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Model Theory and Proof Theory of the Global
Reflection Principle
March 9, 2021
Abstract
The current paper studies the formal properties of the Global Reflection Principle,
to wit the assertion "All theorems of Th are true", where Th is a theory in the language
of arithmetic and the truth predicate satisfies the usual Tarskian inductive conditions
for formulae in the language of arithmetic.
We fix the gap in Kotlarski’s proof from [15], showing that the Global Reflection
Principle for Peano Arithmetic is provable in the theory of compositional truth with
bounded induction only (CT0).
Furthermore, we extend the above result showing that Σ1-uniform reflection over
a theory of uniform Tarski biconditionals (UTB−) is provable in CT0, thus answering
the question of Beklemishev and Pakhomov [2].
Finally, we introduce the notion of a prolongable satisfaction class and use it
to study the structure of models of CT0. In particular, we provide a new model-
theoretical characterization of theories of finite iterations of uniform reflection and
present a new proof characterizing the arithmetical consequences of CT0.
1 Introduction
The Global Reflection Principle (GRP) for a theory Th is the assertion that all theorems
of Th are true. As the statement involves the notion of truth for the language of Th, to
uncover its meaning adequately one shall pass to a proper extension of Th in a richer
language. Minimal such extensions are called axiomatic theories of truth for Th. Each such
theory arises by enriching the language of Th with a single fresh predicate T(x)and
adding a bunch of axioms characterizing T(x)as a truth predicate for the language of
Th. In the paper we focus on one of the most natural such extensions, which comprises
straightforward formalizations of the usual inductive Tarski’s conditions in the language
of Th together with the predicate T(x). Let us denote this theory with CT−.1
The GRP lies at the intersection of at least three, to some extent independent, areas of
research. The first, which was the starting point for the current paper, is the Tarski Bound-
ary project, that seeks to characterize the extensions of CT−+Th which are conservative
over Th. This is non-trivial for two reasons: on the one hand, if Th can develop enough
coding apparatus,2CT−+Th+does not prove any new sentences from the language of
Th. On the other hand, it is an immediate consequence of second Gödels Incompleteness
theorem, that CT−+Th + +GRP is a nonconservative extension of Th. Moreover, the
1The minus sign signalizes the lack of induction for the extended language. Defined as CT in [10] and
CT in [2]
2As shown by [17], this holds whenever Th interprets the elementary arithmetic EA, which, for the pur-
poses of this paper can be taken to be I∆0+ exp.
1
"natural" extensions of CT−+Th that are nonconservative over Th all prove GRP for Th.
Hence, in a sense, GRP is the source of nonconservativity in the realm of truth theories
and it seems highly desirable to know what are the minimal resources needed to prove
it.
The second area is proof theory, especially ordinal analysis as initiated in [21] and
further developed e.g. in [2]. In this approach natural truth-free counterparts of GRP,
Uniform Reflection Principles (REF), play central roles in determining the quantitative
information about the conseqeunces of theories of predicative strength. To get things
right, one adds stratified truth predicates to the picture and studies (partial) uniform
reflection principles over axiomatic theories of truth (this is the method used in [2]).
This is how the GRP enters the scene.
Last but not least, axiomatic theories of truth play an important role in the field of
model theory of Peano Arithmetic (compare [12] and [14]). Any subset S⊆ M such
that (M, S)|=CT−is essentially a full satisfaction class. If (M, S )additionally satisfies
GRP, then Scontains all the theorems of Th, in the sense of M. Satisfaction classes pro-
vide a very powerful tool in constructing interesting models of PA and investigating their
structure3
The current paper contributes to all the three areas. More specifically
1. We prove that ∆0-induction for the truth predicate is enough to prove GRP for PA.
Coupled with the earlier results by Kotlarski [15], this shows that, over CT−+EA,
∆0induction for the truth predicate is equivalent to GRP for PA. This improves on
earlier results from [25] and provides a direct fix to Kotlarski’s argument in [15].4
Additionally, coupled with various developments from the literature, our result
shows that the Global Reflection Principle for PA is a very robust notion, being
equivalent to various others, apparently very different, truth-theoretic principles,
as witnessed by the Many Faces Theorem (Corollary 58).
2. We extend the above result, answering the open problem posed by Beklemishev
and Pakhomov in [2]. We show that not only GRP is provable in CT−+EA in
the presence of ∆0-induction (denote this theory with CT0) but also Σ1-Uniform
Reflection over a weak truth extension of EA, called UTB−+EA (which adds to
the arithmetical part uniquely Uniform Tarski Biconditionals). The result has some
bearings on the analysis performed in [2].
3. We provide a new conservativity proof for CT0. Unlike in the first one from [15] we
are able to show directly that CT0is arithmetically conservative over ω-iterations
of uniform reflection over PA (denote this theory with REFω(PA)5). The proof is
based on an essentially model-theoretic idea of prolonging a (partial) satisfaction
class in an end-extension. This proves to be sufficiently robust to characterize finite
iterations of Uniform Reflection. We show that a model (M, S), where Sis a partial
inductive satisfaction class satisfies niterations of uniform reflection if and only if
a nonstandard restriction of Scan be prolonged ntimes.
The paper is organised as follows: in Section 2 we introduce all the relevant prelimi-
naries and context. In particular we develop handy and uniform conventions regarding
the definable models and satisfaction classes. Section 3 is devoted to the proof of GRP
3For example the construction of a recursively saturated rather classless model of PA by Schmerl [14]
employs them in a crucial way.
4Around 2012 a serious gap in the proof of Theorem 2.2 was discovered Richard Heck and Albert Visser.
5The direct conservativity argument for these theories is presented in [2] as well.
2
in CT0. In particular we describe the history of the problem and comment on flaws in
Kotlarski’s aforementioned proof [15]. The proof is a streamlined version of the one pre-
sented by the author in [18]. In Section 4 we extend the result from the previous section
answering the question of Beklemishev and Pakhomov in [2] in the positive: we give a
proof of Σ1-uniform reflection over UTB−+EA in CT0. The proof makes crucial use of the
Arithmetized Completeness Theorem. Additionally, the section offers some strengthen-
ings of this main result. In Section 5 we give a proof of the conservativity of CT0(PA)over
ω-iterations of uniform reflection over PA. Extending the work of Kaye and Kotlarski [13]
we characterize the theory of n-iterated uniform reflection over PA in terms of models of
the form (M, S)where Sis a partial inductive satisfaction class. Finally we examine the
structure of models of CT0and prove a variation of the main result of Section 4.
To enhance the reading, (as usual) denotes the end of a proof, while means that
the proof is omitted. 4signalizes the end of a definition, remark, convention etc.
Acknowledgements Will be added.
2 Setting the Stage
In this section we gather all the technical preliminaries needed to follow our reasoning
and at the same time develop a useful framework for proving our main results. In particu-
lar most of the results contained in here can be found (sometimes under slightly different
wording) in [12] and [9].
For starters, PA denotes Peano Arithmetic and Ldenotes its language, which we stip-
ulate to contain +,×,0,1,≤as primitive symbols. While studying extensions of PA in a
richer language, we use a handy convention known from [14]: PA∗denotes any theory
in the extended language that admits all instantiations of the induction scheme for the
extended language. Similarly, IΣ∗
ndenotes the extension of PA with induction for Σnfor-
mulae of the extended language. If L0is any language then, L0
Sand L0
Tdenote the result
of extending L0with a single binary predicate Sor a single unary predicate T, respec-
tively. Most of the extensions of PA that we study are formulated either in the language
LSor LT. Last but not least, EA denotes elementary arithmetic, i.e. the extension of I∆0
with a single Π2assertion "exp is total". All the theories we study extend EA, possibly in
a richer language. ∆0(exp) denote the class of bounded formulae in the language with
a symbol for the exponential function exp: it will be used throughout the paper because
various syntactical functions needed to state the axioms for the satisfaction predicate are
in fact ∆0(exp). However, since most of our theories are extensions of PA, the presence of
exp as a primitive symbol do not increase their strength. We explain this in more detail
in Section 3.
To smoothly deal with class sized-objects (such as definable models of arithmetical
theories) various definitions will be stated in the canonical predicative two-sorted ex-
tension of PA i.e. ACA0. Uppercase letters X, Y , Z,X1, Y1, Z1. . . denote second order
variables. In all applications we shall reason about definable classes (perhaps in a richer
language) which will be substituted for the free second order variables. The two sorted
language is denoted L2.
Coding Conventions We assume a standard coding of syntax in PA (defined as in [9]):
primitive symbols of the language are assigned numbers in a recursive way, and then
terms, formulae, sentences etc. are treated as well-formed sequences of such numbers.
3
The notion of sequence is based on the definable Ackermanian membership predicate ∈.
Term(x), ClTerm(x), Var(x), Form(x), Sent(x)denote the arithmetical formulae express-
ing that xis an arithmetical term, a closed term, a variable, an arithmetical formula and
an arithmetical sentence respectively. x∈Subf(y)expresses that xis a subformula of y.
We define x∈Term(y),x∈ClTerm(y)analogously (yis required to be either a formula
or a term). x∈FV(y)expresses that xis a variable which has a free occurrence in (a
formula) y.
The choice of coding apparatus is irrelevant as long as the coding is PA provably
monotone, i.e. the following is provable in PA
∀φ, ψ φ∈Subf(ψ)→φ≤ψ),
We require a similar condition for (the given formalisation) of x∈Term(y). Various
codings which violate this condition are studied in [8] and [11].
Throughout the paper we distinguish between variables of the metalanguage, for
which we reserve the symbols x, y, z, x0, x1, . . . , y0, y1,..., and variables of the arithme-
tized language, which are denoted v, v0, v1, . . . etc. We assume a fixed correspondence
between the first and the second ones. ¯x,¯v,. . . denote sequences of variables. For a
formula φ,pφqdenotes its Gödel code.
Some Model Theory of PA All the definitions and conventions regarding models of PA
are as in [12]. By default M,N,K(possibly with indices) range over nonstandard models
of PA and M,N,Kdenote their respective universes. If Mis any model (possibly for
LS) and φ(¯x)a formula (possibly with parameters from M;¯xdenotes a sequence of
variables), then φMdenotes the set definable by φin M, i.e. {¯a∈M| M |=φ(¯a)}. If
M |=PA and X⊆Mn, then X<b denotes the restriction of Xto all elements smaller
than b. In the case when N ⊆ M,XNdenotes Sb∈MX<b (the restriction of a relation
to the submodel).
Let I⊆M. We write d > I if dis greater than all the elements of I.Iis called an
initial segment of Mif Iis closed downwards with respect to ≤. We say that Iis a cut if I
is an initial segment which is closed under successor, i.e.
∀x x ∈I→x+ 1 ∈I.
If Iis a cut of M, then we call Man end-extension if Iand write I⊆eM(note that Ineed
not be a submodel of M). Any element c∈Msuch that c > ω is called nonstandard.
The following is one of the most basic consequences of induction in models of PA∗:
Lemma 1 (Overspill).If M |=PA∗, then no nontrivial cut of Mis definable.
In particular, if ∅(I(eMis a cut and φ(x)is any formula such that
∀a∈IM |=φ(a),
then there exists a d>Isuch that M |=φ(d).
One last notion which is very tightly linked to the topic of satisfaction classes is recur-
sive saturation:
Definition 2. Fix Mand ¯a∈M. Let p(x)be a set of formulae with at most one variable
xand parameters ¯a. We say that p(x)is recursive (or computable) if so is the set
{pφ(x, ¯y)q|φ(x, ¯a)∈p(x)}.
