Pathologies in satisfaction classes

We study subsets of countable recursively saturated models of $\mathsf{PA}$ which can be defined using pathologies in satisfaction classes. More precisely, we characterize those subsets $X$ such that there is a satisfaction class $S$ where $S$ behaves correctly on an idempotent disjunction of length $c$ if and only if $c \in X$. We generalize this result to characterize several types of pathologies including double negations, blocks of extraneous quantifiers, and binary disjunctions and conjunctions. We find a surprising relationship between the cuts which can be defined in this way and arithmetic saturation: namely, a countable nonstandard model is arithmetically saturated if and only if every cut can be the"idempotent disjunctively correct cut"in some satisfaction class. We describe the relationship between types of pathologies and the closure properties of the cuts defined by these pathologies.


Introduction
Kotlarski-Krajewski-Lachlan famously showed [7] that every countable, recursively saturated model of PA has a full satisfaction class.Enayat-Visser [3] strengthened this result using more typically model-theoretic tools.These results show the conservativity of the theory CT − of compositional truth over the base arithmetic theory PA.Both proofs illustrate the weakness of CT − : not only the theory is a conservative extension of the base theory PA, but also it is consistent with failure of some very basic truth principles such as disjunctive correctness (DC): "A disjunction is true if and only if it has a true disjunct".In particular one can construct models of CT − in which for a nonstandard number a the disjunction 0 = 0 ∨ 0 = 0 ∨ . . .∨ 0 = 0 a times is within the scope of the truth predicate.Thus it is well known how to construct pathological satisfaction classes.
One can easily exclude such pathological behaviour by adding to the theory CT − induction axioms for the extended language.It is well known that the theory CT of an inductive truth predicate is not conservative over PA; indeed, CT proves the Global Reflection Principle for PA, that is the statement (GRP) ∀φ Prov PA (φ) → T (φ) .
In fact, CT 0 , the theory CT − augmented by ∆ 0 -induction for formulas in the language including the truth predicate, is equivalent to (GRP).
Recent work by Enayat and Pakhomov [2] pointed to a deeper connection between non-conservativity and disjunctive correctness.The natural-looking extension of CT − with DC turns out to be equivalent to CT 0 .Ali Enayat (unpublished) separated DC into two principles: DC-out, stating that every true disjunction has a true disjunct, and DC-in, stating that a disjunction with a true disjunct is true.Cieśliński, Le lyk, and Wcis lo [1] show that already CT − + DC-out is equivalent to CT 0 , while CT − + DC-in is conservative over PA.Conservativity of DC-in is shown by proving that every countable model of PA has an elementary extension which is "disjunctively trivial": that is, one in which every disjunction of nonstandard length is evaluated as true.In such disjunctively trivial models of CT − , ω is definable as the cut for which the truth predicate T is "disjunctively correct." In this article, we aim at deepening our understanding of the phenomenon of disjunctive correctness: we consider related questions around which sets can be definable by exploiting pathologies in the satisfaction class.We analyze "local pathologies", along the lines of repeated (idempotent) disjunctions of a single, fixed sentence θ, and non-local pathologies, where, for example, we consider idempotent disjunctions of all sentences.We completely classify the subsets of a model which are definable using local pathologies, and use this to conclude that a countable model of PA is arithmetically saturated if and only if it carries a satisfaction class which makes all disjunctions of nonstandard length true.We also classify the cuts in a model which can be definable using non-local pathologies.
From the definability perspective, our work complements that of [10], where it was shown that for every subset A of a countable recursively saturated model M there is a satisfaction class S such that A is definable in (M, S) as (roughly speaking) the set of those numbers x such that quantifier correctness fails on the x-th formula (in a suitably chosen enumeration).We go in the reverse direction: starting from an idempotent sentential operation F we ask when a set A can be characterized as the set of those numbers x for which the satisfaction class behaves correctly when F is iterated x-times.Unlike in the case of [10] it turns out that in some countable recursively saturated models, not every cut can be defined in this way.
We conclude the paper with several properties about the full disjunctively correct cut.
1.1.Preliminaries.We formulate PA in the usual language L PA = {+, ×, <, 0, 1}.We use script letters M, N , etc to denote models of PA and Roman letters M, N, etc to denote their universes.ElDiag(M) denotes the elementary diagram of the model M. We follow standard definitions and conventions used in the study of models of PA: see [6, Chapter 1].We recall some of these conventions here.
We fix standard coding for finite sets and sequences: for a model M |= PA, a, b ∈ M , • len(a) denotes the length of the sequence coded by a, • (a) b denotes the b-th element of the sequence coded by a, and • we write a ∈ b if a is in the set coded by b.
Definition 1.A model M |= PA is arithmetically saturated iff for every a ∈ M for every type p(x, a) which is arithmetically definable in the type of a, p(x, a) is realized in M.
We note for the reader the equivalence between countable recursively saturated models and countable resplendent models, as well as the equivalence between arithmetically saturated models and recursively saturated models in which ω is a strong cut.The interested reader is again directed to [6] for definitions and other references.
Let M |= PA.By Form M and Sent M we refer to the (definable) sets of (Gödel codes of) formulas and sentences, respectively, in the sense of M. For the rest of this article, we will not distinguish between a formula φ and its Gödel code ⌈φ⌉.We use the following standard abbreviations: • Asn(x, y) is an L PA formula which asserts that y is an assignment for x, which means that it assigns values to all and only those variables which have free occurrences in x (x can be a term or a formula).• s α denotes the value of the term s under the assignment α.
• ∃ denotes the arithmetical operation which given a variable v and a formula φ returns ∃vφ.∨, ¬ and = have analogous meanings.• for any two assignments α, β, we write β ∼ v α iff β differs from α at most on a variable v and the domain of β extends the domain of α at most with v. • for φ ∈ Form L PA , β↾ φ denotes the restriction of β to the variables which have free occurrences in φ.
By convention, when (M, S, X) If S is an X-satsfiaction class on M and A ⊆ M , then by S↾ A we denote the set S ∩ (A × M ).In particular S↾ A is (under the above assumptions) an A ∩ Xsatsifaction class on M. CS − extends CS − ↾ X with an axiom saying that X is the whole universe.When talking about CS − we omit the reference to X.By definition we say that S is a satisfaction class The above axioms of of CS − are weak: in particular, it is known that they are not sufficient to guarantee the correct interaction between quantifiers and term substitutions.For example, it is an exercise to use an Enayat-Visser construction to show that in every countable and recursively saturated model M there is a satisfaction class S such that for some formula φ and assignment α, (∃vφ, α) ∈ S but for no closed term t, φ[t/v], α ∈ S (φ[t/v] denotes the substitution of a closed term t for all occurrences of the variable v.) Because of these and similar problems, it is not known whether in an arbitrary model of (M, S) |= CS − one can define a compositional truth predicate T for the language of arithmetic satisfying the natural axiom where x denotes the canonical numeral naming x.It is known that each standard definition of truth from satisfation (e.g."being satisfied by all assignments" or "being satisfied by an empty assignment") might fail to define a truth predicate in a model of CS − .
To overcome these problems it is customary to extend the above list of axioms of CS − with the regularity axiom (compare [10]).Its full-blown definition is rather involved and we will give it in the Appendix.A satisfaction class which satisfies the regularity axiom is called a regular satisfaction class.Importantly, if S is a regular satisfaction class in M, then terms with the same values can be substituted for free variables in a formula salva veritate, i.e. for every formula φ ∈ Form M , every variable v ∈ Var M , all terms s, t ∈ Term M and all assignments α it holds in (M, S) that One can check, that if S is a regular satisfaction class in M, then the formula Sent(x)∧S(x, ∅) defines in (M, S) a truth predicate which satisfy the above natural axiom for the universal quantifier.
In the Appendix we show how to improve one of our constructions in order to obtain regular satisfaction classes.As a consequence we will be able to construct many pathological truth classes.However, we decided to leave the regularization of all our constructions for further research.
Another basic property of satisfaction classes is satisfying internal induction.Before introducing it let us define one handy abbreviation: if (M, S) |= CS − , and ψ is a formula in the sense of M with exactly one free variable, then T * ψ(x) denotes a L PA ∪ {S}-formula with one free variable x which naturaly expresses "The result of substituting the numeral naming x for the unique free variable of ψ is satisfied by the empty assignment"(see [8], Lemma 3.6.)We say that in (M, S) |= CS − , S satisfies the internal induction iff for every ψ ∈ Form M with a unique free variable, the formula T * ψ(x) satisfies the induction axiom, i.e.
