Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 16, 2004, 3 – 15

© A. Bovykin

Andrey Bovykin
(submitted by Yi Zhang)

ABSTRACT. First, we study a question we encountered while exploring order-types of models of arithmetic. We prove that if MPA is resplendent and the lower cofinality of M \ is uncountable then (M,<) is expandable to a model of any consistent theory T PA whose set of Gödel numbers is arithmetic. This leads to the following characterization of Scott sets closed under jump: a Scott set X is closed under jump if and only if X is the set of all sets of natural numbers definable in some recursively saturated model MPA with lcf(M \ ) > ω. The paper concludes with a generalization of theorems of Kossak, Kotlarski and Kaye on automorphisms moving all nondefinable points: a countable model MPA is arithmetically saturated if and only if there is an automorphism h: M M moving every nondefinable point and such that for all x M, < x < Cl \ , we have h(x) > x.


2000 Mathematical Subject Classification. 03H15, 03C62, 08A35.

Key words and phrases. models of Peano arithmetic, linearly ordered sets, arithmetic saturation, resplendency, automorphisms of models.

The author was supported by a grant of the Swedish Royal Academy of Sciences to stay in Institut Mittag-Leffler and, subsequently, by a NATO-PC Advanced Fellowship via the Scientific and Technical Research Council of Turkey to stay in İstanbul Bilgi University.

1. Introduction

Peano Arithmetic (PA) is the first-order theory in the language LPA = {+,×,<, 0, 1} consisting of the following axioms: associativity and commutativity of + and ×, their neutral elements are 0 and 1 respectively, distributivity, discrete linear order axioms for <, adding 1 gives a successor, and the Induction Scheme:

y¯[ϕ(0,y¯) x(ϕ(x,y¯) ϕ(x + 1,y¯)) xϕ(x,y¯)]

for every LPA-formula ϕ(x,y¯).

Peano Arithmetic is an extremely powerful theory. A folklore knowledge among logicians is that all of classical analysis, number theory and combinatorics can be done within tiny subsystems of Peano Arithmetic. In pre-Gödelean era it was believed that PA comprises an axiomatization of the set of all “truths” about natural numbers and finite sets.

Thus, a model of Peano Arithmetic (that is, a set with operations + and × defined on it so that the above axioms of PA hold) resemble the natural numbers as much as any working mathematician would hope for (all of his concrete mathematics can be conducted inside a model of PA and nobody would notice the difference). As usually for such theories, there are 2λ non-isomorphic models of PA in every infinite cardinality λ.

The structure of models of first-order Peano Arithmetic (PA) has been extensively studied since the 1960s. Due to unclassifyability of the diverse mass of models (even in the countable case) and the elusive nature of completions of PA (especially the ‘true arithmetic’ Th = {ϕϕ LPA,ϕ}), models of PA are among the most difficult to deal with in the whole of model theory.

Certain classes of models were studied that could be to some extent tackled: countable models, ϰ-like models (for a cardinal ϰ), models coding certain sets, realizing certain types. Among the most important notions introduced is recursive saturation. A model M is recursively saturated if it realizes every type (with parameters from M) whose set of Gödel numbers is recursive. Recursively saturated models naturally occur in model theory of arithmetic. For instance, every model of PA obtained from a nonstandard model of PA by an application of the arithmetized completeness theorem is recursively saturated.

A recursively saturated model of PA is uniquely determined by its complete theory and the collection of subsets of definable (coded) in the model: if MPA, NPA are two recursively saturated models of the same completions of PA and code the same subsets of then MN.

Other notions were also introduced, isolating important subclasses of the class of all recursively saturated models: the most important being resplendency and arithmetic saturation. Resplendent models and arithmetically saturated models will be the main objects we study in this paper.

A model M is resplendent if for every a¯ M, and any statement ϕ(a¯) containing additional relation symbols R1,,Rn, if Th(M,a¯) + ϕ(a¯) is consistent then there are relations R1,,Rn on M such that (M,R1,,Rn)ϕ(a¯). Resplendency implies existence of many automorphisms of a model, recursive saturation of a model and many other pleasant properties. Resplendent models are a plentiful class of very ‘regular’ models we can deal with.

