THE STRENGTH OF AN AXIOM OF FINITE CHOICE FOR BRANCHES IN TREES

Abstract

In their logical analysis of theorems about disjoint rays in graphs, Barnes, Shore, and the author (hereafter BGS) introduced a weak choice scheme in second-order arithmetic, called the $\Sigma ^1_1$ axiom of finite choice (hereafter finite choice). This is a special case of the $\Sigma ^1_1$ axiom of choice ($\Sigma ^1_1\text {-}\mathsf {AC}_0$) introduced by Kreisel. BGS showed that $\Sigma ^1_1\text {-}\mathsf {AC}_0$ suffices for proving many of the aforementioned theorems in graph theory. While it is not known if these implications reverse, BGS also showed that those theorems imply finite choice (in some cases, with additional induction assumptions). This motivated us to study the proof-theoretic strength of finite choice. Using a variant of Steel forcing with tagged trees, we show that finite choice is not provable from the $\Delta ^1_1$-comprehension scheme (even over $\omega $-models). We also show that finite choice is a consequence of the arithmetic Bolzano–Weierstrass theorem (introduced by Friedman and studied by Conidis), assuming $\Sigma ^1_1$-induction. Our results were used by BGS to show that several theorems in graph theory cannot be proved using $\Delta ^1_1$-comprehension. Our results also strengthen results of Conidis.

1 Introduction

Reverse mathematics, as initiated by Friedman [5] in the 1970s, is a program in the foundations of mathematics which aims to find the axioms or formal systems needed in order to carry out proofs of particular mathematical theorems. Early results by Friedman, Simpson, and others showed that several basic theorems of analysis, algebra, and combinatorics, when formalized in second-order arithmetic, are equivalent to one of five sets of axioms, now known as the “Big Five.” (The standard reference for subsystems of second-order arithmetic is [15].) These equivalences were proved over the base theory $\mathsf {RCA}_0$ , which formalizes computable mathematics and is the weakest of the Big Five. Later work revealed several exceptions to the Big Five phenomenon, including theorems that lie strictly between consecutive levels of the Big Five, and theorems that are incomparable to some level of the Big Five (for instance, Ramsey’s theorem for pairs). In this paper, we study a weak form of the axiom of choice which lies strictly between consecutive levels of the Big Five, namely $\mathsf {ACA}_0$ and $\mathsf {ATR}_0$ .

The study of choice schemes in second-order arithmetic predates reverse mathematics. In 1962, Kreisel [8] introduced the $\Sigma ^1_1$ axiom of choice ( $\Sigma ^1_1$ - $\mathsf {AC}$ ), which asserts that

$$\begin{align*}\forall n \exists X \varphi(n,X) \rightarrow \exists (X_n)_n \forall n \varphi(n,X_n) \end{align*}$$

for each formula $\varphi (n,X)$ in the language of second-order arithmetic which is arithmetical (i.e., has no set quantifiers). Equivalently, ¹ it asserts that for any sequence of trees $(T_n)_n$ , each of which has a branch (i.e., a ray in the sense of graph theory), one can choose a branch from each $T_n$ .

Kreisel’s interest in $\Sigma ^1_1$ - $\mathsf {AC}$ stemmed at least partially from its intimate connection with the hyperarithmetic sets. (For background on hyperarithmetic theory, see [13].) This connection arises from studying the $\omega $ -models (defined below) of $\mathsf {RCA}_0 + \Sigma ^1_1$ - $\mathsf {AC}$ , or $\Sigma ^1_1\text {-}\mathsf {AC}_0$ for short.

A structure in the language of second-order arithmetic has the form

$$\begin{align*}(N,S,+,\cdot,<,0,1,\in), \end{align*}$$

where $(N,+,\cdot ,<,0,1)$ is a structure in the language of first-order arithmetic (called the first-order part), S is a collection of subsets of N (called the second-order part), and $\in $ is the membership relation between elements of N and elements of S. Such a structure is said to be an $\omega $ -model if its first-order part is the standard natural numbers $\mathbb {N}$ and the symbols $+$ , $\cdot $ , etc. have their standard interpretations. We identify an $\omega $ -model with its second-order part.

Kreisel [8] showed that the $\omega $ -model $\mathrm {HYP}$ consisting of all hyperarithmetical subsets of $\mathbb {N}$ satisfies $\Sigma ^1_1\text {-}\mathsf {AC}_0$ , and furthermore, $\mathrm {HYP}$ is the minimum $\omega $ -model of $\Sigma ^1_1\text {-}\mathsf {AC}_0$ . This is analogous to the well-known facts that $\mathsf {ACA}_0$ has a minimum $\omega $ -model ARITH consisting of all arithmetical subsets of $\mathbb {N}$ , and $\mathsf {RCA}_0$ has a minimum $\omega $ -model REC consisting of all recursive (a.k.a. computable) subsets of $\mathbb {N}$ . In fact, Friedman [5] observed that a subset of $\mathcal {P}(\mathbb {N})$ is a model of $\mathsf {RCA}_0$ ( $\mathsf {ACA}_0$ , respectively) if and only if it is closed under (effective) join $\oplus $ and Turing reduction $\leq _T$ (and the Turing jump operator, respectively). In summary:

One might hope to fill in the final entry of the table with “ $\oplus $ , hyperarithmetical reduction.” Every $\omega $ -model of $\Sigma ^1_1\text {-}\mathsf {AC}_0$ is closed under join and hyperarithmetical reduction [8]. However, not every subset of $\mathcal {P}(\mathbb {N})$ which is closed under join and hyperarithmetical reduction satisfies $\Sigma ^1_1\text {-}\mathsf {AC}_0$ . The issue is not that $\Sigma ^1_1\text {-}\mathsf {AC}_0$ is too strong; rather, there is no theory $\mathsf {T}$ such that the $\omega $ -models of $\mathsf {T}$ are exactly those which are closed under join and hyperarithmetical reduction [17, 2.2.2]. This motivated the following definition:

Definition 1.1 (Steel [16], Montalbán [9]).

A theory $\mathsf {T}$ is a theory of hyperarithmetic analysis (THA) if:

1. – every $\omega $ -model of $\mathsf {T}$ is closed under join and hyperarithmetic reduction, and

2. – $\mathsf {T}$ holds in $\mathrm {HYP}(Y)$ for every $Y \subseteq \mathbb {N}$ (here $\mathrm {HYP}(Y)$ consists of all subsets of $\mathbb {N}$ which are hyperarithmetically reducible to Y).

Before we discuss examples of THAs, note that each THA $\mathsf {T}$ characterizes the notion of hyperarithmetic reduction in the following sense: A set $X \subseteq \mathbb {N}$ is hyperarithmetically reducible to a set $Y \subseteq \mathbb {N}$ if and only if every $\omega $ -model of $\mathsf {T}$ which contains Y also contains X.

The aforementioned results of Kreisel [8] relativize to show that $\Sigma ^1_1\text {-}\mathsf {AC}_0$ is a THA. In the 60s and 70s a number of THAs were identified and proved to be nonequivalent [4, 16, 17]. These THAs are natural fragments of well-studied axiom schemes such as choice schemes or comprehension schemes. In 2006, Montalbań [9] was the first to identify a THA which is a published theorem not stated using concepts from logic (such as first-order formulas). This THA, known as INDEC, is a result of Jullien (see [9]) about indecomposability of linear orderings. Other works on THAs include [2, 10–12].

Recently Barnes, Goh, and Shore [1] (hereafter BGS) identified several theorems of graph theory which are THAs. These theorems are variations of classical results of Halin [6] on disjoint rays in graphs. For simplicity we only discuss the oldest of these results (see [3, Theorem 8.2.5(i)] for its statement), which we call Halin’s theorem. (Much of the discussion below applies to variations of Halin’s theorem as well; see [1].) In their proof that Halin’s theorem is a THA, BGS [1] showed that Halin’s theorem is provable using $\Sigma ^1_1\text {-}\mathsf {AC}_0$ , and conversely, Halin’s theorem implies the following weakening of $\Sigma ^1_1\text {-}\mathsf {AC}_0$ (assuming additional induction axioms on top of what is available in $\mathsf {RCA}_0$ ):

Definition 1.2 (BGS [1]).

Finite- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ consists of $\mathsf {RCA}_0$ and

$$\begin{align*}(\forall n) (\exists \text{ nonzero finitely many }X) \varphi(n,X) \rightarrow (\exists (X_n)_n) (\forall n) \varphi(n,X_n) \end{align*}$$

for each arithmetical formula $\varphi (n,X)$ . Formally, “ $(\exists $ nonzero finitely many $X)\varphi (n,X)$ ” means that there is a nonempty sequence $(X_i)_{i<j}$ such that for each X, $\varphi (n,X)$ holds if and only if $X = X_i$ for some $i < j$ .

Henceforth we will refer to finite- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ as finite choice.

In this paper we present implications and nonimplications between finite choice and other known THAs (Theorems 1.3 and 1.5). First we ought to discuss why finite choice is a THA. To show that a theory is a THA, it suffices to show that it is sandwiched between two THAs. Of course $\Sigma ^1_1\text {-}\mathsf {AC}_0$ implies finite choice. On the other hand, finite choice implies another known THA: unique- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ , which asserts (in addition to $\mathsf {RCA}_0$ ) that

$$\begin{align*}\forall n \exists ! X \varphi(n,X) \rightarrow \exists (X_n)_n \forall n \varphi(n,X_n) \end{align*}$$

for each arithmetical formula $\varphi (n,X)$ . Unique- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ has appeared in the literature with names such as arithmetical replacement $\Pi ^0_{(\omega )}$ - $\mathsf {RA}$ [17, p. 6], $\Pi ^1_0$ -replacement [16, p. 74], and most commonly weak- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ (e.g., [15, VIII.4.12]). A proof that unique- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ is a THA can be found in [15, VIII.4.15–16].

Since finite choice is sandwiched between $\Sigma ^1_1\text {-}\mathsf {AC}_0$ and unique- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ , we can calibrate its proof-theoretic complexity by comparing it with other THAs in this position. One such THA is the $\Delta ^1_1$ -comprehension scheme ( $\Delta ^1_1$ - $\mathsf {CA}_0$ ), which asserts (in addition to $\mathsf {RCA}_0$ ) that

$$\begin{align*}\forall n(\varphi(n) \leftrightarrow \neg\psi(n)) \rightarrow \exists X \forall n (n \in X \leftrightarrow \varphi(n)) \end{align*}$$

for any $\varphi (n)$ and $\psi (n)$ which are $\Sigma ^1_1$ . It is easy to see that $\Sigma ^1_1\text {-}\mathsf {AC}_0 \to \Delta ^1_1$ - $\mathsf {CA}_0 \to $ unique- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ . In order to separate $\Delta ^1_1$ - $\mathsf {CA}_0$ and $\Sigma ^1_1\text {-}\mathsf {AC}_0$ , Steel [16] introduced a new forcing notion and used it to construct an $\omega $ -model which satisfies $\Delta ^1_1$ - $\mathsf {CA}_0$ but not $\Sigma ^1_1\text {-}\mathsf {AC}_0$ . Van Wesep [17] modified Steel’s methods to construct an $\omega $ -model which satisfies unique- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ but not $\Delta ^1_1$ - $\mathsf {CA}_0$ . Subsequently various modifications of Steel forcing have been used to separate THAs [2, 9–12]. Using yet another variant of Steel forcing (Section 3.3), we shall prove the following:

Theorem 1.3. There is an $\omega $ -model which satisfies $\Delta ^1_1$ - $\mathsf {CA}_0$ but not finite choice. Therefore $\Delta ^1_1$ - $\mathsf {CA}_0$ does not imply finite choice, even if additional induction axioms are assumed.