We conjecture that all our constructions can be fine-tuned to yield regular satisfaction classes satisfying internal induction, however we decided to treat this problem on a different occasion.
Remark 3. As shown in [8], Lemma 3.7, if (M, S) |= CS − , S is regular and satisfies internal induction, then for every ψ ∈ Form M with exactly one free variable, if satisfies the full induction schema in the language L PA ∪{X}, where X is interpreted as X ψ .
Suppose φ i : i ≤ c is a coded sequence of elements of Sent M and suppose θ ∈ Sent M .
• i≤c φ i is defined, inductively, so that i≤0 φ i = φ 0 , and Given an M-definable function F : Sent M → Sent M , we define F c (x) by induction on c as follows: , where F ∨ (φ) = φ ∨ φ.These are "binary idempotent disjunctions."Similarly, one can define "binary idempotent conjunctions." , where F ∀ (φ) = ∀xφ.In a model (M, S) |= CS − , we can define the following sets: • for a given θ, the "idempotent disjunctively correct set for θ", • the "idempotent disjunctively correct set": • the "disjunctively correct set": We can similarly define the "conjunctively correct set" for a given θ, the "quantifier correct set" for a given θ (QC θ S ), the "double negations correct set" for a given θ (DNC θ S ), the "binary idempotent disjunctively/conjunctively correct set" (IDC bin,θ S ), or their respective non-local versions (QC S , DNC S , IDC bin S ).Given a set X (often one of the above pathologically definable sets), we introduce the following notation for the longest initial segment of X: This allows us to denote, for example, the idempotent disjunctively correct cut, I(IDC S ).

Separability
In this part, we classify which sets can be IDC 0=1 S for some S. Rather than simply looking at disjunctions, however, we generalize the setting to draw similar conclusions about the conjunctively correct set for 0 = 0, the double negations correct set for any atomic sentence φ, or the binary idempotent disjunctively / conjunctively correct set for φ and much more.
(1) If x, y ∈ Form M , we say x ⊳ y if x is an immediate subformula of y. ( (5) X is finitely generated if there is a finite F ⊆ X that generates it.
We describe a generalization of the idempotent disjunction operation c → i<c θ.
Definition 6. Fix a standard sentence θ.Let Φ(p, q) be a (finite) propositional template, over propositional variables p and q.By this we mean that in Φ we allow all propositional connectives, along with quantifiers (over dummy variables).We insist that Φ has non-zero complexity (that is, Φ(p, q) has at least one propositional connective or quantifier), along with the following properties: • q appears in Φ(p, q), • if M |= θ, then Φ(⊤, q) is equivalent to q, and • if M |= ¬θ, then Φ(⊥, q) is equivalent to q.
Note that for any n ∈ ω and atomic sentence θ, if F is a local idempotent sentential operator for θ and (M, S) |= CS − , then (M, S) |= T (θ) ≡ T (F (n)).In fact, (M, S) |= T (F (x)) ≡ T (F (x + n)), for each x ∈ M .This approach allows us to generalize several examples of local pathologies, for example: can all appear as {F (c) : c ∈ M } for various θ and Φ.We study the question of when, given such a function F , a set X can be the set {x : T (F (x)) ≡ T (θ)} in a model (M, S) |= CS − .We will see that such sets X will require the following property.
Definition 7. Let M |= PA, and A ⊆ D ⊆ M .A is separable from D if for each a such that for every n ∈ ω, (a) n ∈ D, there is c such that for each n ∈ ω, (a) n ∈ A if and only if M |= n ∈ c.We say a set X is separable if it is separable from M .
In Propositions 8, 9, and 10, we refer to definable sets and functions.Here we insist that these are definable in the arithmetic structure of M: that is, they are definable (possibly using parameters) using formulas from L PA .First we notice some basic properties of separability.
Proof.Fix d ∈ D 1 \ A. Assume A is separable from D 1 and fix any a such that for every n, (a) n ∈ D 2 .Let b be defined by Then for every i ∈ ω, (b) i ∈ D 1 , so there is c such that for every i ∈ ω, (b) i ∈ A iff i ∈ c.Then it follows that also for every i ∈ ω, (a Proof.Easy exercise.As stated before, given θ, a local idempotent sentential operator F for θ, and D = {F (x) : x ∈ M }, we wish to classify the subsets A ⊆ D which can be the sets of true sentences in D (equivalently, we wish to classify the sets X such that {F (x) : x ∈ X} is the set of true sentences in D).First we need the following Lemma.
Lemma 11.Let M |= PA.Let θ be an atomic sentence and F a local idempotent sentential operator for θ.Let J 0 , J 1 ⊆ M be closed under predecessors, disjoint, and Then there is a unique X-satisfaction class S such that for each i and x ∈ J i , (F (x), ∅) ∈ S if and only if i = 0.
Proof.Let S 0 = {(F (x), ∅) : x ∈ J 0 }.We extend S 0 to an X-satisfaction class S. Take any φ ∈ X.Then, since J i are closed under predecessors and disjoint, then there is a unique i and minimal x such that φ ∈ Cl(F (x)) and x ∈ J i .Recall that F (x) = Φ(θ, F (x − 1)), and θ is atomic.One notices that the subformulas of Φ(θ, q) must be equivalent to one of q, ¬q, ⊤, or ⊥.Let Ψ(p, q) be the subformula of Φ(p, q) such that Ψ(θ, F (x − 1)) = φ.Again, the presentation of φ as Ψ(θ, F (x − 1)) is unique by induction in M. We put φ, ∅ ∈ S if any of the following hold: • Ψ(θ, q) is equivalent to q and i = 0, • Ψ(θ, q) is equivalent to ¬q and i = 1, or • Ψ(θ, q) is equivalent to ⊤.
One checks that S is an X-satisfaction class.
Theorems 12 and 13 are generalizations of unpublished work by Jim Schmerl 1 .
Theorem 12. Let M |= PA be countable and recursively saturated.Let θ be an atomic sentence and F a local idempotent sentential operator for θ.Let X ⊆ M be separable, closed under successors and predecessors, and for each n ∈ ω, n ∈ X if and only if M |= θ.Then M has an expansion (M, S) Notice that X is separable if and only if M \ X is separable.This means that there is some flexibility in building such satisfaction classes S.
Proof.Let D = {F (x) : x ∈ M } and A = {F (x) : x ∈ X}.Note that A is separable from D. We build sequences F 0 ⊆ F 1 ⊆ . . .and S 0 , S 1 , . . .such that: • each F i is a finitely generated set of formulas such that ∪F i = Form M , • each S i is a full satisfaction class and (M, S i ) is recursively saturated,

and
• for each φ ∈ D ∩ F i , (φ, ∅) ∈ S i if and only if φ ∈ A. Given such a sequence, S = ∪S i ↾ F i would be the required full satisfaction class on M.
Externally fix an enumeration of Form M in order type ω.We can assume, without loss of generality, that θ appears first in this enumeration.Suppose F i and S i have been constructed.Let F i+1 be generated by F i and the least x ∈ Form M \F i in the aforementioned enumeration.Let F ′ = F i ∪ (F i+1 ∩ D).Let a be a sequence such that {F ((a) n ) : n ∈ ω} = F i+1 ∩ D. Note that this is possible since F i+1 is finitely generated.Let c be as in the definition of separability for a.
Since X is closed under successors and predecessors, then if (a) n and (a) m are in the same Z-gap (that is, there is some k ∈ ω such that (a) n and (a) m differ by k), then (a) n ∈ X if and only if (a) m ∈ X.Since X is separable, this means that, if (a) n and (a) m are in the same Z-gap, then n ∈ c if and only if m ∈ c.Let J 0 be the closure under successors and predecessors of {(a) n : n ∈ ω, n ∈ c, and (a) n > ω}, and J 1 be the closure under successors and predecessors of {(a) n : n ∈ ω, n / ∈ c, and (a) n > ω}.By Lemma 11, there is a Cl(F i+1 ∩ D)-satisfaction class S ′ such that for each φ We extend S ′ to a Cl(F ′ ) satisfaction class simply by preserving S i on F i .One notices that if φ ∈ F i ∩ D, then by induction φ, ∅ ∈ S i if and only if φ ∈ A.
Then S ′ is a Cl(F ′ )-satisfaction class, so by [3, Lemma 3.1], M has an elementary extension N carrying a Form M -satisfaction class S agreeing with S ′ on Cl(F ′ ).In particular, this shows the consistency of the recursive theory T h consisting of the following: • S is a full satisfaction class, ) is recursively saturated, by resplendency (M, S i ) has an expansion to T h, and such an expansion is a full satisfaction class agreeing with S ′ on formulas 1 Private communication to Ali Enayat.
from Cl(F ′ ).Recall that countable recursively saturated models are chronically resplendent ([6, Theorem 1.9.3]): by this we mean that such expansions can, themselves, be taken to be resplendent.That is, we can assume that (M, S i , S) is recursively saturated.Let S i+1 = S and continue.