4
We say that p(x)is a type over Mif every finite subset of p(x)is satisfied in M. We say
that Mis realized if there is a b∈Msuch that for every φ(x)∈p(x),M |=φ(b). We say
that Mis recursively saturated (or computably saturated) if every recursive type over Mis
realized in M.4
Some Model Theory in PA Models of theories extending Robinson’s arithmetic are infi-
nite objects, thus inside arithmetic they become essentially second-order objects. In what
follows a set means a second order object and we distinguish it from a coded set (a first or-
der object) and the notion of a ∆nset is explained below. x∈Xshould be understood as
a membership relation between a first and a second order object, whereas x∈ydenotes
the Ackermanian membership (mentioned earlier) between first-order objects.
We recall the notion of a ∆n-set (see [9]): SatΣn(x, y) (SatΠn(x, y)) denotes the arith-
metically definable partial satisfaction predicate for Σn(Πnrespectively) formulae (as in
[12] or [9]) and TrΣn(x) (TrΠn)abbreviates SatΣn(x, ε)(SatΠn(x, ε)). We stress that the
construction of SatΣn(SatΠn)is elementary in n, so it gives rise to a EA-provably total ∆1
map sending nto (the formula) SatΣn.
In PA, a Σnset (Πnset) is any Σn(Πn)formula φ(v)with precisely one free variable.
We define a ∆nset to be a pair of formulae (φ, ψ)such that φis Σn,ψis Πnand
∀xSatΣn(φ, x)≡SatΠn(ψ, x).
The notions of a Σn(Πn,∆n)relation is defined analogously. If Xis a ∆kset given by the
Σkformula φ(v)and a Πkformula ψ(v), then x∈kXabbreviates SatΣk(φ, x). Note that
x∈kXis ∆k.
Observe that a set A⊆Mis definable from parameters in a model Mif and only if, for
some k∈ω, there exists a ∆kset Xsuch that
A:= {x∈M|x∈kX}.
In the paper, except for side remarks, in which case the definitions below can clearly be
adapted, we will only need to talk about models for very specific signatures consisting of
two binary functions +,×, one binary relational symbol S(x, y), reserved for a satisfaction
class and two constants, 0and 1.
We note that since in PA models for theories extending some basic arithmetic (which
we are uniquely interested in) are class-size objects, we do not always have a satisfaction
relation for them. Models without the satisfaction relation will called partial to contrast
them with the full ones for which the truth of an arbitrary sentence can be decided.
Definition 3 (ACA0, Partial model).We say that M= (UM,+M,×M, SM,0M,1M)is a
partial model if
1. UM,SMare sets and SM⊆U2
M,
2. +M,×Mare functions of type U2
M→UM,
3. 0M∈M,1M∈M.
We say that Mis a ∆n-model if it is a ∆nset satisfying the above conditions. 4
Definition 4. If Nis any model of PA then we say that Mis a partial N-definable model if
for some k∈ω,
N |= “Mis a partial ∆kmodel”.
4
5
Note that, according to our convention, “N-definable”means “N-definable with parameters”.
Example 5 (ACA0).For every set S,hv=v, v1+v2=v3, v1·v2=v3, S, 0,1iis a partial
model. We denote it with V[S].v=v, v1+v2=v3, v1·v2=v3denote sets definable with
respective formulae. Vdenotes V[∅].4
Remark 6. If N |=PA and M= (UM,+M,×M, SM,0M,1M)is a partial N-definable
model, then, outside of N, it gives rise to a model for the signature {+,×, S, 0,1}. Indeed,
we may define model Mby putting
M:= ((UM)N,(+M)N,(×M)N,(SM)N,0M,1M).
Such a model will be denoted by (M)N.4
Definition 7. 1. For a natural number n,ndenotes the canonical numeral naming n,
i.e.
1 + (1 + . . . + (1 + 0) . . .)
| {z }
ntimes 1
.
y=xdenotes the formalisation of this relation in PA. We shall often treat xas a
term symbol depending on variable x.
2. An assignment is any function with domain dom(f)⊆Var. αis an M-assignment if
its range is contained in UM. We denote it with α∈Asn(M).αis an M-assignment
for φ, symbolically α∈Asn(φ, M), if αis a M-assignment and FV(φ) = dom(α).
We naturally extend this definition to coded sequences of terms and formulae: if s
is such a sequence, then
α∈Asn(s, M) := ∀i∈dom(s)α∈Asn(si,M).
α∈Asn(s)has an analogous meaning.
3. (ACA0) If αis any assignment and φa formula, then by φ[α]we denote the result
of the simultaneous substitution of α(v)for every free occurrence of v, for every
v∈FV(φ)∩dom(α).t[α]for a term tis defined analogously. If we are interested
in a single substitution in a formula φ, then we write φ[x/v], or φ[x]if vis clear
from context, to mean φ[α]where αis an assignment such that dom(α) = {v}and
α(v) = x. Abusing the notation a little bit, for a term t,φ[t/v]denotes the result of
substituting tfor every free occurrence of v.
Example 8 (PA).If φ=(v0+v1=v2)∧(∃v2v2=v2)and α(v0)=2,α(v2)=3,
then
φ[α] = ((1 + 1 + 0) + x1= (1 + 1 + 1 + 0)) ∧(∃v2v2=v2).
Note that this the same as (2 + x1= 3) ∧(∃v2v2=v2). We shall use both
formats. 4
4. (ACA0) If αis any X-assignment, then αφabbreviates αFV(φ), where fAdenotes
the restriction of a function fto a set A. If φis clear from context, we will write α·
instead of αφ.
5. (ACA0) If tis a term and α∈Asn(t), then tαdenotes the value of tunder α. It is the
same as the value of t[α](t[α]is a closed term).
6
6. (PA)If αand βare any two assignments and vis a variable, then α≤vβexpresses
that βextends αby assigning something to the variable v, i.e. dom(β) = dom(α)∪
{v}and for all w∈dom(α),α(w) = β(w). Note that if α≤vβand v∈dom(α),
then α=β.
7. (PA) If cis any (coded) set of variables and ais a number, then [a]cdenotes the
constant assignment sending everything in cto a. If a variable vis clear from context
then we will omit it writing [a]instead of [a]v.
4
Definition 9 (ACA0).Let Mbe a partial model. An M-evaluation of terms is a partial
function fof type
Term ×Asn(M)→M
such that for every term t,{t} × Asn(t, M)⊆dom(f)and for every terms s, t and every
M-assignment α
1. ht, αi ∈ dom(f)↔FV(t)⊆dom(α),
2. α⊆β∧ ht, αi ∈ dom(f)→f(t, α) = f(t, β),
3. f(0, α)=0M,f(1, α)=1M,
4. f(v, α) = α(v),if v∈dom(α)
undefined, otherwise
5. f((s+t), α) = f(s, α) +Mf(t, α),f((s·t), α) = f(s, α)·Mf(t, α).
4
Observation 10 (ACA0).For every partial model Mthere exists the unique M-evaluation of
terms. We shall denote it with valM. Moreover, if the model is ∆k, then valMcan be taken to be
∆kas well.
Definition 11 (ACA0).Let Mbe a partial model. If Xis a set of formulae closed under
subformulae, then let s(X)denote the set of proper subformulae of formulae from X.S0
is called an X-satisfaction relation for Mif the conditions below holds.
1. X⊆FormLS∧ ∀φ∀ψψ∈Subf(φ)∧φ∈X→ψ∈X.
2. ∀y, z(S0(y, z)→y∈X∧z∈Asn(y, M)
3. ∀s, t∀α∈Asn(s, t, M) (S0(s=t, α)≡valM(s, α) = valM(t, α))
4. ∀s, t∀α∈Asn(s, t, M) (S0(S(s, t), α)≡ hvalM(s, α),valM(t, α)i ∈ SM)
5. ∀φ∈s(X)∀α∈Asn(φ, M) (S0(¬φ, α)≡ ¬S0(φ, α)).
6. ∀φ, ψ ∈s(X)∀α∈Asn(φ, ψ, M) (S0(φ∨ψ, α)≡S0(φ, α·)∨S0(ψ, α·)).
7. ∀φ∈s(X)∀v∀α∈Asn(∃vφ, M)S0(∃vφ, α)≡ ∃β≥vαβ∈Asn(M)∧S0(φ, β·).
Let CS−(X, M, S0)denote the conjunction of the above sentences of L2(we treat M,S0,
Xas second order variables). In the context of S,S(φ, α·)always mean S(φ, αφ).4
Definition 12. [PA; Measures of complexity of formulae]
7
1. The depth of a formula φis the length of the longest path in the syntactic tree of φ.
Equivalently, the depth of φis defined recursively: the depth of an atomic formula
is 0,∃and ¬raises the complexity by one and the depth of the disjunction is the
maximum of the depths of the disjuncts plus one. φ∈dp(x)expresses that the
depth of φis at most x.
2. Let us fix a canonical syntactical transformation, which for a formula φ(¯x)returns
a formula in the Σcform, that is logically equivalent to φ(¯x). Denote with φ(¯x)Σthe
result of applying this transformation. We assume that FV(φ(¯x)) = FV(φ(¯x)Σ). For
a number c, let Σ∗
cdenote the class of formulae φ(x)such that φ(x)Σ∈Σc.
4
Definition 13 (ACA0).For a number c, a c-full model is a tuple (M,SatM)where Mis a
partial model, and SatM⊆FormLS×Asn(M)is a Σ∗
c-satisfaction relation for M. A model
(M,SatM)is a full model if it is a c-full model for every c. Moreover, a tuple (M,SatM)is a
depth-c-full model if SatMis a dp(c)-satisfaction relation for M, i.e. CS−(dp(c),M,SatM)
holds. 4
We stress, that (M,SatM)being full presupposes that Mand SatMsatisfies full in-
duction (treated as additional predicates).
Observe that if for every n∈ω,(M,SatM)is an N-definable n-full model, then we
have two satisfaction classes for Mat our disposal: the metatheoretical one and SatM.
The two relations agree in the following sense: for every φ(x1, . . . , xn)∈ LSand for all
a1, . . . , an∈(UM)N
M |=φ[a1/x1, . . . , an/xn]⇐⇒ N |=SatM(pφ(x0, . . . , xn)q,[a1, . . . , an]).
Convention 14. We reserve calligraphic letters M,N,Kto talk, both internally and ex-
ternally, about models with satisfaction relations, while M,N,Kwill denote arbitrary
partial models.
By the Tarski’s undefinability of truth theorem one obtains that if Mis any model
of PA, then there is no formula SatVwith parameters from Msuch that for every n,
(V,SatV)is an n-full model. However, relativizing the standard partial truth predicates
one obtains the following observation.
Observation 15 (ACA0).If Mis any partial model, then for every kthere are uniquely deter-
mined predicates Satk
Msuch that (M,Satk
M)is a k-full model.
Proposition 16 (ACA0).Let Xbe closed under subformulae. Suppose that SatMis an X-
satisfaction class for M. Let Y⊆Xbe a set of sentences such that M|=SatMY. Then Yis
consistent.
Proof. We reason in ACA0and assume the contrary. Then there is a sequent-calculus
proof of the sequent
Γ⇒0=1,
in the pure first-order logic, where Γ⊆Yis a finite set. By cut-elimination we may
assume that this proof has a subformula property, so every formula occurring in it is a
subformula of a formula from Γ∪ {0 = 1}. By induction on the length of the proof we
can show that for every sequent Θ⇒∆it holds that
∀φ∈ΘM|=SatMφ→ ∃ψ∈∆M|=SatMψ.
This contradicts that the proof ends with 0=1.
8
The above notions of partial and full model lead to the definition of two interpretabil-
ity relations between structures:
Definition 17 (Interpretable models; See [12]).Let Mand Nbe two models of an ex-
tension of PA (not necessarily satisfying PA∗). We say that Minterprets Nif there exists
a partial M-definable model Nsuch that
N= (N)M.
