Saharon Shelah- Choiceless Polynomial Time Logic: Inability to Express

8/3/2019 Saharon Shelah- Choiceless Polynomial Time Logic: Inability to Express

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CHOICELESS POLYNOMIAL TIME

LOGIC: INABILITY TO EXPRESS

Saharon Shelah

Institute of MathematicsThe Hebrew University

Jerusalem, Israel

Rutgers University

Mathematics DepartmentNew Brunswick, NJ USA

Dedicated to my friend Yuri Gurevich

Abstract. We prove for the logic∼

CP T ime (the logic from the title) a sufficientcondition for two models to be equivalent for any set of sentences which is “small”(certainly any finite set ), parallel to the Ehrenfeucht Fraısse games. This enables

us to show that sentences cannot express some properties in the logic∼

CP T imeand prove 0-1 laws for it.

Key words and phrases. Finite model theory; Computer Science; Polynomial time logic;

choiceless; games.

Partially supported by the United States-Israel Binational Science Foundation. Publication634.

I would like to thank Alice Leonhardt for the beautiful typing.

Typeset by AMS -TEX

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Anotated Content

§0 Introduction

§1 The choiceless polynomial time logic presented

[We present this logic from a paper of Blass, Gurevich, Shelah [BGSh533]; where the intention is to phrase a logic expressing exactly the prop-erties which you can compute from a model in polynomial time withoutmaking arbitrary choices (like ordering the model).]

§2 The general systems of partial isomorphisms

[We define a criterion for showing that the logic cannot say too compli-cated things on some models using a family of partial automorphisms(not just real automorphisms) and prove that it works. This is a relative

of the Ehrenfeucht-Fraısse games, and the more recent pebble games.]

§3 The canonical example

[We deal with random enough graphs and conclude that they satisfy the

0-1 law for the logic∼CP T ime thereby proving the logic cannot express

too strong properties.]

§4 Relating the definitions in [BGSh 533] to the ones here

[We show that the definition in [BGSh 533] and the case ι = 7 here areessentially the same (i.e. we can translate at the cost of only in smallincreases in time and space).]

§5 Closing comments

[We present a variant of the criterion (the existence of a simple k-system).We then define a logic which naturally expresses it. We comment ondefining N t[M ] for ordinals.]



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CHOICELESS POLYNOMIAL TIME LOGIC: INABILITY TO EXPRESS 3

§0 Introduction

We deal here with choiceless polynomial time logic, introduced under the

name∼

CP T ime in Blass Gurevich Shelah [BGSh 533]; actually we deal withseveral versions. Knowledge of [BGSh 533], which is phrased with ASM (abstractstate machine), is not required except when we explain in §4 how the definitionshere and there fit. See there more on background, in particular, on ASM-s andlogic capturing P Time. The aim of this logic is to capture statements on a(finite) model M computable in polynomial time and space without arbitrarychoices. So we are not allowed to choose a linear order on M , but if P is a unarypredicate from the vocabulary τ M of M and P M has ≤ log2(M ) elementsthen we are allowed to “create” the family of all subsets of P M , and if e.g.(|P M |)! ≤ M we can create the family permutations of P M . Note that a

statement of the form∼CP T ime captures what can be computed in polynomial

time without arbitrary choices” is a thesis not a theorem. For a given model M ,

we consider the elements of M as urelements, and build inductively N t = N t[M ],with N 0 = M, N t+1 ⊆ N t[M ] ∪P(N t[M ]) but the definition is uniform and thesize of N t[M ] should not be too large, with the major case being: it has apolynomial bound.

So we should have a specific guide Υ telling us how to create N t+1 from N t,hence we should actually write N t[M, Υ]. In the simplest version (we called itpure) essentially Υ = {ψℓ(x, y) : ℓ < m0} and N t is a transitive finite set with M the set of urelements, N 0 = M, N t+1 = N t ∪ {{a : N t |= ψℓ(a, b)} : b ∈ ℓg(y)(N t)and ℓ < m0}; where each ψℓ is a first order formula in the vocabulary of M plus the membership relation ∈, (i.e. τ M ∪ {∈},) and N t has the relations of M and the relation ∈↾ N t. We stop when N t = N t+1, in the “nice” cases afterpolynomial time and space, and then can ask “N |= χ”? getting yes or no.

We consider several versions of the definition of the logic; this should helpus to see “what is the true logic capturing polynomially computable statementswithout making arbitrary choices”.

Our aim here is to deal with finite models and processes, but we dedicate aseparate place at the end to some remarks on infinitary ones and set theory. Wealso comment in this section on classical model theoretic roots; both, of course,can be ignored.

For a logic L it is important to develop methods to analyze what it can say.Usual applications of such methods are to prove that:

(a) no sentence in L expresses some property (which another given logiccan express so we can prove that they are really different)

(b) certain pairs of models are equivalent(c) zero-one laws.

For first order logic we use mostly elimination of quantifiers, E.F. (Ehrenfeucht-Fraısse ) games (see [ CK] or [Ho93] or Ebbinghaus and Flum [EbFl95]) and ? CK ?others; we are most interested in relatives of E.F. games. In finite model theorypeople have worked on the logic Lω,k (usually denoted by Lk

ω,ω) of first order



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4 SAHARON SHELAH

formulas in which every subformula has ≤ k free variables, a relative of the Lλ,κ

logics (see e.g. [ Di]). We can use even such formulas in L∞,ω but for equivalence ? Di ?of finite models this does not matter. For L∞,κ the relatives of E.F. games arecalled pebble games (see [EbFl95]).

The E.F. criterion (see Karp [Ka]) for the L∞,ω-equivalence of two modelsM 1, M 2 of the same vocabulary τ is the existence of a non-empty family F of partial one-to-one functions from M 1 to M 2 such that for every f ∈ F , i ∈ {1, 2}and a ∈ M i there is g ∈ F extending f such that i = 1 ⇒ a ∈ Dom(g)and i = 2 ⇒ a ∈ Rang(g) (a particular case is M = N ; there, of course,M, N are equivalent but the question is when are (M, a), (M, b) equivalent).Note that for finite models and even for countable models, this criterion impliesisomorphism. But if we restrict ourselves to first order sentences of quantifierdepth < k we can replace F by F ℓ : ℓ < k and above we say that for everyf ∈ F ℓ+1, j ∈ {1, 2}, a ∈ M j there is g ∈ F ℓ as there (this is the original E.F.game). For L∞,k this does not work but without loss of generality, f ∈ F &

A ⊆ Dom(f ) ⇒ f ↾ A ∈ F , and above we restrict ourselves to f ∈ F with|Dom(f )| < k. Now probably the simplest models are those with equality only:so every permutation of the (universe of the) model is an automorphism. So

using this group it is proved in [BGSh 533] that∼

CP T ime cannot say much on

such models, thus showing that the∼

CP T ime does not capture P Time logic, infact odd/even is not captured.

But in our case suppose we are given F , a family of partial isomorphismsfrom M 1 to M 2, we have to create such a family for N t[M 1, Υ], N t[M 2, Υ].

We answer here some questions of [BGSh 533]: get a 0-1 law, show that∼

CP T ime+ counting does not capture P Time.

Note that if P M is small enough then we can have e.g. Per(P M ), the group

of permutations of P M

, as a member of N t = N t[M, Υ] for t large enough: justin N 3t we may have the set of partial permutations of P M of cardinality < t.

We thank Yuri Gurevich and Andreas Blass for helpful comments.

Notation:1) Natural numbers are denoted by i,j,k,ℓ,m,n,r,s,t. We identify a naturalnumber t with {s : s < t} so 0 = ∅ and we may allow t = ∞ = the set of naturalnumbers (ω for set theorists). Let [m] = {1, . . . , m}.2) Let τ denote a vocabulary, i.e. a set of predicates P (each predicate witha given arity n(P )); we may also have function symbols, but then it is naturalto interpret them as partial functions, so better to treat them as relations andavoid function symbols. We may attach to each predicate P ∈ τ a group gP

permutations of {0, . . . , n(P ) − 1} telling under what permutations of its argu-ment places the predicate P is supposed to be preserved; if not specified gP is atrivial group.3) Let P,Q,R denote predicate symbols.4) Formulas are denoted by ϕ,ψ,θ,χ; usually θ, χ are sentences.5) Let L denote a logic and L (τ ) denote the resulting language, the set of L -formulas in the vocabulary τ .



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6) Let M, N denote models and let τ M = τ (M ) denote the vocabulary of M and let P M denote the interpretation of the predicate P ∈ τ M . So P M is arelation on M with arity n(P ) and if gP = gτ

P

is not trivial and σ ∈ gP , aℓ :

ℓ < n(P ) ∈ n(P )M , then aℓ : ℓ < n(P ) ∈ P M ⇔ aσ(ℓ) : ℓ < n(P ) ∈ P M .Models are finite if not said otherwise.7) Let |M | be the universe (= set of elements) of M and M the cardinality of |M |, but abusing notation we may say “of M ”.8) Let t, s denote functions from the set of natural numbers to the set of naturalnumbers ≥ 1 and let T denote a family of such functions. We may write t(M )instead of t(M ).9) For a function f and a set A let f ”(A) = {f (x) : x ∈ A and x ∈ Dom(f )}.[Why do we not write f (A)? Because f may act both on A and on its elements;this occurs for functions like G(f ), defined and investigated is §2.]

0.1 Discussion: Would we really like to allow t(M ) to depend not just on M ?

For the definition of the logic we have no problems, also for the criteria of equiva-lence of M 1, M 2 (except saying M 1 = M 1 & t1 = t2 ⇒ t1(M 1) = t2(M 2)).But this is crucial in proving a weak version of the 0-1 law, say for randomgraphs, to overcome this we need to “compensate” with more assumptions.

§1 The Choiceless polynomial time logic presented

Here we present the logic. How does we “compute” when a model M satisfiesa “sentence” θ? Note that the computation should not take too long and usetoo much “space” (in the “main” case: polynomial).

Informally, we start with a model M with each element an atom=ure-element,we successively define N t[M ] and N +t [M ], an expansion of N t[M ], t running on

the stages of the “computation”; to get N t+1[M ] from N t[M ] we add few familiesof subsets of N t, each family for some ψ consist of those defined by a formulaψ(−, a) for some a from N t[M ], and we update few relations or functions, bydefining them from those of the previous stage. Those are P t,ℓ for ℓ < m1. Wemay then check if a target condition holds, then we stop getting an answer:the model satisfies θ or fails by checking if some sentence χ holds; the naturaltarget condition is when we stop to change. Note that each stage increases thesize of N t[M ] by at most a (fixed) power, however in M steps we may haveconstructed a model of size 2M but this is against our intentions. So we shallhave a function t in M , normally polynomial, whose role is that when we havewasted too much resources (e.g. N t[M ] + t) we should stop the computationeven if the target condition has not occurred, in this case we still have to decide

what to do.This involves some parameters. First a logic L telling us which formulas are

allowable

(a) in the inductive step (the ψℓ(−, y) − s)

(b) in stating “the target condition” (we use standard ones: N +t+1 = N +t orP 0 = P 2 or c0 = c2) and the χ telling us if the answer is yes or no.



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Second, a family T of functions t (with domain the family of finite models) whichtell us when we should stop having arrived to the limit of allowable resources(clearly the normal family T is {tk : k < ω} where tk(M ) = M k). So for Υwhich describes the induction step (essentially how to get N t+1[M ] from N t[M ]),χ telling us the answer and t ∈ T we have a sentence θΥ,χ,t. We still have somevariants, getting a logic L Tι [L ∗] for each ι ∈ {1, 2, 3, 4, 5, 6, 7, 11, 22}.

In the case ι = 1 we ignore t, so M |= θΥ,χ,t iff for some t < ∞ the targetcondition N +t = N +t+1 holds, or let t = ∞ and for this t, χ is satisfied by N +t .This is a very smooth definition, but we have lost our main goal. We may restrictourselves to “good” sentences for which we always stop in “reasonable” time andspace.

In the case ι = 2, possibly we stop because of t before the target conditionholds; in this case we say “θΥ,χ,t is undefined for M ”. The case ι = 3 is likeι = 2 but we restrict ourselves to the so called “standard T”, where in N t[M ]we have the natural numbers < t, so we can ignore the “time” as a resource as

always N t[M ] ≥ t. The case ι = 4, is like ι = 3 but instead stopping whenN t[M ] is too large, we stop when at least one of the families of sets added toN t[M ] to form N t+1[M ] is too large. For ι = 5, is like the case ι = 2, but anadditional reason for stopping is t > t(M ). The case ι = 6 is as the case ι = 2separating the bounds on “space” (that is N t[M ]) and “time” (that is t), thecase ι = 7 is similar, not stopping for N +t = N +t+1. The cases ι = 11, ι = 22are like ι = 1, ι = 2 respectively, but using N t[M, Υ, t] (see Definition 1.1(c))instead N t[M, Υ] but for ι = 22 we separate t to two functions.

We treat as our main case ι = 3, see more 1.7.

More formally

1.1 Definition. 1) We are given a model M , with vocabulary τ = τ [0], τ finite

and ∈ not in τ , let τ + = τ [1] = τ ∪{∈}. Considering the elements of M as atoms= urelements, we define V t[M ] by induction of t : V 0[M ] = (M , ∈↾ M ) andclearly with ∈↾ M being empty (as we consider the members of M as atoms =“urelements”). Next V t+1[M ] is the model with universe V t[M ]∪{a : a ⊆ V t[M ]}(by our assumption on “urelements” we have a ⊆ V t[M ] ⇒ a /∈ M ) with thepredicates and individual constants and function symbols of τ interpreted as inM (so function symbols in τ are always interpreted as partial functions) and∈V t+1 [M ] is ∈↾ V t+1[M ].2)

(A) We say Υ = (ψ, ϕ, P ) is an inductive scheme for the logic L f.o. orthe language L f.o.(τ ) (where L f.o. is first order logic) if: letting m0 =ℓg(ψ), m1 = ℓg(ϕ) and τ

[2]= τ

[2][Υ] = τ

[1]∪ {P k : k < m1} we have

(a) P = P k : k < m1 is a sequence (with no repetitions) of predicatesand function symbols not in τ [1] (notationally we treat an n-place

function symbol as (n+1)-predicate); where P k is an mΥ(P k)-placepredicate. Let mΥ

1 be the function giving this information andwhether P k is a predicate or a function symbol (so have domain{0, . . . , m1 − 1})



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(b) ψ = ψℓ : ℓ < m0, ψℓ = ψℓ(x; yℓ) is first order in the vocabularyτ [2],

(c) ϕ = ϕk : k < m1, ϕk = ϕk(xk) is first order in the vocabulary τ 2with ℓg(x) = m(P k), moreover x = xi : i < m(P k).

(B) We say Υ is simple if

(d) each P k is unary predicate and and each ϕk(x) appears among theψℓ’s (with empty yℓ) and for every hereditary model N ∗ ⊆ V∞[M ],and a τ [2]–expansion N + of N ∗ we have {a : N + |= ϕℓ(a)} is amember of N ∗ or is ∅

(C ) We may write mΥ0 , mΥ

1 , ψΥℓ , mΥ, mΥ

1 , ϕΥℓ , ψΥ

ℓ , P Υℓ . We let

ΨΥ = {ψℓ(x, yℓ) : ℓ < mΥ0 }.

(D) We say Υ is predicative if each P k is a predicate; we may restrict ourselvesto this case for the general theorems. We say Υ is pure if m1 = 0

(E ) Υ is monotonic if y ∈ x is ψℓ for some ℓ < mΥ0 (this will cause N t below

to grow); no big loss if we restrict ourselves to such Υ. It is stronglymonotonic if in addition each ϕk(x) has the form P k(x) ∨ ϕ′k(x) (this willcause also each P k to grow)

(F ) Υ is i.c. if each P k is (informally an individual constants scheme) a zeroplace function symbol; in this case if P k is well defined we may write itas ck.

3) For M, τ = τ [0], τ [1], τ [2], a n d Υ = (ψ, ϕ, P ) as above, we shall define by

induction on t a submodel N t = N t[M ] of V t[M ] and P t = P t,k : k < m1and N +t [M ] and Pt,ℓ (for ℓ < m0) as follows; more exactly we are definingN t[M, Υ], P t,k[M, Υ] for k < m1,Pt,k[M, Υ] for k < m0.

We let N +t [M, Υ] = (N t[M, Υ], P t,0[M, Υ], . . . , P t,m1−1[M, Υ]).

Case 1: t = 0: N t[M ] = V 0[M ] and P t,k = ∅ (an mΥ(k)-place relation).

Case 2: t + 1: N t+1[M ] is the submodel of V t+1[M ] with set of elements the

transitive closure of M ∪

ℓ<m0

Pt,ℓ[M ] where we define Pt,k[M ] and P t+1,ℓ by:

Pt,ℓ[M ] =

{a ∈ N t[M ] : N +t [M ] |= ψℓ(a, b)} : b ∈ (ℓg(yi))(N t[M ])

P t,k = {a ∈ m(k)(N t[M ]) : N +t [M ] |= ϕk [a]}

but if P k is a function symbol,

P t,ℓ = {aˆb ∈ N t[M ] : N +t [M ] |= ϕℓ(a, b) & (∃!y)ϕℓ(a, y)}



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(so if Υ is simple, then for t > 1, ψk(x, y) ∈ L f.o.(τ [2]) is actually ψ′k(x, y, ct−1) ∈

L f.o.(τ [1])).

Case 3: t = ∞N t =

s<t

N s, P t,k =s<t

P s,k.

3A) If in addition t ∈ T and Υ is monotonic (see part (2) clause (E)) we defineN t[M, Υ, t], P t,ℓ[M, Υ, t] andPt,ℓ[M, Υ, t] as in part (3) except that the universeof N t+1[M, Υ, t] is the transitive closure of M ∪

{Pt,ℓ[M, Υ, t] : ℓ < m0 and

Pt,ℓ[M, Υ, t] has at most t(M ) members}.4) We say Υ is standard if some ψi guarantees that t ⊆ N t[M ] (so a naturalnumber s belongs to N t[M ] iff s < t; remember that we identify the naturalnumber t with the set {0, 1, . . . , t − 1}).5) Let q.d.(ϕ) be the quantifier depth of the formula ϕ.6) We may replace above first order logic by another logic L . We let L f.o. denote

first order, L card is defined just like first order logic except that we demand thatΥ is standard and defining inductively what is a formula, we allow the formationof formulas the form |{x : θ(x, y)}| = s. We let L card,T (on T see below) bedefined just like L f.o. but for each t ∈ T we allow the quantifier (Qtx)ϕ(x, y)with

N |= (Qtx)ϕ(x, a) iff t(|ure(N )|) < |{b : N |= ϕ(b, a]}|,

where ure(N ) is the set of urelements of N .6A) Lastly, let L f.o.+na be like L f.o. but we add one atomic formula |atoms| = xbeing interpreted as: the number of atoms is x. So this can be expressed inL card,T where T = {id}, id(n) = n.

1.2 Remark. Alternatively to L card: have a quantifier

(Qeqx1, x2)(ϕ2(x1, y1), ϕ2(x2, y2)),

which says that |{x : ϕ2(x, y1)}| = |{x : ϕ2(x, y2)}|.

In the definition below the reader can concentrate on ι = 3. The “t ≥ 2 & . . . ”is not a serious matter.