In the above result, notice that if n ∈ ω, then clearly M |= F (n) if and only if M |= θ.Therefore, ω ⊆ X if and only if M |= θ, and ω ∩ X = ∅ if and only if M |= ¬θ.Moreover, if X = {x : (M, S) |= T (F (x))} then X is necessarily closed under successors and predecessors.The next result shows that separability of X is also necessary.
Proof.Let a ∈ M be such that for each n ∈ ω, (a) n ∈ D. We show that that there is a c so that for all i ∈ ω, (a By a result of Stuart Smith, [9, Theorem 2.19], (M, S) is definably S-saturated.This means that for any coded sequence Let φ j (x) be the formula given by (a) j ≡ (j ∈ x).That is, since (a) j is the code of a sentence, φ j (m) is evaluated as true in a satisfaction class S if the sentence (a) j is evaluated as true and j ∈ m, or (a) j is evaluated as false and j ∈ m.
Let i ∈ ω, and let m ∈ M be such that for all j < i, (a) j ∈ A if and only if j ∈ m.Then, Therefore there is m such that for all i ∈ ω, (M, S) |= T (φ i (m)).In particular, for each n ∈ ω, if (a) n ∈ A, then T ((a) n ) and therefore n ∈ m.Moreover, if n ∈ m, then (M, S) |= ¬T ((a) n ).By assumption this means (a) n ∈ A.

Separable Cuts
In this section, we wish to examine the results of the previous section in case where we have I ⊆ end M a cut.We examine some properties of separable cuts.We conclude this section by showing that a countable model is arithmetically saturated if and only if it has a disjunctively trivial expansion to a model of CS − .Proposition 14.Let M |= PA be nonstandard and I ⊆ end M .The following are equivalent.
(2) There is no a ∈ M such that Compare (3) to the notion of strength: a cut I ⊆ end M is strong if for each a there is c > I such that whenever i ∈ I, (a) n ∈ I if and only if (a) n < c.Clearly, condition (3) is equivalent to strength if I = ω.
We show (1) =⇒ (3): Suppose I is separable and let a ∈ M .We show that there is c ∈ M such that for each n ∈ ω, (a) n ∈ I if and only if (a) n < c.Since I is separable, there is c such that for each n ∈ ω, (a) n ∈ I if and only if n ∈ c.Consider the type This type is finitely satisfiable, so (by restricted saturation of nonstandard models, see [6,Corollary 1.11.4])there is c ′ which satisfies p(x).Now we show (3) =⇒ (1).Let a ∈ M .There is c such that (a) n ∈ I if and only if (a) n < c.Consider the type This type is finitely satisfiable and therefore satisfied by some c ′ ∈ M .Such a c ′ witnesses separability of I.
By Theorem 12, Theorem 13, and Proposition 14, then I is separable if and only if there is S such that (M, S) |= CS − and ))} is also separable?Our next result shows that it is not always the case: if I ⊆ end M has no least Z-gap above it, then there is S such that (M, S) |= CS − and I(IDC 0=1 S ) = I.Later, in Corollary 19, we see that if M is not arithmetically saturated, then such an I need not be separable.
Proposition 15.Let M |= PA be countable and recursively saturated.Suppose I ⊆ end M has no least Z-gap.Then there is S such that (M, S) |= CS − and Proof.First notice that for any c < d in different Z-gaps, for any a ∈ M , there is b such that c < b < d and b ∈ {(a) i : i ∈ ω}.To see this, notice that if a, c, and d are as above, by recursive saturation the type Now we show how to construct the required satisfaction class.Fix a sequence d 0 > d 1 > . . .such that d i+1 is not in the same Z-gap as d i and inf({d i : i ∈ ω}) = I.We proceed similarly to Theorem 12: we build sequences b 0 > b 1 > . .., F 0 ⊆ F 1 ⊆ . . .and S 0 , S 1 , . . .such that: • and Given such a sequence, let S = ∪(S i ↾ F i ).Then S is the required full satisfaction class.To see this, suppose We proceed to the construction.Suppose F i and S i has been constructed satisfying the above.Since F i is finitely generated, there is a coding the lengths of disjunctions of (0 = 1) in F i .By recursive saturation, there is b i such that , and the first formula φ ∈ F i in some externally fixed enumeration of M has an elementary extension N carrying a Form M -satisfaction class S agreeing with S ′ on F ′ .Therefore, the theory T h asserting the following is consistent: • S is a full satisfaction class, • S agrees with S i on formulas from F i , • {S( bi−n (0 = 1), ∅) : n ∈ ω}, and By resplendency, M has a full satisfaction class S satisfying T h; by chronic resplendency, we can assume (M, S) is recursively saturated.Let S i+1 = S and continue.
To find some examples of separable cuts, we recall some definitions from [5].Below, we let Def 0 (a) be the set of elements of M which are ∆ 0 -definable from a in M. Definition 16 ( [5]).Let M |= PA and let I ⊆ end M .
(1) I is coded by ω from below if there is a ∈ M such that (1) I is ω-coded and separable.
(1) =⇒ ( 2): Suppose I is ω-coded, and let a be such that sup({(a) i : i ∈ ω}) = I (the case in which I is coded by ω from above is similar).Suppose also that b ∈ M is such that Def 0 (b) \ I is coinitial in M \ I. Then the following type is realized in M : where t n : n ∈ ω is a recursive enumeration of all ∆ 0 -definable Skolem functions.If c realizes this type, then sup({(c (2) =⇒ ( 1): [5, Proposition 6.2] implies that if I is 0-superrational, then I is ω-coded.To see separability, notice that by 0-superrationality, if Def 0 (a) ∩ I is cofinal in I, then Def 0 (a) \ I is not coinitial in M \ I (and vice versa).
[5, Theorem 6.5] states that ω is a strong cut if and only if every ω-coded cut is 0-superrational.Taken together with the above result, we see that if ω is not strong, then separable cuts are never ω-coded.
Proposition 18.For any M |= PA: (1) If ω is a strong cut, then every cut I which is ω-coded is separable.
(2) If ω is not a strong cut, then every cut I which is ω-coded is not separable. Proof.
(1) is due to [5, Theorem 6.5 (a) =⇒ (c)].We show (2).Suppose ω is not strong.There is a such that inf({(a If I ⊆ end M is a cut which is ω-coded from above, then there is c > I such that I = inf({(c) n : n ∈ ω}).For simplicity assume that the sequence coded by c is a strictly decreasing and its domain consists of all elements smaller than a nonstandard element d.Let b code the sequence defined by (b) i = (c) (a)i .We claim that b witnesses the failure of separability of I.
Indeed, (c) (a)i ∈ I if and only if (c) (a)i < (c) n for each standard n if and only if (a) i > ω.Since the set {(a) i : i ∈ ω} \ ω is coinitial with ω, then {(c) (a)i : i ∈ ω} ∩ I is cofinal with I. Indeed, for any x ∈ I there is a nonstandard number y < d such that x < (c) y ∈ I.However, by the proerties of a there is also a standard number i ∈ ω such that ω < (a) i < y.Since c is strictly decreasing, it follows that for any such The case when I is upward ω-coded is treated similarly.
Corollary 19.Suppose M |= PA is countable, recursively saturated but not arithmetically saturated.Then there are separable sets X such that I(X) is not separable.
Proof.Let c be nonstandard, and I = sup({c + n : n ∈ ω}).Then I has no least Z-gap above it, and so by Proposition 15, there is S such that (M, S) |= CS − and I = I(IDC 0=1 S ).Let X = IDC 0=1 S .Then X is separable by Theorem 13 and I = I(X).Since I is ω-coded, by Proposition 18 (2), if ω is not a strong cut, then I cannot be separable.
Separable cuts always exist in recursively saturated models.We can in fact see more: every recursively saturated model M has a separable cut I ⊆ end M which is not closed under addition.Moreover, M has separable cuts I ⊆ end M that are closed under addition but not multiplication, and ones closed under multiplication but not exponentiation.
To see this, first notice that if (M, I) is recursively saturated and I ⊆ end M , then I is separable.This follows directly from the equivalent definition of separability that says that for each a there is d such that for all i ∈ ω, (a) i ∈ I iff (a) i < d.Now let I ⊆ end M be any cut not closed under addition.By resplendence, there is J ⊆ end M such that (M, J) is recursively saturated and not closed under addition.