A model MPA is arithmetically saturated if it is recursively saturated and the class of subsets of definable in M is closed under jump. Thus, more subsets of are definable in M than is expected from a recursively saturated model. (A recursively saturated model is only expected to code its own complete theory, see Wilmer’s theorem below.) In particular, an arithmetically saturated model codes the sets Πn Th of all true Πn -sentences for all n . In addition, arithmetic saturation implies more homogeneity than just recursive saturation. Recursive saturation implies that the model is homogeneous (if tp(a1,,an) = tp(b1,,bn) then there is an automorphism h: M M such that for all i = 1, 2,,n, h(ai) = bi). Arithmetic saturation gives us extra control over this automorphism. E.g., if ai = bi for all i = 1, 2,n then it can be ensured that this automorphism moves all nondefinable points (i.e., h(x) = x for all x Cl).

Section 3 starts off with an investigation of a problem concerning order-types of resplendent models of Peano arithmetic. The connection with arithmetic saturation is studied in Section 4. Section 5 uses the methods developed in Section 4 to produce a generalization of some well-known results of Kossak, Kotlarski and Kaye.

2. Definitions

If A is a linearly ordered set then lcf(A), the lower cofinality of A, is cf(A), where A is A with the order reversed.

If B MPA then ClM(B) denotes the set of all elements of M definable in M with parameters from B, that is a ClM(B) for some b¯ B and some LPA-formula ϕ(x,y¯),

Mϕ(a,b¯) !xϕ(x,b¯).

A set Y M is definable (with parameters b¯ M) if for some LPA-formula ϕ(x,y¯), Y = {y MMϕ(y,b¯)}.

Definition 1. Let MPA. We define the Standard System of M as SSy(M) = {X there is a definable (with parameters) Y M such that X = Y }. We say that a subset A of is coded in M if A SSy(M).

Definition 2. A Scott set X is a collection of subsets of closed under , , \, complement, relative recursion and such that if T X codes an infinite finitely branching tree then there is B X, B T which codes an infinite path through T.

It is known that for every MPA, SSy(M) is a Scott set. The converse is known to hold for Scott sets of cardinalities ω and ω1 .

Definition 3. A model MPA is called resplendent if for every Σ11-statement Φ(a¯), a¯ M, consistent with Th(M,a¯), we have MΦ(a¯).

Definition 4. A model MPA is called recursively saturated if every recursive type p(x,a¯) (that is, {ϕ(x,y¯))ϕ(x,y¯) p} is recursive) is realized.

It is also known (and we shall often use this fact) that a recursively saturated model M realizes all types that are coded in M, i.e. such that {ϕ(x,y¯)ϕ(x,a¯) p(x,a¯)}SSy(M).

Fact 1. (Kleene). Let L be a finite language and {ϕn(x¯)}n be a recursive set of formulas of L. Then there is a Σ11-formula Φ(x¯) such that in all infinite L-structures M, Mx¯(Φ(x¯) iϕn(x¯)).

It follows from Fact 1 that resplendency implies recursive saturation.

Fact 2. (Wilmers). Let X be a countable Scott set, T PA be a consistent theory, T X. Then there is a countable recursively saturated MT such that SSy(M) = X.

Fact 3. ([4][7][5]). Let MPA be recursively saturated. Then the following are equivalent.

  1. M is arithmetically saturated, i.e., SSy(M) is closed under jump;
  2. for any f M coding a set of pairs determining a function f : M, there is c M \ such that for all n , f(n) > f(n) > c;
  3. for every a M, {θ(x,y)θ(x,y) is an LPA-formula in two variables and min xθ(x,a) Cl}SSy(M);
  4. there is g Aut(M) such that g(a) = a for every a Cl.

3. A question about expanding order-types of resplendent models to models of other theories

A question of H. Friedman [2] asks whether the classes of order-types of uncountable models of T are the same for all T PA. Having embarked on this difficult question, I realized that probably there is some chance of obtaining results in the case of resplendent models. For an up-to-date account of the state of Friedman’s problem, see [1]. Among the results is the following theorem.

Theorem 4. If MPA is resplendent and c M codes a consistent theory T PA then (M,<) can be expanded to a model of T.

The theorem is proved by writing down a Σ1 1-statement expressing the existence of NT, (N,<)(M,<) and noticing that it is realized in every countable submodel of M containing c.

Using a theorem by D.Richard and J.-F.Pabion [8] which says that MPA is ω1 -saturated if and only if (M,<) is ω1 -saturated, we obtain the following corollary.