1.3 Definition. Let T be a set of functions t : N → N∪{∞} and L ∗ be a logic(L f.o. or L f.o.+na or L card usually) and let τ be a vocabulary. If t is constantly∞ we may write ∞.

We define for ι = 1, 2, 3, 4, 5, 6, 7, 11, 22 the logic L Tι [L ∗] below. For all of

those logics the set of sentences for a vocabulary τ called L Tι [L

∗

](τ ) is a subsetof Θ = Θτ = Θτ [L

∗, T] = {θΥ,χ,t : Υ an inductive scheme for L ∗(τ ), χ ∈L ∗(τ ) and t ∈ T}, (equal if not said otherwise). Also for most of those logicswe define the stopping time tι[M, Υ, t] or tι[M, Υ] (if t does not matter). Thesatisfaction relation for L Tι [L ∗](τ ) is denoted by |=ι. Also we write θΥ,χ insteadθΥ,χ,t if t does not matter. (We may let Dom(t) be the set of relevant structures,see 0.1).



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Case 1: ι = 1.

We let

tι[M, Υ] = Min{t : t ≥ 2 and N

+

t [M, Υ] = N

+

t+1[M, Υ]}.(If there is no such t ∈ N we let it be ∞ (i.e., ω for set theorists) and we couldalso have used “undefined”; note that t does not appear).

M |=ι θΥ,χ iff N +t [M, Υ] |= χ for t = tι[M, Υ].

Case 2: ι = 2.

We let tι[M, Υ, t] = Min{t : N t+1[M, Υ] + (t + 1) > t(M ) or t ≥ 2 &N +t [M, Υ] = N +t+1[M, Υ]} and

(a) if for t = tι[M, Υ, t] we have t ≥ 2 & N +t [M, Υ] = N +t+1[M, Υ] then

θΥ,χ,t is true or false in M iff N +t [M, Υ] |= χ or N +t [M, Υ] |= ¬χ re-spectively and we write M |=ι θΥ,χ,t or M |=ι ¬θΥ,χ,t respectively, (so

¬θΥ,χ,t is equivalent to θΥ,¬χ,t).(b) if tι[M, Υ, t] = t(M ) + 1 we say “M |=ι θΥ,χ,t is undefined” and we

say “the truth value of θΥ,χ,t in M is undefined”.

Case 3: ι = 3.

As in Case 2 but we restrict ourselves to standard Υ, see Definition 1.1(4), andlet tι[M, Υ, t] = Min{t : N t+1[M, Υ] > t(M ) or t ≥ 2 & N +t [M, Υ] =N +t+1[M, Υ]} and define |=ι as in Case 2.

Case 4: ι = 4.As in Case 2 but we restrict ourselves to standard Υ and let:

tι[M, Υ, t] = Min{t :for some k < m0 the set Pt+1,k[M, Υ]

has > t(M ) members or

t ≥ 2 & N +t [M, Υ] = N +t+1[M, Υ]}

(so it can be ∞; but by a choice of e.g. ϕ0 we can guaranteePt,0 = {0, . . . , t−1}so that this never happens) and define |=ι as in Case 2.

Case 5: ι = 5.As in Case 2 but we restrict ourselves to standard Υ and tι[M, Υ, t] = Min{t :

t > t(M ) or for some k < m0 the set Pt+1,k[M, Υ] has > t(M ) membersor t > 2 & N +t [M, Υ] = N +t+1[M, Υ, t]}.

Case 6: ι = 6.As in the case ι = 2 but

tι[M, Υ, t] = Min{t : t > ttm(M ) or N t[M, Υ] > tsp(M )

or N +t [M, Υ] = N +t+1[M, Υ]}



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where

ttm(n) = t(2n), tsp(n) = t(2n + 1)

and, of course, tm, sp stand for time and space, respectively. So we may replace tby two functions ttm, tsp and write sentences as θΥ,χ,t′,t′′ (similarly for ι = 7, 22).

Case 7: ι = 7.We define tι[M, Υ, t] = Min{t : t > ttm(M ) or N t[M, Υ] > tsp(M ) or

≥ 2 & N +t [M, Υ] = N +t+1[M, Υ]}

and let:M |=ι θΥ,χ,t iff N t[M, Υ] |= χ for t = tι[M, Υ].

(so unlike cases 2-6, the truth value is always defined)

Case 11: ι = 11.As in the case ι = 1, but we use N t[M, Υ, t], P t[M, Υ, t] (see Definition

1.1(3A)).

Case 22: ι = 22.As in the case ι = 2, but we use N t[M, Υ, twd], P t[M, Υ, twd] where twd ∈ T

is defined by twd(n) = t(2n) (wd for width) and define tι by

tι[M, Υ, t] = Min{t :N t+1[M, Υ, twd] + (t + 1) > tht(M ) or

t ≥ 2 & N +t [M, Υ, twd] = N +t+1[M, Υ, twd]}

where (ht for height)

tht ∈ T is defined by tht(n) = t(2n + 1).

We may write twd, tht instead of t.

1.4 Remark. Alternatively:

Case 10 + ι: For ι = 1, . . . , 7.Like the case ι but we use N t[M, Υ, t], P t[M, Υ, t] and let t10+ι = tι.

Case 20 + ι: ι = 1, . . . , 7.As in the cases ι = 1, . . . , 7, but we use N t[M, Υ, twd], P t[M, Υ, twd], where

twd ∈ T is defined by t(n) = t(2n) (wd for width) and we replace t bytht, tht(n) = t(2n + 1), where for x = tm,sp we derive from tht the functions

tht,x20+ι = (tht)x.



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1.5 Definition. 1) In Definition 1.3 we say “θΥ,χ,t is ι-good” when: for everyfinite model M one of the following occurs:

(a) ι = 1 and tι[M, Υ] < ∞

(b) ι ∈ {2, 3, 4, 5, 6, 7} and in the definition of tι[M, Υ, t] always the lastpossibility occurs and not any of the previous ones

(c) ι = 11 as in the case ι = 1 using N t[M, Υ, t], P t[M, Υ, t]

(d) ι = 22, as for ι = 2 using N t[M, Υ, t], P t[M, Υ, t].

2) Let L Tι (L ∗)good be the logic L Tι (L ∗) restricted to ι-good sentences.3) If in θΥ,χ,t we omit χ we mean P t,0 = the set of atoms.4) We say that M |=ι θΥ,χ,t is in a good way if this case of part (1) holds

1.6 Remark. We can replace in cases ι = 2,3,4 clause (b), the statement N +t [M, Υ]

= N +t+1[M, Υ] by a sentence χ1.

1.7 Discussion: 0) Note that considering several versions should help to see howcanonical is our logic.1) The most smooth variant for our purpose is ι = 4, and the most naturalchoice is L ∗ = L card or L ∗ = Lcard,T, but we are not less interested in thechoice L ∗ = L f.o.,L ∗ = L f.o.+na. From considering the motivation the mostnatural T is {nm : m < ω}, and ι = 3.2) For e.g. ι = 1, 2, 3 some properties of M can be “incidentally” expressed bythe logic, as the stopping time gives us some information concerning cardinality.For example let Y be a complicated set of natural numbers, e.g. non-recursive,and let t∗ ∈ T be: t(M ) is M + 10 if M ∈ Y and t(M ) = M + 6 if

M /∈ Y . We can easily find θ = θΥ,χ,t∗ , with Υ a standard induction schemesuch that it stops exactly for t = 8 and χ saying nothing (or if you like sayingthat there are 8 natural numbers). Clearly for ι = 2, 3 we have M |=ι θΥ,χ,t∗

if M ∈ Y and “not M |=ι θΥ,χ,t∗” if M /∈ Y . Of course, more generally,we could first compute some natural number from M and then compare it witht(M ). This suggests preferring the option |=ι undefined in clause (b) of case2, Definition 1.3 rather than false.3) If you like set theory, you can let t be any ordinal; but this is a side issuehere; see end of §5.

Implicit in 1.3 (and an alternative to 1.3) is (note: an ( M, Υ)-candidate (N, P )is what looks like a possible N t[M, Υ] and a (Υ, t)-successor of it is what looks

like N t+1[M, Υ]):1.8 Definition. Let M, Υ as in Definition 1.3 be given.1) We say (N, P ) is an M -candidate or (M, Υ)-candidate if:

(a) N is a finite transitive submodel of t

V t[M ] which includes M , expanded

by the relations of M (so it is a (τ M )[1]-model)



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(b) P = P k : k < m1, P k an mΥ(k)-relation on N or a partial (mΥ(k) −1)-place function on N when P k is a predicate or a function symbolrespectively.

In fact here the only information from Υ used is mΥ1 of Υ, so we may write

“(M, mΥ1 )-candidate”.

2) We say (N ′, P ′) is the (Υ, t)-successor of (N, P ) if (N ′, P ′), (N, P ) satisfieswhat N +t+1[M, Υ, t], N +t [M, Υ, t] satisfies in Definition 1.1(3A), so

|N ′| = is the transitive closure of M ∪

ℓ<m1

Aℓ[N, Υ, t],

where Aℓ = Aℓ[N, Υ, t] is Pℓ[N, Υ] =

{a : (N, c) |= ψℓ(a, b)} : b ∈ ℓg(gℓ)N

if this family has ≤ t(M ) members or is equal to N and Aℓ is empty otherwise.2A) We say (N ′, P ) is the Υ-successor of (N, P ) if (N ′, P ′), (N, P ) satisfies what

N +t+1[M, Υ], N +t [M, Υ] satisfies in Definition 1.1(3); this means just that (N ′, P ′)is the (Υ, ∞)-successor of (N, P ).2B) If Υ is pure (i.e. mΥ

1 = 0), actually only ψΥ count and we may replace Υby ψΥ.2C) We say that (N, P ) is a (M, Υ)+-candidate if it is an (M, Υ)-candidate andthe sets ∅, |M | (= set of atoms) belongs to N .3) We define N t = N t[M, Υ, t] and P t = P t[M, Υ, t] by induction on t as follows:

for t = 0 it is M (i.e. with P t,k = ∅),

for t + 1, (N t+1, P t+1) is the (Υ, t)–successor of (N t, P t), see below 1.9,

for t = ∞ we take the union.

1.9 Claim. 1) If (N, P ) is an (M, Υ)-candidate, it has exactly one (Υ, t)-successor (and exactly one Υ-successor).2) The pair (N t[M, Υ, ∞], P t[M, Υ, ∞]) defined in Definition 1.8(3), is equal tothe pair (N t[M, Υ], P t[M, Υ]) defined in Definition 1.1(3).3) If Υ is monotonic, (N ′, P ′) the Υ-successor of (N, P ) where both are (M, Υ)+-candidates, then N ⊆ N ′; if Υ is also standard then N ⊂ N ′.4) If Υ is strongly monotonic (see Definition 1.1(2)(E)) and (N ′, P ′) is the Υ-successor of (N, P ) both are (M, Υ)+-candidates, then N ⊆ N ′ and P ℓ ⊆ P ′ℓ for ℓ < mΥ

1 .

There are many obvious inclusions between the variants of logics by natural

translations. We mention the following claim which tells us that there is no realharm if we restrict ourselves to pure Υ’s.

1.10 Claim. 1) Assume the Υ is an inductive scheme in L f.o.(τ +), χ a sen-tence in L f.o.(τ +). Then we can find a pure inductive scheme Υ∗ in L f.o.(τ +)and r∗, r∗∗ and p∗ and sentences θ∗, χ∗1 and formulas ϕ∗(x), ϕ∗k(xk) for k <mΥ

1 , ℓg(xℓ) = mΥ(k) in L f.o.(τ ) such that:



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⊠ for every τ -model and t we have, if t∗ = r∗∗ + r∗t then:

(a) the set N t[M, Υ] is {a ∈ N t∗ [M, Υ∗] : N t∗ [M, Υ∗] |= ϕ∗[a]},

(b) P t,k[M, Υ] = {a ∈m

(N t∗ [M, Υ∗

]) : N t∗ [M, Υ∗

] |= ϕ∗k[a]}, where

m = mΥ(k),

(c) N t[M, Υ] |= χ iff N t∗ [M, Υ∗] |= χ∗,

(d) N +t [M, Υ] = N +t+1[M, Υ] (i.e. it stops) iff N t∗ [M, Υ∗] |= θ∗ iff N t∗ [M, Υ∗] = N t∗+1[M, Υ∗],

(e) N t∗ [M, Υ∗] has exactly p∗(N t[M, Υ]) elements, and N t∗+r[M, Υ∗]has ≤ p∗(N t[M, Υ]) when r < r∗ and p∗ an integer polynomial,

(f ) if Υ is standard then so is Υ∗.

2) Similarly using a logic L ∗ if it is closed under first order operations and substitutions.

Remark. We can similarly deal with N t[M, Υ, t], but then we have to deal withsome form of cardinality quantifiers, etc.

Proof. For simplicity we assume that Υ is standard. Now for every (M, mΥ1 )–

candidate (N, P ) we shall define a M -candidate N ∗ = N ∗(N,P )

. We shall have

N +r∗∗+r∗t[M, Υ∗] is N ∗N +t [M,Υ]

.

The set of natural numbers of N ∗ is {s : s < r∗∗ + r∗t}. The universe of N ∗

is the union of the following sets:

(a) N

(b) {{x, r∗

t} : x ∈ N }(used to define N ),let ax,k,m =: {x, r∗t + 1 + k, r∗t + 1 + mΥ

1 + m} for x ∈ N,k < mΥ1 , m <

mΥ(k)(used to help to code P t,k)

(c) {{axm,k,m : m < mΥ(k)} ∪ {r∗t + 1 + k, r∗t + 1 + mΥ1 + i} :

k < mΥ1 , x0, . . . , xmΥ(k)−1 ∈ N and i ∈ {0, 1} and

i = 1 ⇔ xm : m ∈ mΥ(k) ∈ P t,k}

(d) some more elements to take care of the

“ N t∗ [M, Υ∗] has exactly p∗(N t[M, Υ]) elements ”

(if we agree to N t∗

[M, Υ∗

] has exactly p(N t[M, Υ∗

], t) then this is notnecessary).

The rest should be clear. 1.10

We try to sort out some of the relations between these logics by checking whentwo variants of a sentence say related things, for quite many ι1, ι2.



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1.11 Claim. Let ι1, ι2 ∈ {1, . . . , 7} and θℓ = θΥ,χ,tℓ ∈ L T(L ∗)(τ ) for ℓ = 1, 2and consider

(α) θ2 is ι2-good implies θ1 is ι1-good,(β ) for every (finite) τ -model M, (M |=ι2 θ2) ⇒ (M |=ι1 θ1),

(β )− if θℓ is ιℓ-good for ℓ = 1, 2 then M |=ι2 θ2 ⇒ M |=ι1 θ1.

In the following clauses we list cases which holds under various conditions.

(A)(α) + (β ) if ι1 = ι2 and t1 < t2,

(B)(α) + (β ) if ι1 = 2, ι2 = 3, Υ is standard and t1 ≥ 2t2,

(C )(α) + (β ) if ι1 = 3, ι2 = 2, Υ is standard and t1 ≥ t2,

(D)(α) + (β ) if ι1 = 4, ι2 = 3, Υ is standard and t1 ≥ t2,

(E )(α) + (β )− if ι1 = 3, ι2 = 4, Υ is standard and t1 is large enough,

(F )(α) + (β ) if ι1 = 5, ι2 = 3, Υ is standard and t1 ≥ t2,

(G)(α)+(β ) if ι1 = 3, ι2 = 5, Υ is standard and (∀n)[t1(n) ≥ n+mΥ0 t2(n)t2(n)],

(H )(α) + (β )− if ι1 = 1, ι2 = 2,

(I )(α) + (β )− if ι1 = 2, ι2 = 1 and t1 is large enough,(i.e. t1(n) > Max{tι2 [M, Υ, t2] : M a τ -model with universe [n] and tι2 [M, Υ, t2] < ∞}; note that

Max is taken on a finite set)

(J )(α) + (β ) if ι1 = 6, ι2 = 2 and t1(2n) ≥ t2(n), t1(2n + 1) ≥ t2(n),

(K )(α) + (β ) if ι1 = 2, ι2 = 6 and t2 ≥ t1(2n) + t1(2n + 1),

(L)(α) + (β ) if ι1 = 7, ι2 = 6 and t1 ≥ t2

(note: after “good” stopping, nothing changes) t1 is largeenough

Proof. Straightforward.

1.12 Conclusion. 1) Assume that T satisfies

(∗) (∀s ∈ T)(∀m)(∃t ∈ T)(∀n)(t(n) ≥ n + m(s(n))2).

Then the logics L Tι (L ∗)good for ι = 2, 3, 4, 5, 6 are weakly equivalent whereL 1,L 2 are weakly equivalent if L 1 ≤wk L

2 and L 2 ≤wk L 1; where L 1 ≤wk

L 2 means that for every sentence θ1 ∈ L 1 there is θ2 ∈ L 2 such that for everyM we have1 M |= θ1 implies M |= θ2

2) If in addition T consists of integer polynomials and L ∗

∈ {L f.o.+na

,L card

}we can add ι = 7.

1.13 Remark. In part 1.12(2) we can replace the assumption on t demandingonly that:

1note that if teh truth value of θ1 in M is undefined, then the implication is trivial.



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(∗) for every t ∈ T view t as (ttm, tsp) there is s ∈ T such that we can“compute” ttm(n), tsp(n) is stm(n) time, ssp(n) space in the relevantsense (using the given L ∗, etc.).

1.14 Claim. The family of good sentences in the logic L Tι (L ∗f.o.), that is thelogic L Tι (L ∗f.o.)

good, is closed under the following: the Boolean operation, (∃x),and substitution (that is, up to equivalence) when at least one of the following holds

(∗)1 ι = 1

(∗)2 ι ∈ {2, 3, 5} and T satisfies (∗) of 1.12

(∗)3 ι = 11

(∗)4 ι = 22 and for each t ∈ T, T[twd] =: {sht : s ∈ T, swd = twd} satisfies(∗) of 1.12.

Proof. Straight.

§2 The general systems of partial isomorphisms

Though usually our aim is to compare two models M 1, M 2 we first concentrateon one model M ; this, of course, gives shorter definitions.

Our aim is to have a family F of partial automorphisms as in Ehrenfeucht-Fraıssegames (actually Karp), of the model M we analyze, not total automorphismwhich is too restrictive. But this family has to be lifted to the N t’s. Hencetheir domains (and ranges) may and should sometimes contain an element of high rank. It is natural to extend each f ∈ F to Gt(f ), a partial automorphismof N t. So we should not lose anything when we get up on t. The solution isI ⊆ {A : A ⊆ M } closed downward and F (could have used F ℓ : ℓ < m1),a family of partial automorphisms of M . So every x ∈ N t will have a supportA ∈ I and for f ∈ F , its action on A determines its action on x, (Gt(f )(x)in this section notation). It is not unreasonable to demand that there is thesmallest support, still this demand is somewhat restrictive (or we have to addimaginary elements as in [Sh:a] or [Sh:c], not a very appetizing choice here).But how come we in stage t + 1 succeed to add “all sets X = X ℓ,b” definable by

ψℓ(x, b) for some sequence b ∈ ℓg(b)N t? Let m be such that b = b1, . . . , bm.The parameters b1, . . . , bm each has a support say A1, . . . , Am resp., all in

I ; so when we have enough mappings in the family F , the new set has in

some sense the support A =

mℓ=1

Aℓ, in the sense that suitable partial mappings

act as expected. So if y ∈ N t has support B (BRy in this section notation),f ∈ F , A ∪ B ⊆ Dom(f ) and f ↾ A = idA, then the mapping Gt(f ) which f induces in N t will satisfy y ∈ X ℓ,b ⇔ (G(f ))(y) ∈ X i,b.