We say that Mstrongly interprets Niff there exists Kwitnessing that Minterprets Nand
there exists an M-definable satisfaction predicate SatKmaking Ka full model. Inter-
pretability and strong interpretability will be denoted by Cand CS, respectively. 4
Observe that, as defined neither interpretability, nor strong interpretability is pre-
served under isomorphism, in the sense that from MCNand N ' K we cannot con-
clude that MCK. The next two propositions uncover the important properties of C. The
following routine notion will come in handy:
Definition 18 (ACA0, Relativization).Suppose that Mis a partial model. For every for-
mula φwe define its relativization φMby induction on the complexity of φ:
(s( ¯vs) = t( ¯vt))M:= valM(s, [ ¯vs]) = valM(t, [ ¯vt])
(S(t(¯v)))M:= valM(t, [¯y]) ∈SM
(φ∨ψ)M:= (φ)M∨(ψ)M
(¬φ)M:= ¬(φ)M
(∃xφ)M:= ∃x∈UM(φ)M
Above, valM(s, [ ¯vs]) = yabbreviates the formula
∃αα∈Asn(t, M)∧^
i∈A
α(vi) = xi∧y=valM(t, α).
4
Proposition 19. If MCNand NCK, then MCK.
Proof. Suppose that N= (N)Mand K= (K)N. Suppose further that Nis partial ∆k
model in M. Hence using partial satisfaction predicate for Σkformulae, we can see that
the KN(see Definition 18) makes sense in M, and, in M,KNis a partial ∆kmodel.
Moreover it is easy to observe that
KNM=K,
which ends the proof.
The following proposition will play crucial role in some of our arguments. Its proof
consists in internalizing the argument from Remark 6 and makes use of the arithmetza-
tion of the relativization function introduced above.
Proposition 20 (Enayat-Visser).Suppose that MCSNand NCK, then MCSK.
9
Proof. By Proposition 19 we have MCKand (K)Nis a partial M-definable model wit-
nessing the interpretability. We define the satisfaction relation for (K)Nvia the formula
SatKN(x, y) := FormLS(x)∧y∈Asn(x, KN)∧SatN(xK, y).
In PA we can prove that every consistent theory admits a full model. Since in most
cases both the theory and the model are infinite objects, this is in fact a parametrized
family of theorems:
Theorem 21 (Arithmetized Completeness Theorem).For every n∈ω,PA∗proves the sen-
tence
Every ∆nconsistent theory has a ∆n+1 full model.
Since the proof of this fact (apart from axioms for arithmetical operations) depends
only on the presence of induction, it can be proved also in every extension of PA which
includes full induction scheme (for the extended language). This will be crucial in the
second part of the paper. Let us complete this introductory part with two classical ob-
servations which give us some informations about the structure of interpretable models.
The first one shows that in fact interpretability can be seen as refined end-extendibility.
Definition 22. If Mis a model for a language L0extending L, then MLdenote its L-
reduct. 4
Proposition 23 (Folklore).Let M,Nbe models of PA∗. If Minterprets N, then there exists a
unique M–definable isomorphism between MLand initial segment of NL.
Proof. Suppose N:= hUN,+N,×N, SN,0N,1Niis an Mdefinable partial model witness-
ing the interpretability of Nin M. Let valNbe a valuation function for N. We define the
embedding ι:M→Nvia the formula
ι(x) := valN(x, ε),
where xis a canonical numeral (in the sense of M) naming x. By the earlier remarks
there exists a satisfaction predicate SatNmaking Na1- full model. The fact that ιis an
initial embedding follows since for every xwe can build a quantifier-free sentence (in the
sense of M)
∀vv < x →_
z<x
v=z),
and by induction on xshow that every such sentence is true in Naccording SatN. But
this is equivalent to ιbeing an initial embedding. Now, if ι0is any other Mdefinable
isomorphism between Mand an initial segment of N, then it follows that
M |= “ι0(0) = 0N∧ ∀xι0(x+ 1) = ι0(x) +N1N.”
Then, by induction it follows that, M |=∀xι(x) = ι0(x).
If we strengthen the assumption to strong interpretability, then we can conclude that
the interpreted model is always “longer”.
10
Proposition 24 (Folklore).Let M,Nbe models of PA∗. Suppose that Mstrongly interprets
Nand let ι:ML→ N Lbe the embedding from Proposition 23. Then ιis not an elementary
embedding. In particular, Mis isomorphic to a proper initial segment of N.
Proof. Let M,N,ιbe as above and let Nbe a partial M-definable model such that
N= (N)M. That ιis not elementary follows from Tarski undefinability of truth theo-
rem. Indeed, since ιand SatNare Mdefinable, we can define in Ma predicate S(x, y)
by putting
S(φ, α)≡SatN(φ, ι ◦α).
With such a definition for every formula φ(x1, . . . , xn)and all a1, . . . , an∈Mwe have
M |=S(pφ(x1, . . . , xn)q,[a1, . . . , an]) ⇐⇒ N |=φ(ι(a1), . . . , ι(an)).
Then, if ιwere elementary, then the condition on the righthandside would be equivalent
to M |=φ(a1, . . . , an)which contradicts Tarski’s theorem. The last part follows easily
from the above and Proposition 23.
Let us note one immediate corollary. Recall that Mis κ-like if |M|=κbut every
proper initial segment of Mhas cardinality strictly smaller than κ.
Corollary 25. If Mis κ-like, then Mis not strongly interpreted in any model of PA.
Proof. Obviously, if Ninterprets M, then |N|≥|M|. Moreover, if Mis strongly inter-
pretable in N, then it has a proper initial segment of cardinality |M|.
Moreover, models strongly interpretable in a nonstandard model of PA has to be re-
cursively saturated. This is a corollary to the proposition below (see also [14]):
Proposition 26. Suppose dis a nonstandard element of N. If Mis isomorphic to a depth-d-full
model, then Mis recursively saturated.
Proof. Suppose that M= (M,SatM)is an Ndefinable depth-d-full model, Nand dbeing
as above. Fix an arbitrary recursive type p(x) = {φi(x, a)|i∈ω}with (without loss
of generality) a single parameter aand let σ(x, y)be the ∆1formula representing its
recursive enumeration, i.e. for every i∈ω
PA ` ∀wσ(i, w)↔w=pφi(x, z )q.
Now, since the depth of every φiis less than dand p(x)is a type we have for every n∈ω
N |=∃z∀i≤n∀wσ(i, w)→SatM(w, [x7→ z, y 7→ a])
([x7→ z, y 7→ a]denotes the unique assignment sending the variable xto zand the
variable yto a). Hence, by overspill for some nonstandard cwe have
N |=∃z∀i≤c∀wσ(i, w)→SatM(w, [x7→ z, y 7→ a]),
which shows that p(x)is realised in M.
11
Satisfaction Classes Satisfaction classes provide truth conditions for V.6Usually they
are studied in the context of nonstandard models of PA. Let Mbe such a model.
Definition 27. We say that S⊆M2is a partial satisfaction class on Mif there is a non-
standard csuch that (M, S)|=CS−(dp(c+ 1),V, S ). Equivalently the following holds in
(M, S):
1. ∀x, yS(x, y)→Form(x)∧x∈dp(c)∧y∈Asn(x).
2. ∀s, t∀α∈Asn(s, t) (S(s=t, α)≡sα=tα).
3. ∀φ∈dp(c)∀α∈Asn(φ) (S(¬φ, α)≡ ¬S(φ, α)).
4. ∀φ, ψ ∈dp(c)∀α∈Asn(φ, ψ) (S(φ∨ψ, α)≡S(φ, αφ)∨S(ψ, αψ)).
5. ∀φ∈dp(c)∀v∀α∈Asn(∃vφ)S(∃vφ, α)≡ ∃β≥vαS(φ, βφ).
Henceforth, the conjunction of 1-5will be denoted by CS−(c). If additionally (M, S)|=
PA∗, then Sis called a partial inductive satisfaction class. If (M, S )|=∀xCS−(x), then Sis
called a full satisfaction class. Further define:
UTB−:= {CS−(n)|n∈ω}
UTBn:= UTB−+IΣn(S)
UTB := [
n∈ω
UTBn
Now we define an analogue of arithmetical hierarchy over the theory of a full satis-
faction class.
CS−:= ∀xCS−(x)
CSn:= CS−+IΣn(S)
CS := [
n∈ω
CSn
If Sis a partial satisfaction class on Mand b∈M, then we put
Sb:= {hφ, αi ∈ S|φ∈Σ∗
c}.
Note that Sbis ∆0definable from S, hence for every n, if Sis Σninductive, then so is
Sb.4
Note that if Sis a full satisfaction class, then (M, S)strongly interprets Mand Vis a
∆0partial model witnessing the interpretation. However, (V, S)need not be full, as we
need not have any induction for S. In particular, it does not follow that
(M, S)|=∀φ∈AxPA S(φ, ε),
which, arguably, would mean that Mknows that Vis a model of PA. Let us call such a
satisfaction class PA-correct. It can be shown that for a countable Mthe following condi-
tions are equivalent:
6Obviously one can study satisfaction classes for languages with additional predicates, but we will not
be interested in such objects.
12
1. There exists a full satisfaction class on M.
2. There exists a PA-correct full satisfaction class on M.
3. Mis recursively saturated.
the implication 3.⇒2.has been shown for the first time in [16]. [7] and [17] contain
different proofs. The implication 1.⇒2.is a consequence of Lachlan’s theorem (see
below Theorem 31).
The name UTB−stands for Uniform Tarski Biconditionals7and is normally used for the
theory having as axioms all sentences of the form
∀α∈Asn(φ)S(pφq, α)≡φ((α(x1),...,(α(xn)),
for every φ(x1, . . . , xn)∈FormL.8One can show that, over EA, this set of sentences is
equivalent to the one we’ve officially taken as a definition of UTB−. By Observation 15
each finite portion of UTB−is definable in PA and consequently we obtain the following
proposition (which formalizes in EA)
Proposition 28 (EA).If Th is any extension of PA, then UTB +Th is conservative over Th
Furthermore, observe that (M, S)|=CS−iff Sis a full satisfaction class on Mand
(M, S)|=CS iff Sis a full inductive satisfaction class in the sense of [14]. For further
usage let us observe that the relation of CS to CSnis similar to that between PA and IΣn.
In particular there are definable partial SatΣnsatisfaction predicates for ΣnLSformulae.
Each SatΣnis a ΣnLSformula. As a consequence we obtain
Proposition 29. For every n,EA +CSn+1 `ConEA+CSn.
We note that if Sis a partial satisfaction class on a model M, then for an arbitrary
standard formula φ(x0, . . . , xn)∈ L
(M, S)|=∀α(S(pφq, α)≡φ(α(x1), . . . , α(xn))) .
In particular it follows from Tarski’s theorem that Sis never definable in M(even if we
allow parameters).
Nonstandard satisfaction classes provide a very useful tool for investigating nonstan-
dard models of PA. The first point of interest is that their existence imply recursive satu-
ration. For starters we cite a proposition which directly follows from Proposition 26:
Proposition 30 (Folklore, see [12]).If Sis a partial inductive satisfaction class in M, then
Mis recursively saturated.
Proof. Suppose that (M, S)|=CS(c)for some nonstandard c∈M. Then Mis isomorphic
to a depth-c–full M–definable model. Hence by Proposition 26 is recursively saturated.
Interestingly, with a much more complicated proof one can strengthen the above
proposition lifting the assumption that the chosen satisfaction class is inductive.
Theorem 31 (Lachlan, see [12]).If for some nonstandard c,(M, S)|=CS−(c), then Mis
recursively saturated.
7This theory is defined as UTBin [10] and as UTB in [2].
8In the above αis a bound variable, so φ((α(x1),...,(α(xn)) denotes the formula ∃y1,...,ynVi≤nyi=
(α)i∧φ(y1,...,yn).