In particular, finite choice is strictly stronger than unique- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ . Theorem 1.3 was used by BGS [1] to show that Halin’s theorem and many of its variants cannot be proved using $\Delta ^1_1$ - $\mathsf {CA}_0$ (even if additional induction axioms are assumed).

To show that finite choice is strictly weaker than $\Sigma ^1_1\text {-}\mathsf {AC}_0$ , we will establish a connection between finite choice and a known THA: the arithmetic Bolzano–Weierstrass theorem $\mathsf {ABW}$ introduced by Friedman [5]. Our statement of $\mathsf {ABW}$ below follows Conidis [2, p. 4470].

Definition 1.4. The arithmetic Bolzano–Weierstrass theorem ( $\mathsf {ABW}$ ) states that if $A(X)$ is an arithmetic predicate on $2^N$ , then either $A(X)$ has finitely many solutions, or the set $\{X \in 2^N: A(X)\}$ of A-solutions has an accumulation point.

Friedman asserted that $\mathsf {ABW}$ follows from $\Sigma ^1_1$ - $\mathsf {AC}_0$ . Conidis [2, Theorem 2.1(2)] furnished a proof of that statement. In addition, [2, Theorem 2.1(4)] showed that $\mathsf {ABW}$ implies unique- $\Sigma ^1_1$ - $\mathsf {AC}_0$ over the base theory consisting of $\mathsf {RCA}_0$ and the induction scheme for $\Sigma ^1_1$ formulas (denoted $\mathsf {I}\Sigma ^1_1$ ). (To the best of our knowledge it is not known if $\mathsf {ABW}$ implies unique- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ over just $\mathsf {RCA}_0$ . Nevertheless, as every $\omega $ -model satisfies $\mathsf {I}\Sigma ^1_1$ , Conidis’s result shows that every $\omega $ -model of $\mathsf {RCA}_0+\mathsf {ABW}$ satisfies unique- $\Sigma ^1_1\text {-}\mathsf {AC}_0$ .) We will strengthen Conidis’s result by proving the following (Section 2):

Theorem 1.5. $\mathsf {ABW}$ implies finite choice over $\mathsf {RCA}_0 + \mathsf {I}\Sigma ^1_1$ .

It follows that finite choice is strictly weaker than $\Sigma ^1_1\text {-}\mathsf {AC}_0$ , as $\mathsf {RCA}_0 + \mathsf {I}\Sigma ^1_1 + \mathsf {ABW}$ does not imply $\Sigma ^1_1\text {-}\mathsf {AC}_0$ or even $\Delta ^1_1$ - $\mathsf {CA}_0$ (as witnessed by Van Wesep’s model, see [17, Lemma 1.4] and [2, Theorem 4.7]).

Having established a connection between $\mathsf {ABW}$ and finite choice (Theorem 1.5), our Theorem 1.3 gives an alternate, arguably simpler, proof of Conidis’s [2, Theorem 3.1] result that $\Delta ^1_1$ - $\mathsf {CA}_0$ does not imply $\mathsf {ABW}$ . (See Remark 3.1.)

2 Arithmetic Bolzano–Weierstrass

We adapt a proof of Conidis [2, Theorem 2.1(4)] to prove Theorem 1.5, which asserts that $\mathsf {RCA}_0 + \mathsf {I}\Sigma ^1_1 + \mathsf {ABW} \vdash \mathrm {finite}$ choice.

Proof of Theorem 1.5

Suppose that $\varphi (n,Y)$ is an arithmetical predicate which is an instance of finite choice, i.e., for each n, $\varphi (n,Y)$ has finitely many solutions. Without loss of generality, we may assume that $\varphi (n,\emptyset )$ always fails.

Define an arithmetical predicate A on $2^N$ as follows: $A(\langle X_n \rangle _n)$ holds if

$$\begin{align*}\exists n_0((\forall n \leq n_0)[\varphi(n,X_n)] \land (\forall n>n_0)[X_n = \emptyset]). \end{align*}$$

Using $\mathsf {I}\Sigma ^1_1$ and the assumption that for each n, $\varphi (n,Y)$ has a solution, we can show that for each $n_0$ , $A(\langle X_n \rangle _n)$ has at least $n_0$ distinct solutions. (Note if $X_n$ is $\emptyset $ , then it does not code a solution to $\varphi (n,\cdot )$ .) Hence $A(\langle X_n \rangle _n)$ is an instance of $\mathsf {ABW}$ .

By $\mathsf {ABW}$ , the set $\{\langle X_n \rangle _n: A(\langle X_n \rangle _n)\}$ has an accumulation point Y. Interpret Y as a sequence $\langle Y_n \rangle _n$ . We claim that for all n, $\varphi (n,Y_n)$ holds.

Suppose towards a contradiction that $\varphi (k,Y_k)$ fails. Since $\varphi (k,\cdot )$ has only finitely many solutions, there is some m sufficiently large such that $Y_k\restriction m \neq Y\restriction m$ for every Y such that $\varphi (k,Y)$ holds.

Now, by our choice of $\langle Y_n \rangle _n$ , there are infinitely many $\langle X_n \rangle _n$ satisfying A such that $X_k$ extends $Y_k\restriction m$ . For any such $\langle X_n \rangle _n$ , $\varphi (k,X_k)$ fails, so by definition of A, $X_n = \emptyset $ for all $n \geq k$ .

But for each $n < k$ , $\varphi (n,\cdot )$ has only finitely many solutions, so there cannot be infinitely many $\langle X_n \rangle _n$ satisfying the above conditions. Contradiction. We have showed that $\langle Y_n \rangle _n$ is a finite choice solution to $\varphi (n,Y)$ .

We do not know if finite choice implies $\mathsf {ABW}$ .

3 $\Delta ^1_1$ -comprehension does not imply finite choice

The rest of the paper is devoted to proving Theorem 1.3 using a new variant of Steel’s [16] tagged tree forcing. We assume familiarity with forcing in arithmetic (see, e.g., [14, Section 3]). A helpful reference for forcing in arithmetic over ramified languages is [13, III.4 and IV.3].

We begin by fixing some notation regarding trees. By a tree we mean a subset of $\mathbb {N}^{<\mathbb {N}}$ which is closed under prefix. An element f of $\mathbb {N}^{\mathbb {N}}$ is a branch through a tree T, written $f \in [T]$ , if every prefix of f lies in T. If some $\sigma $ in T is a prefix of a branch through T, we say that $\sigma $ is extendible in T. The empty string is denoted by $\emptyset $ . The length of a string $\sigma $ is denoted by $|\sigma |$ . If $\sigma $ is a prefix of $\tau $ (or a branch f), we write $\sigma \subseteq \tau $ ( $\sigma \subseteq f$ , respectively). If $\sigma $ is a nonempty string, $\sigma ^-$ denotes the prefix of $\sigma $ of length $|\sigma |-1$ . If $\sigma $ is a string and T is a tree, $\sigma \cap T$ denotes the longest prefix of $\sigma $ which lies in T.

3.1 The model

We shall construct an $\omega $ -model $M_\infty \subset \mathcal {P}(\mathbb {N})$ which satisfies $\Delta ^1_1$ - $\mathsf {CA}_0$ but not finite choice. The basic setup follows [10].

To define $M_\infty $ we will construct a generic object

$$\begin{align*}\langle T^G, \{f_i^G: i \in \mathbb{N}\}, h^G\rangle, \end{align*}$$

where $T^G$ is a tree, each $f_i^G$ is a distinct branch through $T^G$ , and $h^G: T^G \to \omega _1^{CK} \cup \{\infty \}$ is the well-founded rank function on $T^G$ , i.e., for all $\sigma $ in the well-founded part of $T^G$ , we have $h^G(\sigma ) = \sup \{h^G(\tau )+1: \tau \in T^G, \, \tau \supseteq \sigma \}$ , while for all $\sigma $ which is extendible in $T^G$ , we have $h^G(\sigma ) = \infty $ . Our convention is that $\infty = \infty +1$ and $\infty> \infty > \alpha $ , so $h^G(\sigma ) = \sup \{h^G(\tau )+1: \tau \in T^G, \, \tau \supseteq \sigma \}$ for all $\sigma $ in $T^G$ .

Then, for each finite $F \subset \mathbb {N}$ (written $F \subset _{\mathrm {f}} \mathbb {N}$ ), we define

$$\begin{align*}M_F = \{X \subseteq \mathbb{N}: \exists \mu < \omega_1^{CK} (X \leq_T (T^G \oplus \langle f_i^G \rangle_{i \in F})^{(\mu)})\}, \end{align*}$$

where $Y^{(\mu )}$ denotes the transfinite iteration of the Turing jump of Y up to level $\mu $ . (For background on transfinite iterations of the Turing jump, see [13]. Later, we will express $M_F$ as the sets which are computable in $H_{\mu ,F}$ for some $\mu < \omega _1^{CK}$ , defined below.) Notice that $\mu $ ranges over the computable ordinals, rather than the ordinals which are computable in $\omega _1^{T^G \oplus \langle f_i^G \rangle _{i \in F}}$ . Nonetheless we will show in Corollary 3.14 that $M_F = \mathrm {HYP}(T^G \oplus \langle f_i^G \rangle _{i \in F})$ .

Finally, our desired $\omega $ -model is defined by

$$\begin{align*}M_\infty = \bigcup_{F \subset_{\mathrm{f}} \mathbb{N}} M_F. \end{align*}$$

Notice that $h^G$ does not appear in the definitions of $M_F$ or $M_\infty $ . Nonetheless it will play a crucial role in showing that $M_\infty $ has the properties we desire.

Let us sketch why finite choice fails in $M_\infty $ . We will show (Lemma 3.9) that for each $F \subset _{\mathrm {f}} \mathbb {N}$ , $M_F \cap [T^G] = \{f_i^G: i \in F\}$ . This implies that the branches through $T^G$ in $M_\infty $ are exactly the branches $f_i^G$ , for $i \in \mathbb {N}$ . This also implies that $M_\infty $ does not contain any infinite sequence of distinct branches $f_i^G$ , for such a sequence has to lie in some $M_F$ , contradicting Lemma 3.9.