Corollary 5. If MPA is resplendent and ω1-saturated then (M,<) can be expanded to a model of any consistent extension of PA.

The hunt for conditions weaker than ω1 -saturation implying expandability of (M,<) to a model of T PA led to the following theorem.

Theorem 6.  
If MPA is resplendent and lcf(M \ ) > ω then for all n ω, (M,<) is expandable to a model of PA + Πn Th .

Proof. For any n ω, let us introduce ΣnDef = the set of all nonstandard definable points of M defined by a Σn-formula.

As lcf(M \ ) > ω, there is a > such that Σ1Def > a. Define A1 = {xϕ(x)ϕ Δ0,Mx < aϕ(x)}. Now, A1 Π1 Th because Δ0M. Also, Π1 Th A1 because if for some ϕ Δ0 such that xϕ(x) there existed x < a¬ϕ(x) then min x¬ϕ(x) would be a nonstandard Σ1-definable point less than a. Hence, A1 = Π1 Th . A1 is definable, hence coded in M.

Suppose at stage n we already know that Πn Th SSy(M). Let b M code Πn Th . Consider the statement

Φn(b) = n,n,n, On, Sn

xy(x nyx < y)

(n,n,n, On, Sn)PA + Πn Th

SSy(M,n,n,n, On, Sn) SSy(M).

Let us show that the last line is expressible by a Σ11-sentence. Let ϕm(x) = (x > m)&zyi < x((z))i = (y)i, where ((z))i means (z)i in the language {n,n,n, On, Sn}. The set {ϕm(x)}m is a recursive set of formulas, hence, by Kleene’s Theorem, there is a Σ1 1-sentence Θ(x) such that in any KPA, Kx( mϕm(x) Θ(x)). Then SSy(M,n,n,n, On, Sn) SSy(M) is implied by the Σ11-sentence xΘ(x). Hence, Φn(b) is a Σ11-sentence.

Φn(b) is consistent because, by Wilmers’ Theorem, as (PA + Πn Th ) SSy(M), there is a countable model

NPA + Πn Th ,SSy(N) = SSy(ClM(b)).

Hence, by resplendency, Φn(b) is already realized in M.

Denote the model (M,n,n,n, On, Sn) by Mn. By construction, MnPA + Πn Th , (Mn,<)(M,<), SSy(Mn) SSy(M).

Let (ΣnDef)Mn > a > . Consider

An+1 = {xϕ(x)ϕ Σn,Mnx < aϕ(x)}.

An+1 Πn+1 Th because if Mnx < aϕ(x) but x¬ϕ(x) then for some k , ¬ϕ(k), which is a Πn-statement. Hence, as MnΠn Th , Mn¬ϕ(k), contradiction.

Πn+1 Th An+1. Let xϕ(x), where ϕ(x) Σn. If Mnx < a¬ϕ(x) then c =: min x¬ϕ(x) is a Σn-definable point less than a. If c then Mn¬ϕ(c), which is a Πn-statement not belonging to Πn Th . Contradiction with MnΠn Th . Hence < c < a contradicting the assumption that (ΣnDef) > a.

Therefore Πn+1 Th = An+1, which is coded in Mn. As SSy(Mn) SSy(M), Πn+1 Th is coded also in M.

Now, by Theorem 4, (M,<) is expandable to a model of PA + Πn Th for every n ω.

Now, let us study a corollary. A theory T is called arithmetic if it has an axiomatization S such that S = {n θ(n)} for some formula θ(x) LPA. Recursive extensions of PA are examples of arithmetic theories. Also, there are complete arithmetic theories by the arithmetized completeness theorem.

Corollary 7.  
For any consistent arithmetic theory T PA, if MPA is resplendent and lcf(M \ ) > ω then there is NT such that (N,<)(M,<).

Since T is arithmetic, T is recursive in the set Πn Th for some n. Hence T is coded in M. Hence, by Theorem 4, (M,<) is expandable to a model of T.

However there is a proof that Πn Th is coded in M different from the one above. Indeed, resplendency implies recursive saturation and for any f : M there is a M such that n (f(n) > f(n) > a) because lcf(M \ ) > ω. Hence, M is arithmetically saturated, thus, applying the machinery of arithmetic saturation (Fact 3), we can conclude that SSy(M) is closed under jump, hence contains Πn Th for all n ω.

In the next section we shall investigate whether recursive saturation and uncountable lower cofinality give us more information about which sets are coded in M than just arithmetic saturation. The answer will be “No”.