13
The converse to Lachlan’s theorem fails, as was shown by Smith.
Theorem 32 (Smith, [22]).If (M, S)|=CS−, then there is S0such that for some nonstandard
c∈M(M, S0)|=CS0(c).
The condition that (M, S0)|=I∆0(S)implies that S0is piecewise coded (in the sense
of [9]) or, using set-theoretical notions, a class on M. Since (M, S0)|=CS−(c)it fol-
lows that S0is not definable (even allowing parameters) in M(or, is a proper class).
Since there are recursively saturated models of PA in which every class is definable (with
parameters; see [14]), Smith’s result shows that there are recursively saturated models
which do not carry a full satisfaction class.
A common strengthening of theorems of Smith’s and Lachlan’s was obtained by Bar-
tosz Wcisło in [25]:
Theorem 33 (Wcisło).If (M, S)|=CS−(c)then there is an S0and a nonstandard csuch that
(M, S0)|=CS(c).
An interesting open problem in the model theory of PA is whether the converse to
the above theorem is true, i.e. whether every M |=PA which admits a partial inductive
satisfaction class admits a full satisfaction class. If one allows to prolong the given model,
then there is a positive answer to this question.
Theorem 34 (Visser).If (M, S)|=CS(c)for some c∈M, then there are M eNand S0
such that (N, S0)|=CS−and S⊆S0.
The proof of this theorem is given in [20], Theorem 43. In Section 5 we shall give an
analogous result for M |=REFω(PA)and CS0instead of CS−.
Remark 35. The theory CS−is a cousin of a compositional truth theory CT−which admits
a unary predicate T. All compositional axioms of CS−can be easily adapted to this new
setting, however in the case of the universal quantifier we have two natural ways to go.
The first candidate is the “numeral”version, i.e.
∀v∀φ(v)T(∀vφ)≡ ∀xT (φ[x/v]).
We stress that φ[x/v]denoted the result of substituting the numeral naming xfor every
free occurrence of the variable v. If such an axiom is adopted, then the resulting theory,
denote it nCT−, can define the satisfaction predicate satisfying CS−via the formula:
S(φ, α) := FormL(φ)∧α∈Asn(φ)∧T(φ[α]).
Let us stress that the above formula is ∆0(exp). The second option is the “term”version
of CT−, denote it tCT−, where the axiom for ∀is the following
∀v∀φ(v)T(∀vφ)≡ ∀t∈Term T(φ[t/v]).
Using Enayat–Visser methods from [7] it can be shown that nCT−and tCT−are indepen-
dent of each other, i.e. neither of them implies the other one (over the remaining axioms
of CS−). Moreover, it is an open problem whether tCT−can define the predicate of CS−.
In this paper CT−will be introduced in Section 3 and will denote the numeral version,
i.e. nCT−.4
14
Reflection Principles Reflection principles are various (families of) statements express-
ing the soundness of a given theory Th in a way which is transparent for Th. In other
words, their aim is to capture the meaning of the metatheoretical assertion
Every theorem of Th is true.
In order to avoid the problem of choosing the presentation for (an abstractly given
theory Th), we will assume that Th is an elementary formula, which, provably in EA
defines the set of sentences. Such a formula will be called a gödelized theory. We use PA
to abbreviate the canonical elementary formula saying "xis an axiom of PA−or an axiom
of induction". Having the satisfaction predicate S(x, y)at our disposal we can express
the above in the form of the Global Reflection Principle
∀φPrTh(φ)→S(φ, ε).(GR(Th))
If Ssatisfies UTB−, then from this one can derive instantiation of the uniform reflection
∀x1, . . . , xnPrTh(pφq[x1, . . . , xn]) →φ(x1, . . . , xn).(REF(Th))
Hence REF(Th)contains all the formulae of the above form for the language of Th.
If Γis a set of formulae of the language of Th, then Γ-REF(Th)denote the restriction of
REF(Th)to formulae from class Γ. Below we will need also its iterated versions:
Γ-REF0(Th) := Th
Γ-REFn+1(Th) := T+ Γ-REF(REFn(Th))
Γ-REFω(Th) := [
n∈ω
Γ-REFn(Th)
In the successor step we tacitly fix the canonical representation of Γ-REFn(Th).
The last definition which is relevant to formalizing soundness claims introduces the
oracle provability predicates.
Definition 36. Let Th be any elementary theory. ProofX
Th(x, y)denotes a ∆0
0(exp) formula
with a second order variable Xwhich canonically formalizes the relation: "yis a proof of
sentence xfrom axioms of Th and sentences belonging to X". PrX
Th is the Σ0
1provability
predicate based on it. 4
The oracle provability predicate defined above enables us to (uniformly) define a clo-
sure conditions on various satisfaction classes. For example we shall often encounter the
assertion
∀φPrS
∅(φ)→S(φ, ε),
which should be read as "Every first-order consequence of true sentences is true", where
"true" abbreviates that S(φ, ε)holds. In the above assertion we simply substitute the de-
finable class {x|S(x, ε)}for the free second-order variable X. Let us also observe that
formally PrX
Th is the same as PrTh∪X, however, for heuristic reasons we prefer to keep the
lower index for absolute definitions and the upper one for arbitrary sets of formulae.
15
Reflection and internal models The theory REF(Th)admits a model-theoretical char-
acterisation in terms of strongly definable models. The following theorem was proved in
[13]. Below KA(M)denote the set of elements of a model Mwhich are definable in M
with parameters from the set A.
Theorem 37. [Kotlarski-Kaye] For an arbitrary recursively saturated Mthe following are equiv-
alent:
1. M |=REF(Th)
2. There exists a full M-definable model N= (N,SatN)such that
(a) M ≡ N and
(b) K∅(M) = K∅(N).
(c) M |=∀φ∈Th SatN(φ, ε).
Thanks to condition (a)imposed on Nin the above, Theorem 37 can be iterated an
arbitrary finite number of times. Let us call the pair (M,N)aKK-pair if it satisfies con-
ditions 1.−3.in the thesis of the above theorem. Thus we obtain
Corollary 38. M |=REF(Th)if and only if there exists {Mi}i∈ωsuch that M0=Mand for
each i,(Mi,Mi+1)is a KK-pair.
In Section 5 we shall offer a similar in spirit model-theoretical characterization of
REFn(Th)and REFω(Th). The main difference will be that we shall work with models
with satisfaction classes.
Reflection and satisfaction classes Full satisfaction classes in non–standard models
embody the conception of a satisfaction relation for V. However, as we have already
remarked, not every satisfaction class provides us with a reasonable truth predicate for
V. One property that one would require from such a truth predicate is the closure under
the internal provability relation. In particular the satisfaction relation for Vshould make
true all the (internal) theorems of first-order logic. This corresponds to the sentence
∀φPr∅(φ)→S(φ, ε)(GR(∅))
being true in a model (M, S). However, as shown for the first time in [4], over CS−the
above sentence implies GR(PA). In particular, in a countable recursively saturated model
Min which there is a proof of inconsistency of PA there is no such a reasonable class for
V(although there are many unreasonable ones.) A characterization of models admitting
a well-behaved satisfaction class was essentially first given by Kotlarski (in [15])9:
Theorem 39 (Kotlarski).Suppose Mis a countable recursively saturated model of PA. Then,
there exists a full satisfaction class Ssuch that
(M, S)|=GR(PA),
if and only if M |=REFω(PA).
9The attribution here is qualified by "essentially", since Kotlarski’s proof that CS0and CS−+GR(PA)are
deductively equivalent contained a significant gap, as we explain in Section 3. Without going through CS0
we think that the proof requires using the result of Smoryński and Cieśliński. This paper contains a different
direct proof of this theorem.
16
A different natural question is how much induction is required to prove the global
reflection for PA. Here a partial answer was given by Wcisło in [25], who showed that
the satisfaction predicate satisfying CS−+GR(PA)is definable in CS0:
Theorem 40 (Wcisło).There exists an LSformula S0(x, y)such that
EA +CS0`[CS−+GR(PA)][S0/S].
In the above φ[S0/S]denotes the result of a uniform substitution of a formula S0(s, t)
for each occurrence of a formula S(s, t)(renaming variables if necessary). In the next
section we improve this result and show that GR(PA)is provable in EA +CS0.
3 Provability of the Global Reflection Principle
In this section we confirm Kotlarski’s claim ([15]) that, over EA, the ∆0induction for a
satisfaction predicate is enough to prove the Global Reflection Principle for PA. We start
by explaining the original strategy and our fix. Unless said otherwise, all theories by
default extend EA. However, it is very easy to see that CS0+EA `PA, since for every
arithmetical formula φ(x),ψ(x) := T(φ[x]) is a ∆0(LT+ exp) and consequently, we have
an induction axiom for it.
Kotlarski’s proof Kotlarski’s proof of GR(PA)in CS0starts by observing that each ∆0
inductive satisfaction class makes all (in the sense of the ground model) the axioms of in-
duction true. He argues as follows: working in CS0fix a formula φ(v)with a free variable
v. Then, S(φ(v),[x]) is a ∆0(exp) formula of x, where [x]denotes the assignment {hv, xi}.
Hence, the following is an axiom of CS0:
S(φ(v),[0]) ∧ ∀xS(φ(v),[x]) →S(φ(v),[x+ 1])−→ ∀xS(φ(v),[x]).
Here Kotlarski’s proof of PA-correctness ends. However, it is not obvious whether the
above is equivalent to S(Ind(φ(v)), ε), where Ind(ψ)denotes the axiom of induction for
a formula ψ. Repeated applications of the compositional clauses to S(Ind(φ(v)), ε)yield
S(φ[0/v], ε)∧ ∀xS(φ(v),[x]) →S(φ[v+ 1/v],[x])−→ ∀xS(φ(v),[x])
Firstly, on the grounds of CS−alone S(φ(v),[0]) neither implies S(φ[0/v], ε)nor is implied
by it. Similarly with S(φ(v),[x+ 1]) and S(φ[v+ 1/v],[x]). To see this one should think of
a nonstandard φ(v)in which voccurs nonstandardly deep in φ(v)(i.e. at a nonstandard
level of φ(v)0s syntactical tree). For example one can take φ(v)to be
0=0∨(0 = 0 ∨(0 = 0 ∨. . .
| {z }
atimes 0=0
∨v=v). . .),
for a nonstandard element a. This is the first problem.
The second problem lies in showing that a ∆0–inductive satisfaction class is closed
under provability, i.e. proving the sentence
∀φPrS
∅(φ)→S(φ, ε).
In the above PrS
∅derives from the oracle provability predicate defined in Definition 36.
Kotlarski’s idea was to work (internally in PA) with a Hilbert–style proof calculus with
17
Modus Ponens as the only rule of reasoning and then using ∆0(exp) induction for a for-
mula
θ(x, p) := “If φis the x-th sentence in p, then S(φ, ε)”.
In θ(x, p)all quantifiers can be bounded by p,10 that can be taken as a parameter. So it is
indeed a ∆0(exp)–formula. In the base step φis either a logical axiom or a true sentence,
so it’s truth is either trivial (the latter case) or seems to follow from the compositional
axioms. In the inductive step, we have to check that if φand φ→ψare true, then so is ψ.
This is indeed guaranteed by the compositional axioms.
However, problems arise while verifying the base step. For example (working in a
nonstandard model) we might encounter the following logical axiom
ψ:= ∀vφ(v)→φ(t)
for some nonstandard formula φand a term t. Then S(ψ, ε)is equivalent to
∀yS(φ(v),[y]) →S(φ(t), ε),
so we encounter problems similar to the ones discussed while dealing with the truth of
induction axioms. Moreover, there are more generic problems: if one does not want to
incorporate the rule of universal generalisation, then one has to accept universal general-
isations of all instances of propositional tautologies as axioms. In particular, (working in
a nonstandard model (M, S)) in the base step one might encounter an axiom of the form
ξ:= ∀v1. . . ∀vaφ∨ ¬φ.