But we are not allowed to increase the family of possible supports and A thougha kind of support is probably too large: in general, I is not closed under unions.



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16 SAHARON SHELAH

But, if we add X = X ℓ,b we have to add all “similar” X ′ = X ℓ,b′. Recall thatnecessarily our strategy is to look for a support A′ ∈ I for X ℓ,b. So we like to findA′ ∈ I which is a support of X , that is such that that if f ∈ F , A∪A′ ⊆ Dom(f ),then f ↾ A induces a mapping of X i,b to some X i,b′ , which when f ↾ A′ =idA′, satisfies that X i,b′ will be equal to X i,b thus justifying the statement

“A′ supports X .” How? We use our bound on the size of the computation.So we need a dichotomy: either there is A′ ∈ I as above or the number of setsX i,b′ defined by ψi(x, b′) varying b′ is too large!!On this dichotomy hangs the proof.

However, we do not like to state this as a condition on N t but rather on M .We do not “know” how ψℓ(x, b′) will act but for any possible A′ this inducesan equivalence relation on the set of images of A′ (for this F has to be largeenough).

Actually, we can ignore the ψℓ’s and develop set theory of elements demandingeach has a support in I through F . Now we break the proof to definition and

claims.We consider several variants of the logic: the usual variant to make preserva-

tion clear, and the case with the cardinality quantifier. We use one F but wecould have used F ℓ : ℓ ≤ k′; in this case actually, for much of the treatmentonly F 0 would count. The relevant basic family of partial automorphisms isdefined in 2.1. Note that the case with cardinality logic, with a stronger as-sumption is clearer, if you like to concentrate on it, ignore 2.1(4) and read inDefinition 2.3 only part (1), ignore 2.9 but read 2.10, ignore 2.17, 2.20 but read2.18, ignore 2.22, 2.24 but read 2.23.

2.1 The Main Definition. 1) We say Y = (M,I,F ) is a k-system if

(A) I is a non empty family of subsets of |M | (the universe of the model M )closed under subsets and each singleton belongs to it[hint: intended as the possible supports of elements N t[M, Υ] and as first approxi-

mation to the possible supports of the partial automorphisms of M , where M is the

model of course; the intention is that M is a finite model]

Let I [m] =: {mℓ=1

Aℓ : Aℓ ∈ I for ℓ = 1, . . . , m}

(B) F is a non empty family of partial automorphisms of M such that f ∈F & A ⊆ Dom(f ) & A ∈ I ⇒ f ”(A) ∈ I (recall f ”(A) = {f (x) :x ∈ A ∩ Dom(f )};F is closed under inverse (i.e. f ∈ F ℓ ⇒ f −1 ∈ F ℓ)and composition and restriction (hence, together with (D) clearly B ∈I [k] ⇒ idB ∈ F )

(C ) if f ∈ F then Dom(f ) is the union of ≤ k members of I (D) if f ∈ F and A1, . . . , Ak−1, Ak ∈ I and ℓ ∈ {1, . . . , k − 1} ⇒ Aℓ ⊆

Dom(f ), then for some g ∈ F we have

f ↾k−1ℓ=1

Aℓ ⊆ g



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Ak ⊆ Dom(g)

2) Assume Y is a k-system and B ∈ I [m], m ≤ k − 2 and A ∈ I

(α) let H Y (B, A) = {g ∈ F : Dom(g) ⊇ B ∪ A and idB ⊆ g}

(β ) E Y (B, A) is the family of equivalence relations E on H Y (B, A) such that:

(∗) if g1, g2, g3, g4 ∈ H Y (B, A) and f ∈ F satisfies idB ⊆ f, g”1(A) ∪g”2(A) ⊆ Dom(f ) and g3 ⊇ f ◦(g1 ↾ (B ∪A)), g4 ⊇ f ◦(g2 ↾ (B ∪A)then g1Eg2 ⇔ g3Eg4.

(this tells you in particular that only g ↾ A matter for determining g/E )

3)

(α) Let H I,m = H Y ,m be the family of functions h from [m] = {1, . . . , m}

to SeqI = {a : a list with no repetitions some A ∈ I ; we can look at aas a one-to-one function from [0, ℓga) onto A}; of course for f ∈ F anda, b ∈ SeqI the meaning of f (a) = b is ℓg(a) = ℓg(b) and f (ai) = bi fori < ℓg(a). Let SeqI,A = {a ∈ SeqI : Rang(a) = A}. For m = 1 we mayidentify h : [m] → SeqI with h(1) so H I,1 is identified with SeqI .

(β ) for h ∈ H I,m and f ∈ F such that

i∈[m]

Rang(h(i)) ⊆ Dom(f ) we define

h′ = f ∗ h as the following function: Dom(h′) = [m] and h′(i) = f ◦ (h(i))

(γ ) let 2r + s ≤ k; for B ∈ I [s] and 1 ≤ m ≤ r let E 0I,B,m = E 0Y ,B,m be the

following 2-place relation on {h : h : [m] → SeqI }:h1E 0I,B,mh2 iff for some f ∈ F , idB ⊆ f and h2 = f ∗ h1.

If B = ∅ we may omit it; similarly m = 1

(δ) for 2m + s ≤ k and B ∈ I [s] let E I,m(B) = E Y ,m(∅) be the family of equivalence relations such that for some h∗ ∈ H I,m , E is an equivalencerelation on the set {h ∈ H I,m : hE 0I,mh∗} which satisfies:

(∗) if h1, h2, h3, h4 ∈ H I,m , f ∈ F , idB ⊆ f , h2 = f ∗ h1, h4 = f ∗ h3

and

{Rang(h1(i)) ∪ Rang(h1(i)) : i ∈ [m]} ⊆ Dom(f ),then h1Eh3 ≡ h2Eh4.If B = ∅ we may omit it. If m = 1 we may omit it.

4) We say that a k-system Y = (M,I,F ) is (t, r)-dichotomical2 when

⊠ if 1 ≤ m ≤ r and E ∈ E I,m(∅), then (β )1 ∨ (β )2 where

(β )1 there is A ∈ I which satisfies:if h1, h2 ∈ H I,m , f ∈ F , idA ⊆ f and h2 = f ∗ h1 then h1Eh2

(β )2 the number of E -equivalence classes is > t(M ).

2note that this is how from “there are not too many” we get “there is a support in I ”



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If we omit r (and write t-dichotomical) we mean r = [k/2], k ≥ 3.

Note that in parts (3), (4) without loss of generality h is one-to-one.

∗ ∗ ∗

2.2 Remark. 1) However, if we shall deal with L ∗ = L card or L ∗ = Lcard,T

we naturally have to require that f ∈ F preserve more. Whereas the (t, r)-dichotomy is used to show that either we try to add too many sets to someN t[M, Υ] or we have support, the “counting (k, r)-system” assure us that thelifting of f preserves the counting quantifier, and the medium (t, r)-dichotomywill be used in the maximal successor, see 2.24.

2) It causes no harm real in 2.1(3)(γ ), (δ) and similarly later, to restrict our-selves to e.g. r + s ≤ k/100, k > 400.

2.3 Definition. 1) We say Y = (M,I,F ) is a counting (or super) (k, r)-system

if: Y is a k-system and

(∗)1 Assume that 0 ≤ m ≤ r and for ℓ = 1, 2 we have Bℓ ∈ I [m] andE ℓ ∈ E Y ,1(Bℓ). If f ∈ F , f maps B1 onto B2 and f maps E 1 to E 2 (see2.4(1)), then |Dom(E 1)/E 1| = |Dom(E 2)/E 2|.

(This should be good for analyzing the model N t[M, Υ, t]). If we omit r (writecounting k-system) we mean r = k − 2, k ≥ 3.2) We say that the k-system Y = (M,I,F ) is medium (t, k , r)-system if

(∗)2 Assume that 1 ≤ m ≤ r and for ℓ = 1, 2 we have Bℓ ∈ I [m] andE ℓ ∈ E Y ,1(Bℓ). If f ∈ F , f maps B1 onto B2 and f maps E 1 to E 2 (see

2.4), then |Dom(E 1)/E 1| = |Dom(E 1)/E 2| or both are > t(M ).3) We omit r if r = k − 2 ≥ 1 (see 2.8 below).

Note that 2.4 is closed to 2.8 and 2.7(2)-(4).

2.4 Observation. Let Y = (M,I,F ) is a k-system.1)

(α) i f 2m + s ≤ k and B ∈ I [s], then E 0Y ,B,m is an equivalence relation

satisfying (∗) of Definition 2.1(3)(δ)

(β ) the following two conditions on B ∈ I [s], m ≤ (k − s)/2, s ≤ k and G areequivalent:

(i) G is an equivalence class of E 0I,B,m

(ii) G is the domain of some E ∈ E I,m(B)

(γ ) if k∗ = 2m + s ≤ k and F ∗ = {f ↾ A : f ∈ F , A ∈ I [k∗]}, andY ∗ = (M,I,F ∗) then Y ∗ is a k∗-system and for each B ∈ I [s] we haveE Y ,m(B) = E Y ∗,m(B) and E 0

Y ,B,m = E 0Y ∗,B,m.



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(δ) if B1, B2 ∈ I [k − 2], f ∈ F , f maps B1 onto B2 and g′1, g′′1 ∈ SeqI ,Rang(g′1) ∪ Rang(g′2) ⊆ Dom(f ), then g′2 =: f ◦ g′1 belongs to SeqI andg′′

2

= f ◦ g′′

1

belongs to SeqI and g′

1

E 0I,B1,1

g′′

1

⇔ g′

2

E 0I,B2,1

g′′

2(ε) If B1 ∈ I [k − 2], A2 ∈ I, f ∈ F , B1 ⊆ Dom(f ), B2 = f ”(B1), then we

can define F eqf,B1,B2

(E ) ∈ E I,1(B2) for E ∈ E I,1(B1) (the image of E by

f ) by: if f ↾ B1 ⊆ g ∈ F , a1, a2 ∈ SeqI , g maps a1 to a∗1 and g maps a2

to a∗2 then a1E a2 ⇔ a∗1F eqf,B1,B2

(E )a∗2.

2) Let 2r + s ≤ k,(δ), (ε), parallely to part (1) with m ≤ r, hℓ ∈ H I,m , Bℓ ∈ I [s].

Proof. Straight, e.g.:

Part (1), Clause (α): We use: F contains idC wherever C ∈ I [k] (for reflexiv-ity), F closed under inverting (for symmetry) and is closed under composition(for transitivity).

2.4

2.5 Discussion 1) In the system Y = (M,I,F ) we deal only with partial au-tomorphisms of M , we need to lift them to the models N t[M, Υ] or actuallyN +t [M, Υ] or N +t [M, Υ, t] appearing in Definition 1.3; this motivates the follow-ing definition 2.6. (We more generally define liftings to (M, Υ)-candidates).2) Note that here probably it is more natural if in the definition of k-system Y ,we replace the relations “f ⊆ g”, “f = g ↾ A”, “f ⊇ g ↾ A” on F by abstractones (so F will be an index set). Also in Definition 2.6 we could demand moreproperties which naturally holds (similarly in Definition 2.13, e.g. if you satisfythe properties of “a set B is Z-support of x” you are one).

2.6 Definition. 1) Let Y = (M,I,F ) be an k-system, M a τ -model, Υ is aninductive scheme for L ∗(τ +) and m1 = mΥ

1 .We say that Z = (N, P , G , R) = (N Z, P Z, GZ, RZ) is a Υ-lifting or m1-lifting

of Y if

(a) (N, P ) is an (M, m1)-candidate so N is a transitive submodel of settheory i.e. of V ∞[M ] with M as its set of urelements and the relationsof M (see Definition 1.4(1))

(b) G is a function with domain F

(c) for f ∈ F

(α) G(f ) is a function with domain ⊆ N, f ⊆ G(f ), moreover f =G(f ) ↾ M and

(β ) G(f ) is a partial automorphism of N

(d) if f ∈ F , g ∈ F , f ⊆ g then G(f ) ⊆ G(g)

(e) R is a two-place relation written xRy such thatxRy ⇒ x ∈ I & y ∈ N [we say: x is a Z-support of y]



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20 SAHARON SHELAH

(f )

(α) if ARy and f ∈ F , A ⊆ Dom(f ), then

y ∈ Dom(G(f )) and f ↾ A = idA ⇒ G(f )(y) = y(β ) if f ∈ F and y ∈ Dom(G(f )) (hence y ∈ N ) then some a Z-support

of y is included in Dom(f )

(g) (∀y ∈ N )(∃A ∈ I )ARy[i.e. every element of N has a Z-support]

(h) if A ∈ I and A ⊆ Dom(f ), y ∈ Dom(G(f )) thenARy ⇔ f ”(A)R(G(f )(y))

(i) for f ∈ F we have G(f −1) = (G(f ))−1

( j) for f 1, f 2 ∈ F , f = f 2 ◦ f 1 we have3 G(f ) ⊆ G(f 2) ◦ G(f 1)

(k) if a ∈m1(ℓ)

(Dom(f )) and f ∈F

and f (a) is well defined, then a ∈ P Z

ℓ ≡f (a) ∈ P Zℓ ; moreover ∅Rcℓ if P Zℓ is the individual constant cℓ = cZℓ whencℓ is well defined (see Definition 1.1(2)(F); this implies that G(f ) is apartial automorphism of (N, P )).

We may write m1 = mZ1 , recall that m1(k) gives the arity of P k and the infor-

mation is it a relation or (possibly partial) function.

2.7 Fact: Let Y = (M,I,F ) be a k-system, and Z be an m1-lifting of Y .1) The 0 − Υ-lifting (in Definition 2.12) exists and is a lifting (see Definition2.6).2) If f 1, f 2 ∈ F and A is a Z-support of y ∈ N then

f 1 ↾ A = f 2 ↾ A & A ⊆ Dom(f 1) ⇒ f 1(y) = f 2(y).

3) From Z we can reconstruct Y ,F , m1; and if I = {A : for some B, A ⊆B and B is a Z-support of some y ∈ N } then we can reconstruct also I (so thewhole Y ).

Proof. 1) Easy [compare with 2.4, 2.8].2) Let y′ = G(f 2)(y), let A1 = A, A2 = f ”2(A1) hence as A is a Z-support of y and A ⊆ Dom(f 1) necessarily y′ is well defined and A2RZy′ (see Definition2.6(1) clause (h)). We know that f −1

2 and f −12 ◦ f 1 belongs to F (see Definition

2.6(1) clauses (i) and (j)). We also know that A2 ⊆ Dom(f −12 ) so as A2RZy′

(see above) we have y′

∈ Dom(G(f −1

2 )) (see Definition 2.6(1), clause (f )(α))and (G(f −1

2 ))(y′) = y (by Definition 2.6(1), clause (i) as (G(f 2))(y) = y′ by thechoice of y′).Clearly A = A1 ⊆ Dom(f −1

2 ◦ f 1) hence (see Definition 2.6(1), clause (f )(β ))

we have y ∈ Dom(G(f −12 ◦ f )).

3so maybe x even has support Aℓ, idAℓ⊆ f ℓ for ℓ = 1, 2 but x has no support ⊆ Dom(f )



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But idA ⊆ f −12 ◦ f 1, so as ARZy we have y = (G(f −1

2 ◦ f ))(y). By Definition2.6(1), clause (j) we have, as the left side is well defined:

(G(f −12 ◦ f 1))(y) = ((G(f −12 )) ◦ (G(f 1)))(y)

and trivially

(G(f −12 )) ◦ (G(f 1)))(y) = (G(f −1

2 ))((G(f 1))(y)).

By the last three equations y = (G(f −12 ◦ f 1))(y) = (G(f −1

2 ))((G(f 1))(y)), but

by the above we note y = (G(f −12 ))(y′). So, as G(f −1

2 ) is one-to-one (having aninverse), we have (G(f 1))(y) = y′, now as y′ was defined as (G(f 2))(y) we aredone.3) Straightforward. 2.7

Note that 2.8 is close to 2.4 and 2.12(2)-(4).

2.8 Definition/Claim. LetY

= (M,I,F

) be a k-system and Z = (N,¯

P , G , R)be an m1-lifting of Y .

1) For B ⊆ N let EB = EY ,ZB be the following 2-place relation on N :

x1EBx2 iff for some f ∈ F we have idB ⊆ f and (G(f ))(x1) = x2.

2) If B ∈ I [k − 2] so k ≥ 3 then

(α) EB is an equivalence relation on N

(β ) if f ∈ F , B ⊆ Dom(f ) then f maps EB to Ef ”(B) which means:

f ↾ B ⊆ g ∈ F &ℓ<2

(G(g))(xℓ) = yℓ ⇒ [x1EBx2 ≡ y1Ef ”(B)y2]

(γ ) if B ⊆ Dom(f ) then there is a one-to-one function F = F f,B from N/EB

onto N/Ef ”(B) such that:

(∗)1 for x1, x2 ∈ N we have: (∃g)(f ↾ B ⊆ g ∈ F & G(g)(x1) = x2) ⇔F (x1/EB) = x2/Ef ”(B)

(δ) if x ∈ N and ARZx and a ∈ SeqI,A then there is an equivalence relationE ∈ E Y ,1(B) with domain {f (a) : f extend idB and A ⊆ Dom(f )} suchthat:

(∗)2 if f ℓ ∈ F , idB ⊆ f ℓ and A ⊆ Dom(f ℓ) for ℓ = 1, 2 then GZ(f 1)(x) =GZ(f 2)(x) ⇔ f 1(a)Ef 2(a)

(∗)3 |x/EB| = |Dom(E )/E |

(ε) if f, F are as in clause (γ ) and x1, x2 ∈ N then

(∗)4 if F (x1/EB) = x2/EB and Y is a counting k-system,then |x1/EB| = |x2/EB|

(∗)5 if F (x1/EB) = x2/EB and Y is a medium (t, k)-system,then |x1/EB| = |x2/EB| or both are > t(M ).



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22 SAHARON SHELAH

Proof of (2).Clause (α):

Reflexivity:For x ∈ N choose a Z-support A ∈ I , so we can find f ∈ F extending idB∪A

hence G(f ) maps x to itself, hence xEBx.

Symmetry:If f ∈ F witnesses xEBy then f −1 ∈ F witnesses yEBx.

Transitivity:If x0EBx1, x1EBx2 let f 1 witness x0EBx2 and let f 2 witness x1EBx2, now

let A0 ⊆ Dom(f 1) be a Z-support of x0, so A1 = f ”1(A0) ⊆ Rang(f 1) is a Z-

witness of x1, now let A∗

1 ⊆ Dom(f 2) be a Z-support of X 1, so B ∪ A∗

1 ∈ I [k − 1]hence without loss of generality A1 ⊆ Dom(f 2) hence A2 = f ”2(A1) is a Z-support of x2, so x1 ∈ Dom(G(f 2 ◦ f 1)) and G(f 2 ◦ f 1) ⊆ G(f 1) ◦ G(f 2) henceG(f 2 ◦ f 1)(x1) = x2 as required.