This allows us to construct a failure of finite choice: For each n, let $T_n$ be the subtree of $T^G$ passing through $\langle n \rangle $ . Our forcing will ensure that each $[T_n]$ contains (nonzero) finitely many $f_i^G$ . Hence $M_\infty $ thinks that $\langle T_n \rangle _n$ is an instance of finite choice. Any finite choice solution to $\langle T_n \rangle _n$ would be an infinite sequence of distinct branches $f_i^G$ (they are distinct because the nth branch has first entry n), and hence would not lie in $M_\infty $ , by the previous paragraph.

In preparation for describing the forcing, we shall give every element of $M_\infty $ a name. Define by recursion on $\mu < \omega _1^{CK}$ :

$$\begin{align*}H_{1,F} = T^G \oplus \langle f^G_i \rangle_{i \in F}, \quad S_{\mu,F,e} = W^{H_{\mu,F}}_e, \quad H_{\mu,F} = \bigoplus_{\nu<\mu,e \in \mathbb{N}} S_{\nu,F,e} \end{align*}$$

for $\mu < \omega _1^{CK}$ , $F \subset _{\mathrm {f}} \mathbb {N}$ , $e \in \mathbb {N}$ . (Here $W^Y_e$ is the eth computably enumerable set relative to the oracle Y.)

3.2 The forcing language

We consider a ramified language $\mathcal {L}_\infty $ which extends the language of second-order arithmetic with constants for each element of $M_\infty $ , and various types of restricted set variables.

In order to define $\mathcal {L}_\infty $ , we first define a language $\mathcal {L}_F$ for each $F \subset _{\mathrm {f}} \mathbb {N}$ . $\mathcal {L}_F$ consists of first-order formulas in the language of second-order arithmetic generated by the following set variables and constants: for each $D \subseteq F$ , there are unranked set variables of the form $X_D$ and ranked set variables of the form $X^\lambda _D$ for each $\lambda < \omega _1^{CK}$ , while the constants are intended to name every element of $M_F$ :

1. – $\mathbf {T}$ , $\mathbf {f}_i$ for $i \in F$ ;

2. – for each $\nu < \omega _1^{CK}$ , $\mathbf {H}_{\nu ,F}$ and $\mathbf {S}_{\nu ,F,e}$ for each $e \in \mathbb {N}$ .

Let $C_F$ denote the above set of constants. If $\textbf {S}$ is of the form $\textbf {S}_{\nu ,F,e}$ , we define $\mathrm {dom}(\textbf {S})$ to be F. We define $C^\mu $ to be the set of all constants of the form $\textbf {H}_{\nu ,F}$ or $\textbf {S}_{\nu ,F,e}$ for some $\nu < \mu $ .

The language $\mathcal {L}_\infty $ consists of $\bigcup _{F \subset _{\mathrm {f}} \mathbb {N}} \mathcal {L}_F$ , unranked set variables of the form X, and ranked set variables of the form $X^\lambda $ for each $\lambda < \omega _1^{CK}$ .

Each formula $\psi $ of $\mathcal {L}_\infty $ can be assigned an ordinal rank as follows:

$$\begin{align*}\mathrm{rk}(\psi) = \omega_1^{CK}\cdot u(\psi) + \omega^2\cdot o(\psi) + \omega \cdot r(\psi) + n(\psi), \end{align*}$$

where:

1. – $u(\psi ) < \omega $ denotes the number of unranked quantifiers in $\psi $ ;

2. – $o(\psi ) < \omega _1^{CK}$ denotes the least upper bound of

   $$ \begin{align*} &\{\nu: \nu \text{ is the superscript of a bound variable in }\psi\} \\ \cup \; &\{\nu+1: \text{ some constant of the form }\textbf{S}_{\nu,F,e} \text{ or } \textbf{H}_{\nu,F} \text{ occurs in }\psi\}; \end{align*} $$

3. – $r(\psi ) < \omega $ denotes the number of ranked set quantifiers in $\psi $ ;

4. – $n(\psi ) < \omega $ denotes the number of connectives.

We say that a formula $\psi $ of $\mathcal {L}_\infty $ is ranked if all of its bound variables are ranked, or equivalently, if $\mathrm {rk}(\psi ) < \omega _1^{CK}$ .

A variable of the form $X^\lambda _D$ or $X_D$ is F-restricted if $D \subseteq F$ . A constant of the form $H_{\mu ,D}$ or $S_{\mu ,D,e}$ is F-restricted if $D \subseteq F$ . A formula of $\mathcal {L}_\infty $ is F-restricted if all of its bound variables and constants are F-restricted.

A formula is $\Sigma $ -over- $\mathcal {L}_F$ if it is built up from ranked F-restricted formulas using $\land $ , $\forall n$ , and $\exists X$ .

For any formula $\psi $ and any $\mu < \omega _1^{CK}$ , we define $\psi ^\mu $ by replacing every unranked set variable in $\psi $ with its ranked counterpart, i.e., X is replaced by $X^\mu $ and $X_F$ is replaced by $X^\mu _F$ .

3.3 The forcing notion

The forcing $\mathbb {P}$ consists of tuples

$$\begin{align*}p = \langle T^p, f^p, h^p \rangle \end{align*}$$

such that:

1. (C1) $T^p \subseteq \mathbb {N}^{<\mathbb {N}}$ is a finite tree;

2. (C2) $f^p$ is a finite partial function from $\mathbb {N}$ to $T^p\backslash \{\emptyset \}$ such that each $\sigma \in T^p$ of length $1$ is an initial segment of some $f^p(i)$

   (we view $f^p$ as a finite collection of finite paths in $T^p$ );

3. (C3) $h^p: T^p \to \omega _1^{CK} \cup \{\infty \}$ satisfies the following:

   1. (a) $h^p$ is a rank function, i.e., if $\tau \subsetneq \sigma $ , then $h^p(\tau )> h^p(\sigma )$

      (by fiat $\infty> \infty > \alpha $ for every $\alpha $ );

   2. (b) for all $i \in \mathrm {dom}(f^p)$ , $h^p(f^p(i)) = \infty $ ;

   3. (c) $h^p(\emptyset ) = \infty $ .

We think of $h^p$ as a labeling or tagging of nodes in $T^p$ . Condition (C2) is novel; ordinary Steel forcing only requires that $f^p$ is a finite partial function from $\mathbb {N}$ to $T^p$ . We impose (C2) in order to ensure that every node of length $1$ is extendible in the generic tree $T^G$ . Note that by (C2) and (C3)(b), for each $\sigma $ in $T^p$ of length $1$ , we must have $h^p(\sigma ) = \infty $ . Therefore, if $T^p$ contains any node of length $1$ , then (C3)(c) follows from (C2) and (C3)(b). There may be nodes $\sigma $ (of length greater than $1$ ) such that $h^p(\sigma ) = \infty $ yet there is no i such that $\sigma \subseteq f^p(i)$ .

We say that q extends p, written $q \leq p$ , if:

1. (E1) $T^q \supseteq T^p$ ;

2. (E2) for all $i \in \mathrm {dom}(f^p)$ , $f^q(i)$ is defined and $f^p(i) = f^q(i) \cap T^p$

   (old paths cannot be extended in the old tree);

3. (E3) for all $i \in \mathrm {dom}(f^q)\backslash \mathrm {dom}(f^p)$ , $f^q(i) \cap T^p = \emptyset $

   (new paths can only intersect the old tree at the root);

4. (E4) $h^q \supseteq h^p$ .

Conditions (E2) and (E3) are needed for us to control the complexity of the forcing relation. Condition (E2) is present in ordinary Steel forcing [16, p. 68]. One could weaken (E2) by considering only conditions p where each $f^p(i)$ is a leaf in $T^p$ (such conditions are dense in the above forcing): In this case we could replace (E2) by “for all $i \in \mathrm {dom}(f^p)$ , $f^q(i)$ is defined and $f^p(i) \subseteq f^q(i)$ .” Condition (E3) is not in [16] but needs to be added to it, as pointed out in [10, Section 2.2].

Remark 3.1. Conidis [2, Section 3.2] used a more complicated variant of Steel forcing (with locks on certain nodes of $T^p$ which can be toggled on and off) to prove that $\Delta ^1_1$ - $\mathsf {CA}_0$ does not imply $\mathsf {ABW}_0$ . Theorems 1.3 and 1.5 together provide an alternate proof of his result.

We will take G to be a sufficiently $\mathbb {P}$ -generic filter (G must decide all $\Sigma $ -over- $\mathcal {L}_F$ formulas for all $F \subset _{\mathrm {f}} \mathbb {N}$ , as well as all formulas of the form $\forall n(\psi (n) \leftrightarrow \neg \varphi (n))$ , where $\varphi $ and $\psi $ are $\Sigma $ -over- $\mathcal {L}_F$ for some $F \subset _{\mathrm {f}} \mathbb {N}$ ). Then we may define $T^G = \bigcup _{p \in G} T^p$ , $f_i^G = \bigcup _{p \in G} f^p(i)$ for $i \in \mathbb {N}$ , and $h^G = \bigcup _{p \in G} h^p$ . By genericity, each $f_i^G$ is an infinite branch in $T^G$ , the branches $f_i^G$ are distinct, and $h^G$ is the well-founded rank function on $T^G$ .

We encode each condition as a number as follows. Fix a computable linear ordering with well-founded part $\omega _1^{CK}$ . (Such orderings are known as pseudo-well-orderings; see [7].) We identify each $\alpha < \omega _1^{CK}$ with its corresponding element in the well-founded part. When we write $\alpha < \beta $ , we always refer to their order as ordinals rather than the natural number ordering. Using the above identification and standard pairing functions, we may encode $\mathbb {P}$ as a $\Pi ^1_1$ subset of $\mathbb {N}$ . For each $\alpha < \omega _1^{CK}$ , define $\mathbb {P}_\alpha $ to be the set of all conditions p such that the range of $h^p$ is contained in $\alpha \cup \{\infty \}$ . We have $\mathbb {P} = \bigcup _{\alpha < \omega _1^{CK}} \mathbb {P}_\alpha $ . Each $\mathbb {P}_\alpha $ is computable, uniformly in $\alpha $ .