If it is true that that for any full satisfaction class Son M,(M, S)|=∀α∈Asn(φ)S(φ∨
¬φ, α), inferring that (M, S)|=S(ξ , ε)requires some argument which cannot be carried
out in CS−alone.
The idea To fill in the gaps in Kotlarski’s reasoning it is sufficient to establish within CS0
a kind of induction on the build-up of formulae. Indeed, what we missed were (inter alia)
the following properties
∀φ∈FormL∀α∈Asn(φ)S(φ, α)≡S(φ[α], ε)
∀φ∈Form≤1
L∀s, t∀α∈Asn(s)∀β∈Asn(t)sα=tβ→S(φ[s/v], α)≡S(φ[t/v], β)
The above hold (provably in CS−) if φis an atomic formula and clearly are preserved by
taking disjunctions, negations and applying existential quantification. However, in order
to secure the step for ∃we need a Π1assumption, saying that the equivalence
S(ψ, α)≡S(ψ[α], ε)
holds under an arbitrary assignment α. It turns out, however, that such assumptions can
be expressed with a ∆0formula. Firstly, the above is clearly equivalent to
∀α∈Asn(ψ)S(ψ≡ψ[α], α).(1)
Secondly, if Scommuted with the blocks of universal quantifiers, the above could have
been further reduced to
S(ucl(ψ≡ψ[α]), ε),(2)
10Or, to be more accurate, by objects of size exponential in p.
18
where ucl(·)is a (definable) function which given a formula returns its universal closure.
Our problem thus reduces to showing the equivalence between conditions of type (1)
and (2). The standard strategy is to use induction on the length of the quantifier prefix.
However, the proof of this once again uses Π0
1induction. To bypass this problem, we shall
first establish commutation with blocks of uniformly bounded universal quantifiers, i.e.
the principle
∀φ∀xS(bucl(φ, x), ε)≡ ∀α≺[x]φα∈Asn(φ)→S(φ, α),(B∀C)
where bucl(φ, x)denotes the universal closure of φ, in which every quantifier in the prefix
is bounded by (the term) xand α≺[x]φsays that dom(α) = FV(φ)and each value of α
is less than x. Having this, we will express ∀αS(φ, α)in a ∆0way via
S(∀vbucl(φ, v), ε),
where vis a variable which do not occur in φ. We now proceed to the details.
The proof One more preparation step will be helpful. We shall expand the language
LSwith symbols for all primitive recursive functions and extend CS0with the defining
equations of them. Let L+
Sdenote the expanded language and CS+
0the extended theory.
Since, trivially, CS+
0is LS-conservative over CS0, it is sufficient to prove (GR(Th)) in
CS+
0. Let us observe that the latter theory proves ∆0induction for the language with
new function symbols.
Proposition 41. CS+
0`I∆0(L+
S).
Proof. Fix a model M |=CS+
0and a ∆0(L+
S)formula with parameters φ(x). Assume
M |=φ(0) ∧ ∀xφ(x)→φ(x+ 1).
Without loss of generality all the terms occurring in φ(x)are of the form f(n, y)where
f(x, y)is a two place p.r. function. Let ψf(x, y, z)be the ∆0formula defining the graph
of f. Fix an arbitrary a∈M, assume that φ(a)holds and let bbe greater than aand all
parameters from φ(x). Without loss of generality assume that bis nonstandard. Let cbe
such that
M |=∀x<b∀y < bf(x, y)< c.
Now let φ0(x)result from φ(x)by recursively changing all subformulae of φ(x)of the
form
R(f(k, x), f (n, y))
into
∃z∃wz < c ∧w < c ∧ψf(n, y, w)∧ψf(k, x, z )∧R(w, z),
where z,ware fresh variables. φ0(x)is a ∆0(LS)formula and clearly we have
M |=∀x<b φ(x)≡φ0(x).
Hence, since a < b and φ(a)holds, φ0(a)holds as well. Since for φ0(x)we have an in-
duction axiom there is the least d < a such that φ0(d)holds. Hence dis the least element
satisfying φ(x).
In the proof below we shall use the following primitive recursive functions (we iden-
tify p.r. relations with their characteristic functions):
19
•≺is a primitive recursive product ordering on functions. i.e. if αand βare two
functions, then α≺βholds if αis smaller than βin the product ordering, i.e.
dom(α) = dom(β)∧ ∀x∈dom(α)α(x)< β(x).
4is the partial ordering based on ≺.
•bucl(φ, x)denotes the universal closure of φ, in which every quantifier in the prefix
is bounded by (the term) x.
•bucl(φ, x, c)returns the formula
∀vi1< x . . . ∀viy< x φ,
where xi1, . . . , xiyare all the elements of the set clisted in the order of decreasing
indices (i.e. iy< iy−1< . . . < i1). In particular xhas to be a term and ca set
of variables. Officially ∀x < tψ, abbreviates ∀xx<t→ψ), hence the above for-
mula is slightly more complicated than it seems to be. Moreover cneed not contain
uniquely variables which are free in φ, hence some of the quantifiers in the prefix
of bucl(φ, x, c)might be dummy.
•[x]creturns the constant function assigning value xto every variable from the set c.
•for a coded set of variables cand a number a,c↑adenotes the set consisting of first
aelements of cand c↓athe set consisting of last aelements of c. For simplicity we
assume that the ordering of variables is given by their indices.
•syntactical relations mentioned at the beginning of Section 2.
Lemma 42 (CS+
0).For every formula φ, every a, every set of variables cand every assignment
αfor bucl(φ, a, c)
S(bucl(φ, a, c), α)≡ ∀β≺[a]cS(φ, α ∪β·).
Proof. Fix φ, a, c and let bbe the cardinality of c. We formalize the standard argument
on the length of the quantifier prefix in bucl(φ, a, c): departing from the assumption that
∀β≺[a]cS(φ, α ∪β·)we check that we can prefix φwith bquantifiers and arrive at
S(bucl(φ, a, c), α). Formally, we use induction on yup to bin the ∆0(L+
S)-formula:
∀β≺[a]c↓b−yS(bucl(φ, a, c↑y), α ∪β·)≡ ∀β≺[a]cS(φ, α ∪β·).
Observe that c↑b=c↓b=c,c↓0=∅. Hence bucl(φ, a, c↑0) = φ,[a]c↓b= [a]c. Moreover, ε
is the unique assignment γsatisfying γ≺[a]∅. Consequently, for y=bthe lefthandside is
equivalent to S(bucl(φ, a, c), α)and for y= 0 both sides of the equivalence are the same.
Assume that the inductive hypothesis holds for dand let idbe the index of the (d+1)-st
variable in c. Observe that
bucl(φ, a, c↑d+1) = ∀vid< a_bucl(φ, a, c↑d).
Hence by compositional axioms, ∀β≺[a]c↓b−(d+1)S(bucl(φ, a, c↑d+1 ), α ∪β·)is equivalent
to
∀β≺[a]c↓b−(d+1) ∀γ≥vidα∪βbucl(φ,a,c↑d+1)γ(vid)< a →S(bucl(φ, a, c↑d), γ ·).(∗)
Observe that the above is equivalent to
∀ββ≺[a]c↓b−d→S(bucl(φ, a, c↑d), α ∪β·),
which, by induction hypothesis, is equivalent to ∀β≺[a]cS(φ, α ∪β·).
20
The above lemma motivates the following abbreviation: we define
cl(φ, c) := ∀v_bucl(φ, v, c).
In the above vis a variable with the least index among those which do not occur in φ.
In particular cl(x, y)is a (partial) primitive recursive function, so we have a symbol for it
in CS+
0. cl(φ)abbreviates cl(φ, FV(φ)).
Now, the following corollary clearly follows from Lemma 42.
Corollary 43 (CS+
0).For all φ,c⊆Var and α∈Asn(cl(φ, c)) it holds that S(cl(φ, c), α)≡
∀β∈Asndom(β) = c→S(φ, α ∪β·).
Proof. Fix φand c. By the compositional axioms and Lemma 42 S(∀vbucl(φ, v, c), α)is
equivalent to ∀x∀β≺[x]cS(φ, α ∪β·), which is clearly equivalent to ∀βdom(β) = c→
S(φ, α ∪β·)), since each assignment for φis dominated by an assignment of the form [a]φ
for some a.
Now, we demonstrate how to use the above corollary for establishing, within CS+
0,
the induction on the build-up of formulae. It will be convenient to isolate a few more
definitions.
Definition 44 (PA).An occurrence of a variable vin a formula φ(term s) is a path in a
syntactic tree of φ(term s) ending with v. A occurrence of a subformula ψof a formula φ
is defined analogously. The fact that ψis a subformula occurrence in φis denoted ψ≤oφ.
The (coded) set of occurrences of variables in a formula φ(term s) will be denoted Occ(φ)
(Occ(s)).
Asubstitution of terms for a formula φis a (coded) function ηsuch that dom(η)⊆Occ(φ)
and rg(η)⊆ClTerm. For a formula φ,φ[η]denotes the result of applying ηto φ.
If ηis a substitution of terms for φand ψis a subformula of φ, then ηnaturally gives
rise to a substitution of terms for ψ(we look only at those paths that pass through ψand
take their suffixes starting from ψ). Such a substitution will be denoted by ηψor simply
η·if it is clear from context which formula should occur in the subscript.
The substiution of terms ηfor a formula φagrees with an assignment α∈Asn(φ)if
whenever p∈dom(η)is an occurrence of a variable vin φ, then (η(p))◦=α(v). Let vp
denote the variable whose occurrence pis.
Let φbe a formula and ψan occurrence of its subformula. Var(φ/ψ)denotes the
(coded) set of variables whose occurrence in ψis free in ψbut bounded in φ.
For a subformula ψof φdefine clφ(ψ) := ∀v_bucl(ψ, v, Var(φ/ψ)) where vis the least
variable not occurring in φ.4
Example 45. Let ψ:= x0=x1∧ ∃x2(x2= ((0 + 1) + 1) ·x0)and φ:= (∀x0ψ)∧x0=x0.
Then
Var(φ/ψ) = {x0}.
Consequently clφ(ψ) = ∀x3∀x0< x3x0=x1∧ ∃x2x2= ((0 + 1) + 1) ·x0.4
Lemma 46 (CS+
0).Assume that φis a formula, ζ∈Asn(φ)and ηis a substitution of terms for
φwhich agrees with ζ. Then it holds that S(φ≡(φ[η]), ζ).
Proof. Fix a formula φ, any assignment ζ∈Asn(φ)and a substitution of terms ηwhich
agrees with ζ. We reason by induction on yin the ∆0(L+
S)-formula
θ(y) := ∀ψ≤oφψ∈dp(y)→S(clφ(ψ≡ψ[ηψ]), ζ·).
21
Let us observe that S(clφ(ψ≡ψ[ηψ]), ζ·)makes sense: each occurrence of a free variable
of ψ≡ψ[ηψ]is either an occurrence of a free variable of φand hence the variable gets
assigned a value by ζψ, or is bounded in φand so, belonging to Var(φ/ψ), gets bounded
by a quantifier occurring in a prefix of clφ(ψ≡ψ[ηψ]).
Let zbe the least variable not occurring in φ. We show that θ(0) holds. The unique
sentences of depth 0are atomic sentences, so let u fix two terms s, t and argue that
Sclφ(s=t≡(s=t[ηs=t])), ζ·
holds. Let c=Var(φ/ψ). Consequently
clφ(s=t≡(s=t[ηs=t])) = ∀z_bucl(s=t≡(s=t[ηs=t]), z, c).
