pi _{E}:\mathcal {M}^{2} \to \mathcal {M}^{2}/E$ be the quotient map. Notice that if $S, T \in \mathcal {P}(\mathbb {I})/\mathcal {I} $ , then $S \subseteq T$ if and only if $\exists a,b,c,d$ in $\mathcal {M}$ such that $\pi _{E}(a,b) = S$ , $\pi _{E}(c,d) = T$ and $\varphi (a,b,c,d)$ . Hence the relation $\subseteq $ is interpretable. Notice that $\pi _{E} = {\mathrm {supp}}_{=}$ which concludes this direction.

$(2) \to (1)$ . By construction, the set $\varphi (x,x',y,y')$ is a definable subset of $\mathcal {M}^{4}$ in $\mathcal {M}^{ {\mathrm {eq}}}$ . Since there are no new definable subsets of $\mathcal {M}^{n}$ in $\mathcal {M}^{ {\mathrm {eq}}}$ , we see that $\varphi (x,x',y,y')$ is already definable in M.

Proof. Assume that $\mathcal {M}= \prod _{\mathcal {I}}\mathcal {M}_i$ and $\mathcal {N}=\prod _{\mathcal {J}}\mathcal {N}_j$ are reduced products of structures from $\mathfrak C$ and let $\Phi \colon \mathcal {M}\to \mathcal {N}$ be an isomorphism. Then $\Phi $ extends to an isomorphism $\mathcal {M}^{ {\mathrm {eq}}}\rightarrow \mathcal {N}^{ {\mathrm {eq}}}$ and by Lemma 3.2, it induces an isomorphism $\alpha : \mathcal {P}(\mathbb {I})/\mathcal {I} \rightarrow \mathcal {P}(\mathbb {J})/\mathcal {J} $ . Then, we can name the parameter S, and $\Phi $ gives rise to an isomorphism between $\mathcal M\restriction S$ and $\mathcal N \restriction {\alpha (S)}$ by Lemma 3.1. Hence the following diagram commutes:

This shows that $\mathfrak C$ recognizes coordinates.

Proof. Only the direct implication requires a proof. Suppose that ZFC and CH together imply $\theta _{\psi }$ . Fix $\mathcal M_n$ , for $n\in {\mathbb N}$ , in $\mathfrak C$ and an ideal ${\mathcal I}$ on ${\mathbb N}$ . Let $\kappa :=\max (\sup _{n\in {\mathbb N}} |\mathcal M_n|, 2^{\aleph _0})$ , and let $\mathbb P$ be the forcing notion whose conditions are functions $p\colon \gamma _p\to \kappa $ , where $\gamma _p$ is a countable ordinal, ordered by the reverse extension (hence $p\leq q$ , i.e., p is a stronger condition than q, if $\gamma _p\geq \gamma _q$ and $p\restriction \gamma _q=q$ ).

This forcing notion is the Lévy collapse of $\kappa $ to $\aleph _1$ . We need two standard facts about forcing with $\mathbb P$ . First, in the generic extension there is a surjection from $\aleph _1$ to $\kappa $ [Reference Jech35, Lemma 15.21]. Second, $\mathbb P$ is $<\aleph _1$ -closed (see the last two lines of the proof of [Reference Jech35, Lemma 15.21]). Therefore $\mathbb P$ does not add any bounded subsets of $\kappa $ [Reference Jech35, Lemma 15.8] and in particular it does not add any new subsets of $\mathcal P({\mathbb N})$ , or any new elements of $\prod _n \mathcal M_n$ .

These two facts imply that CH holds in the forcing extension and that each one of the structures $\prod _{n} \mathcal M_n$ , $\mathcal M$ and $\mathcal M^+$ is unchanged. Moreover, since $\mathfrak C$ is assumed to be closed under elementary equivalence, each $\mathcal M_n$ still belongs to $\mathfrak C$ in the forcing extension. Since ZFC+CH implies $\theta _{\psi }$ , in the forcing extension there exists a formula $\varphi $ such that for all $\bar a$ in $\mathcal M$ we have that $\mathcal M^+\models \psi (\bar a)$ if and only if $\mathcal M\models \varphi (\bar a)$ . Note that $\varphi $ belongs to the ground model, as it is a finite sequence of symbols in ${\mathcal L}$ .

Since $\prod _n \mathcal M_n$ is unchanged by forcing, and since ${\mathcal I}$ is the same (since $\mathcal P({\mathbb N})$ is unchanged by forcing, ${\mathcal I}$ remains an ideal on ${\mathbb N}$ in the forcing extension), $\mathcal M$ is unchanged by forcing. Therefore $\varphi $ and $\psi $ define the same set in the ground model, as required.

Proof. Assume the contrary, that there are an atomless ideal ${\mathcal I}$ on some index set $\mathbb {I}$ and reduced product $\mathcal M:=\prod _{i\in \mathbb {I}} \mathcal M_i/\mathcal {I}$ of structures from $\mathfrak C$ such that no formula $\varphi $ defines $ {\mathrm {supp}}_=$ in $\mathcal M$ . Let $\mathcal F_4$ be the set of all ${\mathcal L}$ -formulas with at most four free variables.

For every formula $\varphi \in \mathcal F_4$ fix $a^\varphi _1,a^\varphi _2,b^\varphi _1,b^\varphi _2$ in $\mathcal M$ such that

(3.1) $$ \begin{align} {\mathrm{supp}}_=(a^\varphi_1,a^\varphi_2)\subseteq {\mathrm{supp}}_=(b^\varphi_1,b^\varphi_2)\qquad \not\Leftrightarrow \qquad \mathcal M\models \varphi(a^\varphi_1,a^\varphi_2,b^\varphi_1,b^\varphi_2). \end{align} $$

Consider the Boolean subalgebra ${\mathcal A}$ of $\mathcal P(\mathbb {I})$ generated by the sets

$$\begin{align*}\zeta^\varphi_\psi:=\{i\in \mathbb{I}\vert \mathcal M_i\models \psi(a^\varphi_1,a^\varphi_2,b^\varphi_1,b^\varphi_2)\} \end{align*}$$

and the sets

$$\begin{align*}{\mathrm{supp}}_=(a^\varphi_1,a^\varphi_2)\setminus {\mathrm{supp}}_=(b^\varphi_1,b^\varphi_2), \end{align*}$$

where $\varphi $ and $\psi $ range over $\mathcal F_4$ .

Let $\theta $ be a regular cardinal sufficiently large to have all of the objects mentioned so far – ${\mathcal L}$ , ${\mathbb I}$ , ${\mathcal I}$ , $\langle \mathcal M_i: i\in {\mathbb I}\rangle $ , $\mathcal F_4$ , the assignment $\varphi \mapsto (a^\varphi _1,a^\varphi _2,b^\varphi _1,b^\varphi _2)$ – belong to the set $H_\theta $ of all sets whose hereditary closure has cardinality less than $\theta $ (see e.g., [Reference Kunen41, Definition I.13.27]).

Let $N\preceq H_\theta $ be a countable elementary submodel that contains all of the objects listed above . Since the Boolean algebra ${\mathcal A}$ is countable, it is included in N. Define

$$ \begin{align*} {\mathbb I}_0&=N\cap {\mathbb I}, \\ {\mathcal I}_0&=\{A\cap {\mathbb I}_0: A\in N\cap {\mathcal I}\}. \end{align*} $$

By elementarity, ${\mathcal I}_0$ is a proper atomless ideal on $\mathbb {I}_0$ . Let $\mathcal M_0:=\prod _{i\in \mathbb {I}_0} \mathcal M_i/{\mathcal I}_0$ . Since $\mathbb {I}_0$ is countable, our assumption implies that there is a formula $\varphi $ such that

$$ \begin{align*} {\mathrm{supp}}_=(a_1,a_2)\subseteq {\mathrm{supp}}_=(b_1,b_2)\qquad \Leftrightarrow \qquad \mathcal M_0\models \varphi(a_1,a_2,b_1,b_2) \end{align*} $$

for all $a_1,a_2,b_1,b_2$ in $\mathcal M$ . In particular, this holds for $a_1=a_1^\varphi \restriction \mathbb {I}_0$ (i.e., $a_1=\pi _{\mathbb {I}_0}(a_1^\varphi )$ ), and analogously defined $a_2,b_1$ and $b_2$ .

By our choice of $\mathbb {I}_0$ and ${\mathcal I}_0$ , we have that

(3.2) $$ \begin{align} {\mathrm{supp}}_=(a_1,a_2)\subseteq {\mathrm{supp}}_=(b_1,b_2)\quad \Leftrightarrow\quad {\mathrm{supp}}_=(a_1^\varphi,a_2^\varphi)\subseteq {\mathrm{supp}}_=(b_1^\varphi,b_2^\varphi). \end{align} $$

By the Feferman–Vaught theorem as stated in [Reference Chang and Keisler9, Theorem 6.3.2], there are a finite list $\psi _j$ , for $1\leq j\leq k$ , of ${\mathcal L}$ -formulas and a formula $\Theta (z_1,\dots , z_k)$ in the language of Boolean algebras such that

$$\begin{align*}\mathcal M\models \varphi(a^\varphi_1,a^\varphi_2,b^\varphi_1,b^\varphi_2)\quad\Leftrightarrow\quad\mathcal P(\mathbb{I})/{\mathcal I}\models \Theta([\zeta^\varphi_{\psi_1}]_{{\mathcal I}},\dots, [\zeta^\varphi_{\psi_k}]_{{\mathcal I}}) \end{align*}$$

and also

$$\begin{align*}\mathcal M_0\models \varphi(a_1,a_2,b_1,b_2)\quad\Leftrightarrow\quad\mathcal P(\mathbb{I}_0)/{\mathcal I}_0\models \Theta([\zeta^\varphi_{\psi_1}\cap \mathbb{I}_0]_{{\mathcal I}_0},\dots, [\zeta^\varphi_{\psi_k}\cap \mathbb{I}_0]_{{\mathcal I}_0}). \end{align*}$$

Both $\mathcal {P}(\mathbb {I})/\mathcal {I} $ and $\mathcal {P}(\mathbb {I}_0)/\mathcal {I}_0$ are atomless Boolean algebras and we may therefore assume $\Theta $ is quantifier-freeFootnote ⁴ . Thus the truth of $\Theta (z_1,\dots , z_k)$ depends only on knowing which of the intersections $\bigcap _{j=1}^k z_j^{\xi (j)}$ are trivial, where , with . The choice of $N\preceq H_\theta $ , the set $\mathbb {I}_0$ and the ideal ${\mathcal I}_0$ assures that for all such $\xi $ we have $\bigcap _{j=1}^k (\zeta ^\varphi _{\psi _k})^{\xi (j)}\in {\mathcal I}$ if and only if $\bigcap _{j=1}^k (\zeta ^\varphi _{\psi _k}\cap \mathbb {I}_0)^{\xi (j)}\in {\mathcal I}_0$ (where the complement is evaluated with respect to $\mathbb {I}_0$ ).

Therefore $ \mathcal M\models \varphi (a^\varphi _1,a^\varphi _2,b^\varphi _1,b^\varphi _2)$ if and only if $\mathcal M_0\models \varphi (a_1,a_2,b_1,b_2)$ . Together with (3.2), this implies the negation of (3.1); contradiction.

Proof of Proposition 3.5.

Suppose that ${\mathcal L}$ is a countable language and $\mathfrak C$ is a full class of $\mathcal {L}$ -structures which recognizes coordinates in ${\mathbb N}$ , and fix a reduced product $\mathcal {M}:= \prod _{i\in \mathbb {I}}\mathcal M_i/\mathcal {I}$ of structures from $\mathfrak C$ . We need to prove that $\mathcal M$ defines the relative support function $ {\mathrm {supp}}_=$ , quotients $\mathcal M_{S}$ and quotient maps $\pi _S\colon \mathcal M\to \mathcal M\restriction S$ , for all $S\in \mathcal {P}(\mathbb {I})/\mathcal {I} $ . By Theorem 2.18 we may assume that the ideal ${\mathcal I}$ is atomless. By Lemma 3.7 it suffices to prove this in the case when $\mathcal M=\prod _{i\in {\mathbb N}}\mathcal M_i/\mathcal {I}$ is a reduced product associated with an atomless ideal $\mathcal {I}$ on ${\mathbb N}$ .

We use Beth’s definability theorem. For $ {\mathrm {supp}}_=$ , it suffices to prove that for every reduced product $\mathcal M$ of structures from $\mathfrak C$ , the support function on $\mathcal M$ is implicitly interpretable. This means that for every elementary extension $\mathcal {N}$ of $\mathcal {M}$ and any two ‘support’ functions $s_1$ and $s_2$ on $\mathcal N$ such that,

$$\begin{align*}(\mathcal{N},s_1) \equiv (\mathcal{M},{\mathrm{supp}}_=) \equiv (\mathcal{N},s_2), \end{align*}$$

we have that for all $a,a',b,b' \in \mathcal {N}$ , we have $s_1(a,a')\subseteq s_1(b,b') $ if and only if $s_2(a,a')\subseteq s_2(b,b')$ .

Fix $\mathcal M$ , $\mathcal N$ , $s_1$ and $s_2$ and let ${\mathcal U}$ be a nonprincipal ultrafilter on ${\mathbb N}$ . If the Continuum Hypothesis (CH) holds and all $\mathcal M_i$ as well as $\mathcal N$ have cardinality no greater than $2^{\aleph _0}$ , then the ultrapowers $\prod _{\mathcal {U}}(\mathcal {N},s_1)$ , $\prod _{\mathcal {U}}(\mathcal {M}, {\mathrm {supp}}_=)$ and $\prod _{\mathcal {U}}(\mathcal {N},s_2)$ have cardinality $2^{\aleph _0}$ and are $\aleph _1$ -saturated. Since they are also elementarily equivalent, we have two isomorphisms $\sigma : \prod _{\mathcal {U}}\mathcal {M} \rightarrow \prod _{\mathcal {U}}\mathcal {N} $ and $\tau : \prod _{\mathcal {U}}\mathcal {M} \rightarrow \prod _{\mathcal {U}}\mathcal {N}$ that moreover respect $ {\mathrm {supp}}_=$ , $s_1$ and $s_2$ so that the following holds:

(3.3) $$ \begin{align} \textstyle (\prod_{\mathcal{U}}\mathcal{N},s_1) \overset{\sigma}{\simeq} (\prod_{\mathcal{U}}\mathcal{M},{\mathrm{supp}}_=)\overset{\tau}{\simeq}(\prod_{\mathcal{U}}\mathcal{N},s_2). \end{align} $$

We will complete the proof under this assumption and then show how it can be removed.

By Proposition 2.19, there is an ideal $\mathcal {J}$ on the index set $\mathbb {J}={\mathbb N}\times {\mathbb N}$ and an enumeration $\mathcal {M}_{ij}$ , for $(i,j)\in {\mathbb N}\times \mathbb {{\mathbb N}}$ (with many repetitions), such that $\prod _{\mathcal {U}}(\mathcal {M}, {\mathrm {supp}}_=) = (\prod _{\mathcal {J}}\mathcal {M}_{ij}, {\mathrm {supp}}_=)$ .

Let $\rho = \tau ^{-1}\circ \sigma $ . This map is an automorphism of $\prod _{\mathcal {J}}\mathcal M_{ij}$ . We need to show that this is an automorphism of $(\prod _{\mathcal {J}}\mathcal M_{ij}, {\mathrm {supp}}_=)$ .

Since $\mathfrak C$ recognizes coordinates, there is an automorphism $\alpha : \mathcal {P}(\mathbb {J})/\mathcal {J} \rightarrow \mathcal {P}(\mathbb {J})/\mathcal {J} $ such that for all $S\in \mathcal {P}(\mathbb {J})/\mathcal {J} $ the following diagram commutes (see Definition 2.5):

To show that $(\rho ,\alpha )$ is an automorphism of $(\prod _{\mathcal {J}} \mathcal M_{ij}, {\mathrm {supp}}_=)$ , we only need to show that it respects the support function. Take a pair $(g,g')$ of elements of $\prod _{\mathcal {J}} \mathcal M_{ij}$ and set $S= {\mathrm {supp}}_=(g,g')$ . Then $\pi _S(g)=\pi _S(g')$ and since the diagram commutes, we have

$$\begin{align*}\pi_{\alpha(S)}(\rho(g))= \pi_{\alpha(S)}(\rho(g')). \end{align*}$$

It follows that $ \alpha ( {\mathrm {supp}}_=(g,g')) \subseteq {\mathrm {supp}}_=(\rho (g),\rho (g'))$ . By considering the isomorphisms $\rho ^{-1}$ and $\alpha ^{-1}$ instead of $\rho $ and $\alpha $ , by the analogous argument we have that

$$\begin{align*}\alpha^{-1}({\mathrm{supp}}_=(\rho(g),\rho(g')) \subseteq {\mathrm{supp}}_=(\rho^{-1}(\rho(g)),\rho^{-1}(\rho(g'))), \end{align*}$$

which gives the other inclusion:

$$\begin{align*}\alpha({\mathrm{supp}}_=(g,g')) \subseteq {\mathrm{supp}}_=(\rho(g),\rho(g')). \end{align*}$$

Therefore, for all $g,g'$ , we have $\alpha ( {\mathrm {supp}}_=(g,g')) = {\mathrm {supp}}_=(\rho (g),\rho (g'))$ and $(\rho ,\alpha )$ is an automorphism of $(\prod _{\mathcal {J}}\mathcal M_{ij}, {\mathrm {supp}}_=)$ .

The rest of the proof is straightforward: let $a,a',b,b' \in \mathcal {N}$ such that $s_1(a,a') \subseteq s_1(b,b')$ . Then in the ultraproduct, we have

$$\begin{align*}\prod_{\mathcal{U}}(\mathcal{N},s_1) \models s_1({\mathrm{diag}}(a),{\mathrm{diag}}(a')) \subseteq s_1({\mathrm{diag}}(b),{\mathrm{diag}}(b'))\end{align*}$$

where $ {\mathrm {diag}}$ is the canonical embedding of $\mathcal {N}$ in the ultrapower. Since $ {\mathrm {Id}} =\sigma ^{-1}\circ \rho \circ \tau $ is a composition of isomorphisms preserving the support, we also have

$$\begin{align*}\prod_{\mathcal{U}}(\mathcal{N},s_2)\models s_2({\mathrm{diag}}(a),{\mathrm{diag}}(a')) \subseteq s_2({\mathrm{diag}}(b),{\mathrm{diag}}(b')), \end{align*}$$

and therefore,

$$\begin{align*}(\mathcal{N},s_2) \models s_2(a,a') \subseteq s_2(b,b'). \end{align*}$$

By the Beth definability theorem, this proves that the relation $ {\mathrm {supp}}_=(x,x')\subseteq {\mathrm {supp}}_=(y,y')$ is first-order definable in $\mathcal M$ if CH holds and all $\mathcal M_n$ and $\mathcal N_n$ have cardinality not greater than the continuum.

The Feferman–Vaught theorem implies that if $\mathcal M_i\equiv \mathcal N_i$ for all i and $ {\mathrm {supp}}_=$ is first-order definable in $\prod _{\mathcal I}\mathcal M_i$ , then $ {\mathrm {supp}}_=$ is first-order definable in $\prod _{\mathcal I} \mathcal N_i$ , by the same definition.Footnote ⁵ Therefore we may apply Lemma 3.6 to the closure of $\mathfrak C$ (under elementary equivalence). By applying it to $\psi $ chosen to be $ {\mathrm {supp}}_=(x_1,x_2)\subseteq {\mathrm {supp}}_=(y_1,y_2)$ , this conclusion follows already in ZFC. Therefore $ {\mathrm {supp}}_=$ is definable in reduced products of structures in $\mathfrak C$ associated with ideals on ${\mathbb N}$ . As pointed out at the beginning of the proof, this implies the general case.

Proof. The implication (1) $\Rightarrow $ (2) is trivial. (2) $\Rightarrow $ (4) is the main implication (see Proposition 3.5); it is also the only implication that uses the assumption that ${\mathcal L}$ is countable. The equivalence between (4) and (5) is Theorem 2.18. As all structures in a full class $\mathfrak C$ have at least three elements, the equivalence of (5) and (6) is Proposition 2.14. The implication (4) $\Rightarrow $ (3) follows from Lemma 3.1. Finally, (3) $\Rightarrow $ (1) is Lemma 3.2 and Lemma 3.3.

In the case when ${\mathcal L}$ is not necessarily countable, it suffices to prove that (1) implies (4). The proof follows the proof of Proposition 3.5 closely. The only difference is that instead of $\aleph _1$ -saturated ultrapowers $\prod _{\mathcal {U}}(\mathcal {N},s_1)$ , $\prod _{\mathcal {U}}(\mathcal {M}, {\mathrm {supp}}_=)$ and $\prod _{\mathcal {U}}(\mathcal {N},s_2)$ one needs to choose saturated elementary extensions of $(\mathcal N,s_1)$ , $(\mathcal M, {\mathrm {supp}}_=)$ and $(\mathcal N,s_2)$ Footnote ⁶ of the same cardinality; the remaining part of the proof is identical.

Proof. This is immediate, as by Theorem 3.8 (5), recognizing coordinates is a property of $ {\mathrm {Th}}(\mathfrak C)$ .

Proof. $(1 \rightarrow 2)$ Assume the relation $a \sqsubseteq b$ defined by $ {\mathrm {supp}}(a)\subseteq {\mathrm {supp}}(b)$ for $a,b\in G $ is definable. This relation is a pre-order, and we denote by $\sim $ the associated equivalence relation: for $a,b\in G$ , $a\sim b$ holds if and only if a and b have the same support. The ordered quotient $(G/\sim , \sqsubseteq )$ is therefore isomorphic to $(\mathcal {P}(\mathbb {I})/\mathcal {I} ,\subseteq )$ and the natural projection is, after identification, equal to the support function $ {\mathrm {supp}}$ . $(2 \rightarrow 3)$ This implication is immediate: for all $a,b\in G$ , we have

$$\begin{align*}{\mathrm{supp}}_=(a,b) = {\mathrm{supp}}(a\cdot b^{-1})^\complement. \end{align*}$$

$(3 \rightarrow 1)$ This is immediate, since for all $a,b\in G$ , $ {\mathrm {supp}}(a)\subseteq {\mathrm {supp}}(b)$ holds if and only if $ {\mathrm {supp}}_=(b,e) \subseteq {\mathrm {supp}}_=(a,e)$ .

Proof. (1) We show that $x=e \rightarrow y=e$ is equivalent to the h-formula:

1. (*) $(\forall z) ~ \left (y\in C(z^G)\rightarrow x\in C(z^G)\right )$ .

To see this, let $\varphi (x,z)$ be the formula $(\forall t)(xtzt^{-1}=tzt^{-1}x)$ , or equivalently, $x\in C_G(z^G)$ . This is an h-formula and, since $\varphi (x,e)$ is true in every group, $(\forall z)(\varphi (y,z)\rightarrow \varphi (x,z))$ is also an h-formula. This formula is clearly equivalent to (*). The assumption that $C_G(a^G)=\{e\}$ for every $G\in \mathcal G$ and every $a\neq e$ in G implies that (*) is equivalent to $x=e\to y=e$ .

The conclusion follows by Theorem 4.1.

(2) Fix a prime p. For a group in $\mathcal {G}_p$ , we have that $x=e \rightarrow y=e$ is equivalent to

(**) $$ \begin{align} (\forall z) ~ \left( [x\in C(z^G)\wedge z^p=e] \rightarrow y\in C(z^G) \right).\end{align} $$

Indeed, notice that for all z of order p or $1$ , if $x= e$ , then $x\in C(z^G)$ . If $x\neq e$ , then $x\in C(z^G)$ only if $z=e$ . Therefore, $x = e \to y = e$ holds if and only if

$$\begin{align*}\{z \mid z^p=e , x\in C(z^G) \} \subseteq \{z \mid z^p=e , y\in C(z^G) \}, \end{align*}$$

and the equivalence with (**) is clear. We observe that (**) is an h-formula as

$$\begin{align*}(\exists z) ~ \left( x\in C(z^G)\wedge z^p=e \right)\end{align*}$$

is always satisfied. The statement holds by Theorem 4.1.

Proof. By Theorem 4.3 (1), the class of all domains recognizes coordinates. By [Reference Cassella, Paolini and Paolini8, Theorem 1.3], every acylindrically hyperbolic group without a nontrivial finite normal subgroup is a domain.

Proof. We proceed as in the proof of Theorem 4.3. We will show that, modulo the theory T, the formula $ x=e \rightarrow y=e $ is equivalent to the following (let $G_{[m]}$ denote the set of elements of G whose order divides m).

(*) $$ \begin{align} (\exists a_1,\dots,a_n \in G_{[m]})& \ [y =a_1 \cdot \dots \cdot a_n \\ \ \wedge ( & \forall a \in G_{[m]}) ~ ( x\in C(a^G)\cap G_{[m]} \rightarrow \bigwedge_{i\leq n}a_i\in C(a^G) )].\nonumber \end{align} $$

Assume that (*) holds, witnessed by $y =a_1 \cdot \dots \cdot a_n$ with $a_i\in G_{[m]}$ , $i\leq n$ . Moreover, assume $x=e$ . Then for every $a\in G_{[m]} \setminus \{e\}$ we have $x\in C(a^G)\cap G_{[m]}$ . By (*), we have for $i\leq n$ that $a_i\in C(a^G)\cap G_{[m]}$ . Since (1) implies that this set is equal to $\{e\}$ , it follows that $a_i=e$ for all i and therefore $y=a_1\cdot \dots \cdot a_n=e$ .

Conversely, assume that $x=e \rightarrow y=e $ holds. If $x=y=e$ , then (*) trivially holds, witnessed by $a_1= \dots = a_n=e$ . Assume $x\neq e$ . By (2) there are $a_1,\dots , a_n\in G_{[m]}$ such that $y=a_1\cdot {}\cdots {} \cdot a_n$ . We check that the consequent of (*) holds for every $a\in G_{[m]}$ . First, (1) implies that $x\notin C(a^G)\cap G_{[m]}$ for every $a\neq e$ . If $a=e$ , then $a_i\in C(a^G)$ for all $i\leq n$ , and therefore, (*) also holds.