3.4 The forcing relation

The forcing relation for formulas in $\mathcal {L}_\infty $ is defined by recursion as follows:

1. (1) for quantifier-free formulas of arithmetic $\psi $ , $p \Vdash \psi $ if and only if $\psi $ is true;

2. (2) $p \Vdash \boldsymbol {\sigma } \in \mathbf {T}$ if either $|\sigma | < 2$ and $\sigma \in T^p$ , or $\sigma ^- \in T^p$ and $h^p(\sigma ^-) \geq 1$ ;

3. (3) $p \Vdash \langle \mathbf {n},\mathbf {m} \rangle \in \textbf {f}_i$ if $i \in \mathrm {dom}(f^p)$ and $f^p(i)(n) = m$ ;

4. (4) $p \Vdash \langle 0,\sigma \rangle \in \textbf {H}_{1,F}$ if $p \Vdash \boldsymbol {\sigma } \in \textbf {T}$ ;

5. (5) $p \Vdash \langle 1,\langle \textbf {i},\langle \textbf {n},\textbf {m} \rangle \rangle \rangle \in \textbf {H}_{1,F}$ if $i \in F$ and $p \Vdash \langle \textbf {n},\textbf {m} \rangle \in \textbf {f}_i$ ;

6. (6) for $\nu> 1$ , $p \Vdash \mathbf {n} \in \mathbf {S}_{\nu ,F,e}$ if $p \Vdash \exists s R(\mathbf {H}_{\nu ,F}; \mathbf {e},s,\mathbf {n})$ where R codes a universal Turing machine;

7. (7) for $\nu> 1$ , $p \Vdash \langle \mathbf {e},\mathbf {n},\mu \rangle \in \mathbf {H}_{\nu ,F}$ if $\mu < \nu $ and $p \Vdash \mathbf {n} \in \mathbf {S}_{\mu ,F,e}$ ;

8. (8) $p \Vdash \forall x\psi (x)$ if for all $n \in \mathbb {N}$ , $p \Vdash \psi (\mathbf {n})$ ;

9. (9) $p \Vdash \forall X^\lambda _F \psi (X^\lambda _F)$ if for all $\nu < \lambda $ , $e \in \mathbb {N}$ , $p \Vdash \psi (\mathbf {S}_{\nu ,F,e})$ ;

10. (10) $p \Vdash \forall X^\lambda \psi (X^\lambda )$ if for all $\nu <\lambda $ , $e \in \mathbb {N}$ , $F \subset _{\mathrm {f}} \mathbb {N}$ , $p \Vdash \psi (\mathbf {S}_{\nu ,F,e})$ ;

11. (11) $p \Vdash \forall X_F \psi (X_F)$ if for all $\nu < \omega _1^{CK}$ , $e \in \mathbb {N}$ , $p \Vdash \psi (\mathbf {S}_{\nu ,F,e})$ ;

12. (12) $p \Vdash \forall X \psi (X)$ if for all $\nu < \omega _1^{CK}$ , $e \in \mathbb {N}$ , $F \subset _{\mathrm {f}} \mathbb {N}$ , $p \Vdash \psi (\mathbf {S}_{\nu ,F,e})$ ;

13. (13) $p \Vdash \varphi \land \psi $ if $p \Vdash \varphi $ and $p \Vdash \psi $ ;

14. (14) $p \Vdash \neg \psi $ if for every $q \leq p$ , $q \not \Vdash \psi $ .

Lemma 3.2. The following facts hold of $\Vdash $ :

1. (1) If $p \Vdash \psi $ , then for every sufficiently generic filter G which contains p, $M_\infty $ satisfies $\psi $ .

2. (2) If G is sufficiently generic and $M_\infty $ satisfies $\psi $ , then there is some $p \in G$ such that $p \Vdash \psi $ .

Analogous facts hold for $\psi \in \mathcal {L}_F$ and each model $M_F$ .

Just as we encoded each forcing condition as a number, we may also encode each formula in $\mathcal {L}_\infty $ as a number. We will show that the forcing relation for certain classes of formulas (encoded as a relation on numbers) can be computed within $M_\infty $ (Lemma 3.12).

3.5 Analyzing the forcing relation for ranked formulas

In order to control the complexity of the forcing relation, we will show that conditions that are sufficiently similar to each other force the same formulas of sufficiently low complexity. The basic notion of similarity in Steel forcing is known as retagging:

Definition 3.3 [16, Definition 4].

Suppose $\mu < \omega _1^{CK}$ and $F \subset _{\mathrm {f}} \mathbb {N}$ . If p and $p^\ast $ are conditions, we say that they are $\mu $ -F-retaggings if:

1. – $T^p = T^{p^\ast }$ and $f^p \restriction F = f^{p^\ast } \restriction F$ ;

2. – $h^p$ and $h^{p^\ast }$ agree on labels strictly smaller than $\mu $ ;

3. – if $h^p(\sigma ) \geq \mu $ , then $h^{p^\ast }(\sigma ) \geq \mu $ .

We make some observations:

1. – $\mu $ -F-retagging is an equivalence relation on conditions.

2. – If p and $p^\ast $ are $\mu $ -F-retaggings, then for any $\mu ' \leq \mu $ and any $F' \subseteq F$ , p and $p^\ast $ are $\mu '$ - $F'$ -retaggings as well.

Central to our analysis will be the ability to replace a quantifier over $\mathbb {P}$ with a quantifier over some subset $\mathbb {P}_\beta $ (recall that $\mathbb {P}$ is $\Pi ^1_1$ while $\mathbb {P}_\beta $ is computable). Once we have established connections between retagging and the forcing relation, the following lemma will be useful for achieving the above:

Lemma 3.4. For any $q \in \mathbb {P}$ and any $\beta < \omega _1^{CK}$ , there is some $q^\ast \in \mathbb {P}_\beta $ such that:

1. – q and $q^\ast $ are $\beta $ -F-retaggings for every $F \subset _{\mathrm {f}} \mathbb {N}$ ;

2. – for every $p \in \mathbb {P}_\beta $ such that $q \leq p$ , we have $q^\ast \leq p$ .

Proof Define $T^{q^\ast } = T^q$ , $f^{q^\ast } = f^q$ , and

$$\begin{align*}h^{q^\ast}(\tau) = \begin{cases} \infty, & h^q(\tau) \geq \beta, \\ h^q(\tau), & h^q(\tau) < \beta. \end{cases} \end{align*}$$

It is easy to see that $q^\ast $ satisfies the desired properties.

A cornerstone of the method of Steel forcing is the following basic retagging lemma.

Lemma 3.5. Let p and $p^\ast $ be $\omega \beta $ -F-retaggings. Then for all $q \leq p$ and all $\gamma < \beta $ , there exists $q^\ast \leq p^\ast $ such that q and $q^\ast $ are $\omega \gamma $ -F-retaggings. Furthermore, if $\mathrm {dom}(f^{p^\ast }) \subseteq F$ (and hence $f^{p^\ast } \subseteq f^p$ ), then we can choose $q^\ast $ such that $f^{q^\ast } \subseteq f^q$ .

The first part of the lemma is identical in form to [10, Lemma 2.5]. The second part has no analog in [10], but its analog would follow immediately from the proof of [10, Lemma 2.5]. For our forcing we will construct $q^\ast $ (specifically $f^{q^\ast }$ ) slightly differently in order to prove the second part. The second part will be used in the proof of Lemma 3.11.

Proof Define $q^\ast $ as follows:

1. – $T^{q^\ast } = T^q$ ;

2. – in general, we define $f^{q^\ast }$ to be the disjoint union of:

   1. – $f^q\restriction F$ ,

   2. – $f^{p^\ast }\restriction (\mathrm {dom}(f^{p^\ast })\backslash (\mathrm {dom}(f^q) \cap F))$ , and

   3. – a function g such that $\mathrm {dom}(g)$ is disjoint from $F \cup \mathrm {dom}(f^{p^\ast })$ and $\mathrm {range}(g) = \{\sigma \in T^{q^\ast }\backslash T^{p^\ast }: |\sigma | = 1\}$ ;

3. – if $\mathrm {dom}(f^{p^\ast }) \subseteq F$ and we want to satisfy the second part of the lemma, we define $f^{q^\ast } = f^q\restriction (F \cup \{i: f^q(i) \cap T^p = \emptyset \})$ ;

4. – $h^{q^\ast }$ is defined by cases:

   $$\begin{align*}h^{q^\ast}(\tau) = \begin{cases} h^{p^\ast}(\tau), & \text{if }\tau \in T^p, \\ \infty, & \text{if }\exists i(\tau \subseteq f^{q^\ast}(i)), \\ h^q(\tau), & \text{if }h^q(\tau) < \omega\gamma, \\ \omega\gamma+|\tau|_Q, &\text{otherwise}, \end{cases} \end{align*}$$
   where $Q = \{\sigma \in T^q: h^q(\sigma ) \geq \omega \gamma \}$ and $|\tau |_Q \in \mathbb {N}$ denotes the rank of $\tau $ in the finite tree Q.

$h^{q^\ast }$ is well-defined. First, we show that the second and third cases in the definition of $h^{q^\ast }$ are mutually exclusive. Suppose $\tau \subseteq f^{q^\ast }(i)$ for some i.

Case 1. If $f^{q^\ast }(i) = f^q(i)$ , then $h^q(\tau ) = \infty> \omega \gamma $ .

Case 2. If $f^{q^\ast }(i) = f^{p^\ast }(i)$ , then $\tau \in T^p$ and $h^{p^\ast }(\tau ) = \infty $ . Since p and $p^\ast $ are $\omega \beta $ -F-retaggings, we have $h^p(\tau ) \geq \omega \beta $ . It follows that $h^q(\tau ) = h^p(\tau ) \geq \omega \beta> \omega \gamma $ .

Case 3. Otherwise, $|f^{q^\ast }(i)| = 1$ . Then $h^q(\tau ) = \infty> \omega \gamma $ .

Second, in order to show that the first and second cases in the definition of $h^{q^\ast }$ do not conflict, we shall show that if $\tau \in T^p$ and $\tau \subseteq f^{q^\ast }(i)$ for some i, then $h^{p^\ast }(\tau ) = \infty $ .

Case 1. Suppose $f^{q^\ast }(i) = f^q(i)$ and $i \in F$ . 1a. If $f^q(i) \cap T^p = \emptyset $ , then $\tau = \emptyset $ so $h^{p^\ast }(\tau ) = \infty $ . 1b. Otherwise, by (E2) and (E3), we have $f^q(i) \cap T^p = f^p(i)$ . Since p and $p^\ast $ are $\omega \beta $ -F-retaggings, we have $f^{p^\ast }(i) = f^p(i)$ . It follows that $\tau \subseteq f^q(i) \cap T^p = f^p(i) = f^{p^\ast }(i)$ , implying $h^{p^\ast }(\tau ) = \infty $ .

Case 2. If $f^{q^\ast }(i) = f^{p^\ast }(i)$ , then $h^{p^\ast }(\tau ) \geq h^{p^\ast }(f^{p^\ast }(i)) = \infty $ .

Case 3. Otherwise, $|f^{q^\ast }(i)| = 1$ . Then $h^{p^\ast }(\tau ) = \infty $ .

To see that the first and third cases in the definition of $h^{q^\ast }$ do not conflict, suppose $\tau \in T^p$ and $h^q(\tau ) < \omega \gamma $ . We have $h^p(\tau ) = h^q(\tau ) < \omega \gamma $ , so by retagging, $h^{p^\ast }(\tau ) = h^p(\tau ) = h^q(\tau )$ as desired. We have shown that $h^{q^\ast }$ is well-defined.

$q^\ast $ satisfies (C2). We address the two definitions of $f^{q^\ast }$ separately. Suppose $\sigma \in T^{q^\ast }$ has length $1$ . We want to show that there is some i such that $\sigma \subseteq f^{q^\ast }(i)$ .

We begin by considering the first definition of $f^{q^\ast }$ . First, consider the case $\sigma \in T^{p^\ast }$ . If there is some $i \in \mathrm {dom}(f^{p^\ast })\backslash (\mathrm {dom}(f^q) \cap F)$ such that $\sigma \subseteq f^{p^\ast }(i)$ , then we are done. Otherwise, since $p^\ast $ satisfies (C2), there is some $i \in \mathrm {dom}(f^{p^\ast }) \cap \mathrm {dom}(f^q) \cap F$ such that $\sigma \subseteq f^{p^\ast }(i)$ . Then $f^{q^\ast }(i) = f^q(i) \supseteq f^p(i) = f^{p^\ast }(i) \supseteq \sigma $ as desired.

Second, if $\sigma \in T^{q^\ast }\backslash T^{p^\ast }$ , then there is some $i \in \mathrm {dom}(g)$ such that $g(i) = \sigma $ . We have $f^{q^\ast }(i) = g(i) = \sigma $ as desired.

Next we consider the second definition of $f^{q^\ast }$ . Assume that $\mathrm {dom}(f^{p^\ast }) \subseteq F$ . If $\sigma \in T^{p^\ast }$ , then since $p^\ast $ satisfies (C2), there is some $i \in \mathrm {dom}(f^{p^\ast })$ such that $\sigma \subseteq f^{p^\ast }(i)$ . Such i lies in F by assumption, so $f^{q^\ast }(i) = f^q(i) \supseteq f^p(i) = f^{p^\ast }(i) \supseteq \sigma $ as desired.

If $\sigma \in T^{q^\ast }\backslash T^{p^\ast }$ , then since q satisfies (C2), there is some $i \in \mathrm {dom}(f^q)$ such that $\sigma \subseteq f^q(i)$ . Since $\sigma \notin T^p$ and $|\sigma | = 1$ , it follows that $f^q(i) \cap T^p = \emptyset $ . So i lies in $\mathrm {dom}(f^{q^\ast })$ . Hence $f^{q^\ast }(i) = f^q(i) \supseteq \sigma $ as desired. This completes the proof that $q^\ast $ satisfies (C2).

To show that $q^\ast $ is a condition, it remains to check (C3). We omit the verification as it is exactly the same as that for ordinary Steel forcing.

$q^\ast $ extends $p^\ast $ . (E1) and (E4) are immediate. To check (E2), consider any $i \in \mathrm {dom}(f^{p^\ast })$ .

Case 1. Suppose $i \in F$ . By retagging, we have $i \in \mathrm {dom}(f^p) \subseteq \mathrm {dom}(f^q)$ and $f^{p^\ast }(i) = f^p(i)$ . In both definitions of $f^{q^\ast }$ we have $f^{q^\ast }(i) = f^q(i)$ , so $f^{q^\ast }(i) \cap T^{p^\ast } = f^q(i) \cap T^p = f^p(i) = f^{p^\ast }(i)$ (the second equality holds by (E2)).

Case 2. If $i \notin F$ , then we have $\mathrm {dom}(f^{p^\ast }) \not \subseteq F$ . So we are only concerned with the first definition of $f^{q^\ast }$ . In this case $f^{q^\ast }(i)$ is defined to be $f^{p^\ast }(i)$ , which lies in $T^{p^\ast }$ , so $f^{q^\ast }(i) \cap T^{p^\ast } = f^{p^\ast }(i)$ .

To check (E3), consider any $i \in \mathrm {dom}(f^{q^\ast })\backslash \mathrm {dom}(f^{p^\ast })$ . We analyze the two definitions of $f^{q^\ast }$ separately.

We begin by considering the first definition of $f^{q^\ast }$ .

Case 1. $i \in \mathrm {dom}(f^q) \cap F$ and $f^{q^\ast }(i) = f^q(i)$ . Since $i \in F\backslash \mathrm {dom}(f^{p^\ast })$ , we have $i \notin \mathrm {dom}(f^p)$ by retagging. Hence by (E3) for $q \leq p$ , we have $f^q(i) \cap T^p = \emptyset $ as desired.

Case 2. Otherwise, $i \in \mathrm {dom}(g)$ and $f^{q^\ast }(i) = g(i)$ . In this case, $g(i) \notin T^{p^\ast }$ and $|g(i)| = 1$ , so $g(i) \cap T^{p^\ast } = \emptyset $ as desired.

Next we consider the second definition of $f^{q^\ast }$ . If $i \in F$ , then the reasoning is identical to Case 1 above. Otherwise, by the definition of $f^{q^\ast }$ , we have $f^{q^\ast }(i) \cap T^p = f^q(i) \cap T^p = \emptyset $ as desired. This completes the proof that $q^\ast $ extends $p^\ast $ .

Finally, it is easy to check that q and $q^\ast $ are $\omega \gamma $ -F-retaggings.

The following consequences of the basic retagging lemma are routine (see [10, Lemma 2.4 and Corollary 2.6]). We include their proofs for completeness.

Lemma 3.6. Let $\psi $ be a ranked formula in $\mathcal {L}_F$ . Suppose that p and $p^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ))$ -F-retaggings. Then $p \Vdash \psi $ if and only if $p^\ast \Vdash \psi $ .

Proof We proceed by induction on the rank of $\psi $ . The only nontrivial case is when $\psi $ is $\neg \varphi $ . Assuming that $p^\ast $ and p are $(\omega \cdot \mathrm {rk}(\neg \varphi ))$ -F-retaggings and that $p^\ast \Vdash \neg \varphi $ , we want to show that $p \Vdash \neg \varphi $ , i.e., for all $q \leq p$ , we have $q \not \Vdash \varphi $ . By Lemma 3.5, there is some $q^\ast \leq p^\ast $ such that $q^\ast $ and q are $(\omega \cdot \mathrm {rk}(\varphi ))$ -F-retaggings. Since $p^\ast \Vdash \neg \varphi $ , we have $q^\ast \not \Vdash \varphi $ . Applying the inductive hypothesis to q and $q^\ast $ shows that $q \not \Vdash \varphi $ as desired.

Corollary 3.7. Suppose $p \in \mathbb {P}_\mu $ and $\psi \in \mathcal {L}_F$ . If there is some $q \leq p$ such that $q \Vdash \psi $ , then there is some $q' \leq p$ in $\mathbb {P}_{\max \{\mu ,\omega \cdot \mathrm {rk}(\psi )\}}$ such that $q' \Vdash \psi $ . Therefore, $p \Vdash \neg \psi $ if and only if for all $q \leq p$ in $\mathbb {P}_{\max \{\mu ,\omega \cdot \mathrm {rk}(\psi )\}}$ , $q \not \Vdash \psi $ .

Proof Apply Lemma 3.4 to obtain some $q' \leq p$ in $\mathbb {P}_{\max \{\mu ,\omega \cdot \mathrm {rk}(\psi )\}}$ such that q and $q'$ are $(\max \{\mu ,\omega \cdot \mathrm {rk}(\psi )\})$ -F-retaggings. By Lemma 3.6, $q' \Vdash \psi $ .

Corollary 3.8. For each condition p and each formula $\psi $ in $\mathcal {L}_F$ , $\emptyset ^{(\mathrm {rk}(\psi ))}$ can decide whether $p \Vdash \psi $ uniformly in p and $\psi $ .

Proof Induction on $\mathrm {rk}(\psi )$ using Corollary 3.7.

We may now analyze which branches on $T^G$ lie in each $M_F$ . The proof below follows [10, Lemma 2.7], with modifications in order to account for (C2).

Lemma 3.9. For each $F \subset _{\mathrm {f}} \mathbb {N}$ , $M_F \cap [T^G] = \{f^G_i: i \in F\}$ . Hence $M_\infty \cap [T^G] = \{f^G_i: i \in \mathbb {N}\}$ , but no infinite sequence of distinct $f^G_i$ lies in $M_\infty $ .

Proof Suppose towards a contradiction that $S = S_{\nu ,F,e} \in [T^G]$ is not $f^G_i$ for any $i \in F$ . Then there is some prefix $\sigma \subset S$ such that $\sigma \not \subset f^G_i$ for any $i \in F$ . For later purposes, fix such $\sigma $ with length greater than $1$ . By Lemma 3.2, fix $p \in G$ such that $p \Vdash \varphi (\textbf {S})$ , where $\varphi (\textbf {S})$ is

$$\begin{align*}\textbf{S} \in [\textbf{T}] \; \land \; \boldsymbol{\sigma} \subset \textbf{S} \; \land \; (\forall i \in F)(\boldsymbol{\sigma} \not\subset \textbf{f}_i). \end{align*}$$

Note $\varphi (\textbf {S})$ is a ranked formula in $\mathcal {L}_F$ .

By genericity, we may assume that $\sigma \in T^p$ . Next, fix $\beta < \omega _1^{CK}$ large enough so that $\beta> \omega \cdot \mathrm {rk}(\varphi (\textbf {S}))$ and $p \in \mathbb {P}_\beta $ .

Since p forces that $\sigma $ is extendible in $T^G$ , we have $h^p(\sigma ) = \infty $ (by Lemma 3.2). We now define $p^\ast $ which is a $\beta $ -F-retagging of p, such that $h^{p^\ast }(\sigma ) \in [\beta ,\omega _1^{CK})$ . Define $T^{p^\ast } = T^p$ . Define $f^{p^\ast }$ to be the union of $f^p\restriction F$ and a function g such that $\mathrm {dom}(g)$ is disjoint from F and $\mathrm {range}(g) = \{\sigma \in T^p: |\sigma |=1\}$ . Finally, define

$$\begin{align*}h^{p^\ast}(\tau) = \begin{cases} \beta+|\tau|_Q, & \text{if }\tau \supseteq \sigma \land h^p(\tau) = \infty, \\ h^p(\tau), & \text{otherwise}, \end{cases} \end{align*}$$

where $Q = \{\tau : \tau \subseteq \sigma \lor (\tau \supseteq \sigma \land h^p(\tau ) = \infty )\}$ .

Since $h^p(\sigma ) = \infty $ and $|\sigma |> 1$ , it is straightforward to check that $h^{p^\ast }$ is a rank function and $h^{p^\ast }(\emptyset ) = \infty $ . In order to show that $p^\ast $ is a condition, it suffices to check that $h^{p^\ast }(f^{p^\ast }(i)) = \infty $ for all $i \in \mathrm {dom}(f^{p^\ast })$ . If $i \in F$ , we have $\sigma \not \subseteq f^p(i) = f^{p^\ast }(i)$ , so $h^{p^\ast }(f^{p^\ast }(i)) = h^p(f^{p^\ast }(i)) = h^p(f^p(i)) = \infty $ . If $i \in \mathrm {dom}(f^{p^\ast })\backslash F$ (that is, $\mathrm {dom}(g)$ ), then $f^{p^\ast }(i)$ has length $1$ , so $h^{p^\ast }(f^{p^\ast }(i)) = h^p(f^{p^\ast }(i)) = \infty $ .

Finally, it is clear that $p^\ast $ is a $\beta $ -F-retagging of p, and hence an $(\omega \cdot \mathrm {rk}(\varphi (\textbf {S})))$ -F-retagging of p. By Lemma 3.6 we have

$$\begin{align*}p^\ast \Vdash \textbf{S} \in [\textbf{T}] \land \boldsymbol{\sigma} \subset \textbf{S}, \end{align*}$$

which is impossible because $h^{p^\ast }(\sigma ) \neq \infty $ .

The following lemma is new; (C2) was designed in order to prove it.

Lemma 3.10. $M_\infty $ does not satisfy finite choice.

Proof Consider the following arithmetical predicate $\varphi (n,X)$ (with parameter $T^G$ ):

$$\begin{align*}X \in [T^G] \; \land \; X(0) = n. \end{align*}$$

First, we claim that for each n, there is some $X \in M_\infty $ such that $\varphi (n,X)$ holds. By genericity, fix some $p \in G$ such that $\langle n \rangle \in T^p$ . By (C2), there is some i such that $\langle n \rangle \subseteq f^p(i)$ . Then $\varphi (n,f^G_i)$ holds.

Second, we claim that for each n, there are at most finitely many $X \in M_\infty $ such that $\varphi (n,X)$ holds. By Lemma 3.9, it suffices to show that there are only finitely many i such that $\varphi (n,f^G_i)$ holds. As above, fix $p \in G$ such that $\langle n \rangle \in T^p$ . For each i, we claim that if $i \notin \mathrm {dom}(f^p)$ , then $\langle n \rangle \not \subseteq f^G_i$ . Suppose to the contrary that there is some $i \notin \mathrm {dom}(f^p)$ such that $\langle n \rangle \subseteq f^G_i$ . Fix some $q \in G$ such that $q \Vdash \langle \textbf {n} \rangle \subseteq \textbf {f}_i$ . Let $r \in G$ be a common refinement of p and q. Then $\langle n \rangle \subseteq f^r(i)$ . But then $f^r(i) \cap T^p \neq \emptyset $ , violating (E3) for $r \leq p$ .

This shows that $\varphi (n,X)$ is an instance of finite choice in $M_\infty $ . By Lemma 3.9, $M_\infty $ does not contain any infinite sequence of distinct branches through $T^G$ , so $M_\infty $ does not contain any finite choice solution to $\varphi (n,X)$ .

3.6 Analyzing the forcing relation for $\Sigma $ -over- $\mathcal {L}_F$ formulas

In order to show that $M_\infty \models \Delta ^1_1$ - $\mathsf {CA}_0$ (Section 3.7), we need to analyze the forcing relation for $\Sigma $ -over- $\mathcal {L}_F$ formulas $\psi $ . We begin by considering formulas $\psi ^\rho $ where $\rho < \omega _1^{CK}$ . Note that such formulas may not lie in any $\mathcal {L}_{F'}$ because they may contain (existential) quantifiers over set variables of the form $X^\rho $ .

To this end we exploit automorphisms of $\mathbb {P}$ , as was done in [16, Lemma 10]. Each permutation $\pi : \mathbb {N} \to \mathbb {N}$ induces an automorphism $\hat {\pi }$ of $\mathbb {P}$ by permuting the finite paths in each condition: $T^{\hat {\pi }(p)} = T^p$ , $h^{\hat {\pi }(p)} = h^p$ , and $f^{\hat {\pi }(p)}(\pi (i)) = f^p(i)$ . Each permutation $\pi $ also induces an automorphism of $\mathcal {L}_\infty $ : For each formula $\psi $ , let $\pi \psi $ denote the formula obtained from $\psi $ by replacing each $\textbf {f}_i$ by $\textbf {f}_{\pi (i)}$ and each $\textbf {S}_{\nu ,F,e}$ by $\textbf {S}_{\nu ,\pi "F,e}$ . (In the proof below, we will denote $\textbf {S}_{\nu ,\pi "F,e}$ by $\pi \textbf {S}_{\nu ,F,e}$ .) By induction on $\mathrm {rk}(\psi )$ , one can show that $p \Vdash \psi $ if and only if $\hat {\pi }(p) \Vdash \pi \psi $ .

Lemma 3.11. Suppose $\psi $ is $\Sigma $ -over- $\mathcal {L}_F$ , $p \Vdash \psi ^\mu $ , and $\mathrm {dom}(f^p) \subseteq F$ . If p and $p^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu )+\omega ^2)$ -F-retaggings, then $p^\ast \Vdash \psi ^\mu $ as well.

Proof We prove by induction on k that if $\psi $ is built from ranked, F-restricted formulas using k steps and p is a condition such that $p \Vdash \psi ^\mu $ , $\mathrm {dom}(f^p) \subseteq F$ , and p and $p^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu )+\omega \cdot 2k)$ -F-retaggings, then $p^\ast \Vdash \psi ^\mu $ . The base case holds by Lemma 3.6.

The only nontrivial inductive step is if $\psi $ has the form $\exists X\varphi (X)$ . We want to prove that for all $q^\ast \leq p^\ast $ , there is some $r^\ast \leq q^\ast $ and some $\textbf {S} \in C^\mu $ such that $r^\ast \Vdash \varphi ^\mu (\textbf {S})$ . Our plan for doing so is illustrated in Figure 1.

Figure 1 Illustration of the proof of Lemma 3.11. Arrows correspond to extension in the forcing. Dotted lines correspond to retaggings.

Given $q^\ast \leq p^\ast $ , there is some $q \leq p$ such that q and $q^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu ) + \omega \cdot (2k+1))$ -F-retaggings. Furthermore, since $\mathrm {dom}(f^p) \subseteq F$ , we can choose such q with $f^q \subseteq f^{q^\ast }$ (by the second part of Lemma 3.5).

Since $p \Vdash \psi ^\mu $ and $q \leq p$ , there is some $r \leq q$ and some $\textbf {S} \in C^\mu $ such that $r \Vdash \varphi ^\mu (\textbf {S})$ . By extending r, we may assume that $F \cup \mathrm {dom}(\textbf {S}) \subseteq \mathrm {dom}(f^r)$ . To complete the inductive step, we want to define $r^\ast \leq q^\ast $ such that r and $r^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu )+\omega \cdot 2k)$ - $\mathrm {dom}(f^r)$ -retaggings. By inductive hypothesis, it would follow that $r^\ast \Vdash \varphi ^\mu (\textbf {S})$ as desired.

The naive definition of $r^\ast $ fails because there may be conflict between $f^r$ and $f^{q^\ast }$ . In order to avoid this conflict, we use an automorphism of $\mathbb {P}$ to permute r and $\textbf {S}$ as follows. Consider a permutation $\pi $ which fixes $F \cup \mathrm {dom}(f^q)$ and moves $\mathrm {dom}(f^r)\backslash (F \cup \mathrm {dom}(f^q))$ to some set which is disjoint with $\mathrm {dom}(f^{q^\ast })$ . Since $\pi $ fixes $\mathrm {dom}(f^q)$ , we have $\hat {\pi } r \leq q$ . Since $\pi $ fixes F and $\varphi $ is $\Sigma $ -over- $\mathcal {L}_F$ , we have $\pi \varphi ^\mu = \varphi ^\mu $ . So $\hat {\pi } r \Vdash \varphi ^\mu (\pi \textbf {S})$ . Therefore, by replacing r and $\textbf {S}$ with $\hat {\pi } r$ and $\pi \textbf {S}$ , we may assume that $\mathrm {dom}(f^r)\backslash (F \cup \mathrm {dom}(f^q))$ and $\mathrm {dom}(f^{q^\ast })$ are disjoint.

We claim that q and $q^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu ) + \omega \cdot (2k+1))$ - $\mathrm {dom}(f^r)$ -retaggings. Since $f^q \subseteq f^{q^\ast }$ , this reduces to showing that $\mathrm {dom}(f^{q^\ast }) \cap \mathrm {dom}(f^r) \subseteq \mathrm {dom}(f^q)$ . Suppose $i \in \mathrm {dom}(f^{q^\ast }) \cap \mathrm {dom}(f^r)$ . By assumption on r, it follows that $i \in F \cup \mathrm {dom}(f^q)$ . If $i \in \mathrm {dom}(f^q)$ , then we are done. Otherwise $i \in F$ , but since $i \in \mathrm {dom}(f^{q^\ast })$ and q and $q^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu ) + \omega \cdot (2k+1))$ -F-retaggings, it follows that $i \in \mathrm {dom}(f^q)$ as well. This proves the claim.

By Lemma 3.5, there is some $r^\ast \leq q^\ast $ such that r and $r^\ast $ are $(\omega \cdot \mathrm {rk}(\psi ^\mu )+\omega \cdot 2k)$ - $\mathrm {dom}(f^r)$ -retaggings. Since $F \cup \mathrm {dom}(\textbf {S}) \subseteq \mathrm {dom}(f^r)$ , we may apply the inductive hypothesis and the fact that $r \Vdash \varphi ^\mu (\textbf {S})$ to conclude that $r^\ast \Vdash \varphi ^\mu (\textbf {S})$ as desired.

The following lemma is not stated in [10] but some form of it is used in the proof of [10, Lemma 2.9(1)].

Lemma 3.12. Given a condition p, a formula $\psi $ which is $\Sigma $ -over- $\mathcal {L}_F$ , and some $\rho < \omega _1^{CK}$ , $\emptyset ^{(\mathrm {rk}(\psi ^\rho ))}$ can decide whether $p \Vdash \psi ^\rho $ uniformly in p, $\psi $ , and $\rho $ .

Proof We proceed by induction on the number of steps it takes to construct $\psi $ from ranked F-restricted formulas. The base case holds by Corollary 3.8.

For the inductive step, the only nontrivial case is when $\psi $ has the form $\exists X\varphi (X)$ . Fix any $\mu \geq \omega \cdot \mathrm {rk}(\varphi ^\rho (\textbf {S})) + \omega ^2 + \omega $ such that $p \in \mathbb {P}_\mu $ . We claim that $p \Vdash \exists X^\rho \psi ^\rho (X^\rho )$ if and only if for every $q \leq p$ in $\mathbb {P}_\mu $ , there is some $r \leq q$ in $\mathbb {P}_\mu $ and some $\textbf {S} \in C^\rho $ such that $r \Vdash \varphi ^\rho (\textbf {S})$ . This claim suffices for proving this case of the inductive step.

To prove the forward direction, suppose we are given $q \leq p$ in $\mathbb {P}_\mu $ . By assumption, there is some $r \leq q$ and some $\textbf {S} \in C^\rho $ such that $r \Vdash \varphi ^\rho (\textbf {S})$ . We can retag r to some $r^\ast \leq q$ in $\mathbb {P}_\mu $ such that r and $r^\ast $ are $\mu $ - $(F \cup \mathrm {dom}(f^r))$ -retaggings (Lemma 3.4). Since $\mu \geq \omega \cdot \mathrm {rk}(\varphi ^\rho (\textbf {S})) + \omega ^2$ , it follows from Lemma 3.11 that $r^\ast \Vdash \varphi ^\rho (\textbf {S})$ as desired.

To prove the backward direction, suppose we are given $q \leq p$ . In order to construct some $r \leq q$ and some $\textbf {S} \in C^\rho $ such that $r \Vdash \varphi ^\rho (\textbf {S})$ , we follow Figure 2. We begin by retagging q to some $q^\ast \leq p$ in $\mathbb {P}_\mu $ such that q and $q^\ast $ are $\mu $ - $F'$ -retaggings for every $F' \subset _f \mathbb {N}$ (Lemma 3.4). By assumption, there is some $r^\ast \leq q^\ast $ in $\mathbb {P}_\mu $ and some $\textbf {S} \in C^\rho $ such that $r^\ast \Vdash \varphi ^\rho (\textbf {S})$ .

Figure 2 Illustration of the proofs of Lemmas 3.12 and 3.13. Arrows correspond to extension in the forcing. Dotted lines correspond to retaggings.

To complete the proof of the backward direction, we want to produce some $r \leq q$ which is a retagging of $r^\ast $ . Define $F'$ to be the union of F, $\mathrm {dom}(f^{r^\ast })$ and $\mathrm {dom}(\textbf {S})$ (recall that $\mathrm {dom}(\textbf {S}_{\nu ,D,e})$ is defined to be D.) Then $\varphi ^\rho (\textbf {S})$ is $\Sigma $ -over- $\mathcal {L}_{F'}$ . Since $\mu \geq \omega \cdot \mathrm {rk}(\varphi ^\rho (\textbf {S})) + \omega ^2 + \omega $ , there is some $r \leq q$ such that r and $r^\ast $ are $(\omega \cdot \mathrm {rk}(\varphi ^\rho (\textbf {S})) + \omega ^2)$ - $F'$ -retaggings (Lemma 3.5). It follows from Lemma 3.11 that $r \Vdash \varphi ^\rho (\textbf {S})$ as desired.

Next, we shall show that in order to understand the forcing relation for $\Sigma $ -over- $\mathcal {L}_F$ formulas $\varphi $ , it suffices to consider formulas $\varphi ^\rho $ where $\rho < \omega _1^{CK}$ . This will be useful for showing that $M_\infty \models \Delta ^1_1$ - $\mathsf {CA}_0$ .

Lemma 3.13. Suppose $\psi $ is $\Sigma $ -over- $\mathcal {L}_F$ . If $p \Vdash \psi $ , then there is some $\mu < \omega _1^{CK}$ such that for all $\rho \in [\mu ,\omega _1^{CK})$ , $p \Vdash \psi ^\rho $ .

Proof We proceed by induction on the number of steps it takes to construct $\psi $ from ranked F-restricted formulas. The base case is trivial. For the inductive step, the only nontrivial case is when $\psi $ has the form $\exists X\varphi (X)$ . Since $p \Vdash \exists X \varphi (X)$ , for every $q \leq p$ , there exist $r \leq q$ and $\textbf {S}$ such that $r \Vdash \varphi (\textbf {S})$ . By the inductive hypothesis, $r \Vdash \varphi ^\rho (\textbf {S})$ for all sufficiently large $\rho < \omega _1^{CK}$ .

Hence, for each $q \leq p$ , there exist $\gamma _q < \omega _1^{CK}$ , a condition $r \leq q$ in $\mathbb {P}_{\gamma _q}$ , and $\textbf {S} \in C^{\gamma _q}$ such that $r \Vdash \varphi ^{\gamma _q}(\textbf {S})$ . By Lemma 3.12, we can hyperarithmetically search for the least such $\gamma _q$ , r and $\textbf {S}$ .

By boundedness, for each $\beta < \omega _1^{CK}$ , there is some $\gamma < \omega _1^{CK}$ such that for every $q \leq p$ in $\mathbb {P}_\beta $ , there exists $\gamma _q < \gamma $ as above. For later purposes, we will choose $\gamma < \omega _1^{CK}$ sufficiently large such that $\omega \cdot \mathrm {rk}(\varphi ^{\gamma _q}(\textbf {S}))+\omega ^2+\omega \leq \gamma $ for every $q \leq p$ in $\mathbb {P}_\beta $ .

Since we can find $\gamma $ from $\beta $ hyperarithmetically, by boundedness, there is some limit $\mu < \omega _1^{CK}$ such that (1) $p \in \mathbb {P}_\mu $ ; (2) for every $\beta < \mu $ , there is some $\gamma < \mu $ satisfying the previous paragraph.

We shall show that $p \Vdash \exists X^\mu \varphi ^\mu (X^\mu )$ . Given $q \leq p$ , we want to construct $r \leq q$ and $\textbf {S} \in C^\mu $ such that $r \Vdash \varphi ^\mu (\textbf {S})$ . Our plan for doing so is illustrated in Figure 2.

We begin by retagging q to some $q^\ast \leq p$ in $\mathbb {P}_\mu $ such that q and $q^\ast $ are $\mu $ - $F'$ -retaggings for every $F' \subset _f \mathbb {N}$ (Lemma 3.4). Since $\mu $ is a limit and $\mathrm {dom}(h^{q^\ast }) = T^{q^\ast }$ is finite, we have $q^\ast \in \mathbb {P}_\beta $ for some $\beta < \mu $ . By construction of $\mu $ , there exist $\gamma _{q^\ast } < \gamma < \mu $ , a condition $r^\ast \leq q^\ast $ in $\mathbb {P}_{\gamma _{q^\ast }}$ , and $\textbf {S} \in C^{\gamma _{q^\ast }}$ such that $\omega \cdot \mathrm {rk}(\varphi ^{\gamma _{q^\ast }}(\textbf {S}))+\omega ^2+\omega \leq \gamma $ and $r^\ast \Vdash \varphi ^{\gamma _{q^\ast }}(\textbf {S})$ .

Define $F'$ to be the union of F, $\mathrm {dom}(f^{r^\ast })$ , and $\mathrm {dom}(\textbf {S})$ . Since $\omega \cdot \mathrm {rk}(\varphi ^{\gamma _{q^\ast }}(\textbf {S}))+\omega ^2+\omega \leq \gamma < \mu $ and q and $q^\ast $ are $\mu $ - $F'$ -retaggings, by Lemma 3.5, there is some $r \leq q$ such that r and $r^\ast $ are $(\omega \cdot \mathrm {rk}(\varphi ^{\gamma _{q^\ast }}(\textbf {S}))+\omega ^2)$ - $F'$ -retaggings. By Lemma 3.11, we have $r \Vdash \varphi ^{\gamma _{q^\ast }}(\textbf {S})$ and so $r \Vdash \varphi ^\mu (\textbf {S})$ as desired.

Following [10, Lemma 2.9(2)], we conclude the following corollary.

Corollary 3.14. For each $F \subset _f \mathbb {N}$ , $M_F \models \Sigma ^1_1$ - $\mathsf {AC}_0$ . It follows that $M_F = \mathrm {HYP}(T^G \oplus \langle f_i^G \rangle _{i \in F})$ .

Proof Suppose that $M_F \models \forall n\exists X\varphi (n,X)$ , where $\varphi (n,X)$ is arithmetic with parameters from $M_F$ . Fix some $p \in G$ such that $p \Vdash \forall n \exists X\varphi (n,X)$ . By Lemma 3.13, fix some $\mu < \omega _1^{CK}$ (greater than the rank of every constant symbol in $\varphi $ ) such that $p \Vdash \forall n \exists X^\mu \varphi (n,X^\mu )$ . It follows that $M_F \models \forall n \exists X^\mu \varphi (n,X^\mu )$ .

For each n, let $e_n$ be least such that $M_F \models \varphi (n,S_{\mu ,F,e_n})$ . For each e, $H_{\mu +\omega ,F}$ computes whether $M_F \models \varphi (n,S_{\mu ,F,e})$ , so $\langle S_{\mu ,F,e_n} \rangle _n$ is a $\Sigma ^1_1\text {-}\mathsf {AC}_0$ solution which lies in $M_F$ .

We are ready to prove that $M_\infty $ satisfies $\Delta ^1_1$ - $\mathsf {CA}_0$ . The overall strategy follows [16, Lemma 12], with modifications in order to account for (C2).

3.7 Proof that $M_\infty $ satisfies $\Delta ^1_1$ -comprehension

Suppose $\varphi (n)$ and $\psi (n)$ are $\Sigma $ -over- $\mathcal {L}_F$ with only n free and $M_\infty \models \forall n(\psi (n) \leftrightarrow \neg \varphi (n))$ . We want to define $D \in M_\infty $ such that

$$\begin{align*}M_\infty \models \forall n(\psi(n) \leftrightarrow n \in D). \end{align*}$$

A naive attempt is to consider

$$\begin{align*}\{n \in \mathbb{N}: \exists q \in G(q \Vdash \psi(\mathbf{n}))\}, \end{align*}$$

but there are two obstacles preventing us from showing that the above set lies in $M_\infty $ :

1. – $M_\infty $ does not contain G so we cannot search over $q \in G$ ;

2. – deciding whether $q \Vdash \psi (\mathbf {n})$ is not hyperarithmetic in general.

The second obstacle is overcome using Lemma 3.13. To overcome the first obstacle, we shall use retagging to change the scope of our search to a class of conditions which look like they might lie in G, based on information from $T^G$ and finitely many branches $f^G_i$ (all of which lie in $M_\infty $ , unlike G).

Recall that (by genericity) $h^G$ is the well-founded rank function of $T^G$ . We say that a rank function $h: T \to \omega _1^{CK} \cup \{\infty \}$ is $\nu $ -good if $T \subset T^G$ and for all $\sigma \in T$ :

$$ \begin{align*} h^G(\sigma) < \nu \quad &\Rightarrow \quad h(\sigma) = h^G(\sigma), \\ h^G(\sigma) \geq \nu \quad &\Leftrightarrow \quad h(\sigma) \geq \nu. \end{align*} $$

We will use the notion of $\nu $ -goodness in our definition of D.

We begin the proof by fixing $p \in G$ such that $p \Vdash \forall n(\psi (n) \leftrightarrow \neg \varphi (n))$ . By genericity and by expanding F if necessary, we may assume that $F = \mathrm {dom}(f^p)$ . Since $p \Vdash \forall n(\psi (n) \lor \varphi (n))$ (which is $\Sigma $ -over- $\mathcal {L}_F$ ), by Lemma 3.13, we may fix $\mu < \omega _1^{CK}$ large enough such that $p \Vdash \forall n(\psi ^\mu (n) \lor \varphi ^\mu (n))$ and $\mu $ is greater than the rank of any constant in $\varphi $ and $\psi $ . Since $p \Vdash \forall n(\neg \psi (n) \lor \neg \varphi (n))$ and $\mu $ is large enough, we also have $p \Vdash \forall n(\neg \psi ^\mu (n) \lor \neg \varphi ^\mu (n))$ and so $p \Vdash \forall n(\psi ^\mu (n) \leftrightarrow \varphi ^\mu (n))$ .

Next, fix $\nu < \omega _1^{CK}$ large enough such that $p \in \mathbb {P}_\nu $ and $\mathrm {rk}(\varphi ^\mu (\mathbf {d}) \lor \psi ^\mu (\mathbf {d})) < \nu $ for all $d \in \mathbb {N}$ . Now we define $D \subseteq \mathbb {N}$ as follows: $d \in D$ if and only if there is some $q \in \mathbb {P}_{\omega \nu +\omega ^2+\omega }$ extending p such that:

1. (1) $q \Vdash \psi ^\mu (\mathbf {d})$ ;

2. (2) $T^q \subset T^G$ (this is implied by (3) but we include it for emphasis);

3. (3) $h^q$ is $(\omega \nu +\omega ^2+\omega )$ -good;

4. (4) $\forall i \in F(f^q(i) = f^G_i \cap T^q)$ .

Note that if $q \in G$ , then q satisfies (2) (by (E1)) and (4) (by (E2)). (3) must hold as well, but even more is true: we must have $h^q \subseteq h^G$ . Since $M_\infty $ does not have access to $h^G$ , we make do with sufficiently good approximations. Observe that D is hyperarithmetic in $T^G \oplus \langle f^G_i \rangle _{i \in F}$ , so $D \in M_F \subseteq M_\infty $ . It remains to show that $M_\infty \models \psi (\mathbf {d})$ if and only if $d \in D$ .

The forward direction is straightforward: If $M_\infty \models \psi (\mathbf {d})$ , then $M_\infty \models \psi ^\mu (\mathbf {d})$ , so we may fix $q' \in G$ extending p which forces $\psi ^\mu (\mathbf {d})$ . Using $q'$ , we define q as we did in the proof of Lemma 3.4: $T^q = T^{q'}$ , $f^q = f^{q'}$ , and

$$\begin{align*}h^q(\sigma) = \begin{cases} \infty, & \text{if }h^{q'}(\sigma) \geq \omega\nu+\omega^2+\omega, \\ h^{q'}(\sigma), & \text{otherwise}. \end{cases} \end{align*}$$

It is straightforward to check that q witnesses $d \in D$ (using Lemma 3.11 to show that $q \Vdash \psi ^\mu (\textbf {d})$ ).

To prove the backward direction, suppose $M_\infty \models \varphi (\mathbf {d})$ . Then $M_\infty \models \varphi ^\mu (\mathbf {d})$ , so we may fix $r \in G$ extending p which forces $\varphi ^\mu (\mathbf {d})$ . Suppose towards a contradiction that $d \in D$ , as witnessed by some q. Our plan is to construct conditions $q^\ast $ extending q, $r^\ast $ extending r, and $s^\ast $ extending p, as illustrated in Figure 3.

We begin by using an automorphism of $\mathbb {P}$ to avoid conflict between $f^q$ and $f^r$ . Consider a permutation $\pi $ which fixes F and moves $\mathrm {dom}(f^q)\backslash F$ to some set which is disjoint with $\mathrm {dom}(f^r)$ . Since $F = \mathrm {dom}(f^p)$ , we have $\hat {\pi } q \leq p$ . It is straightforward to check that $\hat {\pi } q$ witnesses $d \in D$ . Therefore, by replacing q with $\hat {\pi } q$ , we may assume that $\mathrm {dom}(f^q)\backslash F$ and $\mathrm {dom}(f^r)$ are disjoint. It follows that $\mathrm {dom}(f^q) \cap \mathrm {dom}(f^r) = F$ .

Next, fix finite partial functions $g_{q^\ast }$ and $g_{r^\ast }$ on $\mathbb {N}$ such that:

1. – $\mathrm {dom}(g_{q^\ast })$ , $\mathrm {dom}(g_{r^\ast })$ , and $\mathrm {dom}(f^q) \cup \mathrm {dom}(f^r)$ are pairwise disjoint;

2. – $\mathrm {range}(g_{q^\ast }) = \{\sigma \in T^r\backslash T^q: |\sigma | = 1\}$ ;

3. – $\mathrm {range}(g_{r^\ast }) = \{\sigma \in T^q\backslash T^r: |\sigma | = 1\}$ .

We may now define $q^\ast $ as follows:

1. – $T^{q^\ast } = T^q \cup T^r$ ;

2. – $f^{q^\ast }(i) = \begin {cases} f^q(i), & \text {if }i \in \mathrm {dom}(f^q)\backslash F; \\ f^G_i \cap T^{q^\ast }, & \text {if }i \in F; \\ g_{q^\ast }(i), &\text {if }i \in \mathrm {dom}(g_{q^\ast }); \end {cases}$

3. – $h^{q^\ast }$ is defined by cases:

   $$\begin{align*}h^{q^\ast}(\tau) = \begin{cases} h^q(\tau), & \text{if }\tau \in T^q, \\ \infty, & \text{if }\exists i(\tau \subseteq f^{q^\ast}(i)), \\ h^r(\tau), & \text{if }h^r(\tau) < \omega\nu+\omega^2, \\ \omega\nu+\omega^2+|\tau|_Q, &\text{otherwise}, \end{cases} \end{align*}$$
   where $Q = \{\sigma \in T^r\backslash T^q: h^r(\sigma ) \geq \omega \nu +\omega ^2\}$ .
   

   Figure 3 p and r lie in G, while q “looks like” it lies in G.

Next, define $r^\ast $ as follows:

1. – $T^{r^\ast } = T^q \cup T^r$ ;

2. – $f^{r^\ast }(i) = \begin {cases} f^r(i), & \text {if }i \in \mathrm {dom}(f^r)\backslash F; \\ f^G_i \cap T^{r^\ast }, & \text {if }i \in F; \\ g_{r^\ast }(i), & \text {if } i \in \mathrm {dom}(g_{r^\ast }); \end {cases}$

3. – $h^{r^\ast } = h^G\restriction T^{r^\ast }$ .

Finally, define $s^\ast $ as follows:

1. – $T^{s^\ast } = T^q \cup T^r$ ;

2. – $f^{s^\ast } = f^{q^\ast } \cup f^{r^\ast }$ ;

3. – $h^{s^\ast }(\sigma ) = \begin {cases} h^G(\sigma ), &\text {if } h^G(\sigma ) < \omega \nu + \omega ^2; \\ \infty , &\text {otherwise.} \end {cases}$

One may verify that:

1. – $q^\ast $ is a condition which extends q;

2. – $r^\ast $ is a condition which extends r;

3. – $s^\ast $ is a condition which extends p;

4. – $q^\ast $ and $s^\ast $ are $(\omega \nu +\omega ^2)$ - $\mathrm {dom}(f^{q^\ast })$ -retaggings;

5. – $r^\ast $ and $s^\ast $ are $(\omega \nu +\omega ^2)$ - $\mathrm {dom}(f^{r^\ast })$ -retaggings.

The verification is the same as that for ordinary Steel forcing, with minor additions in order to handle the new paths contributed by $g_{q^\ast }$ and $g_{r^\ast }$ . We only discuss $g_{q^\ast }$ here; the argument for $g_{r^\ast }$ is similar. First, note that $g_{q^\ast }$ was defined to ensure that $q^\ast $ satisfies (C2). Second, (E3) is satisfied for $q^\ast \leq q$ because for each $i \in \mathrm {dom}(f^{q^\ast })\backslash \mathrm {dom}(f^q)$ ( $=\mathrm {dom}(g_{q^\ast })$ ), we have $g_{q^\ast } \notin T^q$ and $|g_{q^\ast }(i)| = 1$ .

With the above facts, we may complete the proof that if $M_\infty \models \varphi (\textbf {d})$ , then $d \notin D$ . Since $q^\ast \leq q$ and $q \Vdash \psi ^\mu (\textbf {d})$ , we have $q^\ast \Vdash \psi ^\mu (\textbf {d})$ . It follows from Lemma 3.11 that $s^\ast \Vdash \psi ^\mu (\textbf {d})$ (note $F \subseteq \mathrm {dom}(f^{q^\ast })$ so $\psi ^\mu (\mathbf {d})$ is $\Sigma $ -over- $\mathcal {L}_{\mathrm {dom}(f^{q^\ast })}$ ). Similarly, we have $s^\ast \Vdash \varphi ^\mu (\mathbf {d})$ . But since $M_\infty \models \forall n(\psi ^\mu (n) \to \psi (n))$ and $M_\infty \models \forall n(\varphi ^\mu (n) \to \varphi (n))$ , it follows that $s^\ast \Vdash \psi (\mathbf {d}) \land \varphi (\mathbf {d})$ . This contradicts the fact that $p \Vdash \forall n(\psi (n) \leftrightarrow \neg \varphi (n))$ and $s^\ast \leq p$ .

This completes the proof that $M_\infty $ satisfies $\Delta ^1_1$ - $\mathsf {CA}_0$ .

Remark 3.15. We are unable to extend our proof to show that $M_\infty $ satisfies the $\Pi ^1_1$ -separation principle (studied by Montalbán [10]). We do not know if $\Pi ^1_1$ -separation implies finite choice. If there is a model of $\Pi ^1_1$ -separation and $\mathsf {I}\Sigma ^1_1$ (such as an $\omega $ -model of $\Pi ^1_1$ -separation) which does not satisfy finite choice, then we would obtain a negative answer to [2, Question 1.11] (using Theorem 1.5).

Acknowledgments

We thank Richard A. Shore, Antonio Montalbán, Linda Brown Westrick, James Barnes, and the anonymous referees for many useful discussions and suggestions.