Call the above sentence on the righthandside ξ. By Corollary 43 we know that S(ξ, ζ·)
is equivalent to
∀α∈Asndom(α) = c→S(s=t≡(s=t[ηs=t]), ζξ∪α).
The last sentence holds, since, by the compositional conditions for atomic sentences and
connectives, it is equivalent to the assertion that for every α∈Asn such that dom(α) = c
(sζξ∪α=tζξ∪α)≡(s[ηs=t]ζs=t[ηs=t]∪α=t[ηs=t]ζs=t[ηs=t]∪α).($)
To avoid double restrictions let us abbreviate ζξwith ζξ. Observe that ζξs=t=ζξ. To
prove ($) it is sufficient to show that each occurrence of a variable in either sor tgets
assigned the same value on both sides. This holds since if o∈Occ(s), then either o∈
dom(ηs=t)or not. In the former case by our assumption ζξ(vo) = ζ(vo) = (η(o))◦. If
o /∈dom(ηs=t), then o /∈dom(η)and vo∈dom(ζξs=t[ηs=t]∪α)⊆dom(ζξ∪α), hence o
get assigned the same value on both sides of the equivalence. Consequently
s[ηs=t]ζξs=t[ηs=t]∪α=sζξ∪α.
Now assume that the thesis holds for yand consider (an occurrence of) ψof depth y+ 1.
We shall do the case of ψ=ψ1∨ψ2and ψ=∃vψ3. We treat them simultaneously.
As previously put ξ:= clφψ≡(ψ[ηψ]),ξi:= clφψi≡(ψi[ηψi]),c=Var(φ/ψ)
and ci=Var(φ/ψi). Applying the inductive assumption and Corollary 43 we have for
i∈ {1,2,3}
∀α∈Asndom(α) = ci→S(ψi≡(ψi[ηψi]), ζξi∪α).
Now observe that ζξiψi=ζξiand if dom(α) = ci, then αψi=α. By this and com-
positional conditions, for arbitrary αsuch that dom(α) = ci, the succedent of the above
implication is equivalent to
S(ψi, ζξi∪α)≡S(ψi[ηψi], ζξiψi[ηψi]∪αψi[ηψi]).(3)
In the case ψ=ψ1∨ψ2we have c=c1∪c2. Now, by compositional conditions, the
following are equivalent for an arbitrary assignment αsuch that dom(α) = c:
S(ψ1∨ψ2≡(ψ1∨ψ2[ηψ1∨ψ2]), ζξ∪α)(4)
_
i∈{0,1}
S(ψi, ζξψi∪αψi)
≡
_
i∈{0,1}
S(ψi[ηψi], ζξψi[ηψi]∪αψi[ηψi])
.(5)
22
Now observe that ζξψi=ζξiψi=ζξi. Indeed, vis a free variable in clφ(ψ≡ψ[ηψ])
(=ξ) and a free variable in ψi, if and only if vis a free variable in clφ(ψi≡ψi[ηψi])
(=ξi) and in ψi. Hence dom(ζξψi) = dom(ζξiψi)and this completes our claim. The
same reasoning shows also that ζξψi[ηψi]=ζξiψi[ηψi]. Finally, if dom(α) = c, then
dom(αψi) = ci. It follows that (3) implies the above condition (5) and the case of ∨is
done.
In the case of ∃, we observe that
c3=c∪ {v},if v∈FV(ψ3)
cotherwise.
The following are equivalent for every α∈Asn such that dom(α) = c:
S(∃vψ3≡(∃vψ3[η∃vψ3]), ζ ξ∪α)(6)
∃β≥v(α∪ζξ)∃vψ3S(ψ3, βψ3)≡∃β≥v(α∪ζξ)∃vψ3[η∃vψ3]S(ψ3[ηψ3], β ψ3[ηψ3])
(7)
∃β≥v(α∪ζξ)S(ψ3, βψ3)≡∃β≥v(α∪ζξ)∃vψ3[η∃vψ3]S(ψ3[ηψ3], βψ3[ηψ3])
(8)
Observe that v /∈dom(ζξ)∪dom(ζξ3), because every occurrence of vin ∃vψ3is bounded
in φ. Hence ζξ=ζξ3, and consequently ζξψ3=ζξ3ψ3=ζξ3. With this observation
the proof in the case v /∈FV(ψ3)is straightforward, for (8) immediately reduces to (for
all αsuch that dom(α) = c3)
S(ψ3, α ∪ζξ3)≡S(ψ3[ηψ3],(α∪ζξ3)ψ3[ηψ3])
The above is the same as our induction assumption. So we may assume that v∈FV(ψ3).
Now fix αsuch that dom(α) = c. Suppose first that β≥v(α∪ζξ3)is such that S(ψ3, βψ3)
holds. Let us observe that in this case, βψ3=β. Moreover dom(α)∩dom(ζξ3) = ∅,
hence for some α0such that dom(α0) = c3,β=α0∪ζξ3. Hence S(ψ3[ηψ3], βψ3[ηψ3])fol-
lows by induction assumption (3). It is left to show that βψ3[η∃vψ3]≥v(α∪ζξ)∃vψ3[η∃vψ3].
This holds since no occurrence of vis in dom(ηψ3) = dom(η∃vψ3)and β≥v(α∪ζξ3).
So now assume that for some β≥v(α∪ζξ3)∃vψ3[η∃vψ3]S(ψ3[ηψ3], βψ3[ηψ3])holds. As
no occurrence of vis in dom(ηψ3), we can infer that S(ψ3[ηψ3], β)holds. Extend αto α0
such that dom(α0) = c3and α0(v) = β(v). Then (α0∪ζξ3)ψ3[ηψ3]=βand by (3) we
obtain S(ψ3, α0∪ζξ3). Hence there exists β0such that β0≥vα∪ζξ3and S(ψ3, β 0). This
ends the whole proof.
Corollary 47 (CS+
0).For every φ,ψ,α∈Asn(φ),β∈Asn(ψ), if φ[α] = ψ[β], then S(φ, α)≡
S(ψ, β).
Proof. By Lemma 46 S(φ, α)is equivalent to S(φ[α], ε)and S(ψ, β)is equivalent to S(ψ[β], ε).
Corollary 48. CS+
0` ∀φ(v)S(Ind(φ(v)), ε).
Proof. By the previous considerations at the beginning of this section, for a fixed φ(v),
S(Ind(φ(v)), ε)is equivalent to
S(φ[0/v], ε)∧ ∀xS(φ(v),[x]) →S(φ[v+ 1/v],[x])−→ ∀xS(φ(v),[x]).
23
By Lemma 46 and Corollary 47 we have
S(φ[0/v], ε)≡S(φ(v),[0])
∀xS(φ[v+ 1/v],[x]) ≡S(φ(v),[x+ 1]).
Hence, finally S(Ind(φ(v)), ε)is equivalent to the following axiom of ∆0(L+
S)-induction:
S(φ(v),[0]) ∧ ∀xS(φ(v),[x]) →S(φ(v),[x+ 1])−→ ∀xS(φ(v),[x]).
Corollary 49 (CS+
0).For every formula φ(v), term t(possibly having some variables), which is
substitutable for vin φ(v)and every α∈Asn(φ(t/v)), if S(∀vφ(v), α·), then S(φ(t/v), α).
Proof. Fix φ(v),t,α, as above and suppose S(∀vφ(v), α·). By compositional conditions
we know that for every βsuch that β≥vα∀vφ(v),S(φ(v), β·)holds. Define βsuch that
β∀vφ(x)=α∀vφ(x)and β(v) = tα. Hence, by our assumption we have S(φ(v), β ). Let
η0be a substitution of t[α]for every occurrence of vin φ(v). Them η0agrees with β, so
by Lemma 46 we have S(φ(v)[η0], β.). Observe that βφ(v)[η0]=αφ(v)[η0], hence we have
S(φ(v)[η0], α.). Let πbe a (coded) set of occurrences of the free variables from φ(t/v)
that are within the new occurrences of t(observe that there might be some occurences
of tin φ(v)). Let ηbe a substitution of numerals such that for every occurrence p∈π,
η(p) = α(vp)(recall that vpis the variable whose occurrence pis). By the definition of η,
we have φ(v)[η0] = φ(t/v)[η], hence we can conclude that S(φ(t/v)[η], α.)holds. Since η
agrees with α, Lemma 46 yields S(φ(t/v), α).
Recall that φ(t/v)denotes the substitution of tfor all (free) occurrences of vin φ(v).
Theorem 50. CS0` ∀φPrS
∅(φ)→S(φ, ε).
Proof. We reason in CS+
0. We fix a sequent calculus for the first order-logic with equality,
as in [24] (this choice is just a matter of convenience11). We fix a proof pof a sentence φ
and by induction on it’s length argue that whenever a sequent Γ⇒∆occurs in p, then
S(cl ^Γ→_∆, ε)(9)
holds, where VΓand WΓdenote (the canonically parenthesized) conjunction and dis-
junction over sentences from sets Γand ∆, respectively. To simplify the notation VΓ→
W∆will be abbreviated using the sequent notation as Γ⇒∆. In the course of the induc-
tion we rely on the fact that the following sentence is provable in CS0:
∀α∈Asn(Γ)S_Γ, α≡ ∃φ∈ΓS(φ, αφ).(DC)
The proof of (DC) in CT0consists in a straghtforward induction on the size of Γand a
similar argument can be given in the case of Vyielding a dual equivalence (see also [25]
and [3] for precise arguments). We go back to the main induction on the length of the
fixed proof p. In the base step we have to establish that all initial sequents satisfy (9).
These include initial sequents for equality and all sequents of the form φ⇒φ. In both
cases the proof follows the same pattern: first using Corollary (43) we get rid of the quan-
tifier prefix and then verify that the formula following it is satisfied by every assignment.
11Strictly speaking this calculus is formulated only for the language with both ∨and ∧and both ∃,∀, but
we can always extend the language by defining the missing symbols and adding axioms defining them
24
In the case of the initial sequents for equality we use the conditions for atomic sentences
from CS−, in case of φ⇒φwe use the axioms for ¬and ∨.
In the induction step the cases of quantifier rules are the unique non-obvious ones.
Let us consider the dictum de omni rule:
Γ, φ(t)⇒∆
Γ,∀vφ(v)⇒∆
where φ(v)is a formula and tis free for vin φ(v). We may safely assume that vis a free
variable in φ. For simplicity abbreviate φ(t/v)with simply φ(t). So suppose for every
α∈Asn Γ + φ(t)⇒∆it holds that
SΓ + φ(t)⇒∆, α.(10)
Fix an arbitrary α∈Asn Γ + ∀vφ(v)⇒∆and assume that for every θ∈Γ∪{∀vφ(v)},
S(θ, α·)holds. Consider any β∈Asn Γ + φ(t)⇒∆such that βΓ+∀vφ(v)⇒∆=
α. Since S(∀vφ(v), β·), then by Lemma 49, S(φ(t), β ·)as well. Hence for every θ∈
Γ∪ {φ(t)}S(θ, β·). By the induction assumption, for some ψ∈∆,S(ψ, β·)and since
βψ=αψ, this ends the proof.
Now consider the rule of universal generalisation
Γ⇒∆, φ(v)
Γ⇒∆,∀vφ(v),
where vdoes not occur free in the lower sequent. As previously assume that for all α∈
Asn (Γ ⇒(∆ + φ(v)))
SΓ⇒∆ + φ(v), α.
Fix an arbitrary β∈Asn Γ⇒∆ + ∀vφ(v)and arbitrary γ≥vβand assume that for
all ψ∈Γ,S(ψ, γ·). Since γΓ=βΓ(vis not a free variable in Γ) it follows that there
is ψ∈∆∪ {φ(v)}such that S(ψ, γ ·). If ψ∈∆, then we are done, since γ∆=β∆.
Otherwise S(φ(v), γ·)and it follows that S(∀vφ(v), β·), since γwas arbitrary.
Now, the thesis of the theorem follows, since, for arbitrary φ, if PrS
∅(φ)holds, then
there is a sequent calculus proof of Γ⇒φ, where for every ψ∈Γwe have S(ψ, ε).
Hence, by the proof above we obtain
Scl ^Γ→φ, ε
and since VΓ→φdoes not admit any free variables, then we have S(VΓ→φ, ε). The
thesis follows by the conjunctive correctness and compositional conditions: since for ev-
ery ψ∈Γwe have S(ψ , ε), then S(VΓ, ε)holds and an application of Modus Ponens
yields S(φ, ε).
Corollary 51. For every gödelized theory Th, we have
CS0+∀xTh(x)→S(x, ε)` ∀φPrTh(φ)→S(φ, ε).
Hence,
CS0+∀xTh(x)→S(x, ε)`REFω(Th).
Proof. The first part follows directly from Theorem 50. Having it, we prove the second
one: by induction on nwe prove that
CS0+∀xTh(x)→S(x, ε)` ∀φPrREFn(Th)(φ)→S(φ, ε).
25
For n= 0 this follows from the first part. Fix nand assume the thesis holds for it. By the
compositional clauses for CS−we have
CS0+∀xTh(x)→S(x, ε)` ∀φS(pPrREFn(Th)(φ)→φq, ε).
Consequently, reapplying the first part of this corollary for REFn(Th)substituted for Th
we get the induction thesis for n+ 1.
The corollary below easily follows from the corollary above and Corollary 48.
Corollary 52. CS0` ∀φPrPA (φ)→S(φ, ε).
Corollary 53. CS0`REFω(PA).
3.1 Corollaries
Compositional satisfaction vs. compositional truth The above results immediately
transfer to the setting of the following theory of truth:
Definition 54. CT−is the L∪{T}theory extending EA with the following axioms:
1. ∀xT(x)→Sent(x).
2. ∀s, t ∈ClTermT(s=t)≡(s)◦= (t)◦.
3. ∀φ, ψ ∈SentT(φ∨ψ)≡(T(φ)∨T(ψ)).
4. ∀φ∈SentT(¬φ)≡ ¬T(φ).
5. ∀φ(v)∈Form≤1T(∃vφ(v)) ≡ ∃xT (φ[x].
As usual CTndenotes the result of extending the following theory with induction axioms
for Σnformulae of the extended language. 4
Let L+
Tdenote the extension of LTwith function symbols for all p.r. recursive func-
tions and CT+
0denote the extension of CT0with all defining axioms for fresh functions
symbols in L+
T. Then we have an analogue of Proposition 41:
Proposition 55. CT+
0`I∆0(L+
T)
Now we show that the result on the provability of (GR(Th)) in CS0transfers to the
setting with the truth predicate.
Corollary 56. CT0` ∀φ∈SentPrPA (φ)→T(φ).
Proof. Work in CT+
0and put
S(x, y) := Form(x)∧y∈Asn(x)∧T(x[y]).
The above is a ∆0(L+
T)formula, so obviously we have I∆0(L+
S). Now, we show that
S(x, y)behaves compositionally. We focus on the ∃-axiom. Pick a formula φ(v)and
α∈Asn(∃vφ(v)). Observe that the following equivalences hold:
S(∃vφ(v), α)≡T(∃vφ(v)[α]).
≡ ∃xT (φ[α][x])).
≡ ∃β≥vαT (φ[β]).
≡ ∃β≥vαS(φ, β ).
26
In the second and the third equivalence we use the fact that v /∈dom(α). Hence (in CT0)
by Theorem 50 we have
∀φ∈SentPrPA(φ)→S(φ, ε).
Translating it back to the language with the truth predicate, we get our thesis.
The proof of the above corollary proceeds by defining the satisfaction predicate sat-
isfying CS+
0in CT+
0. In fact, the same translation works also in the context of the non-
inductive versions of both theories, CS−and CT−. However, it is not known, whether
the reverse is true in the context of these theories, i.e. whether CS−can define the truth
predicate of CT−. Using the Enayat–Visser method ([7]) of constructing pathological
models for CS−one can show that standard methods of defining truth from satisfaction
do not work. However, the results from the previous section witness that ∆0induction
is sufficient to overcome these deficiencies of CS−.
Proposition 57. The truth predicate satisfying CT0is definable in CS0.
Proof. Working in CS0, put
T(x) := S(x, ε).
As previously, T(x)is a ∆0(LS)formula, so CS0`I∆0(LT). Since sentences are the
unique formulae for which εis an assignment, so axiom 1.of CS−implies the correspond-
ing axiom of CT−. Once again we focus on the compositional axioms for ∃. Working in
CS0fix φand without loss of generality assume that v∈FV(φ). Observe that the follow-
ing equivalences hold:
T(∃vφ)≡S(∃vφ, ε)
≡ ∃β≥vεS(φ, β ·)
≡ ∃xS(φ, [x])
≡ ∃xS(φ[x], ε)
≡ ∃xT (φ[x]).
The proof of the fourth equivalence involves the crucial use of Lemma 46.
Many Faces Theorem Corollary 56 coupled with some known results from the litera-
ture, shows that the Global Reflection Principle is a very robust notion. Not only it is
equivalent to bounded induction but is immune to, apparently significant, variations.
This is summarized in the corollary below (we state it for the theory of compositional
truth, however all the equivalences should transfer to the setting of a satisfaction predi-
cate without significant changes). PrT
Sent(φ)asserts that φis provable from the set of true
sentences in pure sentential logic, while DC is a truth variant of the principles from the
proof of 50.
Corollary 58. Over CT−+EA the following theories are equivalent
1. I∆0(LT);
2. ∀φPrPA(φ)→T(φ);
3. ∀φPr∅(φ)→T(φ);
4. ∀φPrT
∅(φ)→T(φ);
27
5. ∀φPrT
Sent(φ)→T(φ);
6. DC.
The implication 3.⇒4. is established in [5]. The equivalence between 5. and 1. is
demonstrated in [3]. Much later it was significantly fine-tuned in [6], yielding the impli-
cation 6.⇒1.
Fullness The following is one of the most useful properties of CS0. It implies that every
model of CS0is full, a theorem first demonstrated by Bartosz Wcisło and presented in
[25]. The proof below is an observation also due to Wcisło which crucially uses the prov-
ability of (GR(Th)). It’s proof is included also in [20] (Fact 33) but we give it here for
completeness. In the definition below we fix a canonical primitive recursive translation
transforming a given formula φinto one in the Σnform. We assume that it formalizes in
PA.
Recall (Definition 12) that φ(x)Σdenotes the canonical Σcform of φ(x)and Σ∗
c:=
{φ|φΣ∈Σc}. Moreover recall that Scdenotes the restriction of Sto all formulae which
are equivalent to sentences of Σccomplexity (in the sense of M). More precisely
Sc:= {hφ, αi | (M, S)|=φ∈Σ∗
c∧S(φΣ, α)}.
Theorem 59. Suppose that (M, S)|=CS−+GR(PA). Then for every c,(M, Sc)|=CS(Σ∗
c).
Proof. Fix a model (M, S)|=CS−+GR(PA)and c∈M. It will be easier to switch to the
truth predicate, so put T:= S(x, ε). Since for every formula φ∈Σ∗
cand every α∈Asn(φ)
we have
(M, S)|=S(φ, α)≡T(φ[α]),
it is sufficient to show that (M, Tc)|=Ind(LT), where Tcis the restriction of Tto the
formulae of Σ∗
ccomplexity, i.e.
Tc:= (Σ∗
c)M∩(T)(M,S).
From now on we work in (M, T ). By the classical metamathematics of PA, for every
cthere is a formula SatΣcsuch that for every φ∈Σ∗
cand every α∈Asn(φ)we have
PrPA φ[α]≡SatΣc(φΣ, α).
Hence, by GR(PA) we conclude that for every sentence φ∈Σ∗
c
T(φ)≡T(SatΣc(φΣ, ε)).
Put ξ(v) := SatΣc(v, ε)and T0(x) := T(ξ[x]) ∧x∈Σ∗
c. Then we see, that
Tc= (T0)(M,T )∩(Σ∗
c)M
Consequently, T0satisfies the compositional axioms of CT−for formulae from the Σ∗
c
class. We shall now show (M, T 0)|=Ind(LT). Thus let η[T0]be an arbitrary axiom of
induction for a formula with T0(we mark all occurrences of T0in η). We may assume
that η[T0]is in the semirelational form (as defined in [19]). Since, using the notation of
[19], T0is of the form T∗ξ, by Lemma 25 in [19] we have
(M, T )|=η[T0]≡T(η[ξ]).
However, η[ξ]is an axiom of induction (in the sense of M) for an arithmetical formula
η[ξ], hence T(η[ξ]) by GR(PA).
28
Remark 60. A very similar reasoning was used in Kotlarski in [15]. However various
parts of this paper are negatively influenced by the significant gaps already discussed at
the beginning of this section. We decided to reprove it in a rigorous way. Essentially the
same proof is given in [20]. 4
Corollary 61. For any (M, S)|=CS−the following conditions are equivalent:
1. (M, S)|=GR(PA).
2. For every c∈M,(M, S)|=CS(Σ∗
c).
3. (M, S)|=CS0.
Proof. We show the remaining implication 2⇒3. By the classical fact in the metamath-
ematics of PA, a subset of the model satisfies ∆0-induction if and only if it is piecewise
coded, i.e. it is sufficient to show
(M, S)|=∀c∃d∀x<cS((x)0,(x)1)≡x∈d,
where (x)idenotes the i-th projection of x. Working in (M, S)fix an arbitrary c. Obvi-
ously, if a formula φ<cthen φ∈Σ∗
c. Hence, it is sufficient to find a dsuch that
(M, Sc)|=∀x<cS((x)0,(x)1)≡x∈d.
This clearly can be done as Scsatisfies full induction.
4 Consequences of the Global Reflection Principle
In this section we focus on the ∆0-inductive truth predicate. We remind the Reader that
by default all theories extend EA. We extend the result from the previous section and
prove the following theorem
Theorem 62. For every φ(x)∈Σ1(LT)and every n∈ωthe following sentence is provable in
CT0
∀xPrT
UTBn(φ[x]) →φ(x).
Corollary 63. CT0`Σ1(LT)-REF(UTB−).
The above affirmatively answers the question of Lev Beklemishev and Fedor Pakho-
mov from [2].
Convention 64. Working in an extension of UTB−, it makes sense to treat the predicate
Tas a theory composed of all true sentences. Thus, we shall often write (for a gödelized
theory Th)
Γ-REF(Th +T)
to denote the theory consisting of all sentences
∀xPrT
Th(φ[x]) →φ(x),
for φ(x)∈Γ.4
29
Let us start by explaining that Theorem 62 really improves on the results from the
previous section. Let Th be a gödelized theory extending EA in a language L0and let
UTB−(Th)denote the extension of Th with UTB−axioms. It is sufficient to observe that
over EA
∆0(LUTB−(Th))-REF(UTB−(Th)) `GR(Th).
Indeed, working in EA + ∆0(LUTB−(Th))-REF(UTB−(Th)) fix an arbitrary LTh sentence φ
and assume PrTh(φ). By the axioms of UTB−(Th)we immediately obtain PrUTB−(Th)(pT(φ)q).
Therefore, by the ∆0reflection for UTB−(Th)we obtain, T(φ).
Proof of Theorem 62 starts with a lemma12:
Lemma 65. For every φ∈∆0(LT)
CT0` ∀xPrT
UTB(φ[x]) →φ(x).
Proof. Fix (M, T )|=CT0,φ(x)∈∆0(LT)and a∈Mand any proof psuch that (M, T )|=
ProofT
UTB(p, φ[a]). Since in φall the quantifiers are bounded, then there is a b∈Msuch
that for every (M0, T 0)satisfying (1) M ⊆eM0and (2) T0<b =T<b we have
(M, T )|=φ(a)⇐⇒ (M0, T 0)|=φ(a).(A)
Recall that for X⊆M,X<b denotes the set of elements of Xbelow b. Fix any such b.
In short, bis big enough so that any end-extension of (M, T )which agrees with Tup to
bis absolute with respect to φ(a). Without loss of generality assume that p < b and let
cbe big enough so that any formula (with code) smaller than bbelongs to (Σ∗
c−1)M. By
Theorem 59, (M, Tc)|=CT(Σ∗
c). Now we work in (M, Tc). Consider the theory
Th := PA +{φ∈Σ∗
c|T(φ)}.
By GR(PA) (Corollary 52) Th is consistent and by the trivial conservativity proof for UTB
it follows that UTB +Th is consistent as well. Hence, by the Arithmetized Completeness
Theorem (for PA∗) there is a full model ((N, T 0),Sat(N,T 0))such that
(M, Tc)|=(N, T 0)|=Sat(N,T 0)UTB +Th.(B)
Since (N, T 0)is strongly interpretable in (M, Tc), then, by Proposition 24, M ⊆eN. Since
internally in (M, Tc),(N, T 0)is a model of UTB +Th, then every sentence occurring in
pis true in (N, T 0). So we see that (N, T 0)|=φ(a). We show that T0<b =T<b. Fix any
sentence ψsuch that ψ < b. Then, by the choice of c,ψ∈Σ∗
c−1and hence the following
conditions are equivalent
1. ψ∈T<b
2. ψ∈Th
3. (M, Tc)|=N |=Sat(N,T 0)ψ
4. (N, T 0)|=T(ψ)
5. ψ∈T0<b.
12This lemma can be obtained by the methods of [2] as well. Indeed, the result (for UTB−instead of UTB
and without the oracle T) is mentioned in an open question at the end of p. 15.
30
The equivalence between 3.and 4.follows by (B). Hence it follows by (A) that (M, T )|=
φ(a), which suffices to prove the claim.
Remark 66. The above proof generalises to the case in which PA is replaced with a (for-
malized) theory Th in an expanded (at most countable) language L0(we assume a fixed
Gödel coding of L0) such that Th `IndL0. This allows us to obtain:
Lemma 67. For every φ∈∆0(L0
T)
CT0+∀φTh(φ)→T(φ)` ∀xPrT
UTB(L0)+Th(φ[x]) →φ(x).
In order to bypass the problems with infinitely many additional predicates in L0it is
sufficient to work with a M-bounded fragment of Th and consider only the fragment of
L0consisting of predicates which occur in a formula in the fixed proof p.4
The following lemma suffices to complete the proof of Theorem 62.
Lemma 68 (Bounding lemma).For every formula φ(x)∈∆0(LT)and every n∈ω, the
following implication is provable in CT0:
PrT
UTBn(∃vφ(v)) → ∃yPrT
UTB(∃v < yφ(v)).
Before we prove it, let us show a proposition which was the motivation for the proof
of the above lemma:
Proposition 69 (Σ1-completeness for UTB−).For every ∆0(LT)formula φ(x), if (N,Th(N)) |=
∃xφ(x), then for some n∈ωTh(N) + UTB−` ∃x < nφ(x).
Proof. Suppose that for every n, UTB−+Th(N) + ∀x<n¬φ(x)is consistent. A trivial
compactness argument then shows that Th(N) + UTB−+{∀x < nφ(x)|n∈ω}is
consistent as well. So let us take (M, T )|=Th(N)+UTB−+{∀x < nφ(x)|n∈ω}and look
at (N, T N). Since NeM,(N, T N)|=UTB−, and, consequently TN=Th(N), because
in Nthere is only one interpretation for the UTB−-truth predicate. Since φ(x)∈∆0(LT),
then (N,Th(N)) |=∀x¬φ(x), which suffices to end the proof.
Remark 70. Essentially the same proof shows that already TB−(a non-uniform version
of UTB−based solely on Tarski biconditionals for arithmetical sentences) is Σ1-complete
in the above sense. 4
Proof of Lemma 68. Fix n∈ω,φ(x)∈∆0(LT)and a model (M, T )|=CT0. Assume that
for all a∈M,(M, T )|=¬PrT
UTB(∀v<a¬φ(v)). By formalizing in PA the trivial compact-
ness argument, we see that for a (M, T )-definable theory
Th := UTB +{φ|T(φ)}+{∀v < a¬φ(v)|a∈M},
(M, T )|=ConTh. However, contrary to what happened in the proof of Σ1-completeness
for UTB−(Proposition 69), there need not be an (M, T )-definable full model of Th, since
we may not have Σ1-induction for LT. As a remedy, we shall work with restrictions of
T. For an arbitrary cwe shall show that ¬PrTc
UTBnc(∀v¬φ(v)), which suffices to prove the
claim by a trivial compactness argument (UTBncdenotes the first c−1axioms of UTBn).
Fix cand let c < b be big enough so that every formula in Tcand every axiom of UTB−c
belongs to Σ∗
b−1. Working in (M, Tb)consider
Thb:= UTB +{φ|T(φ)∧φ∈Σ∗
b}+{∀v < a¬φ(v)|a∈M}.
31
Since Thb⊆Th, then Thbis consistent. Thbis definable in (M, Tb)|=PA∗hence we can
fix a full model ((N, T 0),Sat(N,T 0))such that
(M, Tb)|=(N, T 0)|=Sat(N,T 0)Thb.
Consider a model (M, T 0M)(i.e. we restrict T0to model M). Since (M, T 0M)⊆e(N, T 0)
and φ(x)∈∆0(LT), then (M, T 0M)|=∀x¬φ(x). We check that (M, T 0M)|=CS−(Σ∗
b).
To this end it is sufficient to show that for every φ∈Σ∗
bwe have
φ∈T0M⇐⇒ φ∈T. (∗)
So fix an arbitrary such φ. Now the following conditions are equivalent
1. φ∈T0M;
2. (N, T 0)|=T(φ);
3. (M, Tb)|=(N, T 0)|=Sat(N,T 0)φ;
4. (M, T )|=T(φ).
Now, the equivalence between (2) and (3) is by the fact that (N, T 0)is an (M, Tb)-definable
full model of UTB. The equivalence between (3) and (4) holds because Nsatisfies all sen-
tences from Σ∗
b, which Tdeems true in M. Since T0Mis (M, Tb)-definable it satisfies full
LTinduction. To sum up, we have just concluded that
(M, T 0M)|=CS(Σ∗
b) + ∀x¬φ(x).
We work in (M, T 0M). First observe that T0
bmakes V(i.e. the class of all numbers, see
Example 5) a b- full model for the arithmetical vocabulary (because T0
b=Tb, by (∗)). For a
fixed k∈ωlet Xkconsists of all boolean combinations of sentences from Σ∗
b(L)∪Σ∗
k(LT).
Observe that for all k∈ω, all the uniform Tarski biconditionals are in Xkand this class
is closed under subformulae. Now for every k∈ω, there exists an (M, T 0M)definable
satisfaction class for V[T]and sentences from Xk. Denote such a satisfaction predicate
with SatT0
b
Σk. Then SatT0
b
Σn+2 makes V[T]an Xn-model for the definable restricted theory
{φ∈ LPA |Tc(φ)}+UTBnc+∀x¬φ(x).
By Proposition 16 this is enough to show that this theory is consistent.
Remark 71. The proof of Lemma 65 can be modified to yield a new proof of Theorem 1
from [2] (for finite languages). Let us restate it and prove it here:
Theorem 72 (Beklemishev-Pakhomov).Let L0be an arbitrary finite language with a fixed
Gödel numbering and Th be a gödelized L0theory. Then,
UTB−(L0)+∆0(L0
T)-REF(UTB−(L0) + Th)
is L0-conservative over REF(Th).
Proof. Pick any (M, S)|=REF(Th) + CS(Σ∗
c(L0)) (the satisfaction class is defined for L0).
Using overspill, as in the proof of Lemma 75 we may pick a nonstandard d<csuch that
(M, S)|=∀φ∈Σ∗
d(L0)PrSd
Th(φ)→S(φ, ε).
32
Then, in (M, S)the L0
S-theory Th +UTB−(L0) + {φ∈Σ∗
d(L0)|S(φ, ε)}is consistent, so
let (N, S0)be its full model. We claim that
(M, S0M)|=UTB−(L0)+∆0(L0
T)-REF(UTB−(L0) + Th).
UTB−(L0)holds in (M, S0M)since, in M,Sand S0Mcoincide on all formulae from Σ∗
d.
To show ∆0reflection, we use the fact that (N, S0)is strongly interpretable in (M, S). Fix
a∆0(L0
T)-formula φ(x)and working in (M, S)assume that pis a proof of φ(a)(for some
element a) from the axioms of UTB−(L0) + Th. Since
(M, S)|=(N, S 0)|=Sat(N,S 0)UTB−+Th,
then (N, S0)|=φ(a)and since (M, S0M)⊆e(N, S 0),φ(a)holds in (M, S0M)by abso-
luteness of bounded formulae.
4
5 The Arithmetical Part of CT0
In this section we reprove the following result of Kotlarski ([15])13
Theorem 73 (Kotlarski, Smoryński).CT0is arithmetically conservative over REFω(PA).
The idea of Kotlarski’s argument is to mimic the Henkin proof of Completeness Theo-
rem in a countable recursively saturated model. Also, [2] gives a different, syntactic proof
of Theorem 73. We choose a still different path and our main ingredient is the following:
Theorem 74. Let Th be any gödelized Ltheory. Suppose that M |=REFω(Th)and Sis a
satisfaction class for Msuch that
(M, S)|=CS(Σ∗
c(L0))
for some nonstandard c. Then there exists a nonstandard d∈Mand (N, S0)|=CS0+∀xTh(x)→
S(x, ε)such that M(eNand Sd⊆S0.
Observe that the conditions M(eNand Sd⊆S0jointly imply that MeN(assum-
ing Sdand S0are satisfaction classes). The construction of (N, S0)proceeds in ω-stages
and is motivated by Corollary 61: Nwill admit a cofinal ω-sequence a0, . . . , an, . . . and
at the n-th stage of the construction we shall build a satisfaction class for all large enough
formulae. The following lemma makes this idea more precise. Henceforth Th is any
gödelized theory in L14.
Lemma 75. Suppose that M |=REFω(Th),cis a nonstandard element of Mand (M, S)|=
CS(Σ∗
c(L0)). Then there exists a nonstandard d∈Mand a sequence {(Mn, Sn, cn)}n∈ωsuch
that (M0, S0, c0)=(M, Sd, d)and for each n
1. MneMn+1 and Sn⊆Sn+1.
2. (Mn+1, Sn+1)|=CS(Σ∗
cn+1 ) + ∀φ∈Σ∗
cn+1 Th(φ)→S(φ, ε).
13In order to get the arithmetical theory right one should compose Kotlarski’s result with Smoryński’s
observation from [23].
14All the results generalize to the setting where Lis substituted with an arbitrary finite language. We have
decided for the reduced version to keep the definition simpler.
33
3. cn+1 ∈Mn+1 \Mn.
Thus {(Mn, Sn, cn)}n∈ωconsists of proper end-extensions and each satisfaction class
Sn+1 in the sequence decides all formulae in the sense of Mn. Let us show that Lemma
75 immediately implies Theorem 74.
Proof of Theorem 74 modulo Lemma 75. Fix M |=REFω(Th)and Ssuch that (M, S)|=
CS(Σ∗
c). Fix a chain {(Mn, