Proof. We build $\mathcal {A}$ to satisfy the requirements $R_e$ for all e.

1. R^e: $\varphi _e^K$ does not assign to each tuple $\bar {a}$ an existential formula defining the orbit.

To satisfy $R_e$ , we designate two elements $a_e,b_e$ —these will represent different connected components. We make sure that if $\varphi _e^K(a_e) = d$ , where d is an existential formula true of $a_e$ , then d is also true of $b_e$ but the two are not automorphic. To start off, we make $a_e$ and $b_e$ each part of a cycle of size $p_e$ (different values for different e). If we guess that $\varphi _e^K(a_e)\downarrow = d$ , where d is an existential formula true of $a_e$ , and also of $b_e$ , then we extend so that $a_e,b_e$ are not automorphic, making $a_e$ part of a cycle of length r and making $b_e$ part of a cycle of size $s\not = r$ (choosing r and s larger than the stage and the size of any other cycles already in the graph).

If, later, we guess that $\varphi _e^K(a_e)\downarrow = d'$ , where $d'$ is an existential formula true of $a_e$ , we first extend so that the connected components of $a_e$ and $b_e$ match— $d'$ is then true of both. Then we extend further, making $a_e$ part of a cycle of length $r'$ and $b_e$ part of a cycle of length $s'\not = r'$ as before. We continue in this way. For each new guess that $\varphi _e^K(a_e)\downarrow = d_{e,n}$ , where this is an existential formula true of $a_e$ , we first extend the connected components of $a_e$ and $b_e$ to make them match, so that $d_{e,n}$ is also true of $b_e$ . We then make $a_e$ part of a new cycle of length $r_{e,n}$ and we make $b_e$ part of a cycle of length $s_{e,n}\not = r_{e,n}$ .

If $\varphi _e^K(a_e)$ converges to d, where d is an existential formula true of $a_e$ , then after finitely many steps, we will computably know the part of K and $K^c$ used in the computation. In this case d is true of both $a_e$ and $b_e$ , but the two are not in the same orbit. The connected components of $a_e$ and $b_e$ are finite and not isomorphic. If $\varphi _e^K(a_e)$ actually diverges, but infinitely often we guess that it converges to an existential formula true of $a_e$ , then $a_e$ and $b_e$ are in the same orbit, which is defined by the formula describing the original cycle of size $p_e$ and one extra cycle that each is a part of. The connected components of $a_e$ and $b_e$ are both infinite and isomorphic.

To produce a computable $\Pi _2$ Scott sentence for $\mathcal {A}$ , we first consider the set S of all existential sentences true in $\mathcal {A}$ and the set T of all universal sentences true in $\mathcal {A}$ . Since $\mathcal {A}$ is computable, the set S is c.e., and the set T is co-c.e. Let $\{\psi _n: n \in \omega \}$ be a computable list of all finitary universal sentences in the language of graphs. Then by a standard technique, employed, for instance, in [Reference Ash and Knight4], there is a sequence of infinitary computable $\Pi _1$ sentences $\tau _n$ , each built entirely from the logical constants for truth and falsity, so that $\tau _n$ is logically equivalent to truth iff $\psi _n \in T$ ; and $\tau _n$ is logically equivalent to falsity iff $\psi _n \notin T$ . Therefore, the conjunction $\left (\bigwedge _{\varphi \in S}\varphi \right ) \wedge \left (\bigwedge _{n} (\tau _n \rightarrow \psi _n)\right )$ is a computable $\Pi _2$ sentence logically equivalent to the conjunction of the finitary existential and universal sentences true in $\mathcal {A}$ .

Next, consider the computable $\Pi _2$ sentence $\gamma $ stating that every finite tuple $\bar {x}$ that forms a finite linear chain of distinct elements is part of one or two cycles; i.e., for each such $\bar {x}$ , either there is $\bar {y}$ so that $\bar {x}\bar {y}$ forms a cycle; or there are distinct $x, \bar {x}', \bar {x}," \bar {y}$ , and $\bar {z}$ so that $\bar {x} = \bar {x}'x\bar {x}"$ and $x\bar {x}'\bar {y}$ forms a cycle, and $x\bar {x}"\bar {z}$ forms a cycle. Next, consider the computable $\Pi _2$ sentence $\rho $ saying that for each x, there exists y such that for some e, y lies on a cycle of length $p_e$ and there is a finite path from x to y.

Let $\varphi = \left [\left (\bigwedge _{\varphi \in S}\varphi \right ) \wedge \left (\bigwedge _{n} (\tau _n \rightarrow \psi _n)\right )\right ] \wedge \gamma \wedge \rho $ . Clearly, $\mathcal {A} \models \varphi $ . Now consider a countable graph $\mathcal {B}$ so that $\mathcal {B} \models \varphi $ . The conjuncts $\gamma $ and $\rho $ imply that each connected component of $\mathcal {A}$ and $\mathcal {B}$ consists entirely of finite cycles, one of which is a cycle of length $p_e$ for some e. The fact that $\mathcal {A}$ and $\mathcal {B}$ satisfy the same existential and universal sentences implies a number of basic facts. First, each connected component contains exactly one element that is a part of multiple cycles. Moreover, since $\gamma $ says that each finite chain is itself part of a cycle, each connected component actually contains one element that is a “center,” i.e., part of every cycle in that component. Second, for each e, there are exactly two cycles of length $p_e$ , and these are not connected to each other; and for $e \not =e'$ , a cycle of length $p_e$ is not connected to a cycle of length $p_e'$ .

By the construction of $\mathcal {A}$ , for each e, consider the two connected components of $\mathcal {A}$ where each contains a cycle of length $p_e$ ; either they are isomorphic, or each contains a cycle of a length that the other does not. Because $\mathcal {B}$ satisfies the same existential and universal sentences that $\mathcal {A}$ does, the two connected components of $\mathcal {B}$ that contain a cycle of length $p_e$ must look exactly the same as those in $\mathcal {A}$ . Because $\rho $ implies that $\mathcal {A}$ and $\mathcal {B}$ consist entirely of these kinds of connected components, we conclude $\mathcal {A} \cong \mathcal {B}$ .

Proof. To say of a given existential formula $d(\bar {x})$ that it defines an orbit, it is enough to say that it is satisfied, and for all $\bar {y}$ that satisfy $d(\bar {y})$ , the $\forall $ -formulas true of $\bar {x}$ are true of $\bar {y}$ . This is computable $\Pi _2$ , uniformly in $d(\bar {x})$ . From this, it is clear that we have a $\Delta ^0_3$ function taking each $\bar {a}$ to the first orbit-defining $\exists $ -formula $d(\bar {x})$ that is true of $\bar {a}$ .

Notation. For a structure $\mathcal {A}$ , $NS(\mathcal {A})$ is the set of existential formulas $d(\bar {x})$ that are satisfied in $\mathcal {A}$ and do not split in $\mathcal {A}$ .

Proof. We write $NS$ for the common set $NS(\mathcal {A}) = NS(\mathcal {B})$ . We may suppose that each formula $d(\bar {x})\in NS$ implies $\pm \alpha (\bar {x})$ for $\alpha $ among the first n atomic formulas in variables in the n-tuple $\bar {x}$ —we drop from $NS$ , the formulas that do not have this feature. Let $D_{\top }$ be the set of formulas $d(x)\in NS$ in one variable. For each formula $d(\bar {u})\in NS$ , let $D_{d(\bar {u})}$ be the set of formulas $d'(\bar {u},x)\in NS$ such that $d(\bar {u})\ \&\ d'(\bar {u},x)\in NS$ . In both $\mathcal {A}$ and $\mathcal {B}$ , the formulas in $D_{\top }$ define the orbits of single elements. The formulas in $D_{d(\bar {u})}$ define the orbits of one-element extensions of tuples in the orbit that is defined by $d(\bar {u})$ .

Proceeding much as Scott did, we arrive at a common Scott sentence $\varphi $ that is the conjunction of the following:

1. 1. the sentence saying $(\forall x)\bigvee _{d\in D_{\top }}d(x)\ \&\ \bigwedge _{d\in D_{\top }}(\exists x) d(x)$ , and

2. 2. a sentence for each $d(\bar {u})\in NS$ saying that for all $\bar {u}$ satisfying $d(\bar {u})$ , $(\forall x)\bigvee _{d'\in D_{d(\bar {u})}}d'(\bar {u},x)\ \& \ \bigwedge _{d'\in D_{d(\bar {u})}} (\exists x)d'(\bar {u},x)$ .

Scott built a Scott sentence for a structure, focusing on tuples in the structure, and choosing for each tuple a particular formula that defines the orbit. We focus directly on formulas that define the orbits. We use the fact that for all tuples in the orbit defined by a formula $d(\bar {u})$ (in either structure), the formulas $d'(\bar {u},x)$ that define the orbits of $1$ -element extensions are the same.

Proof of Proposition 3.4

Recall that S is the $\forall \exists $ -theory of $\mathcal {A}$ , which is assumed to be $\Sigma _2^0$ .

Define the sentence $\Theta = [\bigwedge _{\overline {x}} \forall \overline {x} \bigvee _{i \in \omega } \varphi _{i}(\overline {x})]\ \&\ \bigwedge _{\psi \in S} \psi $ . The first conjunct is computable $\Pi _2$ , so consider the second. Since the set S is $\Sigma _2^0$ , we have the standard trick of creating a computable sequence of computable $\Sigma _2^0$ sentences $\tau _n$ built out of $\top $ and $\bot $ to define membership in S of the nth finitary $\forall \exists $ -sentence $\psi _n$ . Therefore, the second conjunct is logically equivalent to the computable $\Pi _2$ sentence $\bigwedge _{n \in \omega } (\tau _n \rightarrow \psi _n)$ . Thus, $\Theta $ is computable $\Pi _2$ .

Proof of Claim

Let $\mathcal {B}$ be a countable model of $\Theta $ . We show that $\mathcal {A}\cong \mathcal {B}$ using the lemma above.

1. 1. $\mathcal {A}$ has a $\Pi _2$ Scott sentence.

   Each tuple $\bar {a}$ in $\mathcal {A}$ satisfies some $\varphi _i(\bar {x})$ . Then the orbit is defined by one of finitely many $\exists $ -formulas $\gamma _j(\bar {x})$ .

2. 2. $NS(\mathcal {A}) \subseteq NS(\mathcal {B})$ .

   Let $d(\bar {x})$ be a formula in $NS(\mathcal {A})$ . Since $\mathcal {B}$ satisfies the $\forall \exists $ -theory of $\mathcal {A}$ , $d(\bar {x})$ is satisfied in $\mathcal {B}$ . If $d(\bar {x})$ split in $\mathcal {B}$ , we would have an $\exists $ -formula $d'(\bar {x})$ and tuples $\bar {b}_1$ , $\bar {b}_2$ , such that $\bar {b}_1$ satisfies $d(\bar {x})\ \&\ d'(\bar {x})$ and $\bar {b}_2$ satisfies $d(\bar {x})\ \&\ \neg {d'(\bar {x})}$ . However, the $\forall \exists $ sentence saying

   $$\begin{align*}(\forall\bar{x},\bar{x}')[(d(\bar{x})\ \&\ d(\bar{x}')\ \&\ d'(\bar{x}))\rightarrow d'(\bar{x}')]\end{align*}$$
   is true in $\mathcal {A}$ , so it is also true in $\mathcal {B}$ .

3. 3. $\mathcal {B}$ has a $\Pi _2$ Scott sentence.

   Each tuple $\bar {b}$ in $\mathcal {B}$ satisfies some $\varphi _i(\bar {x})$ , so it satisfies one of the $\exists $ -formulas $\gamma _j(\bar {x})$ . Since $\gamma _j(\bar {x})$ defines the orbit of some tuple in $\mathcal {A}$ , it is in $NS(\mathcal {A})$ , so it is also in $NS(\mathcal {B})$ . Then by Corollary 1.3, $\mathcal {B}$ has a $\Pi _2$ Scott sentence.

4. 4. $NS(\mathcal {B})\subseteq NS(\mathcal {A})$ .

   Let $d(\bar {x})\in NS(\mathcal {B})$ . Then $d(\bar {x})$ is satisfied by some $\bar {a}$ in $\mathcal {A}$ . Say that $\gamma (\bar {x})$ is an $\exists $ -formula that defines the orbit of $\bar {a}$ . Suppose that $d(\bar {x})$ splits in $\mathcal {A}$ . We have some $\bar {a}'$ , not automorphic to $\bar {a}$ , but also satisfying $d(\bar {x})$ . Say that the orbit of $\bar {a}'$ is defined by the $\exists $ -formula $\gamma '(\bar {x})$ . Then the $\exists $ -sentence $(\exists \bar {x})[d(\bar {x})\ \&\ \gamma '(\bar {x})]$ is true in $\mathcal {A}$ but false in $\mathcal {B}$ , a contradiction.

We have shown that the hypotheses of Lemma 3.5 are satisfied. Therefore, $\mathcal {A}\cong \mathcal {B}$ . This completes the proof that $\Theta $ is a computable $\Pi _2$ Scott sentence for $\mathcal {A}$ .

Proof. The structure $\mathcal {A}$ is, essentially, a countable family of unary structures, based on a uniformly computable family of trees $T_e$ with no dead ends. For each e, $T_e$ will be either the set of finite sequences of $0$ ’s, or the set of sequences of form $\sigma _e\tau $ , where $\sigma _e$ is a finite sequence of $0$ ’s of a special length and $\tau $ has all of its $1$ ’s before any of its $0$ ’s. In the first case, $T_e$ has just one path. In the second, $T_e$ has infinitely many paths. All but one are isolated. The language of $\mathcal {A}$ consists of unary relations $U_e$ for $e\in \omega $ and $V_{e.n}$ for $e,n\in \omega $ . In $\mathcal {A}$ , the relations $U_e$ partition the universe into infinite sets, and the relations $V_{e,n}$ are subsets of $U_e$ . The elements of $U_e$ represent paths through $T_e$ via their membership (or not) in the $V_{e,n}$ . More specifically, each node $\sigma \in T_e$ corresponds to a quantifier-free formula $d_\sigma $ saying that $x\in U_e$ and $x\in V_{e,n}$ if $\sigma (n) = 1$ and $x\notin V_{e,n}$ if $\sigma (n) = 0$ . In $\mathcal {A}$ , each element a is in exactly one $U_e$ , where it represents some isolated path in $T_e$ . Some $\sigma $ isolates the path and then $d_\sigma $ defines the orbit. Moreover, each isolated path of $T_e$ is represented in $\mathcal {A}$ by infinitely many elements.

To construct $T_e$ , we start putting in at step s just $0^s$ . If for some s, when we have put into $T_e 0^n$ for all $n < s$ , we see that $W_{e,s}$ has a quantifier-free formula $d(x)$ satisfied by an element of $U_e$ that is not in any $V_{e,n}$ , then $T_e$ is the result of attaching to the sequence $0^s$ the sequences $\sigma $ of form $1^j0^k$ . We assume that at stage s, $W_{e,s}$ can only contain formulas with symbols $V_{e,n}$ for $n < s$ . If there is no such d in $W_e$ , then $T_e$ has just the path $0^*$ . Thus, either $T_e$ has just the path $0^*$ , and the elements representing that path do not satisfy any formula $d\in W_e$ , or else $T_e$ has a formula d satisfied by elements representing all but the non-isolated path in $T_e$ , infinitely many.

We now describe the explicit construction of $\mathcal {A}$ from the uniformly computable family of trees $T_e$ . The universe of $\mathcal {A}$ is divided into infinitely many infinite, computable sets $A_e$ . For each stage s that $T_e$ just looks like the set of nodes of all $0$ ’s, we assign the sth element a of $A_e$ to this path, so $U_e(a)$ and $\neg V_{e, n}(a)$ for all n. If we see at some stage s that $T_e$ is not just the set of nodes of all $0$ ’s, then we know exactly what the paths in $T_e$ look like: (1) the path of all $0$ ’s; (2) paths that begin with $0^s$ and continue with a $1^j$ for some j and then have all $0$ ’s; and (3) the sole non-isolated path of $0^s$ followed by all $1$ ’s. We divide the remaining elements of $A_e$ into infinitely many infinite, computable sets and assign infinitely many elements to each of the isolated paths in $T_e$ , i.e., those of the form (1) or (2).

For a Scott sentence $\varphi $ , we take the conjunction of the following:

1. 1. the existential and universal sentences true in $\mathcal {A}$ —the conjunction of the existential sentences is computable $\Pi _2$ , and the conjunction of the universal sentences is also equivalent to a computable $\Pi _2$ sentence;

2. 2. the computable $\Pi _2$ sentence $(\forall x)\bigvee _e U_ex$ ;

3. 3. the computable $\Pi _2$ sentences $(\exists x)V_{e,n}x\rightarrow (\forall x)\bigvee _{n'> n}\neg {V_{e,n'}x}$ , for all e and n.

Let $\mathcal {B}$ be some model of $\varphi $ . By (2), in $\mathcal {B}$ , each element is in some $U_e$ . Also, because the existential theories match, if $\sigma $ is an isolating node, then there are infinitely many elements satisfying $d_\sigma $ . There are universal sentences $(\forall x)(U_e(x)\rightarrow \bigvee _{\sigma \in T_{e,n}}d_\sigma (x))$ , saying that each element of $U_e$ in $\mathcal {B}$ satisfies $d_\sigma $ for some $\sigma $ at level n in $T_e$ . These sentences guarantee that each element of $U_e$ represents some path in $T_e$ . We must show that no element of $\mathcal {B}$ represents a non-isolated path. We suppose that $W_{e,s}$ has a quantifier-free formula $d_e$ true of elements representing the all- $0$ path in $T_e$ . We may suppose that $d_e$ mentions only $V_{e,n}$ for $n < s$ . The non-isolated path in $T_e$ has $s 0$ ’s, followed by all $1$ ’s. By (3), for $e,s$ , this path is not represented in $\mathcal {B}$ . For an isomorphism between $\mathcal {A}$ and $\mathcal {B}$ , we just map elements of $\mathcal {A}$ in $U_e$ and representing a given path in $T_e$ to elements of $\mathcal {B}$ in $U_e$ and representing the same path.

We need to show that if every element of $\mathcal {A}$ satisfies some existential formula in $W_e$ , then some existential formula $d\in W_e$ is satisfied by elements in infinitely many orbits. We will show something stronger. The theory of $\mathcal {A}$ has effective elimination of quantifiers, so we may replace each existential formula by a quantifier-free formula that is equivalent over the theory. We may suppose that if d is enumerated into $W_e$ at stage s, then d mentions only $V_{e,n}$ for $n < s$ , not for $n\geq s$ . Let $c_e$ be an element of $U_e$ representing the path $p^*$ in $T_e$ . If $W_e$ has an existential/quantifier-free formula d satisfied by $c_e$ , then $T_e$ has infinitely many paths, and d is satisfied by all elements of $U_e$ —infinitely many representing each of the isolated paths.

Proof of Theorem 4.1

We split the proof into some lemmas. The first lemma gives the easier implication.

Proof. Since $\mathcal {A}$ is unary, $Th(\mathcal {A})$ is generated by its $\forall \cup \exists $ -sentences. Let $\beta $ be a computable $\Pi _2$ sentence equivalent to the conjunction of these. (We use that $Th(\mathcal {A})$ is computable here.) Then $\varphi = \beta \ \&\ (\forall x)\psi (x)$ is also computable $\Pi _2$ . We claim that $\varphi $ is a Scott sentence for $\mathcal {A}$ .

Clearly, $\mathcal {A}\models \beta $ . Since $\mathcal {A}$ has a $\Pi _2^0$ Scott sentence, $\mathcal {A}$ is atomic. So, $\mathcal {A}\models (\forall x)\psi (x)$ and thus $\mathcal {A}\models \varphi $ . Let $\mathcal {B}$ be another model of $\varphi $ . Satisfying $\beta $ implies that a model is a unary structure with $Th(\mathcal {A})$ ; satisfying $(\forall x)\psi (x)$ implies that the model is atomic. (More specifically, the formulas $\sigma (x)$ , for $\sigma \in 2^{<\omega }$ , that are satisfied and do not split are the same for all models of $\beta $ . The quantifier-free type generated over $Th(\mathcal {A})$ by such a $\sigma (x)$ is determined by $\beta $ and $\sigma $ , and $\beta $ guarantees that $\sigma (x)$ is satisfied by infinitely many elements.) Since $\mathcal {A}$ and $\mathcal {B}$ are countable atomic models of the same theory, $\mathcal {A}\cong \mathcal {B}$ .

Proof. We may suppose that $\varphi = \bigwedge _i (\forall \bar {u}_i)\psi _i(\bar {u}_i)$ , where for each i, $\psi _i(\bar {u}_i)$ is computable $\Sigma _1$ . In $Th_T$ , we have quantifier elimination. We may suppose that $\psi _i(\bar {u}_i)$ has the form $\bigvee _j\alpha _{i,j}(\bar {u}_i)$ , where $\alpha _{i,j}(\bar {u}_i)$ is quantifier-free. In addition, we may suppose that $\alpha _{i,j}(\bar {u}_i)$ gives the complete quantifier-free type of $\bar {u}_i$ in some finite language $L_s = \{U_n:n < s\}$ , saying for each pair of variables $u,v\in \bar {u}_i$ whether $u = v$ , and saying for each single variable v, that $\sigma (v)$ holds for some $\sigma \in 2^s$ .

For the models $\mathcal {B}_p = \mathcal {A}\cup \{b\}$ , we may fix $\mathcal {A}$ , say with universe the odd numbers, and we may let $b = 0$ . For each i, we consider the possible assignments t mapping $\bar {u}_i$ to $\mathcal {A}\cup \{b\}$ . For an assignment t taking $\bar {u}_i$ to $\mathcal {A}\cup \{b\}$ , let $\bar {x}_{i,t} = t^{-1}(b)$ and let $\bar {y}_{i,t} = t^{-1}(\mathcal {A})$ . For the least s such that $\alpha _{i,j}(\bar {u}_i)$ is in the language $L_s$ , we may write $\alpha _{i,j}(\bar {u}_i)$ as $\rho _{i,j}(\bar {u}_i)\ \&\ \beta _{i,t,j}(\bar {x}_{i,t})\ \&\ \gamma _{i,t,j}(\bar {y}_{i,t})$ , where $\rho _{i,j}(\bar {u}_i)$ gives the equality relation on $\bar {u}_i$ , $\beta _{i,t,j}(\bar {x}_{i,t})$ assigns some $\sigma _x\in 2^s$ to each $x\in \bar {x}_{i,t}$ , and $\gamma (\bar {y}_{i,t})$ assigns some $\sigma _y\in 2^s$ to each $y\in \bar {y}_{i,t}$ . Say that $\sigma _u$ is the element of T assigned to the variable u in $\alpha _{i,j}(\bar {u}_i)$ .

We write $\alpha _{i,j}(t(\bar {u}_i))$ for the result of plugging in $t(u)$ for the variable u in the formula $\alpha _{i,j}(\bar {u}_i)$ . The formula is satisfied in $\mathcal {B}_p$ by the assignment just in case all three parts are satisfied. We have $\mathcal {B}_p\models \rho _{i,j}(t(\bar {u}_i))$ when $t(u) = t(v)$ iff $u = v$ is a conjunct of $\rho _{i,j}$ . We can check this effectively, just looking at $\rho _{i,j}(\bar {u}_i)$ and t. We have $\mathcal {B}_p\models \gamma _{i,t,j}(t(\bar {y}_{i,t}))$ iff $\mathcal {A}\models \bigwedge _{y\in \bar {y}_{i,t}}\sigma _y(t(y))$ . Finally, $\mathcal {B}_p\models \beta (t(\bar {x}_{i, t}))$ iff $\sigma _x\subseteq p$ for all $x\in \bar {x}_{i,t}$ .

For each i and each assignment t mapping $\bar {u}_i$ to the universe of $\mathcal {B}_p$ , let $S(i,t) = \{j: \mathcal {B}_p\models \rho _{i,j}(t(\bar {u}_i))$ and $\mathcal {B}_p\models \gamma _{i,t,j}(t(\bar {y}_{i,t}))$ }. Our $\psi (x)$ should say that for all i and all $t:\bar {u}_i\rightarrow \mathcal {A}\cup \{b\}$ , there is some $j\in S(i,t)$ such that t makes $\beta _{i,t,j}(\bar {x}_{i,t})$ true.

In case $\mathcal {A}$ is computable, then for each $i,t$ , the set $S(i,t)$ is c.e., depending only on $\mathcal {A}$ and the c.e. set of disjuncts from the original $\varphi $ . We can then let $\psi (x) = \bigwedge _{i,t}\bigvee _{j\in S(i,t)}\beta ^*_{i,t,j}(x)$ , where $\beta ^*_{i,t,j}(x)$ is the result of replacing each variable in $\bar {x}_{i,t}$ by x. This is computable $\Pi _2$ , with the desired meaning.

In case $\mathcal {A}$ is not computable, the construction of $\psi (x)$ is a little more complicated. We first recall the set $NS(\mathcal {A})$ , defined in Section 3, which is actually the same for all models of $Th_T$ , and is easily seen to be a $\Pi _1^0$ set, since T is computable. Next, for each i and each assignment t, and each quantifier-free formula $d(\bar {y}_{i, t})$ , let $S(i, t, d) = \{j: \mathcal {B}_p\models \rho _{i,j}(t(\bar {u}_i))$ and $Th_T\vdash d(\bar {y}_{i, t})\rightarrow \gamma _{i,t,j}(\bar {y}_{i,t})\}$ . Note that $S(i, t, d)$ is computable; it doesn’t, in fact, even depend on where the assignment t sends $\bar {y}_{i, t}$ in $\mathcal {A}$ beyond whether the elements are equal or not.

Now we are almost ready to construct our modified $\psi (x)$ . First, for each i consider the set $C_i$ consisting of the conjuncts $\lambda _t$ , one for each assignment t, from the $\psi (x)$ we defined in case $\mathcal {A}$ was computable; i.e., $\lambda _t = \bigvee _{j \in S(i, t)} \beta ^*_{i, t, j}(x)$ . For each quantifier-free formula $d(\bar {y}_{i, t})$ , let $\lambda _{t, d} = \bigvee _{j \in S(i, t, d)} \beta ^*_{i, t, j}(x)$ . Now let $C_i^* = \{\lambda _{t, d}: d(\bar {y}_{i, t}) \in NS(\mathcal {A})\}$ .

Proof of Claim

Let $\lambda _t \in C_i$ , and let $\bar {a} = t(\bar {y}_{i, t})$ . Since $\mathcal {A}$ has a $\Pi _2$ Scott sentence, there is $d(\bar {y}_{i, t}) \in NS(\mathcal {A})$ with $\mathcal {A} \models d(\bar {a})$ . Thus, for all j, we have $j \in S(i, t)$ iff $j \in S(i, t, d)$ . Therefore, $\lambda _t = \lambda _{t, d} \in C^*_i$ . Conversely, let $\lambda _{t, d} \in C^*_i$ . Let $\bar {a} \in \mathcal {A}$ with $\mathcal {A} \models d(\bar {a})$ . Let $t'$ be the assignment with $t'(\bar {x}_{i, t}) = t(\bar {x}_{i, t})$ , and $t'(\bar {y}_{i, t}) = \bar {a}$ . Thus, for all j, we have $j \in S(i, t')$ iff $j \in S(i, t, d)$ . Therefore, $\lambda _{t, d} = \lambda _{t'} \in C_i$ .

Proof. Let $\mathcal {B}$ be a model of $Th_T$ , and let b be an element. Since $\mathcal {A}$ is the prime model, we may suppose that $\mathcal {A}\subseteq \mathcal {B}$ . Moreover, since $\mathcal {A}$ has infinitely many elements representing any given isolated path, we may suppose that $b\notin \mathcal {A}$ . Since the formula $\psi (x)$ is quantifier-free, with just the variable x, $\mathcal {B}\models \psi (b)$ iff $\mathcal {A}\cup \{b\}\models \psi (b)$ .