Clause (β ):So assume f ↾ B ⊆ g ∈ F and (G(g))(xℓ) = yℓ, for ℓ = 1, 2 and we should

prove x1EBx2 ≡ y1Ef ”(B)y2. It suffices to prove x1EBx2 ⇒ y1Ef ”(B)y2 (as

applying it to B′ = f ”(B), f ′ = f −1, g′ = g−1, y1, y2, x1, x2 we get the otherimplication). As x1EBx2 we can find a witness h, i.e., idB ⊆ h ∈ F and(G(h))(x1) = x2. Let A1 ⊆ Dom(h) be a Z-support of x1 and let A2 = h”(A1),so A2 is a Z-support of x2.

If B ∈ I [k − 4] without loss of generality A1, A2 ⊆ Dom(g), and let A∗1 =

f ”(A1), A∗2 = f ”(A2), lastly let g∗ = g ◦ h ◦ g−1. Now (G(g∗))(y1) = y2,

idf ”(B) ⊆ g∗ so g∗ witnesses y1Ef ”(B)y2 as required.But maybe B /∈ I [k −4], still B ∈ I [k −2]; now for ℓ = 1, 2, as (G(f ))(xℓ) = yℓ

there is a Z-support C ℓ of xℓ such that C ℓ ⊆ Dom(g) and let C ′ℓ = g”(C ℓ). Sowe can find, for ℓ = 1, 2 a function g1 ∈ F such that g ↾ (B ∪ C 1) ⊆ g1

and A1 ⊆ Dom(g2) and also there is h1 ∈ F such that h ↾ (B ∪ A1) ⊆ h1 andC 1 ⊆ Dom(h1). So h1◦g−1

1 extends (g ↾ B)−1 and C ′1 ⊆ Dom(g−11 ), (g−1

1 )”(C ′1) =

C 1 ⊆ Dom(h1) but C ′1 is a Z-support of y1. Hence (G(h◦g−11 ))(y1) is well defined

and equal to (G(h) ◦ G(g−11 ))(y1) = (G(h))((G(g−1

1 ))(y1)) = (G(h1))(x1) = x2.

Let g2 ∈ F be such that g ↾ (B ∪ A2) ⊆ g2 and (h1 ◦ g−11 )”(C ′1) ⊆ Dom(g2) and

similarly we get g−12 ◦ h1 ◦ g−1

1 extends idg”(B) and (G(g−12 ◦ h1 ◦ g−1

1 ))(y1) = y2,so we are done.

Clause (γ ), (δ), (ε):Should be clear. 2.8

Quite naturally for such a m1-lifting Z of Y the family {G(f ) : f ∈ F Y }helps us to understand first order logic on (N Z, P Z).



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2.9 Claim. Assume Y = (M,I,F ) is a k-system and Z = (N, P , G , R) is an m1-lifting of Y .Then

(∗) Assume ϕ(x) is first order and k ≥ quantifier depth (ϕ(x)) + ℓg(x), or just ϕ(x) ∈ L ∞,k which means:every subformula of ϕ(x), (e.g. ϕ itself) has ≤ k free variables.

If a ∈ ℓg(x)N, AℓRaℓ and Aℓ ⊆ Dom(f ) (hence aℓ ∈ Dom (G(f ))) for ℓ < ℓg(x)and f ∈ F then

(N, P ) |= “ϕ[. . . , aℓ, . . . ]ℓ<ℓg(x)” ⇔ (N, P ) |= “ϕ[. . . , G(f )(aℓ), . . . ]ℓ<ℓg(x)”

Proof. We prove this by induction on the quantifier depth of ϕ. Let m = ℓg(x)so x = xℓ : ℓ < m and without loss of generality m ≤ k.

Case 1: ϕ atomic.As G(f ) is a partial automorphism of N and even (N, P ) this should be clear.

Case 2: ϕ = ¬ψ or ϕ = ψ1 ∧ ψ2 or ϕ = ϕ1 ∨ ϕ2.Straight.

Case 3: ϕ = ϕ(x) = (∃y)ψ(y, x).Without loss of generality y is not a dummy variable in ψ, that is has a free

occurrence. Let x = x0, . . . , xm−1 so m < k.As G(f −1) = G(f )−1 it is enough to prove N |= “ϕ[a0, . . . , am−1]” ⇒ N |=

“ϕ[G(f )(a0), . . . , G(f )(am−1)]”.So we assume the left side, i.e.

(∗)1 N |= ϕ[a0, . . . , am−1],hence for some a∗ ∈ N we have

(∗)2 N |= ψ[a∗, a0, . . . , am−1].

Necessarily a∗ has a Z-support A∗ ∈ I .Now k ≥ m + 1 and Y is a k-system hence there is f ∗ ∈ F such that f ↾

(ℓ<m

Aℓ) ⊆ f ∗ and A∗ ⊆ Dom(f ∗). So each of a∗, a0, . . . , am−1 has a Z-

support included in Dom(f ∗) hence by the induction hypothesis applied toψ[a∗, a0, . . . , am1

] and (∗)2 we have

(∗)3 N |= ψ[G(f ∗)(a∗), G(f ∗)(a0), . . . , G(f ∗)(am−1)].

So by the definition of |= we get

(∗)4 N |= (∃y)ψ(y, G(f ∗)(a0), . . . , G(f ∗)(am−1).

But for ℓ < m, the set Aℓ is a Z-support of aℓ and f ∗ ↾ Aℓ = f ↾ Aℓ hence(G(f ∗))(aℓ) = G(f )(aℓ) so

(∗)5 N |= (∃y)ψ(y, G(f )(a0), . . . , G(f )(am−1)].



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But ϕ(x) = (∃y)ψ(y, x) so we are done. 2.9

Having dealt with first order logic we should deal with cardinality logic (actually

any of the variants we mention). Here we use the counting version, really natu-rally the medium version suffices but for it we have to use more “bookkeeping”of the various things used, and the reader can use only this smoother case.Note that if we like to add cardinality quantifiers on pairs we need s ≥ 2, etc.,but we may create the set of pairs in N t so not so necessary.

2.10 Claim. Assume that Y is a counting k-system (see Def 2.3(1)) and Z =(N, P , G , R) is an m1-lifting of Y .Then

(∗) assume ϕ(x) ∈ L card,T or ϕ(x) ∈ L card (can have both kinds quantifiers;recall that L f.o. + na is included in a special case of L card,T) and every

subformula of ϕ(x), including ϕ(x) itself, has < k free variables and m = ℓg(x) < k

(α)ϕ(x) if f ∈ F and Aℓ is a Z-support of aℓ and Aℓ ⊆ Dom (f ) for ℓ < mthen:

(N, P ) |= “ϕ[a0, . . . , am−1]” iff (N, P ) |= “ϕ[G(f )(a0), . . . , G(f )(am−1)]”

(β )ϕ(x) if f ∈ F and Aℓ is a Z-support of aℓ for ℓ = 1, . . . , m− 1 and Aℓ ⊆Dom (f ) then the sets {b ∈ N : (N, P ) |= “ϕ[b, a1, . . . , am−1]”},and {b ∈ N : (N, P ) |= “ϕ[b, G(f )(a1), . . . , G(f )(am−1)]”} have thesame number of elements.

Proof. We prove by induction on the quantifier depth of ϕ.We first show that

⊠ (α)ϕ(x) ⇒ (β )ϕ(x)

Why? So assume (α)ϕ(x), and a1, . . . , am−1 ∈ N be given (where ℓg(x) = m) andalso f ∈ F and A1, . . . , Am−1 ∈ N, Aℓ is a Z-support of aℓ for ℓ = 1, . . . , m − 1such that Aℓ ⊆ Dom(f ), and we should prove the equality in (β )ϕ(x). Let ai

ℓ

be aℓ if i = 1 and (G(f ))(aℓ) if i = 2. Let Aiℓ be Aℓ if i = 1 and (G(f ))”(Aℓ) if

ℓ = 2.

Let Bi =m−1

ℓ=1

Aiℓ so Bi ∈ I [m − 1] and f maps B1 onto B2.

By Definition 2.8(1) we know that EBℓis an equivalence relation on N and, see

Definition 2.8(2), clause (γ ), (δ), the function F = F f,B1satisfies

(i) F is a one-to-one function from N/EB1onto N/EB2

(ii) f ↾ B1 ⊆ g ∈ F & (G(g))(x1) = x2 ⇒ (F (x1)/EB1) = x/EB2



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(iii) for every x1 ∈ N, A1 ∈ I such that A1RZx1, and a1 ∈ SeqI,A1for some

E = E x1 ∈ E Y ,1(B1) we have Dom(E ) = x1/EB1and

(∗) if [idB1 ⊆ f ℓ & A1 ⊆ Dom(f ℓ)] for ℓ = 1, 2 then f 1(a1)E xf 2(a2) ⇔(G(f 1))(x1) = (G(f 2))(x1)

(iv) if x1 ∈ N, x2 ∈ N, (G(g))(x1) = x2 for some g ∈ F such that f ↾ B1 ⊆ g,then |(x1/EB1

)/E x1| = |(x2/EB2/E x2 |.

Hence it suffices to prove, assuming F (x1/EA1) = x2/EA2

that

N |= “ϕ[x1, a1, . . . , am−1]” ⇔ N |= “ϕ[x2, G(f )(a1), . . . , G(f )(am−1)]”.

As we can replace f by a suitable extension of f ↾ B1, without loss of generalitythere are y1 ∈ x1/EB1

and y2 ∈ x2/EB2such that (G(f ))(y1) = y2. We can find

C 1 ⊆ Dom(f ) which is aZ

-support of y1.As x1EB1y1 we can find f 1 ∈ F such that idB1

⊆ f 1 and (G(f 1))(x1) = y1;as m < k, without lose of generality C 1 ⊆ Rang(f 1) and let C 0 = (f −1

1 )”(C 1),let C 2 = f ”(C 1). As x2EB1

y2 we can find f 2 ∈ F such that idB2⊆ f 2 and

(G(f 2))(y2) = x2; as m < k without lose of generality C 2 ⊆ Dom(f 2) and letC 3 = f ”2(C 2). So f ′ = f 2 ◦ f ◦ f 1 belongs to F and extends f ↾ B1 and it mapsC 0 onto C 3 hence y1 ∈ Dom(G(f ′)), and clearly (G(f 1))(x1) = y1, (G(f ))(y1) =y2, (G(f 2))(y2) = x2 hence (G(f ′))(x1) = x2 and, of course, Bℓ ⊆ Dom(f ′)hence applying (α)ϕ(x) to f ′, x1, a1, . . . , am−1 we get

(N, P ) |= “ϕ[x1, a1, . . . , am−1]” ⇔ (N, P ) |= “ϕ[x2, G(f )(a1), . . . , G(f )(am−1)]”

recalling x2 = (G(f ′))(x1). So ⊠ holds.

Now the inductive proof of (α)ϕ is separated to cases. The case ϕ atomic,ϕ = ¬ϕ, ϕ = ψ1 ∧ ψ2, ϕ = (∃y)(ψ(y, x)) works as in the proof of 2.9. The newcases hold because (β )ψ hold by the induction hypothesis + ⊠. 2.10

2.11 Claim. We can weaken in 2.10 the assumption on Y to “medium (t, k)-dichotomical” provided that:

⊠ if B ∈ I [m], 1 ≤ m ≤ r, then every equivalence class of EY ,ZB (so a subset

of N Z, see Definition 2.8(1)) has ≤ t(M Y ) members.


2.12 Definition. For Y = (M,I,F ) a k-system, the 0−Υ-lifting or the 0−m1-lifting if m1 = mΥ

1 is (M, P , G , R) where

(a) G is the identity on F

(b) ARy ⇔ A ∈ I & y ∈ A

(c) each P ℓ is the empty relation.



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Clearly our intention requires us for a k-system Y , to move from an Υ-liftingZt = (N t[M, Υ, t], P t[M, Υ, t], Gt, Rt) to a Υ-lifting

Zt+1 = (N t+1[M, Υ], P t+1[M, Υ], Gt+1, Rt+1).

Toward this aim naturally in Definition 2.13 below we define for Υ-lifting Z somesuccessors, and in Claim 2.16 we prove what they satisfy.In Definition 2.13(6) we can define “Z′ is the full t-successor of Z” (both m1-liftings of a k-system Y ).

2.13 Definition. Let Y = (M,I,F ) be a k-system and Z = (N, P , G , R) bean m1-lifting of Y .1) We say X is good or (Y , Z)-good if

(a) X a subset of N

(b) for some A ∈ I we have “A supports X ” (for our Y and Z) which means:if f ∈ F ,BRy (so B ∈ I, y ∈ N ) and A∪B ⊆ Dom(f ), and f ↾ A = idA

then y ∈ X ⇔ (G(f ))(y) ∈ X (note: (G(f ))(y) is well defined by clause (f) of 2.6)

(c) X /∈ N .

2) Let P = PY ,Z be the family of good subsets of N , let R = R Y ,Z be thetwo-place relation defined by: A R X iff A supports X , i.e. (b) of part (1) holds.3) For f ∈ F we define a function G+(f ) = G+

Y ,Z(f ) with domain ⊆ PY ,Z ∪ N

as follows, (well, now (G+(f ))(X 1) = X 2 is just a relation, but by 2.14(1) clause(ii) below it is a function)

(α) For good X such that A R X, A ∈ I when A ⊆ Dom(f ) we let (G+(f ))(X )= {y ∈ N : for some g ∈ F and y′ ∈ X we have f ↾ A ⊆ g and G(g)(y′) =y}

(β ) G+(f ) ↾ N = G(f ).Note that no contradition arises between clauses (α) and (β ) because of clause (c) in part (1).

4) We define E = E Y ,Z as the following two place relation: X 1EX 2 iff X 1, X 2are good subsets of N Z and for some f ∈ F we have (G+(f ))(X 1) = X 2; this isan equivalence relation (see 2.15(2) below).5) Z′ = (N ′, P ′, G′, R′) is a successor of Z if:

(a) N ⊆ N ′ ⊆ N ∪PY ,Z

(b) X 1E Y ,ZX 2 & X 1 ∈ PY ,Z & X 2 ∈ PY ,Z ⇒ [X 1 ∈ N ′ ↔ X 2 ∈ N ′](c) G′ is a function with domain F and for f ∈ F the function G′(f ) is

defined as G+(f ) from part (3) restricted to N ′

(d) R′ is R ∪ [ R ↾ (I × N ′)]

(e) the pair (N ′, P ′) is an (M, m1)-candidate; so P ′ℓ is an m1(ℓ)-ary relationor function as dictated by m1



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6) We say Z′ is a ϕ-reasonable successor of Z if it is a successor and

(e)′ P ′ℓ = {b ∈ N : (N, P ) |= ϕℓ(b)} but when Υ is i.c. (see Definition 1.1(2),

clause (F)) this is so only if P ′ℓ ∈ N ′ and P ′ℓ = ∅ otherwise (for eachℓ < m1 (so we demand P ′ℓ ∈ N ′).

We may omit ϕ if clear from the content.7) We say that Z′ = (N ′, P ′, G′, R′) is a full successor of Z if it is a successorof Z and N ′ = N ∪PY ,Z. We say it is the full successor if in addition P ′ℓ = ∅and it is the full reasonable successor (or reasonable full successor) if it is a fullsuccessor which is a reasonable successor.8) Z′ = (N ′, P ′, G′, R′) is a full t-successor of Z if it is a successor of Z and

(a)∗8 N ′ = N ∪ {X ∈ PY ,Z : |X/E Y ,Z| ≤ t(M )}.

So if we omit t we mean t(N ) = ∞.

9) Z′ = (N ′, P ′, G′, R′) is the true (Υ, t)-successor of Z if m1 = mΥ1 and Z′ is areasonable successor of Z and:

(a)∗9 (N ′, P ′) is the (Υ, t)-successor of (N, P ), (see Definition 1.8(2)).

10) Z′ = (N ′, P ′, G′, R′) is the true Υ-successor of Z if it is a ϕΥ-reasonablesuccessor of Z and:

(a)∗10 (N ′, P ′) is the Υ-successor of (N, P ), (see Definition 1.8(2A))[this just means the true (Υ, ∞)-successor of Z].

Note that the names above indicate our intentions, but we have to prove that“Z′ is a successor of Z” implies that “Z is an m1-lifting of Y ” (done in 2.16),

the true (Υ, t)-successor of Z is a t-successor of Z (done in 2.17, 2.18, 2.20) andsimilarly without the t.

2.14 Claim. Assume Y is a k-system, Υ an inductive scheme (so τ is common)and Z is an Υ-lifting of Y .1) In Definition 2.13(3), if k ≥ 3 then for f ∈ F and (Y , Z)-good X we have:

(i) if the relation X 2 = (G+(f ))(X 1) holds and (G(f ))(x1) = x2 then x1 ∈X 1 ≡ x2 ∈ X 2

(ii) the value (G+(f ))(X ) does not depend on A, so G+(f ) is well defined.(iii) if the relation X 2 = (G+(f ))(X 1) holds then X 2 is a (Y , Z)-good

2) There is a unique object Z′ which is the full successor of Z. ; there is a uniqueobject Z′ which is the reasonable full successor of Z and there is a unique object which is the reasonable t-full successor of Z.3) If the Z′ is the true (Υ, t)-successor of Z, then Z′ is a reasonable successor of Z which implies Z′ is a successor of Z.4) There is at most one true successor Z′ of Z.

Proof. Easy, using 2.16(2) below for part (2); e.g.1) Clause (i) So assume that X 1 is (Y , Z)-good, A1 is a Z-support of X 1,



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A1 ⊆ Dom(f ), f ∈ F and X 2 = {(G(g))(x) : x ∈ Dom(g), f ↾ A ⊆ g} ⊆ N and(G(f ))(x1) = x2. First x1 ∈ X 1 ⇒ x2 ∈ X 2 by the definition of X 2. Secondassume that x2 ∈ X 2, so for some y ∈ X 1 and g ∈ F we have f ↾ A1 ⊆ gand (G(g))(y) = x2. Let B2 ⊆ Dom(g) be a Z-support of y. Let B1 ⊆ Dom(f )be a Z-support of x1; as k ≥ 3 there is f 1 ∈ F such that f ↾ (A1 ∪ B1) ⊆ f 1and g”(B2) ⊆ Rang(f 1) so (G(f 1))(x1) = x2. Now (G(g−1 ◦ f 1))(x1) is welldefined as B2 ⊆ Dom(g−1 ◦ f 1) hence is equal to ((G(g−1) ◦ (G(f 1)))(x1) = yand (g−1 ◦ f 1) ↾ A1 = idA1

hence x1 ∈ X 1 ≡ y ∈ X 1 but y ∈ X 1 hence x1 ∈ X 1as required.

Clause (ii) So assume that f ∈ F , X is good and for ℓ = 1, 2 the set Aℓ ∈ I is asupport of X and Aℓ ⊆ Dom(f ). For ℓ = 1, 2 let X ℓ =: {y ∈ N : for some g ∈ F and y′ ∈ X we have f ↾ Aℓ ⊆ g and (G(g))(y′) = y}.By the symmetry it is enough to show that y ∈ X 1 ⇒ y ∈ X 2. So assume y ∈ X 1hence there are g ∈ F and y′ ∈ X such that f ↾ A1 ⊆ g and (G(g))(y′) = y.As y′ ∈ X ⊆ N , by Definition 2.6(1), clause (g) there is B ∈ I which is a

Z-support of y′. As (G(g))(y′) = y without loss of generality B is such thatB ⊆ Dom(g) (see Definition 2.6(1), clause (f)). As Y is a k-system and k ≥ 3there is f ∗ ∈ F such that f ↾ (A1 ∪ A2) ⊆ f ∗ and g”(B) ⊆ Rang(f ∗) hencefor some B∗ ⊆ Dom(f ∗) we have (f ∗)”(B∗) = g”(B) so for some y∗ ∈ N wehave (G(f ∗))(y∗) = y. By clause (i) applied to A2, f ∗, y∗, y we have y∗ ∈ X ≡y ∈ X 2, so it is enough to prove that y∗ ∈ X . Now easily g−1 ◦ f ∗ ∈ F ,B∗ ⊆ Dom(g−1 ◦ f ∗), idA1

⊆ g−1 ◦ f ∗ and so y∗ ∈ Dom(G(g−1 ◦ f ∗)) and(G(g−1 ◦ f ∗))(y∗) = y′. Hence, as X is good (see Definition 2.13(1) clause (b)),we have y∗ ∈ X ≡ y′ ∈ X , but y′ ∈ X by its choice, so we are done.

Clause (iii) Easy, or see the proof of ⊠(∗)1 inside the proof of 2.16 below.2.14

Now for the definition of successor for liftings of Y , we naturally ask whetherthere is any.

2.15 Claim. Assume Y is a k-system k ≥ 3, Υ an inductive scheme and Z isan m1-lifting of Y and t ∈ T.1) There is a ϕΥ-reasonable full t-successor of Z (and it is unique), similarly without t.2) E Y ,Z, defined in 2.13(4) is an equivalence relation on PY,Z (see Definition 2.13) and for every f ∈ F , the function G+(f ) : PY ,Z → PY ,Z preverse theE Y ,Z-equivalence class.

Proof. 1) All is straight modulo part (2) (recalling 2.14).2) Let Z∗ be the ϕΥ-reasonable full successor of Z which exists by 2.14(2), and

is a successor of Z by 2.14, and is an m1-lifting by 2.16 below.Why is E Y ,Z an equivalence relation on P Y ,Z = N Z

∗

\N Z? In short, by theproperties of Z; in details:

E Y ,Z is reflexive:Let X ∈ P Y ,Z, so for some A ∈ I we have A R Z

∗

X (or equivalently A R X )(see Definition 2.13(2)) and there is f ∈ F which is the identity on A, hence



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(see Definition 2.13), X ∈ Dom(GZ∗

(f )), and (GZ∗

(f ))(X ) = X (as clause (f))of Definition 2.6(1) holds as Z∗ is an m1-lifting of Y .

E Y ,Z is symmetric:Use GZ

∗

(f −1) = (GZ∗

(f ))−1.

E Y ,Z is transitive:Use GZ

∗

(f 1 ◦ f 2) ⊆ GZ∗

(f 1) ◦ GZ∗

(f 2).[This is similar to 2.4.]

By the definition of E Y ,Z (see Definition 2.13(4)), clearly for f ∈ F the

mapping G+(f ) = GZ∗

↾ PY ,Z preserves the E Y ,Z-equivalence class or use2.16. 2.15

Clearly for “reasonable” cases, everything can be interpreted in N ZfullY ,t , see later.

We now prove that Definition 2.13(1)-(5) works as intended, i.e. any successorof Z is an m1-lifting of Y . In particular, we have to show that the functionsdefined are functions with the right domain and range and the E ’s are equivalencerelations. This is included in the proof of 2.16.

2.16 Claim. Assume Y is a k-system and Z is an m1-lifting of Y (see Defi-nition 2.1(2),Definition 2.6) and k ≥ 3.Any successor Z′ of Z is an m1-lifting of Y .

Proof. We check the clauses in Definition 2.6. Let G+, G′, R′, N ′, P ′ be as inDefinition 2.13.

Clause (a): As N is transitive with M its set of urelements, and X ∈ N ′\N ⇒X ∈ PY

,Z ⇒ X ⊆ N also N ′ is transitive with M its set of urelements. Clearly

N has the right vocabulary τ + = τ M ∪ {∈} and Q ∈ τ M ⇒ QN ′ = QM . So N ′

is as required. Also P ′ = P ′ℓ : ℓ < m1, each P ′ℓ as required by m1.

Clause (b): By Definition 2.13(5)(c) we have that G′ is G+ ↾ N ′ where thefunction G+ is defined in part (3) of Definition 2.13 and f ∈ F implies G+(f )is a partial function with domain ⊆ N ∪PY ,Z (see 2.14(1)). So G′ is a functionwith domain F and G′(f ) by its definition is a partial function with domain⊆ N ′.

Clause (c):

Subclause (α):For f ∈ F we know that G(f ) is a function, G(f ) ↾ M = f (see Definition

2.6(1)), clause (c)(i)) and G(f ) = (G+(f )) ↾ N (see Definition 2.13(3) par-ticularly subclause (β ), remembering that “X good ⇒ X /∈ N ” by Definition2.13(1), clause (c)). As M ⊆ N ⊆ N ′ and G′(f ) = G+(f ) ↾ N ′, together we getf = (G+(f )) ↾ M = (G′(f )) ↾ M .

Subclause (β ): Let f ∈ F , G(f ) = f 1, G′(f ) = f 2 and let x, y ∈ N ′ belongs tothe domain of f ′ and we should prove



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(α) f 2(x) ∈ N ′

(β ) x ∈ N ′\N ⇒ f 2(x) ∈ N ′\N

(γ ) if x = y are from N ′

then f 2

(x) = f 2

(y)(δ) for every predicate Q ∈ τ M , f 2 preserve Q and ¬Q

(ε) N ′ |= “y ∈ x” ⇔ N ′ |= “f 2(y) ∈ f 2(x)”(we shall do more toward proving clause (g) of Definition 2.6(1) below).

Note that for clause (α), as f 2 ↾ N = G(f ′) ↾ N = G(f ) = f 1, it is enoughto check it for x ∈ N ′\N , which is done in ⊠, (∗)4 + (∗)1 + (∗)6 below (as agood subset of N does not belong to N ). Clause (β ) also follows from ⊠, (∗)4 +(∗)1 + (∗)6 below. As for clause (γ ), if x, y ∈ N use G(f ′) ↾ N = G(f ); if x ∈ N & y ∈ N ′\N note that G(f ′)(x) = (G(f ))(x) ∈ N and (G(f ′))(y) /∈ N by clause (β ); similarly if x ∈ N ′\N & y ∈ N ; lastly if x, y ∈ N ′\N we useclause (ε) proved below and N ′ being transitive (as f ′(x), f ′(y) are subsets of N so /∈ M ). Now clause (δ) is easy as G′(f ) ↾ M = f and f being a partialautomorphism of M being from F .

Lastly, we consider clause (ε), so we let x, y ∈ N ′. If x ∈ N , then f 2(x) isnecessarily in N too, but N is transitive, hence N ′ |= “y ∈ x” ⇒ y ∈ N andN ′ |= “z ∈ f 2(x)” ⇒ z ∈ N , so as f 2 ↾ N = f 1 we are done. So we can assume

x ∈ N ′\N , so x is a good subset of N , so for some A0 ∈ I, A0RZ′

x. We define

z =:

b ∈ N :for some g ∈ F and b′ ∈ x we have⊗1

f ↾ A0 ⊆ g and G(g)(b′) = b

.

We need the following, and it suffices

⊠ assume x ∈ N ′\N and z is defined as in ⊗1.Then

(∗)1 z is a good subset of N with A1 =: f ”(A0) a support of z

(∗)2 x = {b′ ∈ N : for some g ∈ F and b ∈ z we havef −1 ↾ (f ”(A0)) ⊆ g and G(g)(b) = b′}

(∗)3 z does not belong to N

(∗)4 z = f 2(x)

(∗)5 if B is another Z-support of x, then z′ = z when z′ = {b ∈ N :for some g ∈ F and a ∈ x we have f ↾ A ⊆ g and G(g)(a) = b}.

(∗)6 z ∈ N ′

Proof of (∗)1. We should check clauses (a),(b),(c) of Definition 2.13(1). Now

clause (a) is trivial and clause (c) is dealt with in (∗)3 which we prove below(and we do not use it till then, so no vicious circle). So we concentrate on provingclause (b). So suppose:

(i) a, b ∈ N and

(ii) g1 ∈ F satisfies A1 ⊆ Dom(g1) and g1 ↾ A1 is the identity and

(iii) a ∈ Dom[G(g1)] and b = G(g1)(a).



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Now we should prove that a ∈ z ⇔ b ∈ z. It is enough to prove ⇒ as applyingit to g−1

1 we get the other implication. As b = G(g1)(a) necessarily by clause (i)of Definition 2.6 for some Z-support B1 of a we have B1 ⊆ Dom(g1).

Assume a ∈ z then by the definition of z we can find g ∈ F and a′ ∈X such that f ↾ A0 ⊆ g and G(g)(a′) = a. By Definition 2.6(1), clause(f )(β ) there is B2 ∈ I such that B2 is a Z-support of a′ and B2 ⊆ Dom(g).As k ≥ 3 and as we can replace g by any g∗ such that g ↾ (A0 ∪ B2) ⊆g∗ ∈ F ,without loss of generality B1 ⊆ Rang(g). So, possibly changing B2

without loss of generalityB1 = g”(B2) (see clause (b) of Definition 2.6(1)).Let g′ = g1 ◦ g, so A0 ∪ B2 ⊆ Dom(g′), g′ ↾ A0 = g ↾ A0 = f ↾ A0.

[Why? As g ↾ A0 = f ↾ A0, f ”(A0) = A1 and g1 ↾ A1 = idA1; also B2 ⊆

Dom(g) and g”(B) is equal to B1 which is ⊆ Dom(g1)]. Hence a′ ∈ Dom(G(g′))and so G(g′)(a′) = (G(g1)) (G(g)(a′)) = (G(g1)) (a) = b. (See Definition 2.6(1),clause (j).)

So g′, a′ witness b ∈ z; so b ∈ z has been proved under the assumption a ∈ z.

So by symmetry we have proved a ∈ z ⇔ b ∈ z.

Proof of (∗)2.Call the set in the right side x′.First assume that a ∈ x, so a has a Z-support B1 hence for some g1 ∈ F

we have A0 ∪ B1 ⊆ Dom(g1) and f ↾ A0 ⊆ g1, hence a ∈ Dom(G(g1)) and letb =: (G(g))(a), so b ∈ z by the definition of z, also b has Z-support B2 =: g”(B1).Let g2 = g−1

1 so g2 ∈ F and G(g2) = G(g1)−1 hence (G(g2))(b) = a. Lastly, asf ↾ A0 ⊆ g1 clearly f −1 ↾ (f ”(A0)) ⊆ g2. Together g2, b witness that a ∈ x′. Sowe have proved a ∈ x ⇒ a ∈ x′.

Second, assume that a ∈ x′, so we have witnesses g, b for this, i.e. g ∈F , b ∈ z, f −1 ↾ (f ”(A0)) ⊆ g and (G(g))(b) = a. So we can find B1 ⊆

Dom(g) a Z-support of b, so B0 = g”(B1) is a Z-support of a. As b ∈ zthere are witnesses for it, that is, there are g1 ∈ F and b′ ∈ x such thatf ↾ A0 ⊆ g1 and (G(g1))(b′) = b, hence g−1

1 ∈ F , G(g−11 ) = (G(g1))−1 so

without loss of generalityB1 ⊆ Rang(g1) and let B2 = (g−11 )”(B1), but B1 is

a Z-support of b hence B2 is a Z-support of b′. Let g′ = g ◦ g1 ∈ F . NowA0 ⊆ Dom(g1) and g”1(A0) = f ”(A0) ⊆ Dom(g) hence A0 ⊆ Dom(g′), andas g1 ↾ A0 = f ↾ A0, and g ↾ f ”(A0) = f −1 ↾ f ”(A0) clearly g′ ↾ A0 = idA0

.Also B2 ⊆ Dom(g1), B1 = g”(B2) ⊆ Dom(g), hence B2 ⊆ Dom(g′) and(G(g′))(b′) = (G(g ◦ g1))(b′) = G(g)((G(g1))(b′) ) = (G(g))(b) = a, but asg′ ↾ A0 = idA0

and (∗)1 we have b′ ∈ x ⇔ (G(g′))(b′) ∈ x which meansb′ ∈ x ⇔ a ∈ x. But we have chosen b′ ∈ x hence a ∈ x. So we have proved thata ∈ x′ ⇒ a ∈ x. Thus finishing the proof of (∗)2.

Proof of (∗)3. If z ∈ N there is A∗ ∈ I such that A∗ is a Z-support of z.Now there is f 1 ∈ F , f ↾ A0 ⊆ f 1 such that A∗ ⊆ Rang(f 1). So z1 =

G(f −11 )(z) is well defined and by (∗)2 we can check that {b ∈ N : b ∈ z1} = x;

contradiction to “x /∈ N ”.

Proof of (∗)4.



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Should be clear.

Proof of (∗)5.By 2.14(1).

Proof of (∗)6.By clause (b) of 2.13(5).

We continue checking the clauses in Definition 2.6.

Clause (d):Easy as for f, g ∈ F we have f ⊆ g ⇒ G(f ) ⊆ G(g).

Clause (i):By the symmetry it is enough to show that G′(f −1) ⊆ G′(f )−1.So let (G′(f −1))(x) = z.

Now we know that both G(f ) and G(f −1) maps N to N and N ′\N to N ′\N (that is when defined), so if x ∈ N we have z ∈ N and we use “Z satisfiesDefinition 2.6 (1), clause (i)” to get (G(f )−1)(z) = x as required. So assumex ∈ N ′\N hence z ∈ N ′\N . By ⊠(∗)4 + ⊗1 and ⊠(∗)2 we are done.

Clause (j):Assume x0 ∈ Dom(G′(f )), hence x0 has a Z′-support A0 ⊆ Dom(f ), so by

the definition of f 2 ◦ f 1 = f we have A0 ⊆ Dom(f 1) and f ”1(A0) ⊆ Dom(f 2).So we have x0 ∈ Dom(G′(f 1)) and x1 =: (G′(f 1))(x0) has Z′-support A1 =:f ”2(A0). Similarly x2 =: (G′(f 2))(x1) is well defined and has Z′-support A2 =:

f ”1(A1) which is ⊆ Rang(f 2 ◦ f 1) = Rang(f ). Now we would like to show thatx2 = (G′(f ))(x0); if x0 ∈ N this should be clear so assume that x0 ∈ N ′\N hence x1, x2 ∈ N ′\N . Let x′2 = (G′(f ))(x0), it is well defined as x0 has Z′-support A0, A0 ⊆ Dom(f ) and it suffices to prove that x2 = x′2. So let y ∈ N and we shall prove that (y ∈ x2) ≡ (y ∈ x′2).Let B2 be a Z′-support, equivalently Z-support of y and let y2 = y. We canfind f ′2 ∈ F such that f ′2 ↾ A1 = f 2 ↾ A1 and B2 ⊆ Rang(f ′2) so as A1

is a Z′-support of x1, clearly (G′(f ′2))(x1) = (G′(f 2))(x1) = x2. Let B1 =((f ′2)−1)”(B2). Also we can find f ′1 ∈ F such that f ′1 ↾ A0 = f 1 ↾ A0 andB1 ⊆ Rang(f ′1) so as A0 is a Z′-support of x0 clearly (G′(f ′2))(x0) = x1 and letB0 = ((f ′1)−1)”(B1). Let y1 =: (G′((f ′2)−1))(y2), so y1 ∈ N has Z-support B1,and let y0 = (G′((f ′1)−1))(y1), so y0 ∈ N has Z-support B0. As G′(f ′2) maps x1 to

x2 and y1 to y2 we have by clause (c)(β ) on Z′

which we have already proved that(y2 ∈ x2) = (y1 ∈ x1). Similarly as G′(f ′1) maps x0 to x1 and y0 to y1 we have byclause (c)(β ) that (y1 ∈ x1) ≡ (y0 ∈ x0) so together (y2 ∈ x2) ≡ (y0 ∈ x0). Nowf ′ = f ′2 ◦ f ′1 ∈ F and its domain include A0 ∪ B0 and G′(f ′) maps y0 to y2 (byclause (j) for Z!); also as x0 is in its domain (as A0 ⊆ Dom(f ′) is a Z′-supportof x0) and as f ↾ A0 = f ′ ↾ A0 we have (G′(f ′))(x0) = (G′(f ))(x0) but thelater is x′2. So (G′(f ′))(x0) = x′2, so as y0 ∈ Dom(G′(f ′)) by clause (c) we have



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y0 ∈ x0 ≡ (G′(f ′))(y0)) ∈ x′2 but (G′(f ′))(y0) = y2 so (y0 ∈ x0) ≡ (y2 ∈ x′2). Asearlier we have gotten (y2 ∈ x2) ≡ (y0 ∈ x0) together (y2 ∈ x2) ≡ (y2 ∈ x′2) buty2 = y so we are done.

Clause (e): See Definition of RZ′

in Definition 2.13(5), clause (d).

Clause (f):

Subclause (α):So assume ARZ

′

y, f ∈ F and A ⊆ Dom(f ). First, if y ∈ N then we useG′(f ) ↾ N = G(f ) and Z satisfying Definition 2.6(1), clause (f )(α). Second, if

y ∈ N ′\N then ARZ′

y means A R y and clearly y is a good subset of N and bythe definition of G′(f )(= G+(f )), necessarily y ∈ Dom(G′(f )). If in additionf ↾ A = idA, we should prove that (G′(f ))(y) = y. Now by ⊠(∗)4 apply to y, Ainstead x, A0 we have

(G′(f ))(y) = {b ∈ N :for some g ∈ F and b′ ∈ y we have

f ↾ A ⊆ g (i.e. idA ⊆ g) and G(g)(b′) = b}.

But as A R y we have:

idA ⊆ g & b ∈ Dom(G(g)) ⇒ [b ∈ y ≡ G(g)(b) ∈ y]

which means that (G′(f ))(y) = y, as required.

Subclause (β )(of (f )):So assume f ∈ F and y ∈ Dom(G′(f )) (hence y ∈ N ′). First, if y ∈ N ,

recall that G′(f ) ↾ N = G(f ) and use Z satisfying clause (f )(β ) of Definition

2.6(1) and

⊠2 R′ ↾ (I × N ) = R.

Second, if y ∈ N ′\N see the definition of G′(f ) = G+(f ) and R′.

Clause (g):See the choice of R′, R .

Clause (h):The new case is: assume A ⊆ Dom(f ), A ∈ I, X ∈ Dom(G′(f )), X ∈ N ′\N .

We have to show ARZ′

X ⇔ f ”(A)RZ′

(G′(f ))(X ); now by by clause (i) it isenough to prove the implication ⇐.

Let A∗

=: f ”(A) and X ∗

=: (G′

(f ))(X ), so we know that A, A∗

∈ I andX, X ∗ ∈ N ′\N and A∗ R X ∗. We have to show that A R X . If ¬A R X , then wecan find g ∈ F , g ↾ A = idA, and z0 ∈ Dom(G(g)), z1 = G(g)(z0), such thatz0 ∈ X ≡ z1 /∈ X . We can find B0 ∈ I such that B0Rz0 and B0 ⊆ Dom(g) andlet B′

1 =: g”(B0). We can find f 1, f ↾ A ⊆ f 1, B0 ∪ B1 ⊆ Dom(f 1), f 1 ∈ F andwithout loss of generalityf 1 = f and Dom(g) = A∪B0. Let g∗ = f ◦g◦f −1, B∗

0 =f ”(B0) and B∗

1 = f ”(B1). Clearly B∗0 , B∗

1 ∈ I and f −1, g ◦ f −1, g∗ = f ◦ g ◦



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34 SAHARON SHELAH

f −1 ∈ F . Also B∗0 ⊆ Dom(f −1), (f −1)”(B∗

0 ) = B0 ⊆ Dom(g), g”(B0) = B1

and f ”(B1) = B∗1 hence together g∗(B∗

0 ) = B∗1 . Let z∗0 =: (G(f ))(z0), z∗1 =

(G(f ))(z1), so (G(f −1))(z∗

0

) = z0, (G(g))(z0) = z1, (G(f ))(z1) = z∗

1

, and as B∗

0is Z-support of z∗0 , B∗0 ⊆ Dom(g∗) necessarily (G(g∗))(z∗0 ) = z∗1 .

We can also show that (z0 ∈ X ) ≡ (z∗0 ∈ X ∗) and (z1 ∈ X ) ≡ (z∗1 ∈ X ∗) byclause (c)(β ) which we already proved so remembering (z0 ∈ X ) ≡ (z1 /∈ X ) weget a contradiction to “A R X ∗” which we have assumed.

Clause (k):Trivial. 2.16

Well we have Υ-successors of candidates (in Definition 1.8, implicitly in Defini-tion 1.1) and we have successors of m1-liftings Z of Y = (M,I,F ) where Z hasin it a candidate (N Z, P Z).

Of course, we like to connect then, specifically show that true (Υ, t)-successorof Z exists. This is not always true, as Definition 1.8 can lead us to elements of N ′\N with no support in I . In Definition 2.13 we restrict ourselves to elementswith support in I , and we can change the definition in 1.1, 1.8 to have it, but itseems to me not so convincing for a logic. Rather we show that the dichotomyassumptions (as in 2.1(3), 2.3) help.

When we use L ∗ = L f.o., then dealing with Υ-successor is easier, we haveto look less carefully at cardinalities, still we need a dichotomy property (seeDefinition 2.6) in order to get a Z-support to every member.

2.17 Claim. Assume that:

(a) Y = (M,I,F ) is a k-system, k ≥ 3

(b) Υ is an inductive scheme for L f.o.(τ +M )

(c) for every ℓ < mΥ0 any subformula of ψΥ

ℓ has ≤ k free variables; also for every ℓ < mΥ

1 any subformula of ϕΥℓ has ≤ k free variables

(d) t ∈ T

(e) Y is a t-dichotomical k-system

(f ) for every ℓ < mΥ0 the formula ψΥ

ℓ has ≤ k/2 free variables.

Then

(α) if Z is an mΥ1 -lifting of Y then Z has a true (Υ, t)-successor (see Defi-

nition 2.13(9))

(β ) for every t ≤ ∞ there is an mΥ1 -lifting Zt of Y such that recalling Defi-

nition 1.1(3A), we have (N Zt

, P Zt

) = (N t[M, Υ, t], P

t[M, Υ, t])

(γ ) if t ≤ tι[M, Υ, t] and ι ∈ {1, . . . , 7} or t ≤ tι[M, Υ, t] & ι = 11, . . . , 17

then there is an mΥ1 -lifting Zt of Y such that (N Z

t

, P Zt

) is equal to

(N t[M, Υ], P t[M, Υ]) or (N t[M, Υ, t], P t[M, Υ, t]),

respectively; similalry for ι = 21, . . . , 27.



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Proof. Clearly clause (β ) follows from clause (α), just prove by induction on t.Also clause (γ ) follows from (β ) and clause (α), so we deal with clause (α). LetZ = (N, P , G , R).

The main point is to prove

⊠ assume ℓ < mΥ0 and for a ∈ ℓg(y)N and X = X a = {b ∈ N : (N, P ) =

ψℓ[b, a]}. Then A R X for some A ∈ I or{{b ∈ N : (N, c) |= ψℓ[b, a′]} : a′ ∈ ℓg(y)N }| > t(M ).

Let m = ℓg(y), so 2(m + 1) ≤ k and a = aℓ : ℓ < m, let Aℓ ∈ I be a Z-supportof aℓ and let bℓ be a list of Aℓ without repetitions, and define a function h∗ withdomain [m] with h∗(ℓ) = bℓ. We define a 2-place relation E on h∗/E 0I,m, see

Definition 2.1(3)(γ ):

⊕1 if for j = 1, 2, f j ∈ F and

ℓ∈[m]

Rang(h∗(ℓ)) ⊆ Dom(f j) and hj = f j∗h∗

then h1Eh2 ⇔ X G(f 1)(a) = X G(f 2)(a)

(where G(f j)(aℓ : ℓ < m) = G(f j)(aℓ) : ℓ < m.

Now

⊕2 E belongs to E I,m(∅),see Def 2.1(3)(δ).

Why? by traslating this means that:

(∗) if a1, a2, a3, a4 ∈ {(G(f ))(a) :

ℓ<m

Aℓ ⊆ Dom(f ) and f ∈ F }, and

(G(f ))(a1â2) = a3â4, then N |= ψ(a1, a2) ≡ ψ(a3, a4)

where ψ(y1, y2) =: (∀x)[ϕ(x, y1) ≡ ϕ(x, y2)].Now every subformula of ψ(y1, y2) has at most 2m + 1 < k free variables or

is a subformula of ϕ(x, y) hence has ≤ k free variables, hence by 2.9 we haveN |= ψ(a1, a2) iff N |= ψ(a3, a4), so we are done showing ⊕2.

Now m ≤ k/2 and we are assuming that Y is a t-dichotomical k-system, so(β )1 or (β )2 of Definition 2.1(4) holds. Now (β )2 gives the desirable first possibleconclusion in ⊠ and (β )1 gives that |{b : N |= ψℓ(b, a′) : a′ ∈ mN }| > t(M ),hence second possible conclusion in ⊠. 2.17

2.18 Claim. 1) Assume that

(a) Y is a counting k-system with k ≥ 3(b) Υ is an inductive scheme, in L f.o. or in L f.o. + na or L card or L card,T

(c) for ℓ < mΥ0 , every subformula of ψℓ(y, x) has at most k − 1 free variables

and for ℓ < mΥ1 every subformula of ϕℓ has at most k − 1 free variables

(d) t ∈ T

(e) Y is a t-dichotomical k-system



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36 SAHARON SHELAH

(f ) for ℓ < mΥ0 the formula ψℓ has ≤ k/2 free variables.

Then

(α) if Z is an mΥ

1 -lifting, then Z has a true (t, Υ)-successor (β ) for every t there is an mΥ

1 -lifting Zt such that

(N Zt

, P Zt

) = (N t[M, Υ, t], P t[M, Υ, t]).

Proof. The proof is as in 2.17 , but we know by 2.10 that the partial automor-phism G(f ) preserves also ψ1(y, x) and even (∃ky)ψ1(y, x) when every subfor-mula of ψ1 has < k free variables; (note that only now having 2m + 1 < k ratherthan 2m + 1 ≤ k seem helpful or repeat the proof of 2.10 as we can use the < k just for subformulas of ϕ). 2.18

2.19 Remark. Why do we still need in 2.18 the “t-dichotomical”? Just to guar-

antee that the true (t, Υ)-successor is included in the full one.

2.20 Claim. Assume

(a) Y is a k-system with k ≥ 3

(b) Υ is an inductive scheme in L f.o.

(c) for ℓ < mΥ0 , every subformula of ψℓ(y, x) has at most k − 1 free variables

and for ℓ < mΥ1 every subformula of ϕℓ has at most k − 1 free variables

(d) t ∈ T

(e) Y is medium t-dichotomical

(f ) for ℓ < mΥ0 the formula ψℓ has ≤ k/2 free variables.

Then: the conclusion of 2.18 holds, so if t ≤ tι(M, Υ, t), then for some mΥ1 -lifting Zt we have (N Z

t

, P Zt

) = (N t[M, Υ, t], P t[M, Υ, t]).

Proof. Like the proof of 2.17, 2.18.

2.21 Definition. 1) We say that H is a witness to the k-equivalence of Y 1 andY 2 if

(a) for ℓ = 1, 2 we have Y ℓ = (M ℓ, I ℓ,F ℓ) is a k-system

(b) H is a family of partial isomorphisms from M 1 into M 2

(c) for every g ∈ H , we have Dom(g) ∈ I 1, Rang(g) ∈ I 2

(d) if g ∈ H and f 1 ∈ F 1 then g ◦ f 1 ∈ H (e) if g ∈ H and f 2 ∈ F 2 then f 2 ◦ g ∈ H

(f ) if g ∈ H and A ∈ I 1[k − 1] and B ∈ I 1, then for some g1 ∈ H we haveg ↾ A ⊆ g1 and B ⊆ Dom(g1)

(g) if g ∈ H and A ∈ I 2[k − 1] and B ∈ I 2, then for some g1 ∈ H we haveg−1 ↾ A ⊆ g−1

1 and B ⊆ Rang(g1).



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2) We say that H is a witness to the dichotomical (k, r)-equivalence of (Y 1, t1)and (Y 2, t2) if

(i) Y ℓ is a (tℓ, k , r)-dichotomical k-system for ℓ = 1, 2(ii) H is a witness to the k-equivalence of Y 1 and Y 2

(iii) each g ∈ H preserved the possibility chosen in the definition of (t, k , r)-dichotomical.

If we omit r, we mean s = [k/2].3) We say that H is a witness to the counting k-equivalence of Y 1 and Y 2 if

(i) Y ℓ is a counting (t, k , r)-system for ℓ = 1, 2

(ii) H is a witness to the k-equivalence of Y 1 and Y 2

(iii) each g ∈ H preserve the cardinalities involved in the definition of “count-ing (t, k , r)-system”.

4) Similarly with “medium dichotomy”.

2.22 Main Conclusion: Assume

(a) Y ℓ = (M ℓ, I ℓ,F ℓ) is a tℓ-dichotomical k-system, and τ (M ℓ) = τ forℓ = 1, 2

(b) H is a witness to the k-equivalence of Y 1 and Y 2

(c) χ ∈ L f.o.(τ +), i.e. a first order sentence in the vocabulary τ + = τ ∪{∈},

(d) Υ is an inductive scheme for L f.o.

(e) every subformula of χ and of ψΥℓ and of ϕΥ

ℓ has at most < k free variables

(f ) t1, t2 ∈ T

(g) every formula ψΥ

ℓ has ≤ k/2 free variables and k ≥ 3 of course.Then

(α) let4 ι ∈ {2, 3, 4, 5}; the truth value of θΥ,χ,t1 in M 1 and θΥ,χ,t2 in M 2under |=ι are equal except possibly when: for some ℓ ∈ {1, 2} we havethe truth value of θΥ,χ,tℓ in M ℓ is undefined whereas that of θΥ,χ,t3−ℓ

inM 3−ℓ is well defined and tι[M ℓ, Υ, tℓ] < tι[M 3−ℓ, Υ, t3−ℓ].

(β ) For any t, if N ℓ = N t[M ℓ, Υ, tℓ] is well defined for ℓ = 1, 2, then for everysentence θ ∈ L f.o.(τ +) such that every subformula has at most k freevariables, we have N 1 |= θ ⇔ N 2 |= θ

(γ ) if ι ∈ {2, 3, 4, 5} and θℓ = θΥ,χ,tℓ ∈ L Tι (L f.o.(τ )) is ι-good for ℓ = 1, 2:

then M 1 |=ι “θ1” iff M 2 |=ι “θ2”

(δ) for any t, if N ℓ = N t[M ℓ, Υ, tℓ] is well defined for ℓ = 1, 2 and θ =θ(x1, . . . , xn) ∈ L f.o.(τ +) is a formula such that every subformula has atmost k free variables then:

⊕ let a1, . . . , am ∈ N 1 and g ∈ H , thenN 1 |=ι θ[a1, . . . , am] iff N 2 |=ι θ[(G(f ))(a1), . . . , (G(f ))(am)].

4here and below, for ι ∈ {6, 7} the conclusion is similar but expressed more cumbersomely



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Proof. First we can prove clause (δ) by induction on the quantifier depth of θ,

as in 2.9.Second, note that clause (α) follows from clause (β ).Third, note that clause (β ) follows from clause (δ)and the definition of satisfaction 1.1.Lastly, concerning clause (γ ) follows the definition of L -good (and clause

(β )).2.22

2.23 Conclusion. Assume

(a) Y ℓ = (M ℓ, I ℓ,F ℓ) is a counting k-system, τ (M ℓ) = τ for ℓ = 1, 2

(b) H is a witness for the counting k-equivalence of Y 1 and Y 2

(c) χ ∈ Lcard,T

(d) Υ is an inductive scheme for L ∗ = L card,T (τ +) or L card(τ +)

(e) every subformula of χ and of ψΥℓ (for ℓ < mΥ

0 ) and of ϕΥℓ (for ℓ < mΥ

1 )has at most k − 1 free variables

(f ) t ∈ T

(g) if ℓ < mΥ0 then ψΥ

ℓ has ≤ k/2 free variables and k ≥ 3 of course.

Then

(α) if θ = θΥ,χ,t and ι ∈ {2, 3, 4, 5, 6, 7, 11, 22} then M 1 |=ι θ ⇔ M 2 |=ι θ andM 1 |=ι ¬θ ⇔ M 2 |=ι ¬θ

(β ) For any t if N ℓ = N t[M ℓ, Υ, t] is well defined for ℓ = 1, 2, then for everysentence θ ∈ L f.o.(τ +) such that every subformula has at most k − 1 freevariables, we have N 1 |=ι θ ⇔ N 2 |=ι θ

(γ ) for any t, if N ℓ = N t[M ℓ, Υ, t] are well defined (for ℓ = 1, 2), and θ =θ(x1, . . . , xn) ∈ L f.o.(τ +) is a formula such that every subformula has atmost k − 1 free variables we have:

⊕ if a1, . . . , am ∈ N 1 and g ∈ H , each (G(f ))(aℓ) is well defined thenN 1 |=ι θ[a1, . . . , am] iff N 2 |=ι θ[(G(f ))(a1), . . . , (G(f ))(am)].

Proof. Straight.

Now

2.24 Conclusion. Assume

(a) Y ℓ = (M ℓ, I ℓ,F ℓ) is a medium t-dichotomical k-system, τ (M ℓ) = τ forℓ = 1, 2



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(b) H is a witness for the medium t-dichotomical k-equivalence of Y 1 andY 2

(c) χ ∈ L f.o.(τ +

)(d) Υ is an inductive scheme for L ∗ = L f.o.(τ +)

(e) every subformula of χ and of ψΥℓ (for ℓ < mΥ

0 ) and of ϕΥℓ (for ℓ < mΥ

1 )has at most k − 1 free variables

(f ) tℓ ∈ T

(g) if ℓ < mΥ0 then ψΥ

ℓ has ≤ k/2 free variables and k ≥ 3 of course.

Then

(α) if θ = θΥ,χ,t and ι ∈ {1, . . . , 5, 6, 7, 11, 22} then M 1 |=ι θ ⇒ M 2 |=ι θ andM 1 |=ι ¬θ ⇔ M 2 |=ι ¬θ

(β ) For any t if N ℓ

= N t[M ℓ, Υ, tℓ] is well defined for ℓ = 1, 2, then forevery sentence θ ∈ L f.o.(τ +) such that every subformula has at most(k − 1)-free variables, we have N 1 |= θ ⇔ N 2 |= θ.


2.25 Discussion We consider now some variants.1) We have to consider the stopping times. If L ∗ = L car,T or L card,T thisis natural, (and they are stronger logics than the earlier variants). If we stillwould like to analyze in particular for the others, we should be careful howmuch information can be gotten by the time.

2) We can modify Υ such that in N t+1 we can reconstruct the sequence (N ℓ,¯

P ℓ) :ℓ ≤ s (see §4).3) We can change our presentation: first proving the equivalences for N Z[Y ℓ,tℓ] forℓ = 1, 2, (see Definition 5.5) and then proving that (N t[M Y ℓ , Υ, t], P t[M, Υ, t])

is interpretable in N ZfullY ℓ,t uniformly.

§3 The canonical example

We apply §2 to the canonical example: random enough graph.

3.1 Definition. Let τ be a fixed (finite) vocabulary consisting of predicatesonly. We say M is a (s, k)-random τ -model if every quantifier free 1-type over

A ⊆ M, |A| < k (not explicitly inconsistent) is realized in M by at least s(M )elements. If s is constantly ∞ we may write k-random.

Remark. We can restrict the set of allowable quantifier free types if it is niceenough e.g. R two-place symmetric irreflexive. More generally see e.g. [BlSh528].



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3.2 Definition. Tpol is TQ, where for a set Q ⊆ R containing an unboundedset of reals > 0 let TQ be {f q : q ∈ Q,q > 0} where f q : ω → ω is f q(n) = nq, ormore exactly, ⌈nq⌉ the least integer ≥ nq.

3.3 Claim. Assume

(a) q∗, k are integers > 1 and k∗ = q∗k

(b) s ≤ k,s > 0 integer

(c) M ℓ is (sℓ, 3k∗)-random τ -model for ℓ = 1, 2

(d) tℓ(M ℓ) < (sℓ(M ℓ))q∗+1/(q∗ + 1)! and sℓ(M ℓ) > q∗

(e) Υ is an inductive scheme for L f.o.(τ [2]), χ a sentence in L f.o.(τ [2]) and

each subformula of any formula among ψΥℓ (ℓ < mΥ

0 ), ϕΥm(m < mΥ

1 ) and χ has at most s free variables

Then

⊕ if ι ∈ {2, 3, 4, 5, 6} and θΥ,χ,tℓ is ι-good (at least for M ℓ that is, the truth values below are defined) for ℓ = 1, 2, then M 1 |=ι θΥ,χ,t1 ⇔ M 2 |=ι

θΥ,χ,t2 .

3.4 Remark. 1) Compare clause (β ) with 2.21(2).2) Why in 3.3 do we use clause (d)? As there we use N t[M ℓ, Υ, tℓ] so for some t wemay add the sets in Pt[M 1, Υ, t1] to N t[M 1, Υ, t1] (in defining N t+1[M 1, Υ, t1]but do not add Pℓ[M 2, Υ, t2] to N t[M 2, Υ, t2] in defining N t+1[M 2, Υ, t2].3) We concentrate on ι-good sentences (or local versions) in order to have neatresults. Otherwise we have really to be more careful, e.g. about the cardinalities

of the N t[M, Υ, t]’s. This is very reasonable for counting logic.4) We have ignored in this claim, and others in this section, the cases 10 < ι.We can deal with them, if we note the following required changes. We have tonote that the function t is split to two functions; one, tsp, for telling us howto increase N t to N t+1, that is which additional families of subsets of N t areallowed to be added, and for this function the parallel of clause (d) should bedemanded. Secondly we have to consider what families are added in each stage(so for the counting and the medium analog our situation may be better).

Proof. We let I ℓ = {A ⊆ M ℓ : |A| ≤ q∗} and

F ℓ = {f :f is a partial automorphism of M ℓ

and Dom(f ) has ≤ q

∗

k elements}

(∗)1 Y ℓ = (M ℓ, I ℓ,F ℓ) is a k-system[why? the least obvious clause in Definition 2.1(1) is clause (D) whichholds by Definition 3.1 above.]

(∗)2 Y ℓ = (M ℓ, I ℓ,F ℓ) is (tℓ, s)-dichotomical, (see Def 2.1(4)).



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Why? The proof of (∗)2 takes most of the proof. Let m ∈ N be 1 ≤ m ≤ s andlet E be an equivalence relation from E 0I,m(∅) so it is an equivalence relation on

h∗/E 0Y ℓ,m

where h∗ : [m] → SeqI ℓ

, E satisfying (∗) of clause (δ) of Definition

2.1(3). For h ∈ h∗/E 0Y ℓ,m

let bh be the concatenation of h(1), h(2), . . . , h(m).

Without loss of generality h∗ is one-to-one and even bh∗ is with no repetitions.Let t∗ = ℓg(bh∗) so t∗ ≤ q∗k ≤ k∗. By clause (c) and the definition of F thereis a quantifier free formula ϕ(x, y) ∈ L (τ ) with ℓg(x) = ℓg(y) = t∗ such thath1Eh2 iff M ℓ |= ϕ[bh1 , bh2 ].

Clearly without loss of generalityϕ(x, y) tells us that the quantifier free type of x and of y (same as that of bh∗ = h∗(1)ˆh∗(2)ˆ . . . ˆh∗(m)), call it p(x) and let

p[M ] = {a ∈ t∗M : a realizes p(x)}, so we can look at E restricted to this set.Can there be a0, a1 ∈ t∗(M ℓ) realizing the same quantifier free type p(x) (overthe empty set) which are not E -equivalent? If not, then | p[M ℓ]/E | = 1 and weare done so assume there are. We can find a2 ∈ t∗(M ℓ) realizing the same quanti-

fier free type p(x) and disjoint to a0â1 (use “M ℓ is (sℓ, 3k∗

)-random”), so a2, aj

are not E -equivalent for some j ∈ {0, 1}; so without loss of generality a0, a1 aredisjoint. Now we ask “are there disjoint b0, b1 ∈ t∗(M ℓ) realizing p(x) which areE -equivalent”? If yes, we easily get a contradiction to “E an equivalence rela-tion” (by finding b′, a sequence from M ℓ realizing p(x) such that both b′â0, b′â1

realize the same quantifier free type as b0ˆb1; contradiction). So: no disjointb0, b1 ∈ p[M ℓ] are E -equivalent. Next we claim that

for some u ⊆ [0, t∗) and subgroup g of the group of permutations of u,moreover of g∗ = g∗u = {σ ∈ Per(u): if a ∈ t∗(M ℓ) realizes p(x) thenaσ(i) : i ∈ u realizes the same quantifier type as a ↾ u}, we have:

E ↾ {a ∈ t∗(M ℓ) : a realizes p(x)} = E g ↾ {a ∈ t∗(M ℓ) : a realizes p(x)}

where for any subgroup g′ of g∗, E g′ = E t∗,M ℓ

g

′ is defined by: for a, b ∈t∗(M ℓ), aE g′ b ⇔ (∃σ ∈ g)(aσ(t) : t ∈ u = br : t ∈ u) (this is anequivalence relation).

[Why? It is enough to show: assume a, b, c ∈ t∗(M ℓ) realize p(x) and a, b areE -equivalent, r∗ < t∗ and ar∗ /∈ {bt : t < t∗} then c/E does not depend on cr∗

(i.e. if c′ ∈ t∗(M ℓ) realizes p(x) and r′ < t∗ & r′ = r∗ ⇒ cr′ = c′r′ then cE c′.)Toward this end, let σ be the partial function from [0, t∗) to [0, t∗) such thatσ(r1) = r2 ⇔ ar1 = br2 . Clearly σ is one to one and r∗ /∈ Dom(σ).

We choose by induction on j∗ ≤ t! a sequence aj ∈ p[M ℓ] such that a0 = a, a1 =b, the quantifier free type of ajâj+1 in M ℓ is the same as the quantifier freetype of a0â1 = aˆb in M ℓ and (∀r)[aj+1

r /∈ {ajr : r < t∗} ⇒ aj

r /∈ {ar : r < t∗}].Clearly j < t∗! ⇒ ajE aj+1, hence j ≤ t∗! ⇒ aE aj . Let σj be the partial function

from [0, t∗) to [0, t∗) defined by σj(r1) = r2 ⇔ ar1 = ajr2 . Clearly σ0 = id[0,t∗)

and σj+1 = σ ◦ σj . Clearly σt∗! is the identity function on some subset of [0 , t∗)

and at∗!r2

= a0r1

(= ar1) ⇔ r1 = r2 & σt∗!(r1) = r1. Now given c′ as above we

can find b ∈ p[M 1] such that cˆb and c′ˆb realizes the same quantifier free typeas a0ât!, hence cE b & c′E b hence cE c′. Easily we are done proving

⊠ p[M ℓ]/E has cardinality ≥ (sℓ(M ℓ)|u| · (|g∗|/|g|).



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[Why? Clear by ⊠ and Definition 3.1.]This number is ≥ (sℓ(M ℓ)|u|/|u|!. Hence if |u| > q∗ as sℓ(M ) > k∗ = q∗k ≥|u| > q∗ (see assumption (d)) the number is ≥ (sℓ(M ℓ))q

∗+1/(q∗ + 1)! whichby assumption (d) is > tℓ(M ℓ) and we get one of the allowable answers inDefinition 2.1(4). So we can assume that |u| ≤ q∗ and this gives the secondpossibility so we have finished proving (∗)2.

Let

H =

f :f is a partial embedding of M 1 into M 2

with Dom(f ) having ≤ q∗k members

.

(∗)3 H is a (k, s)-witness to the equivalence of (Y 1, t1) and (Y 2, t2)[why? straight.]

So we can apply 2.22 and get the desired result. 3.3

Lastly, we can conclude the answer for the question in [BGSh 533]:

3.5 Theorem. 1) Assume ι ∈ {2, 3, 4, 5}, τ = {R}, R binary symmetric ir-reflexive, p ∈ (0, 1) and T are given and each t ∈ T is bounded by a poly-nomial. The logic L Tι (L f.o.)(τ ) satisfies the undecided 0-1 law for finite ran-dom enough model, that is graph with a fix probability p ∈ (0, 1) which means;if θ1 = θΥ,χ,t ∈ L Tι (L f.o.)(τ ) and θ0 = θΥ,¬χ,t then Min {Prob(M n |= θ0),Prob(M n |= θ1)} : n < ω converge to zero5, where M n is G(n, p), the random graph on n with edge probability p.2) Moreover also the undecided + 0 − 1-law hold; which means:

if θ1 = θΥ,χ,t ∈ L Tι (L f.o.)(τ ) and θ0 = θΥ,¬χ,t then for some ℓ ∈ {0, 1} the

sequence {Prob(M n |= θℓ) : n < ω converges to zero,3) Similarly for any fixed (finite) vocabulary τ consisting of predicates only p =pR : R ∈ τ , pR ∈ (0, 1)R.

Proof. 1) Let θ0, θ1, Υ, χ, t be as above and ε > 0. Let s be large enough suchthat assumption (e) of 3.3 holds, choose k = 3s so assumption (b) there holds.We choose sℓ as s(n) = (n − k) × (Min{pk/2, (1 − p)k/2}) and let q∗ be integer> 0 such that for n large enough s(n)q

∗+1 ≥ t(n)(q∗ + 1)! and let k∗ = q∗k.Let n be large enough and M 1n ,M 2n be random enough (for G(n, p)). We wouldlike to apply Claim 3.3 with M 1 = M 1n , M 2 = M 2n and Υ, χ as in the definitionof θ0, θ1 and t1 = t2 = t and s1 = s2 large enough. This is straight, noting thatthe case of the truth value of θ1 in M n is undefined, i.e. we run out of resources, just help us.2) Similarly, this time for n is large enough, n1 ≥ n, n2 ≥ n and M 1n1 ,M 2n2 arerandom enough (for G(n1, p), G(n2, p) respectively).3) Similarly 3.5

5we do not ask that for some ℓ < 2 the probability for the satisfaction of θℓ converges to1, as the decision when to stop may be complicated. If we e.g. use an inside “clock” to tell uswhen to stop, this disappears



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3.6 Discussion: 1) It is reasonable to consider the undecided law if we knowthat the (N t[M ℓ, Υ, tℓ], P t[M ℓ, Υ, tℓ]) for ℓ = 1, 2 are quite equivalent for everyt, when M 1 = M 2, but we do not have information otherwise.2) We might prefer to have the usual zero-one law. There are some avenuesto get (at some price), see also 3.7,3.8 below; we may consider all sentences,ι ∈ {2, . . . , 5} and the usual 0-1 law.

We have to try to use t which tries to diagonalize the right sets. That is, using 3.3for t1 = t2 = t, we can get strong enough equivalence of N +t [M 1, Υ], N +t [M 2, Υ],which is fine if M 1 = M 2 < M 2, so it is enough if N +t1 [M ℓ, Υ], N +t2 [M ℓ, Υ]with t1 = tι[M 1, Υ, t1], t2 = tι[M 2, Υ, t1] and choose t such that they are quiteequivalent. As in N +t [M ℓ, Υ] we can define N ↾ {0, . . . , t − 1}, this requirest(M ℓ) to be quite large compare to M ℓ. So we can get our desired 0-1 lawsand all possible ι’s, but for a logic remote from our intention.

On the other hand, we may restrict our family of sentences (here)

3.7 Theorem. If in Theorem 3.5 we restrict ourselves to the good sentences,i.e. the logic is L Tι (L f.o.)

good and ι ∈ {2, 3, 4, 5}, then the usual 0-1 law holds.

Proof. Similar. 3.7

3.8 Theorem. 1) Assume ι ∈ {2, 3, 4, 5}, τ = {R}, R binary symmetric ir-reflexive predicate, p ∈ (0, 1)R and T are given and for each t ∈ T for someinteger r and ε ∈ (0, 1/2)R we have 0 = lim t(n)/nr+1−ε : n ∈ N and ∞ = lim t(n)/nr+ε : n ∈ N. Then the logic L Tι (L f.o.)(τ ) satisfies the re-sults in 3.3 - 3.7.2) Similarly for any fixed (finite vocabulary τ consisting of predicates only,

¯ p = pR : R ∈ τ , pR ∈ (0, 1)R.

Proof. 1) Suppose that M ∗ |= θΥ,χ,t and this because in stage t∗ the run stop,i.e., in a good way; and assume further that M is random enough graph (forour given Υ and χ). We can find E ℓ for ℓ < ℓ∗ such that E ℓ is a quantifierfree formula with 2mℓ variable defining an equivalence relation on pℓ(M ) forevery random enough graph M , pℓ(x) a complete quantifier free type with mℓ

variables said to be pairwise distinct. We can find non negative integers rtℓ for

ℓ < ℓ∗, t ≤ t∗ such that: if M is random enough graph and N t = N t[M, Υ]then N t = Σℓ<ℓ∗rtℓ|(mℓM )/E ℓ|. Now the expected value of |(mℓM )/E ℓ| is of the form p′ × binomial(n, mℓ) for some constant p′. The distribution is similar

enough to normal (see [Sh 550]) to ensure that the run on M n will not stop fort ≤ t∗ for over using resources.2) Similarly 3.8

What we have done for random graphs we can do to unary predicate. Thepoint is to replace claim 3.3 by a parallel one (the rest will follow).



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3.9 Claim. 1) Assume that the vocabulary τ is {P }. P a unary predicate and

(a) q+, q−, k are integers > 1 and h is a decreasing (not necessarily strictly)

function from {0, 1, . . . , q+} to {0, 1, . . . , q−},(b) s is an integer ≤ k but > 0

(c) M ℓ = (|M ℓ|, P M ℓ) for ℓ = 1, 2

(d) if q ≤ q+ then

(i) tℓ(M ℓ) is at least |P M ℓ |q × |(M ℓ \ P M ℓ)|h(q)

(ii) tℓ(M ℓ) is strictly smaller than

binomial(|P M ℓ |, q + 1) × binomial(|M ℓ \ P M ℓ |, h(q)),

(iii) tℓ(M ℓ) is strictly smaller than

binomial(|P M ℓ |, q) × binomial(|M ℓ \ P M ℓ |, h(q) + 1).

2) The parallel of 3.5- 3.8 holds

Proof. Similar but letting I ℓ = {A ⊆ M ℓ: for some q ≤ q+ we have |A∩P M ℓ | ≤ qand |A \ P M ℓ | ≤ h(q)}. 3.9

3.10 Definition. 1) We say M is a τ -model with k-elimination of quantifiersif for every subsets A0, A1 of M , |A0| = |A1| < k and an isomorphism f fromM ↾ A0 onto M ↾ A1 and a0 ∈ M there is a1 ∈ M such that f = f ∪ {a0, a1}is an isomorphism from M ↾ (A0 ∪ {a0}) onto M ↾ (A1 ∪ {a1}).2) We replace “quantifiers” by “quantifier and counting” if we add: and thetwo sets {a′0 ∈ M : a′0, a0 realize the same quantifier free type over A0} and

{a′1 ∈ M : a′1, a1 realize the same quantifier free type over A1} has the samenumber of elements (we can then get it to equivalence relations on m-tuples).

3.11 Claim. 1) Assume (a), (b), (e) in 3.3 replacing first order by counting logic and

(c)− (α) M ℓ are τ -models which has k-elimination of quantifier for ℓ = 1, 2

(β ) if ϕ(x, y) is a quantifier free formula defining an equivalencerelation and ℓg(x) = ℓg(y) ≤ k∗ then the number of classes is> tℓ(M ℓ) for ℓ = 1, 2 or for some u ⊆ [0, ℓg(x)) with ≤ q∗

elements, some ϕ′(x ↾ u, y ↾ u) defines the equivalence relation in M 1 and in M 2

(γ ) if 2r + s ≤ k and for ℓ = 1, 2, aℓ ∈s

(M ℓ), xj

= xj

i : i < s, for j = 1, 2,ϕℓ = ϕℓ(x1, x2, , aℓ) is first order and defines in M 0an equivalence relation E ℓ and ϕ1 = ϕ2 and the quantifier free types of a1 in M 1 and a2 in M 2 are equal,then |r(M 1)/E ℓ|, |r(M 2)/E 2| are equal or

|r(M 1)/E 1| > t1(M 1) & |r(M 2)/E 2| > t2(M 2).



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Then

⊕ if ι ∈ {1 − 7, 11 − 17, 21 − 27} then M 1 |=ι θΥ,χ,t1 ⇔ M 2 |=ι θΥ,χ,t2 .

2) We have a theorem like 3.3, 3.8 (for ι as above) using 3.11(1) instead of 3.3.3) We can in parts (1), (2) and in 3.3, 3.5, 3.7, 3.8 replace L f.o. by L f.o. + na.

Proof. Straight.

3.12 Claim. 1) Choiceless polynomial time does not capture counting logic.2) Similarly for the pair (L Tι (L f.o.),L Tι (L f.o.+na)), the pair (L Tι (L f.o.+na),L Tι (L card)) and the pair (L Tι (L f.o.+na),L Tι (L card,T)).3) We can apply 3.11 to show that the pairs (M 1n, M 2n) of models from [GuSh 526] are not distinguished by our logics (for a sentence θ for n large enough.

Proof. 1) Use 2.22, 3.9 on the question: |P M | ≥ M /2 with τ = {P }.

2),3) Also easy 3.12

3.13 Remark. : Y. Gurevich asks for 0-1 laws, as in 3.3 - 3.8, for the generalframework of §2. The answer is quite straight by 3.14, 3.15, when we use constantI .

3.14 Definition. Let τ be a fixed vocabulary consisting of predicates only. Wesay M is (s, k)-random model if: every quantifier free 1-type over A ⊆ M, |A| < k(not explicitly inconsistent) is realized in M by at least s(M ) elements.

We are, of course, using

3.15 Claim. Let k , s > 0 be integers, let τ = {R}, R symmetric irreflexiveand p ∈ (0, 1

2 ]R. The probability that: for M n a G(n; p) random graph (soset of vertices in [n]), (M n, I ) is not (s, k)-random is ≤ Σℓ<kbinomial(n, ℓ)×Prob(flipping n − ℓ times a coin with probability p/2ℓ for a head we get < sheads).

§4 Relating the definitions in [BGSh 533] to the one here

4.1 Discussion If we just like to replace the creation of N t[M, Υ, t] by ASM,we can note that we can straightforwardly code the actions of the ASM by amonotone Υ; the waste is small except that we are not allowed to omit oldelements so for fine measurements this make a difference. But we can justreplace N t[M, Υ, t] by a situation of the ASM with no lose and no real differencein the proofs. Still, the reader may instead of just accepting or understandingthis observation choose to read the formal translation below. Though this seemtrivial, writing the details of a translation is tedious.

4.2 Discussion: How do we relate between the definitions above and [BGSh533]?



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(i) an infinite structure I there corresponds to a τ -model here

(ii) a state A there corresponds to a model of the form N = N t[M, Υ] in 1.3

and (N,¯

P ) in 1.8(iii) dynamic function there corresponds to P t,ℓ here

(iv) an object x is active at A in 5.1 there, corresponds to x ∈ N

(v) a program 5.7 there corresponds to an Υ in 1.1(2) here (mainly the firstorder formulas used);

(vi) the counting function in 5.5 there corresponds to the cardinality quanti-fier (1.1(6)) here

(vii) the polynomial functions p, q in 5.1 there corresponds to tsp, ttm ∈ There

(viii) the logic there corresponds to L Tι (L ∗),L ∗ ∈ {L f.o.,L r.o.+ na,L card}here.

If we insist on P ℓ being individual constants this still can be done with a price.The P t+1,ℓ can in the usual set theory manner be actually 7-place function fromN t to N t or 7-place relation on N t, or be the universe of N t. Understandingthis to interpret the successor step there to here we need that all parts of theprogram are expressible in L f.o. (or L card). For the other direction we need toshow f.o. operations can be expressed by the programs of ASM there (see 6.1there), no problem (and not needed to show our results solved problems there).

4.3 Lemma. 1) Let π be a program concerned with τ -models (in [BGSh 533,4.7]’s sense). We can find a natural number r∗ ≥ 1 and Υ such that:

⊠1(a) Υ is an inductive scheme in L f.o. with τ Υ = τ π

(b) for every integer polynomials p(n), q(n) and τ -model M such that p(n) ≥

2, q(n) > n + 2 we have M |= θΥ,p∗

(n),q∗

(n) (in the sense of Definition 1.3) iff M |= π (in the sense of [BGSh 533]) when

(∗) p∗(n) = 3 + r∗ p(n), q∗(n) = 2 + 2q(n).

2) If π is as in [BGSh 533, §11], that is with being able to compute the cardinality of M (= set of atoms) then a similar result holds but Υ is in L f.o. + na.3) If π is as in [BGSh 533] for cardinality logic (see §4 there), then a similar result holds but Υ is in L card and spaces do not use θΥ,3+n+r∗( p),2+n+2q(n).4) We can replace p, q by arbitrary function from T and get similar results.

Before proving 4.3:

4.4 Observation. 1) If we identify truth, false with the sets ∅, |M | and m1 is asin 1.1, then the state (for π, i.e. Υ there, etc.) of [BGSh 533] are the same(M, m1)+-candidate here when we identify a state there with its set of activeelements (O.K., as they carry the same information). Sometimes we use anytransition set ⊆ V ∞(M ) containing the active member.

The main point is



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4.5 Claim. Let π be a program concerned with τ -models and the states corre-spond to (M, m1)-candidates.1) For every term σ for some natural number r(σ) (actually at most its depth)and pure inductive scheme Υ which is monotonic and ψ0 = [y = x0] we have:

⊠1 if (N i, P i) is a (M, m1)-candidate for i = 0, . . . , r(σ) and (N i+1, P i+1) isthe (M, Υ)-successor of (N i, P i) and ζ is a variable assignment of x, a sequence listing the free variables of σ into N σ, then the interpretation from σ under ζ in N i satisfies d =: val N i,ζ(σ) ∈ N r(σ).

2) Moreover in (1) we can find also formulas ψℓ,r(x, yℓ, z) : ℓ < mΥ1 ,r < r(σ)

such that

⊠2 in ⊠1 above we can add:if r ≤ r(σ) and let N ′r = N 0∪(T C (d)∩N r) where T C is transitive closure,then N ′r+1 = N r ∪ {a : for some b ∈ ℓg(y)N r, (N ′r, P ) |= ψℓ,2(a, b, ζ )}

(identifying ζ with a sequence of members of N 0).

3) For every rule R (see [BGSh 533, 4.5]) there are r(R) ∈ N and an inductivescheme Υ in L f.o. such that ( P 0 is zero place relation):

⊠3 if (N i, P i) is a (M, Υ)-candidate for i ≤ r(R) and (N i+1, P i+1) is theΥ-successor of (N i, P i) for i < r(R) and P i,0 = truth, then i < r(R) ⇒N i ⊆ N i+1, the stationary (N i(R), P −

i(R)) is the R-udpate of (N 0, P −0 ),

P i,r(R) = truth where P − = P ↾ [1, mΥ1 ).

4) In (3) we can even have

⊠4 N r = N 0 ∪ {x : x active in N r∗}.

Proof. 1) By induction on the term.2) As N r+2 can be (uniformly) interpreted in N r.3) By induction on the rule R. 4.5

Proof of 4.3.We describe what is an Υ-successor rather than let r∗ = r(Rπ) + 1, Rπ is the

rule which π is. Then say formally what is Υ.Now the predicates (and function symbols) {P Υk : k < mΥ

0 } serve some pur-poses:

kind 1: The dynamic predicate and function symbols of π, say P k for k ∈ w(1),

say P k(1,0) will denote ∅, P k(1,a) will denote the set of atoms.For notation simplicity

kind 2: P k(2,0) unary predicate will serve to denote the set of active elements;and

kind 3: P k(3,1), . . . , P k(3,r∗) will be zero place relations, they will denote the timemodulo r∗, say for t = 0 they are all false; for t = 1 we get true, true false ...,



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for t = 2 we get true, true, true, false... (without loss of generalityr∗ ≥ 3 fort = 3 + r∗s + r, r < r∗ we have P k(3,r′) ≡ r′ = r). The reason is that in ourtranslation one step for π will be translated to r∗ steps in the construction of theN t’s and the translation begins only with t = 1. Now we can describe almost atranslation.

Now Υ is such that:

(a) N 1[M, Υ] is M ∪ {M, ∅},N 2[M, Υ] is M 1[M, Υ] ∪ {1}, recall 1 = {∅}N 3[N, Υ] is N 1[M, Υ] ∪ {1, 2},

(b) if t∗ = 3 + r∗s, then P t∗,k(3,0) = true, P t∗,k(3,1), . . . , ct∗,k(3,r∗) are falseand we take care that P t∗+r,k(3,r′) = r′ = 0 mod r∗ for r′ ≤ r∗ and

(N t∗+2, P t∗+r) : i ≤ r(R∗) is as in 4.4(3)+(4). Moreover P t∗+r∗ =P t∗+r(R∗) and N t∗+r∗ is the set of active members of (N t∗+r(R∗), P t∗+r(R∗).

Well, as in θ = θΥ,1+r∗ p,q and ι = 1, the stopping decision for time will be thesame, but we still have to deal with the space (up to now using r + r∗ p wouldbe O.K.). However, between t∗ and t∗ + r∗ we have to preserve N t∗ till creatingN t∗+r(R∗) and only then can we omit the elements of N t∗ no longer necessary.

So we will have

kind 4: individual constants P k(4,1), P k(4,2)

(c) if t∗ = 3 + r∗s then: P t,k(4,0) < P t,k(4,1) < P t,k(4.2) are the three lastordinals, P t,k(4,0) is an active ordinal but not P t,k(4,1), P t,k(4,2) and x ∈N t∗ is active iff P t,k(4,2) ∪ x ∈ N t∗ .

So N t∗ = 2 + 2{x ∈ N t∗ : x active}.

Now starting with N t∗

in deciding N t∗

we omit and non-active elements exceptthe two last ordinals and then do as earlier by 4.4(3)+(4) for r = 1, . . . , r(Rπ)taking care to have the right natural numbers.

So defining N i∗+2∗ , we take care of the “doubling”.2), 3), 4) Similarly. 4.3

Less critical is the inverse relation

4.6 Lemma. 1) Let Υ be an inductive scheme for L f.o., ι = 7, τ = τ Υ, χ ∈L f.o.(τ ).Then we can find r∗ ≥ 1 and a program π for the same vocabulary as in [BGSh 533, §4] such that for every integer polynomials p(n), q(n) and τ -model M such that:

M |=ι θΥ,χ,p,q (see Definition 1.3) iff M |= π where π = (π, p∗, q∗)and |= is as in [BGSh 533, §4] where: p∗(n) = r∗, p∗(n) + r∗, q∗(n) =q(n) + 2



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4.7 Lemma. 1) If π is as in [BGSh 533, §11], that is with being able to computethe cardinality of M (= set of atoms) then a similar result holds but Υ is in L f.o. + na.2) If π is as in [BGSh 533] for cardinality logic (see §4 there), then a similar result holds but Υ is in L card and spaces do not use θΥ,3+n+r∗( p),2+n+2q(n).3) We can replace p, q by arbitrary function from T and get similar results.

Proof. 1) We ignore too short runs for simplicity. The “q(n) = q(n) + 2” comesfrom [BGSh 533] starting with two extra elements so during the computation wepreserve having two entry elements (except when we notice we are to stop-seebelow).

Now every step of the computation for π is translated to r∗ step during thecomputation of the N t[M, Υ]’s.What do we do in those r∗ steps? First we compute the relations on N t definablefirst order subformulas of the ψℓ. We also translate χ to be equivalent to whatshould be in the next state and then add the new elements (so N t+1 |= χ wascomputed in N t, as in 4.4).2), 3), 4) Similar to part (1) + 4.3.

[How do we “compute” the first order formulas? Where P ϕ(x) code ϕ(x), we of course represent all subformulas and do it inductively.]

Atomic are given

negation is by “if P ϕ(x)(a) = truth then P ϕ(x)(a) = false, else P ϕ(a) is truth(and appropriate “for all” also adding dummy variables is possible by “for all”.

For conjunctions ϕ(x) = ϕ1(x) use “if P ϕ(x)(a) = truth then P ϕ(x)(a) = P ϕ2(x)(a)else P ϕ(x)(a) = false.

For existential quantifier, ϕ(x) = (∃y)ψ(y, x) use“if P ψ(x,bary)(a) = truth then P ϕ(x)(a) = truth else do nothing”. 4.3



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§5 Closing comments

We may consider a context is (K,I ) such that logics related to our proof in

§2. The first version in 5.1(3) changes the satisfaction the second (in 5.1(4),(4A))changes also the syntax.

5.1 Definition. 1) A context is a pair (K,I ) such that

(a) K be a class of models with vocabulary (= the set of predicates) τ

(b) I is a function

(c) Dom(I ) = K

(d) I (M ) is a family of subsets of M , whose union is |M |, and closed undersubsets.

1A) We call (K,I ) invariant if

(e) I is preserved by isomorphisms, i.e. if f is an isomorphism from M 1 ∈ K onto M 2 ∈ K and A ∈ I (M 1) then f ”(A) ∈ I (M 2).

1B) We say “f is an I -isomorphism from M 1 ∈ K onto M 2 ∈ K ” if f is anisomorphism from M 1 onto M 2 such that I (M 2) = {f ”(A) : A ∈ I (M 1)}.

Recall that I (M 2)[k] = {ℓ<k

Aℓ : Aℓ ∈ I [M ]}.

2) In 1) let

SeqαI (M ) = {a : a a sequence of members of M of length α, Rang(a) ∈ I (M )}.

3) Let L = L f.o. or L = L λ,κ; recalling that the logic L λ,κ,α for λ, κ cardinals,

is defined like first order logic but we allow conjunctions of i<α

, for α < λ and

existential quantifier (∃x) with x a sequence of variables of length < κ, and

the depth of the formulas is < α. We define L [k] = L [k]λ,κ,α, logics with the

same syntax but with a difference in the definition of the satisfaction relation,

M |=[k]I

ϕ[a] or (M,I ) |=[k] ϕ(a) is defined inductively on α as usual, exceptthat

(∗) we demand Rang(a) ∈ I [M ][k] (otherwise not defined) , that is

M |=[k]I

(∃x)ϕ(x, b) iff

Rang(b) ∈ I [M ][k − 1] and for some a ∈ ℓg(x)M withRang(a) ∈ I (M ) we have M |=k

I ϕ[a, b].

Let L λ,κ,α;card,L λ,κ;+na,L λ,κ;card,T be defined similarly (so

M |=[k]I ∃!µxϕ(x, b) iff (b ∈ ℓg(b))M ), Rang(b) ∈ I (M )[k − 1]

and µ = |{a : {a} ∈ I and M |=[k]I ϕ[a, b]}|.



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Omitting α means some α.If λ,κ,α are omitted they are ℵ0 (so Lλ,κ,α is first order).

4) We define a logicL

[k]

λ,κ,α. Let us define the formulas inL

[k]

λ,κ,α by inductionon α, each formula ϕ has the form ϕ(x0, x1, . . . , xk1−1), k1 ≤ k, where the xℓ’sare pairwise disjoint sequences of variables of length < κ (so if κ = ℵ0, finitesequences) and every variable appearing freely in ϕ appear in one of those se-quences (so any formula is coupled with such x0, . . . , xk1−1, probably some notactually appearing).

α = 0: quantifier free formula; i.e. any Boolean combination of atomic ones(with the right variables, of course).

α + 1: α non-limit ϕ(x0, . . . , xk1−1) is fromL [k]λ,κ,α or is a Boolean combination of

formulas of the form (∃y)ψ(xi0 , . . . , xik2−2, y) where k2 ≤ k, ψ(xi0 , . . . , xik2−2

, y) ∈

L

[k]λ,κ,α.

α limit: L [k]λ,κ,α =

β<α

L [k]λ,κ,β.

α + 1, α limit: L [k]λ,κ,α+1 is the set of ϕ ∈ L [

k]λ,κ,α or ϕ a Boolean combinations of

members of L ∗λ,κ,α of the right variables.

Let L kλ,κ =α

L [k]λ,κ,α and L

[∗]λ,κ,α =

k<ω

L [k]λ,κ,α and L

[k]α = L[k]

ℵ0,ℵ0,αand

L [k]<α =

β<α

L [k]β and L [k] = L

[k]ℵ0,ℵ0

and L[∗] =k<ω

Lk.

4A) We now define a satisfaction relation M |= ϕ(a0, . . . , ak1−1) where k1 ≤ k(depending on I ).

I.e. we define by induction on α, for ϕ(x0, . . . , xk1−1) ∈ L [k]λ,κ,α, aℓ ∈ Seq

ℓg(xℓ)I

(M ),

when does M |= ϕ[a0, . . . , ak1−1] and when M |= ¬ϕ[a0, . . . , ak1−1]. Thisis done naturally, in particular M |= (∃y)ϕ(a0, . . . , ak2−2, y) iff for some b ∈

Seqℓg(y)I

(M ), (so Rang(b) ∈ I (M )) we have M |= ϕ[a0, . . . , ak2−2, b].

5) We can define for L one of the above, (L )card similarly adding the quanti-fiers [ϕ′(x; z)/ϕ′′(x, y, z)] saying: ϕ′′(. . . , . . . ; z) define an equivalence relationon {x : ϕ′(x)} with exactly s equivalence classes.6) We can above replace models M by pairs (M, I ), I ⊆ P(M ) closed undersubsets.

5.2 Discussion: We may replace M by M +, adding elements coding each A ∈I (M ), with decoding by functions, but

(a) this does not capture |=[k] and

(b) for |=[k] this requires infinitely many functions, we need to actually codeany sequence listing each A ∈ I (M ).



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52 SAHARON SHELAH

Still this framework seems to work quite smoothly for its purposes. We couldhave made it more central (use 5.5 below).

Note that

5.3 Observation. For any k-system Y and Z and mΥ1 -lifting of Y , letting M Y =

M we have

Y Z = (N Z, {A : (∃f ∈ F )[Dom(f ) ∈ I F & A ⊆ Dom(G(f ))]}, {G(f ) : f ∈ F })

is a k-system.

Now easily (and it makes a connection with §1, §2):5.4 Observation: 1) Let (K,I ) be a context and M 1, M 2 ∈ K be (finite as

always), τ -models, τ a vocabulary, and k < ω .The following are equivalent:

(A) there are k-systemsY ℓ = (M ℓ,I (M ℓ),F ℓ) for ℓ = 1, 2 and H as in Definition 2.21(1),(2)

(B) for every sentence ψ ∈ L [k]

f.o.(τ ) we have

M 1 |=[k]I

ψ ⇔ M 2 |=[k]I

ψ

(C ) for infinite λ,κ,α, for every sentence ψ ∈ L [k]λ,κ,α(τ ) we have

M 1 |=[k]I

ψ ⇔ M 2 |=[k]I

ψ

(D) for every t and infinite λ,κ,α for every sentence ψ ∈ L [k]λ,κ,α(τ ) we have

N Zfull

Y 1,t |=[k]I Y 1t

ψ ⇔ N Zfull

Y 2,t |=[k]I Y 2t

ψ where N Zfull

Y ℓ,t is defined in 2.13(7) and

I Y t = {Dom(f ) : f ∈ GZfullY ,t}.

5.4

Proof. Straight.

5.5 Definition. 1) We say Υ (from Definition 1.1) is pure if m1[Υ] = 0 so noP ℓ.2) Let Y = (M,I,F ) be a k-system; let “Z∗ is the full t-successor of Z” beas defined in 2.13. We define by induction on t; Zt = Zt[Y , t] as follows: for

t = 0, N Zt

= M, P Zt

ℓ is the empty set, GZt

is the identity on F and RZ

={(A, x) : A ∈ I, x ∈ A}; for t = s + 1 let Zt be the full t-successor of Zs. LetI Z = I Z,Y = {B : B ⊆ S Z(A) for some A ∈ I } where S Z(A) = S Z,A = {x ∈N Z : A is a Z-support of x}.

We can also see:



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5.6 Claim. 1) Assume

(a) Y ℓ = (M ℓ,I ℓ,F ℓ) is a t -dichotomical k-system for ℓ = 1, 2

(b) Zℓt = Zt[M ℓ, I ℓ], so Zℓt+1 is the full successor of Zℓt.

Then the following are equivalent:

(α) (M 1, I 1), (M 2, I 2) are L [k]-equivalent

(β ) for every t < ∞ the pairs (M Zt1 , I Zt), (M Zt2 , I Z2) are L [k]-equivalent.

2) For any given (M, I ) there in a sentence ψ ∈ L [k]

f.o.(τ M ) satisfied by (M, I )

and implying any other sentence ψ′ ∈ L[k]∞,∞(τ µ) satisfied by (M, I ).

5.7 Remark. 1) So in part (2) we can apply 2.22, 2.23, 2.24.2) Note that by 5.3, L [k] satisfies addition theorems.

5.8 Fact: For any Υ we can find Υ′

which is equivalent if we use in Definition1.3 the case ι = 4 (well when t(M ℓ) always is ≥ 2). In fact, we can reconstruct(i.e. define by a formula in L ∗) the sequence of P t,ℓ : t′ < t in N t.

5.9 Conclusion 1) Assume Y is counting k-system (see 2.3). Then we can defineRt, Gt for every t (N t the “computation” in time t) such that

(M, P 0, G0, R0) is 0-lifting

(N t+1, P t+1, Gt+1, Rt+1) is a lifting, successor of (N t, ct, Gt, Rt).

2) So the formula the ϕ defines is preserved by f ∈ F 0.

Proof. Straight.

∗ ∗ ∗

5.10 Discussion: 1) In §2 and in 5.5 we can allow infinite models M and defineN t[M ] = N t[M, Υ, t] for every ordinal t, for this better assume Υ is standard,monotonic and pure (or strongly monotonic, i.e. demand P ℓ,t[M, Υ, t] is in-creasing with t; for t limit we take union and so V [M, Υ, t] = ∪{N α[M, Υ, t] :α an ordinal}, see below. Now as in the case ι = 4, the analysis in §2 works forthis but it is not clear if we can get any interesting things.Note that those definitions remind us of Godel’s construction of L, particularly of Lα+1 from Lα, and the Frankel-Mostowski models (which use automorphisms).

May can this give interesting proofs of consistency for set theory with no choicebut with urelements? It seems they can be reduced to the classical case.2) We can prove various equivalence and 0-1 laws by 5.5, by proving that therelevant model N t can be interpreted in Zt[M, I ] from 5.5, using f.o. logic whichsuffices.

Proof. Straight.



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REFERENCES.

[BlSh 528] John T. Baldwin and Saharon Shelah. Randomness and Semi-

genericity. Transactions of the American Mathematical Society ,349:1359–1376, 1997. math.LO/9607226.

[BGSh 533] Andreas Blass, Yuri Gurevich, and Saharon Shelah. Choiceless Poly-nomial Time. Annals of Pure and Applied Logic, 100:141–187, 1999.math.LO/9705225.

[EbFl95] Heinz-Dieter Ebbinghaus and Jorg Flum. Finite model theory . Per-spectives in Mathematical Logic. Springer-Verlag, Berlin, 1995.

[GuSh 526] Yuri Gurevich and Saharon Shelah. On finite rigid structures. Jour-nal of Symbolic Logic, 61:549–562, 1996. math.LO/9411236.

[Ho93] Wilfrid Hodges. Model theory , volume 42 of Encyclopedia of Math-

ematics and its Applications. Cambridge University Press, Cam-bridge, 1993.

[Ka] Carol R. Karp. Finite quantifier equivalence. In J. W. Addison,L.A. Henkin, and A. Tarski, editors, The Theory of Models, pages407–412. North–Holland Publ. Co, 1965.

[Sh 550] Saharon Shelah. 0–1 laws. Preprint . math.LO/9804154.

[Sh:a] Saharon Shelah. Classification theory and the number of noniso-morphic models, volume 92 of Studies in Logic and the Foundationsof Mathematics. North-Holland Publishing Co., Amsterdam-NewYork, xvi+544 pp, $62.25, 1978.

[Sh:c] Saharon Shelah. Classification theory and the number of noniso-morphic models, volume 92 of Studies in Logic and the Founda-tions of Mathematics. North-Holland Publishing Co., Amsterdam,xxxiv+705 pp, 1990.

Documents

Saharon Shelah- Choiceless Polynomial Time Logic: Inability to Express