Again, notice that this proof generalizes to show that if f and g are increasing definable functions such that there is any cut I ⊆ end M closed under f but not g, then there is I ⊆ end M that is separable and closed under f but not g.Hence there are separable cuts which are closed under addition but not multiplication, and cuts which are closed under multiplication but not exponentiation.
3.1.Arithmetic Saturation.In [1, Lemma 26], we see that there exist disjunctively trivial models: models (M, T ) |= CT − such that for all sequences φ i : i < c of sentences such that c > ω, (M, T ) |= T ( i<c φ i ).That is, models such that all disjunctions of nonstandard length are evaluated as true.In this part we see that disjunctive triviality implies arithmetic saturation.We construct sequences F 0 ⊆ F 1 . . . of finitely generated sets of formulas such that ∪F i = Form M and full satisfaction classes S 0 , S 1 , . ... Suppose S i is a full satisfaction class such that (M, S i ) is recursively saturated and if φ ∈ F i ∩ Sent M is disjunction of nonstandard length, then S i (φ, ∅).
Let a code the lengths of all disjunctions in F i+1 .That is, suppose (b) n is the n-th element of F i+1 , and (a) n is the maximum c such that there is a sequence φ j : j < c such that (b) n = j<c φ j .Since ω is strong, there is d > ω such that for each n ∈ ω, (a) n ∈ ω if and only if (a) n < d.By [1, Lemma 26], the theory T h asserting the following is consistent: • ElDiag(M), Since T h is a consistent, recursive theory and (M, S i ) is recursively saturated, by resplendence, (M, S i ) has an expansion (M, S i , S i+1 ) |= T h.Continue as before, obtaining S = ∪S i ↾ F i , a full satisfaction class which is disjunctively trivial.
We observe that, for each n, there is an arithmetical sentence θ n := "There exists a ∆ n full model of CS − which is disjunctively trivial above ω".Here by "ω" we mean the image of the canonical embedding of the ground model onto an initial segment of the model and a "full model" means a model with a satisfaction relation satisfying the usual Tarski's truth condition.Corollary below shows that each such sentence is false.It follows from the above corollary that the construction of the disjunctively trivial model of CT − does not formalize in any true arithmetical theory, in particular it does not formalize in PA.Hence one cannot hope to interpret CT − + DC − in in PA by using the construction of a disjunctively trivial model internally in PA.This is unlike in the case of a standard Enayat-Visser construction: [4] shows how to formalize the model theoretical argument from [3] in PA in order to conclude that CT − is feasibly reducible to PA and, in consequence, it does not have speed-up over PA.

Non-local Pathologies
In previous sections, we have considered a single, fixed θ and functions F such that F (x) is the x-th iterate of θ in some sense.We described sets defined by certain correctness properties with respect to this θ.In other words, we explored "local" pathologies (pathologies that are local to a fixed θ).In this section we address several sets defined using non-local pathologies: for example, instead of fixing a θ and looking at the idempotent disjunctions of θ, we look at all idempotent disjunctions (of any sentence).These sets can include IDC S , QC S , IDC bin S , DNC S , among others.Notice that QC S is necessarily closed under addition, since if, for each φ, T ((∀x) c φ) ≡ T (φ), then let θ = (∀x) c φ, and so Since (∀x) c θ = (∀x) 2c φ, we conclude that c ∈ QC S if and only if 2c ∈ QC S .Suppose that QC S is not a cut, and let c 0 < c 1 be such that c 0 / ∈ QC S and c 1 ∈ QC S .Then there is φ such that ¬[T ((∀x ∈ QC S , since T ((∀x) c0+c1 φ) ≡ T ((∀x) c0 φ).Let I ⊆ end J 0 ⊆ end J 1 ⊆ end M be separable cuts closed under addition such that c 0 ∈ J 0 and c 1 ∈ J 1 \ J 0 .Then X = I ∪ (J 1 \ J 0 ) is separable, but by the above argument, there can be no S such that (M, S) |= CS − and QC S = X.
This remark shows that there are complications that occur with sets defined by these non-local pathologies.For the remainder of this section, we look instead at the cuts defined by these pathologies.
We again generalize the setting to draw conclusions about I(IDC S ), I(QC S ) and I(IDC bin S ).To formalize this notion, we again look at finite propositional templates Φ(p, q) (recall this notion from the beginning of Section 2).We restrict our attention to Φ with the following properties: • Φ(p, q) is not equivalent to p, • the complexity of Φ(p, q) is non-zero, • q must appear in Φ(p, q), • p ∧ q ⊢ Φ(p, q), and Definition 24.Suppose Φ has the above properties.Then F : M × Sent → Sent defined as follows: • F (0, φ) = φ for all φ ∈ Sent M , and . is called an idempotent sentential operator.We say that Φ is a template for F .
Notice that for any θ, the function F (•, θ) is one to one.
Lemma 25.Let Φ be a template for F , and F an idempotent sentential operator.If p does not appear in Φ(p, q), then for all x, y ∈ M and φ ∈ Sent M , M |= F (x + y, φ) = F (x, F (y, φ)).
Proof.If p does not appear in Φ(p, q), then there is a propositional function Ψ(q) such that Φ(p, q) = Ψ(q).Let G : Sent M → Sent M be defined by G(φ) = Ψ(φ).Then, Since F and G are M-definable, by induction, one observes that for all x, F (x, φ) = G x (φ), the x-th iterate of G. Therefore, As before, notice that if p appears in Φ(p, q), then for each φ and x, φ ∈ Cl(F (x, φ)).For this reason, if p appears in Φ(p, q), we refer to F as accessible.If not, then because of Lemma 25, we say F is additive.
Definition 26.Let F be an idempotent sentential operator.
• The F -length of φ is the maximum x such that there is θ such that F (x, θ) = φ.• The F -root of φ is the unique θ such that F (x, θ) = φ, where x is the F -length of φ.
Remark 27.By working through the possible truth tables for Φ(p, q), one notices that if Φ(p, q) has the required properties, then it is logically equivalent to one of the following propositional formulae: We say that Φ(p, q) is q-monotone if it is logically equivalent to either p ∨ q or to q.
Lemma 28.Let F be an idempotent sentential operator.
(1) If F is accessible, then for all x, y > 0, φ, ψ ∈ Sent M , if F (x, φ) = F (y, ψ), then x = y and φ = ψ.In other words, when x > 0, the F -root of , then the F -root of φ and F -root of ψ are the same.
Next we show (2).Suppose F is additive and θ is the F -root of φ.Then F (x, θ) = φ and x is the F -length of φ.If θ is not F -irreducible, then there is y > 0 and ψ such that F (y, ψ) = θ.Then φ = F (x, θ) = F (x, F (y, ψ)) = F (x + y, ψ), the last equality holding by additivity.Since x + y > x, this contradicts that x is the F -length of φ.
To show the "moreover" part of (2), let x, y > 0, φ, ψ ∈ Sent M , and Notice that G is one to one.Since G is one to one, then if x = y, G x (φ) = G y (ψ) implies, by induction in M, that φ = ψ.Suppose x > y.Then again by induction in M, we have that M |= G x−y (φ) = ψ.Let θ be the F -root of φ, so that there is a such that G a (θ) = φ.Then G a+(x−y) (θ) = ψ, so θ is the F -root of ψ.
• Φ(p, q) = ¬¬q.Then F (x, φ) = (¬¬) x φ.The goal of this section is to characterize those cuts I such that This would allow us to characterize I(IDC S ), I(IDC bin S ), and I(QC S ), among others.For IDC bin S and QC S the relevant F functions are additive, while for IDC S , F is accessible.For the most part we will restrict our attention to Φ with syntactic depth 1.This covers most of the above cases, with the notable exception of ¬¬q; we treat this case separately.Given a cut I ⊆ end M , we say I is F -closed if either F is accessible or F is additive and I is closed under addition.We say I has no least F -gap if one of the following holds: • F is accessible and if x > I, then there is a y such that for each n ∈ ω, x − n > y > I, or • F is additive and if x > I, there is a y such that for each n ∈ ω, ⌊ x n ⌋ > y > I. Our next main result shows that if I is F closed and either separable or has no least F -gap, then there is S such that (M, S) |= CS − and Our method of proof will be similar to our previous results: we build sequences of finitely generated sets F 0 ⊆ F 1 ⊆ . . .and full satisfaction classes S 0 , S 1 , . . .with particular properties.We first prove two important lemmas which we use in the inductive step of our construction.
For the rest of this section, we modify Definition 5 so that we say φ ⊳ ψ if either φ is an immediate subformula of ψ or φ is the F -root of ψ.Similarly modify the definitions of closed sets and finitely generated sets so that such sets are closed under F -roots.Note that by Lemma 28, if F is accessible, this changes nothing about finitely generated and/or closed sets, but this does have an effect for additive F .
Definition 30.Let F be an idempotent sentential operator with template Φ(p, q).Let I ⊆ end M , X ⊆ Form M closed, and S a full satisfaction class.
(1) S is F -correct on I for formulae in X if for each φ ∈ X and x ∈ M , whenever F (x, φ) ∈ X and x ∈ I, then S(F (x, φ), α) if and only if S(φ, α) for all assignments α of φ.
(2) S is F -trivial above I for formulae in X if for each φ ∈ X and x ∈ M , whenever F (x, φ) ∈ X and x > I, then either Φ(p, q) is q-monotone and S(F (x, φ), α) for all assignments α, or Φ(p, q) is not q-monotone and ¬S(F (x, φ), α) for all assignments α of φ.
Lemma 31.Let M |= PA be countable and recursively saturated.Let F be an idempotent sentential operator with template Φ(p, q), and assume Φ(p, q) has syntactic depth 1.Let I ⊆ end M be F -closed and separable.Suppose X ⊆ Form M is finitely generated, S is a full satisfaction class, (M, S) is recursively saturated, S is F -correct on I for sentences in X and F -trivial above I for sentences in X.Then for any finitely generated X ′ ⊃ X, there is a full satisfaction class S ′ such that (M, S ′ ) is recursively saturated, S ′ ↾ X = S ↾ X, S ′ is F -correct on I for sentences in X ′ , and S ′ is F -trivial above I for sentences in X ′ .
Proof.Suppose we have X, X ′ and S as given in the statement of the Lemma.Let a, b, c code enumerations such that {(c) n : n ∈ ω} enumerates Sent M ∩X ′ , (b) n is the F -root of (c) n , and F ((a) n , (b) n ) = (c) n .By separability, there is d such that for each n ∈ ω, (a) n ∈ I if and only if (a) n < d.
To obtain S ′ , we proceed in two stages.First, we find an X ′ -satisfaction class S 1 with the following properties:

and
• if ψ = F (x, φ) ∈ X ′ , and x > I, then if Φ(p, q) is q-monotone, then (ψ, ∅) ∈ S 1 , and if Φ(p, q) is not q-monotone, then (ψ, ∅) ∈ S 1 .As before, by [3, Lemma 3.1], M has an elementary extension N with a full satisfaction class S N agreeing with S 1 on X ′ .Moreover, if U c = {x : c, x ∈ S 1 } for c ∈ X ′ , then (M, U c ) c∈X ′ is resplendent, so we can assume that there is a full satisfaction class S ′ on M itself and that (M, S ′ ]) is recursively saturated.
To find such an X ′ -satisfaction class S 1 , we use an Enayat-Visser argument.Let T h be the theory in the language L PA ∪ {S, S 1 } asserting the following: (1) the elementary diagram of M, • if Φ(p, q) is q-monotone, then We refer to either of the conditions in (5) as F -triviality.
Before we show that T h is consistent, we first show that if (M, S, S 1 ) |= T h is recursively saturated, then S 1 satisfies the required properties.First, since S 1 is compositional for formulas in X ′ , S 1 is an X ′ -satisfaction class.Moreover, the preservation scheme implies that S 1 ↾ X = S ↾ X.To show the other properties, suppose θ = F (x, φ) ∈ X ′ and φ ∈ X ′ .Then there are n, m ∈ ω such that By F -triviality, if Φ is q-monotone, we have (θ, ∅) ∈ S 1 , and if Φ is not q-monotone, we have (θ, ∅) / ∈ S 1 .Now we return to showing that T h is consistent.Let T 0 ⊆ T h be a finite subtheory.Let C be the set of formulas such that the instances of their compositionality, preservation, F -correctness and F -triviality appear in T 0 .Then C is finite, so the modified subformula relation, ⊳, is well-founded on C. We define S inductively on this relation: Suppose φ is minimal in C. If α is an assignment for φ, we put (φ, α) ∈ S 1 if any of the following hold: (1) φ ∈ X and (φ, α) ∈ S, (2) φ is atomic, α is an assignment for φ and M |= φ[α], or (3) Φ(p, q) is q-monotone, φ = F ((a) n , (b) n ), α = ∅ and (a) n > d.
Define φ of higher rank using compositionality if possible.If it is not possible, meaning that no immediate subformula of φ is in C, then there must be ψ ∈ C such that ψ is the F -root of φ.Let φ = F ((a) n , (b) n ), where (b) n = ψ.In this case, put (φ, α) ∈ S 1 if either (ψ, α) ∈ S 1 or (a) n > d and Φ is q-monotone.
We show that (M, S, S 1 ) |= T 0 .Clearly, (M, S, S 1 ) satisfies the elementary diagram of M, and by definition, (M, S, S 1 ) satisfies all compositional axioms in T 0 .
We show that (M, S, S 1 ) satisfies the preservation scheme.Suppose φ ∈ X.Then if φ is minimal in C in the subformula relation, then S 1 (φ, α) if and only if S(φ, α) by construction.If φ is not minimal, then S 1 (φ, α) if and only if S(φ, α) follows by compositionality along with F -correctness and F -triviality of S on sentences from X.
Next we show F -triviality.Suppose φ = F ((a) n , (b) n ) ∈ C and (a) n > d.We assume Φ(p, q) is q-monotone; the other case is similar.If φ is minimal in C, then by construction, (φ, If ψ ∈ C, then by our construction, (φ, ∅) ∈ S 1 if and only if either ((b) n , ∅) ∈ S 1 or (a) n > d (and Φ is q-monotone).Since (a) n < d, then (φ, ∅) ∈ S 1 if and only if Since T h is consistent, there is a model (M ′ , S ′ , S ′ 1 ) |= T h.By resplendency of (M, S), (M, S) has an expansion (M, S, S 1 ) |= T h.This S 1 is the required X ′ -satisfaction class.
We shall now prove an analogous lemma with a different assumption about I: instead of separability we shall require that there is no least F -gap above I.In the proof we shall need one more notion, which we shall now define: Definition 32.Let M |= PA, and let F be an idempotent sentential operator.Assume that F is additive.For Z ⊆ Form M and d ∈ M , let Z d be the set of those formulae of the form F (c, φ), for which there are n ∈ N, a ∈ M , such that • φ is the root of F (a, φ).For uniformity of our proofs, when F is accessible, we take Z d to be just the closure of Z (under immediate subformulae and taking F -roots).
Proposition 33.Let M |= PA, F an idempotent sentential operator, and Z ⊆ Form M .Then, for every Proof.This is clear if F is accessible, so assume F is additive.Fix an arbitrary c, φ such that F (c, φ) ∈ (Z d ) d .Choose a, n such that F (a, φ) ∈ Z d and 0 < a − c < n • d.By definition it follows that for some c ′ , n ′ , φ ′ and a ′ , F (a, φ) = F (c ′ , φ ′ ), F (a ′ , φ ′ ) ∈ Z and 0 < a ′ − c ′ < n ′ • d.Since F is additive this means that φ = φ ′ (since both of them are roots) and a = c ′ , hence Lemma 34.Let M |= PA be countable and recursively saturated.Let F be an idempotent sentential operator with template Φ(p, q), and assume Φ(p, q) has syntactic depth 1.Let I ⊆ end M be F -closed and has no least F -gap.Suppose S is a full satisfaction class, (M, S) is recursively saturated, d > I and S is F -correct on [0, d).Suppose further that X ⊆ Form M is finitely generated.Then for any formula φ ∈ Form M , there are I < d 0 < d 1 < d, a finitely generated X ′ ⊇ X and a full satisfaction class S ′ such that p hi ∈ X ′ , (M, S ′ ) is recursively saturated, Proof.Fix M, I, S, X, d and φ as in the assumptions.Let ⊙ denote + if F is accessible and • if F is additive.Let d 1 , d 0 be any numbers above I such that for every n, k ∈ ω, d 0 ⊙ n < d 1 ⊙ k < d.Suppose that every formula in X ∪ { φ} has complexity smaller than r ∈ M .Let θ := (¬) 2r 0 = 0 if F is not q-monotone and θ := ¬(¬) 2r 0 = 0 in the other case.We note that θ is the We shall start our construction by extending S to a Y ∪ Y d0 -satisfaction class on Additionally Y ∪ Y d0 is closed under roots and under immediate subformulae.We argue that X d0 ∩ Cl(F (d 1 , θ)) d0 = ∅.To this end observe that if ψ ∈ Cl(F (d 1 , θ)) d0 , then either ψ is in Cl(θ), and hence the complexity of ψ is greater than 2r − n for some standard n, or ψ contains θ as a subformula.In both cases the complexity of ψ is at least 2r − n for some standard n.Consequently, if ψ ∈ Cl(F (d 1 , θ)) d0 , then ψ does not belong to X d0 , because each formula in X d0 is a subformula of a formula in X, and hence its complexity is not greater than r.Moreover, if φ, F (b, φ) are both in Y ∪Y d0 and b < d 0 , then φ ∈ X d0 ⇐⇒ F (b, φ) ∈ X d0 .Indeed, from right to left this follows since (X d0 ) d0 ⊆ X d0 .From left to right this is so, since if The first case is impossible since each formula in Cl(θ) d0 starts with a negation which does not occur in Φ.In the latter case it follows that θ is a subformula of φ (because θ is F -irreducible) and hence φ / ∈ X d0 .Let us put Y ′ = Y ∪ Y d0 .We extend S↾ X to a Y ′ -satisfaction class S 0 , which is compositional and d 0 -correct for formulae in Y ′ .For every φ ∈ Y ′ and every α: , then (φ is a sentence and) S 0 (φ, α) iff α = ∅ and F is q-monotone.• if φ is in the closure of F (d 1 , θ), then, since Φ(p, q) has syntactic depth 1, φ is either in Cl({θ}) or φ = F (d 1 − n, θ) for some n ∈ ω.We have already taken care of the former case.In the latter case we let the value of φ on α be the same as that of F (d 1 , θ) on α.
. This can happen only if F is additive.Since Y is closed under roots, ψ ∈ Y , hence for each α the value of ψ on α has already been defined.We stipulate that the value of F (a − b, ψ) on α is the same as that of F (a, ψ) on α.We observe that this is independent of the choice of F (a, ψ) ∈ Y : if F (a, ψ) and F (a ′ , ψ ′ ) both satisfy the above conditions, then either both F (a, ψ), F (a ′ , ψ ′ ) belong to X or both of them belong to Cl(F (d 1 , θ)).If the former holds our claim follows because S is F -correct on [0, d).If the latter holds, it must be the case that ψ = ψ ′ = θ and |a − a ′ | is standard, so our claim follows by construction.
We check that S 0 is F -correct on [0, d 0 ) for sentences in Y ′ .If F is accessible, this easily follows from our construction.Assume that F is additive.Assume 0 < b < d 0 and fix an arbitrary φ.By previous considerations either both φ, F (b, φ) belong to X d0 or they both belong to Cl(F (d 1 , θ)) d0 .In the latter case both φ and F (b, φ) are of the form In particular, for an arbitrary α, φ and F (b, φ) get the same value on α (by construction).
We check that with so defined S, (M, S) |= T 0 .That the compositional clauses hold is clear from the construction.We check that S is F -correct on [0, d 0 ) for sentences in E. By induction on n we prove that for all φ, F (a, φ) ∈ E, rk(φ) + rk(F (a, φ)) = n, a < d 0 , then for every α, S(φ, α) ⇐⇒ S(F (a, φ), α).Since rk(φ)+ rk(F (a, φ)) = 0 only if a = 0, the base step is trivial.Assume rk(φ) + rk(F (a, φ)) is positive.Then certainly rk(F (a, φ)) is positive.If all immediate subformulae of F (a, φ) belong to E, then at least one of them is of the form F (a − 1, φ) and the thesis follows by inductive hypothesis and idempotency of Φ, since F (a − 1, φ) has lower rank than F (a, φ).Otherwise, for some ψ ∈ E and b < d 0 such that F (a, φ) = F (b, ψ) and we decided that for every α, the values of F (b, ψ) and ψ are the same.By Lemma 28, for some b ′ , either φ = F (b ′ , ψ) or ψ = F (b ′ , φ).Hence the thesis follows by the inductive assumption.Now we argue for the preservation axioms.By induction on the rank of φ we prove that if φ ∈ Y ′ , then for every α, S(φ, α) iff S 0 (φ, α).This is immediate for formulae of rank 0. In the induction step we use the definition of S and the closure properties of Y ′ .
For the step induction step we first consider extend S n ↾ Mn to the set Form M ∪ (Form M ) d0 ⊆ Form Mn+1 .Then we argue as in the first step.
Theorem 35.Let M |= PA be countable and recursively saturated and I ⊆ end M .Let F be an idempotent sentential operator with template Φ(p, q), and assume Φ(p, q) has syntactic depth 1. Suppose I is F -closed.Then if I is separable or has no least F -gap above it, there is S such that (M, S) |= CS − and Proof.We construct sequences F 0 ⊆ F 1 ⊆ . . .and S 0 , S 1 , . . . of sets such that: (1) F i is finitely generated and ∪F i = Form M , (2) S i is a full satisfaction class and (M, S i ) is recursively saturated S i is F -correct on I for sentences from F i , and (5) for each x > I, there is I < y < x, i ∈ ω and φ ∈ F i such that F (y, φ) ∈ F i and ¬(S i (F (y, φ), α) ≡ S i (φ, α)) for all assignments α.
If I is separable, we also ensure that S i is F -trivial above I for sentences in F i .
Prior to starting the construction, if I has no least F -gap above it, we externally fix a sequence d 0 > d 1 > . . .such that inf{d i : i ∈ ω} = I and for each i, d i and d i+1 are in different F -gaps.Additionally, we externally fix an enumeration of Form M (in order type ω).
Suppose we have constructed F i and S i .Let φ be the least formula in our enumeration that is not in F i .If I is separable, let F i+1 be generated by F i and φ, and apply Lemma 31 to obtain S i+1 .Otherwise, we suppose S i is F -correct on [0, d i ) and apply Lemma 34 to obtain F i+1 , S i+1 , and I < c 0 < c 1 < d i such that S i+1 is F -correct on [0, c 0 ) but not on [0, c 1 ).(In fact, there is θ ∈ F i+1 that witnesses the failure of F -correctness on [0, c 1 ).)Without loss of generality, we can replace d i+1 with the minimum of {c 0 , d i+1 }, so that we can assume S i+1 is F -correct on [0, d i+1 ) and continue.
Having constructed these sequences, let S = ∪S i ↾ F i .Then it follows that S is F -correct on I and for each x > I, there is φ such that ¬(T (φ) ≡ T (F (x, φ))).
Remark 36.It is easy to see that in fact a tiny modification of our proof of Theorem 35 shows something more: we can perform our construction in such a way that S is F correct on I not only on all sentences but on all formulae.Hence, given M, I and F as in the assumptions of Theorem 35 we can find a satisfaction class S such that We assume that Φ has depth 1 in the previous results because the more general case is quite complicated.In particular, if Φ has depth at least 2, then it might not be possible to ensure that S is F -trivial above I as we do in Lemma 31.For example, suppose Φ(p, q) = (¬¬)q, φ = (0 = 0) and ψ = ¬(0 = 0).Then, for any x and any satisfaction class S, T ((¬¬) x φ) ≡ ¬T ((¬¬) x ψ).However, we show in our next result that we can still ensure that, if I is separable and closed under addition, there is S such that I is the (¬¬)-correct cut.
To show that T h is consistent, let T 0 ⊆ T h be finite, and let C be the set of formulas whose instances of compositionality, preservation, double negation correctness and double negation incorrectness are in T 0 .Since C is finite, then the modified subformula relation ⊳ is well-founded on C, and we define S ′ inductively on this relation.
Suppose φ is minimal in C. If α is an assignment for φ, we put (φ, α) ∈ S ′ if either φ is atomic and M |= φ[α], or φ ∈ X and (φ, α) ∈ S. We define φ of higher rank using compositionality if possible.If this is not possible, then it must be the case that there is n ∈ ω such that φ = (¬¬) We verify that (M, S, S ′ ) |= T 0 .Clearly it satisfies the diagram and compositionality axioms by construction.Suppose φ ∈ X is such that ∀α(S ′ (φ, α) ≡ S(φ, α)) ∈ T 0 .If φ is of minimal rank, then this is true by construction.If not, we can assume, by induction, that whenever ψ ⊳ φ is such that ψ ∈ C, then ∀α(S ′ (ψ, α) ≡ S(ψ, α)).If φ is determined via compositionality, then the result for φ follows from the fact that both S and S ′ are compositional for formulas in X.Otherwise, the result for φ follows from either double negation correctness up to I, or double negation incorrectness above I.A similar argument shows double negation incorrectness in the case that (a) n > d.
By Theorem 35, if I is either separable or has no least Z-gap above it, there is T such that (M, S) |= CS − and I(IDC S ) = I.In fact, if ω is a strong cut, then by Proposition 18 every cut I is either separable or has no least Z-gap, and therefore every cut I can be I(IDC S ) for some satisfaction class S. Similarly, if ω is strong, then every additively closed cut I is either separable or has no least additive gap above it, and therefore each additively closed cut can be I(IDC bin S ).To complete the picture, we can show that if F is an idempotent sentential operator and I is the F -correct cut, then either I has no least F -gap above it or is separable.Therefore, if M is not arithmetically saturated, then there are cuts I which cannot be realized as I(IDC S ) for any T .
Proposition 38.Let F be an accessible idempotent sentential operator.Suppose (M, S) |= CS − and I ⊆ end M is such that Then either there is no least Z-gap above I or I is separable.This completes the picture for accessible F .In particular, we have a complete picture for which cuts can be I(IDC S ).If ω is strong, then every cut can be I(IDC S ) for some S, and if ω is not strong, then only those cuts which have no least Z-gap above it can be I(IDC S ).What about for cuts which are F -correct for additive F , like I(QC S )?
Proof.Consider the following sequence of formulae The above conjunction is of the form (φ s0 ∧ (φ s1 ∧ (. ..) . ..)where {s i } i<2 n is an enumeration of all binary sequences of length ≤ n according to the length-first lexicographic ordering.By Smith's result [9,Theorem 2.19] there is a ∈ M such that for all n ∈ ω, T (φ n (a)) holds.Hence {s ∈ {0, 1} <ω | s ∈ a} is an infnite finitely branching tree, so it has a coded infinite branch, b.Since b ⊆ a, for every i ∈ ω we have (M, S) |= T (b(i)).
Proposition 40.Let F be an additive idempotent sentential operator.Suppose (M, S) |= CS − and I ⊆ end M is such that Then either there is no least +-closed gap above I or I is separable.
Proof.Suppose there is a least +-closed gap above I and let a code a sequence such that (a) n+1 = ⌊ (a)n 2 ⌋ and inf n∈ω (a) n = I.Let c be the length of a. Observe that sup(I ∩ im(a)) = I, so by Proposition 10 it is sufficient to show that I ∩ im(a) is separable.Fix φ such that (M, S) |= ¬T (F ((a) 0 , φ) ≡ φ).Then for every n it holds that Define the labelling t of a full binary tree of height c by recursion as follows: In the above, x * is the unique sentence ψ such that there is θ such that x = ¬(θ ≡ ψ).By our assumption, t ↾ T has arbitrarily long branches, so there is an infinite coded branch b of t such that for every i ∈ ω (M, S) |= T (b(i)).Moreover, by the construction of t, for every i ∈ dom(b), It follows that the set A = {ψ ∈ im(b) : T (¬ψ)} is separable.Observe that for every i < len(b) we have where G is the definable function (a) i → b(i).By Proposition 9 this ends the proof.
Note that the implication (2) =⇒ (1) holds in more generality: it does not rely on Φ having syntactic depth 1.
Proof.We show (1) =⇒ (2).Suppose ω is a strong cut.Let a ⊙ n be a − n, if F is accessible, and ⌊ a n ⌋, if F is additive.By Proposition 18, if I is not separable, then I is not ω-coded, and so there is no a > I such that inf({a ⊙ n : n ∈ ω}) = I.Therefore, every F -closed cut I is either separable or has no least F -gap above it.The result follows from Theorem 35.
Conversely, if M is not arithmetically saturated, let I ⊆ end M be any cut with a least F -gap above it.For example, fix a nonstandard c and let I = inf({c ⊙ n : n ∈ ω}).Since ω is not strong, by Proposition 18, I is not separable.It follows by Proposition 38 for accessible F , and by Proposition 40 for additive F , that there is no S such that (M, S) |= CS − and I = {x : ∀y ≤ x∀φ(T (F (y, φ)) ≡ T (φ))}.

Disjunctively Correct Cut
We proceed to the strongest correctness property, that of full disjunctive correctness (DC S ).As usual we shall focus on I(DC S ).The first proposition states that the intuitive strength of full disjunctive correctness is reflected in the closure properties of I(DC S ): Proposition 42.For every (M, S), I(DC S ) is closed under multiplication.
Proof.We shall use a result from [1]: define the sequential induction cut SInd S to be the set of those c such that the following is true in (M, S) : Then the proof of [1,Theorem 8] directly shows that DC S ⊆ SInd S .Now we proceed to the main argument: fix any c ∈ DC S and let b ≤ c 2 .Fix any d, r such that b = dc + r and r < c.Fix any φ i : i ≤ b and assume first that T ( i≤b φ i ) and, aiming at a contradiction that for every i ≤ b, T (¬φ i ).Define the auxiliary sequence of length d: for each i ≤ d let θ i = j≤ic φ j and let θ d+1 = φ b .We show that for every i < d + 1, T (¬θ i ) → T (¬θ i+1 ).Fix any i and assume T (¬θ i ).Let c ′ be c if i < d and r if i = d.Consider the sequence ψ k = j≤ic+k φ j .We claim that for any k < c ′ T (¬ψ k ) → T (¬ψ k+1 ).Indeed, fix any k < c ′ and assume T (¬ψ k ).Observe that by the definition of T , the definition of ψ k+1 and the compositional axioms we have The last sentence is clearly true by our assumptions.Hence, since c ′ ∈ SInd S , we conclude that T (¬ψ c ′ ).Since by definition ψ c ′ = θ i+1 , we established that for any i < d T (¬θ i ) → T (¬θ i+1 ).Since d + 1 ∈ SInd S , we conclude that T (¬θ d+1 ).By definition, we obtain that T (¬ i≤b φ i ), which contradicts our assumption.Now assume that for some e ≤ b, T (φ e ) holds.In particular, it holds that T ( i≤e φ i ).Let us fix d ′ , r ′ such that b − e = d ′ c + r ′ and for j ≤ d ′ define θ j = i≤e+jc φ i and θ d ′ +1 = i≤b φ i .As in the above proof we show that for each j ≤ d ′ , T (θ j ) → T (θ j+1 ) and obtain T ( i≤b φ j ), which concludes the proof of the reverse implication and the whole argument.
We conclude with a limitative result which shows that methods used to prove the main results of previous sections are unsufficient for obtaining the analogous results in the context of DC S .This is because, as conjectured, our methods show that, in an arithmetically saturated model, any cut can be characterized as I(IDC S ) for some regular satisfaction class which satisfies the internal induction axiom, For such a satisfaction class S, S(φ, ∅) behaves like a truth predicate satisfying the axioms of CT − and we have the following small insight.Below Con PA (x) is a formula with a free variable x which canonically expresses that there is no proof of 0 = 1 in PA whose code is smaller than x.
Proposition 43.Suppose that (M, S) |= CS − , S is regular and satisfies the internal induction axiom.Then, for every a ∈ DC S , M |= Con PA (a).
Sketch.Let CT − (x) denote a formula of the language L PA ∪{T } with a free variable x which expresses "T (x) satisfies Tarski's inductive truth conditions for sentences of logical depth at most x".By inspection of the proof of Theorem 3.1 from [11] one sees that if a ∈ DC S , then there is (typically nonstandard) ψ ∈ M with a unique free variable such that M |= Form L PA (ψ(x)) and (M, S) |= CT − (a)[T * ψ(x)/T (x)].T * ψ(x) denotes a formula with a free variable x which expresses "The result of substituting the numeral of x for the unique free variable in ψ is true" (we use the notation from [8], Lemma 3.6) and CT − (a)[T * ψ(x)/T (x)] is the formula obtained by substituting T * ψ(x) for T (x).As in [8], Lemma 3.7 we conclude that T * ψ(x) satisfies full induction scheme in (M, S).It follows that no proof with a code less than a can be the proof of 0 = 1 from the axioms of PA, because each formula in this proof is of complexity at most a and all the premises are made true by T * ψ(x).

Appendix
In this Appendix we indicate how to modify the proof of Theorem 12 in order to obtain a much better-behaved satisfaction class.In particular we would like the constructed satisfaction classes to define a truth predicate.We start with introducing the notion of regularity.The definition is taken from [10]: In other words, α φ assigns to a variable v, the value of the term γ(v) under the assignment α.For illustration assume that θ is either a true atomic sentence or the negation of a true atomic sentence and F is a local idempotent operator for θ with a template Φ(p, q) (as in Definition 6).Then for any x, F (x) can differ from F (x) only in that • F (x) may use different free and bound variables; • each element of [θ] F (x) is of the form v i = v j for some variables v i and v j (if θ is a true atomic sentence) or each element of [θ] F (x) is of the form ¬v i = v j for some variables v i and v j (if θ is the negation of a true atomic sentence.Moreover all the variables in F (x) occur only in formulae from [θ] F (x) .In particular F (x) is not a sentence.
Moreover, observe that, since F (x) is a sentence, then ∅ is the unique assignment for F (x). Hence, if θ is either s = t or ¬s = t, where s and t are closed terms whose value is a, then ∅ F (x) is constantly equal to a.
The above described situation of a local idempotent operator for θ will be the only one which we shall consider in this section.
We are now proceeding to strengthening Theorem 12.For notational reasons we write M, α |= φ instead of M |= φ[α] to mean that a formula φ is satisfied in M by an assignment α.Definition 50.Fix M |= PA, X ⊆ M , θ, F and Φ such that F is local idempotent sentential operator for θ with syntactic template Φ(p, q).
(1) We say that a formula φ is an F -intermediate formula if for some x, F (x) is a subformula of φ (not necessarily direct or proper) and φ is a subformula (not necessarily direct or proper) of F (x + 1).( 2) For an intermediate formula φ, the F -length of φ is the maximal x such that F (x) is a subformula of φ. (3) Recall that compl(φ) denotes the complexity of a formula φ (defined in Preliminaries).For an F -intermediate formula φ, assignment α for φ and x such that for some n ∈ ω, compl(φ) = compl(F (x)) + n, we say that α (X, x)-satisfies φ if M, α |= φ[A/F (x)] where A is 0 = 0 if x ∈ X and 0 = 1 otherwise and φ[A/F (x)] denotes the result of replacing in φ every occurrence of F (x) φ with A. We say that α, X-satisfies φ if α (X, x)−satisfies φ where x is the F -length of φ.
We note that the above definition makes sense, since φ[A/F (x)] is a formula of standard complexity (possibly with variables with nonstandard indices).
Proposition 51.Fix any M |= PA and X ⊆ M which is closed predecessor.For an arbitrary intermediate formula φ of nonstandard complexity and assignment α for φ the following are equivalent: (1) α X-satisfies φ.
Proof.Follows immediately from the definition of F and the fact that θ, Φ are chosen so that Φ(θ, q) is equivalent to q.
Theorem 52.Let θ be either a true atomic sentence or a negation of a true atomic sentence and F a local idempotent sentential operator for θ with template Φ(p, q).Let X ⊆ M be separable, closed under successors and predecessors, and for each n ∈ ω, n ∈ X if and only if M |= θ.Then M has an expansion (M, S) |= CS − such that X = {x ∈ M : (M, S) |= T (F (x)) ≡ T (θ)} and S is a regular satisfaction class.
Proof.The initial structure of the argument is very similar to that used in proving Theorem 12. Let D = {F (x) : x ∈ M } and A = {F (x) : x ∈ X}.Note that A is separable from D. We build sequences F 0 ⊆ F 1 ⊆ . . .and S 0 , S 1 , . . .such that: • each F i is a finitely generated set of formulas such that ∪F i = Form M , • each S i is a regular full satisfaction class and (M, S i ) is recursively saturated, • S i+1 ↾ F i = S i ↾ F i , and • for each φ ∈ D ∩ F i , (φ, ∅) ∈ S i if and only if φ ∈ A.
Given such a sequence, S = ∪(S i ∩ F i × M ) would be the required full satisfaction class on M.
Externally fix an enumeration of Form M in order type ω.We can assume, without loss of generality, that θ appears first in this enumeration.We let F 0 be {θ} and S 0 be any regular full satisfaction class which satisfies internal induction.Let F i+1 be generated by F i and the least x ∈ Form M \F i in the aforementioned enumeration.Let F ′ = F i ∪ (F i+1 ∩ D).Let a be a sequence such that {F ((a) n ) : n ∈ ω} = F i+1 ∩D.Note that such a sequence exists since F i+1 is finitely generated.Let c be as in the definition of separability for a.
We shall now show how to construct an elementary extension N of M and a full regular satisfaction class S ′ on N such that • For each n ∈ ω, (F ((a) n ), ∅) ∈ S ′ ⇐⇒ M |= n ∈ c.By a straightforward resplendence argument one can then copy S ′ to M. This last step crucially uses the fact that (M, S i ) is recursively saturated and the facts that (1) F i+1 is finitely generated and (2) that we can code the membership in F i+1 ∩ X via the parameter c.The construction of S ′ follows the lines of a standard Enayat-Visser construction (as presented in [3]): we build a sequence of models M = M 0 M 1 M 2 , . . .and sets S ′ 1 , S ′ 2 , . . .such that (1) (3) for each n ∈ ω, (F ((a) n ), ∅) ∈ S ′ i ⇐⇒ M |= n ∈ c (4) for every φ ∈ Form Mi and α ∈ M i+1 , if α is an assignment for φ, then (φ, α) ∈ S ′ i+1 ⇐⇒ ( φ, α φ ) ∈ S ′ i+1 .
Then one easily checks that for N = i M i and S ′ = i S ′ i+1 ∩ (Form Mi ×M i+1 ), (N , S ′ ) satisfy the conditions A,B,C above and S ′ is a full regular satisfaction class.We note that condition 4 does not contradict the fact that S ′ i+1 is defined only for By the inductive assumption, the last condition is equivalent to: ( * * ) ∅ ψ0 X − satisfies ψ 0 or ∅ ψ1 X − satisfies ψ 1 .
Let κ 0 be the occurrence of ψ 0 in φ as the left disjunct, and κ 1 be the occurrence of ψ 1 in φ as the right disjunct.Then (κ 0 ) φ differs from ψ only up to the bound variables renaming and a permutation of free variables.Let σ be the permutation of free variables such that σ[(κ 0 ) φ ] is (up to bounded variables renaming) the same as ψ 0 .By unraveling the definitions it follows that ∅ φ ↾ (κ 0 ) φ = ∅ ψ0 • σ.The same holds for the pair (κ 1 ) φ and ψ 1 .So we conclude that ( * * ) is equivalent to The above however is clearly equivalent to the right-hand side of ( * ).

Proposition 10 .
Let M |= PA.Let I ⊆ e M and A ⊆ M be an M-definable set such that sup(A ∩ I) = I and A ∩ I is separable.Then I is separable.Proof.Define the function f by f (x) = µy.{y∈ A : x ≤ y} if such y exists 0 otherwise Then, by the assumptions, I = f −1 [A ∩ I].The result follows by Proposition 9.
is either coded by ω from below or from above.(2) I is 0-superrational if there is a ∈ M such that either Def 0 (a) ∩ I is cofinal in I and for all b ∈ M , Def 0 (b) \ I is not coinitial in M \ I, or Def 0 (a) \ I is coinitial in M \ I and for all b ∈ M , Def 0 (b) ∩ I is not cofinal in I. Theorem 17.Let M |= PA and I ⊆ end M .Then the following are equivalent:

Definition 20 .
Let (M, S) |= CS − and I ⊆ end M .(1) If, for every c > I and every sequence of sentences (in the sense of M) φ i : i < c , (M, S) |= T ( i<c φ i ), then we say (M, S) is disjunctively trivial above I.If (M, S) is disjunctively trivial above ω, we simply say (M, S) is disjunctively trivial.(2) If, for every c ∈ I and every sequence of sentences (in the sense of M) φ i : i < c , (M, S) |= T ( i<c φ i ) ≡ ∃i < c T (φ i ), we say that (M, S) is disjunctively correct on I. Corollary 21.Suppose (M, S) |= CS − and I ⊆ end M .If (M, S) is disjunctively trivial above I and disjunctively correct on I, then I is separable.In particular, if (M, S) is disjunctively trivial above ω, then M is arithmetically saturated.Conversely, if M is arithmetically saturated, there is S such that (M, S) |= CS − and is disjunctively trivial above ω.Proof.If (M, S) |= CS − is disjunctively trivial above I and correct on I, then I = {c : (M, S) |= ¬T ( c (0 = 1))}.Therefore I is separable by Theorem 13.If I = ω, then (by Proposition 14) ω is a strong cut in M and therefore M is arithmetically saturated.Conversely, suppose M is arithmetically saturated.
(a)n ((b) n ) and (b) n ∈ C has lower rank than φ.We put (φ, α) ∈ S ′ if either (a) n < d and at an earlier stage we decided ((b) n , α) ∈ S ′ , or if (a) n > d and, at an earlier stage we decided ((b) n , α) ∈ S ′ .