We can also reformulate this question as follows. If M is recursively saturated and lcf(M \ ) > ω then SSy(M) is closed under jump. Does every countable Scott set closed under jump occur in this way? The answer will be “Yes”.

4. Do recursive saturation and uncountable lower cofinality say more about coding than arithmetic saturation?

Lemma 8. Let MPA be recursively saturated. Then M is arithmetically saturated if and only if for all a M , Cl(a) \ is bounded below in M \ .

Proof. Suppose, for all a M, Cl(a) \ is bounded below. Let f M code a function whose domain contains . For every n , f(n) Cl(f). If b M \ is such that Cl(f) > b then for all n , (f(n) > f(n) > b).

Let M be arithmetically saturated, c > . The type which says: F M codes a function F : [0,c]M with F(θ) = tθ(a) (where θ ranges over all formulas of LPA with two variables and tθ is the Skolem term defined by θ) is recursive, hence realized. But if Cl(a) \ is unbounded below then {F(θ)} (M \ ) is not separated from contradicting arithmetic saturation.

Let E = {x M there are no nonstandard definable points below x}. If a M \ Cl, Ea = {x M for all c Cl,c < x c < a}. By Lemma 8, E = and for any a such that < a < Cl \ , Ea = E. The following lemma establishes some homogeneity properties of Ea which will be important in the rest of this section.

Lemma 9. Let M be recursively saturated, a M \ Cl.

  1. If p(x,b¯) is realized by c Ea, c > Cl(b¯) Ea then for all x Ea there is y > x such that p(y,b¯).
  2. If p(x,b¯) is realized by c Ea, c < Cl(b¯) Ea then for all x Ea there is y < x such that p(y,b¯).

Proof. 1. Let Aupper = {x My Cl,a < y < x}. For an arbitrary e Ea, let us find d > e such that p(d,b¯). Let us show that for all θ(x,b¯) p(x,b¯),Mθ(y,b¯) for unboundedly-many y Ea. Consider the two cases. If A = {x AupperMθ(x,b¯)} is unbounded below then Mθ(y,b¯) for unboundedly-many y Ea by overspill. Otherwise, let k Cl, a < k < A. Define g = max x < kθ(x,b¯). We observe that g Cl(b¯), while c g < Aupper, which is a contradiction.

Thus for any e E, p(x,b¯) {x > e} is finitely satisfied. By recursive saturation, p(x,b¯) {x > e} is coded, hence realized.
2. Analogous proof.

Lemma 10. Let MPA be a countable arithmetically saturated model, < e < Cl \ . Then there is an elementary embedding h : M M such that for all x > , h(x) > e.

A forth-argument. Let us enumerate M as {a1,a2,,ai,}i<ω and build inductively a sequence {b1,b2,,bi,}i<ω with tp(b1,,bi) = tp(a1,,ai) and bi > e bi > for all i and define h(ai) = bi.

Suppose at stage i we already have tp(a1,,ai) = tp(b1,bi),e < Cl(b1,,bi) \ . By Lemma 8, Cl(a1,,ai+1) \ is bounded below. Let c < Cl(a1,,ai+1) \ . By Lemma 9 (2),

p(x,a1,,ai,c) = {θ(a1,,ai,x)Mθ(b1,,bi,e)}{x < c}

is satisfied, say, by e E.

As tp(a1,,ai,e) = tp(b 1,,bi,e), by recursive saturation, there is an elementary embedding (actually, an automorphism) h: M M such that h(a1) = b1,,h(ai) = bi,h(e) = e. Put bi+1 = h(ai+1). By construction, e < Cl(b1,,bi+1).

Hence, M has an elementary extension N M, NM and there is a N \ M such that < a < M. Since a union of an elementary chain of recursively saturated models is recursively saturated, we can repeat this extension ω1 times to obtain the following Theorem, which was promised earlier.

Theorem 11.  
Let X be a countable Scott Set. Then X is closed under jump if and only if there is a recursively saturated MPA, lcf(M \ ) > ω, SSy(M) = X.

The countability assumption cannot be dropped yet because for Scott sets X with cardX > ω1, the existence of a model MPA such that SSy(M) = X is still an open problem.

5. Automorphisms moving all nondefinable points

Now, as we are discussing arithmetic saturation, let us turn to automorphism groups where arithmetic saturation has profound consequences. We shall employ lemmas and methods of the previous section.

Fact 12. (Kaye, Kossak, Kotlarski [3]) If MPA is countable and arithmetically saturated then M has an automorphism which moves every nondefinable point.

Fact 13. (Kossak [6]) If MTh is countable and arithmetically saturated then there exists h Aut(M) such that for all x > , h(x) > x, i.e. h moves every nonstandard point upwards.

Fact 14. (Kossak, Schmerl [7]) If M is countable and arithmetically saturated then there is an automorphism f of M such that for every x > Cl, f(x) < x.

Notice that Fact 14 generalizes Fact 13. We are going to prove a theorem generalizing Fact 13 in a different direction and at the same time fusing it somehow with Fact 12.

Recall the notation E = {x M there are no nonstandard elements of Cl below x}. In general, if Th(M) = Th , there exists no h Aut(M) such that for all x Cl, h(x) > x.

Proof. Let a < b < e, e Cl \ ,h(a) = b. Then e a > e b, hence h(e a) > h(e b), hence e b > h(e b).

But what we can expect is the following Theorem.

Theorem 15.  
If MPA is countable and arithmetically saturated then there is h Aut(M) such that for all x E, h(x) > x and h moves every nondefinable point.

Since in the case of Th(M) = Th , we have E = M \ , this Theorem generalizes Fact 13. The proof uses Kossak’s method and the following two lemmas.

Lemma 16. (Kaye, Kotlarski) If M is arithmetically saturated, tp(a¯) = tp(b¯) and for any Skolem term t,

t(a¯) = t(b¯) t(a¯) Cl

then for any c M there is d such that tp(a¯,c) = tp(b¯,d) and for any Skolem term t,

t(a¯,c) = t(b¯,d) t(a¯,c) Cl.

Notice that Fact 12 follows from Lemma 16 by a back-and-forth argument.

Lemma 17.   Let MPA be recursively saturated.

  1. If c < Cl(a¯) \ then for any b there is b such that tp(a¯,b) = tp(a¯,b),
    c < Cl(a¯,b) \ .
  2. If c > Cl(a¯) E then for any b there is b such that tp(a¯,b) = tp(a¯,b),
    c > Cl(a¯,b) E.

Proof. 1) By Lemma 9, (1), there is d < Cl(a¯,b) such that tp(a¯,c) = tp(a¯,d). By recursive saturation, there is h : M M,h(a¯) = a¯,h(d) = c. Denote h(b) by b. As h is elementary, tp(a¯,b) = tp(a¯,b). As d < Cl(a¯,b), c < Cl(a¯,b).
2) Similar proof.

Proof. We shall construct a string of points {di}i unbounded above and below in E such that our future automorphism h takes di to di+1 which will guarantee that each point of E moves upwards: if a (di,di+1) then h(a) (hdi,hdi+1) = (di+1,di+2). Also, it obviously follows that there will be no h-fixed initial segment in E other than sup E and .

By Lemma 16 there are c0,c1 E such that tp(c0) = tp(c1) and (t(c0) = t(c1) t(c0) Cl), hence, considering the type {ϕ(x)Mϕ(c1)}{t(x) = t(c1)t(c1) Cl}, we deduce, using Lemma 9, that there are d0,d1 E such that

tp(d0) = tp(d1),

t(d0) = t(d1) t(d0) Cl,

Cl(d0) E < d1,

Cl(d1) \ > d0.

Let {si}iω be an enumeration of the whole of M \ Cl. By stage n we shall already have:

a¯ = a0,,a2n1,

b¯ = b0,,b2n1,

d¯ = dn,dn+1,,dn,dn+1

satisfying the following conditions:

tp(a¯,dn,,dn) = tp(b¯,dn+1,,dn+1),

dn < Cl(b¯,dn+1,,dn+1) E,

dn+1 > Cl(a¯,dn,,dn) E,

t(a¯,dn,,dn) = t(b¯,dn+1,,dn+1) t(a¯,dn,,dn) Cl.

(At stage n = 0, a¯ and b¯ are empty.)
Let b2n = sn. Let e < Cl(b2n,b¯,dn,,dn+1). By Lemma 16 (applied to the tuples (b¯,dn+1,,dn+1) and (a¯,dn,,dn) and the new point dn), the set of formulas

p(x) = {ϕ(a¯,x,dn,,dn)Mϕ(b¯,dn,dn+1,,dn+1)}

{t(a¯,x,dn,,dn) = t(b¯,dn,dn+1,,dn+1)

t(b¯,dn,dn+1,,dn+1) Cl}

is realized, hence, by Lemma 9 (2), is realized by a point less than e, hence, by Lemma 17 (2), is realized by a point dn1 < e such that

Cl(a¯,dn1,,dn) E < dn+1.

Let q(x) = {ϕ(a¯,dn1,,dn,x)Mϕ(b¯,dn,,dn+1,b2n}

{t(a¯,dn1,,dn,x) = t(b¯,dn,,dn+1,b2n)

t(b¯,dn,,dn+1,b2n) Cl}.

By Lemma 16, q(x) is realized, hence, by Lemma 17 (2), is realized by some point a2n such that

Cl(a¯,dn1,,dn,a2n) E < dn+1.

By construction,

t(a¯,dn1,,dn,a2n) = t(b¯,dn,,dn+1,b2n)

t(b¯,dn,,dn+1,b2n) Cl,

i.e., every nondefinable point of Cl(b¯,dn,,dn+1,b2n) moves. Let us show that if t(b ¯ ,dn,,dn+1,b2n) E then

t(a¯,dn1,,dn,a2n) < t(b¯,dn,,dn+1,b2n).

If t(b¯,dn,,dn+1,b2n) > dn+1 then t(a¯,dn1,,dn,a2n) (dn,dn+1) because Cl(a¯,dn1,,dn,a2n) E < dn+1. If t(b¯,dn,,dn+1,b2n) (di,di+1),i = n,,n, then t(a¯,dn1,,dn,a2n) (di1,di).
If t(b¯,dn,,dn+1,b2n) < dn then, by construction of dn1,
t(b ¯ ,dn,,dn+1,b2n) (dn1,dn), hence t(a¯,dn1,,dn,a2n) < dn1.
Let a2n+1 = sn. Using Lemmas 16, 9 (1), 17 (1), we choose dn+2 such that

tp(b¯,b2n,dn,,dn+2) = tp(a¯,a2n,dn1,,dn+1),

t(b¯,b2n,dn,,dn+2) = t(a¯,a2n,dn1,,dn+1)

t(a¯,a2n,dn1,,dn+1 Cl,

dn+2 > Cl(a¯,a2n,a2n+1,dn1,,dn+1) E,

dn1 < Cl(b¯,b2n,dn,,dn+2) E.

Now, using Lemmas 16 and 17 (1), we choose b2n+1 such that

tp(b¯,b2n,b2n+1,dn,,dn+2) = tp(a¯,a2n,a2n+1,dn1,,dn+1),

t(b¯,b2n,b2n+1,dn,,dn+2) = t(a¯,a2n,a2n+1,dn1,,dn+1)

t(a¯,a2n,a2n+1,dn1,,dn+1) Cl,

 and  dn1 < Cl(b¯,b2n,b2n+1,dn,,dn+2).

Having obtained the points ai, bi for all i ω, we observe that h: M M defined as h(ai) = bi for all i ω is an elementary isomorphism possessing the required properties.


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[2]   Friedman H. (1975). One hundred and two problems in mathematical logic. Journal of Symbolic Logic, 40, pp. 113-129.

[3]   Automorphisms of First-Order Structures. (1994). Ed. Kaye and Macpherson, Oxford University Press.

[4]   Kirby, L.A.S. and Paris, J.B. (1977). Initial segments of models of Peano’s axioms. Set Theory and Hierarchy Theory, V. Bierutowice, Poland.

[5]   Kaye, R., Kossak, R., Kotlarski, H. (1991). Automorphisms of recursively saturated models of arithmetic. Annals of Pure and Applied Logic, 55 pp. 67-91.

[6]   Kossak, R. (1991). Exercises in ‘back-and-forth’. Proceedings of the Nineth Easter Conference on Model Theory, Gosen.

[7]   Kossak, R., Schmerl, J. (1995). Arithmetically saturated models of arithmetic. Notre Dame Journal of Formal Logic, 36.

[8]   Pabion, J.F. (1982). Saturated models of Peano Arithmetic. Journal of Symbolic Logic, 47, pp. 625-637.


E-mail address: andrey@logic.pdmi.ras.ru

Received January 16, 2004; Revised version May 14, 2004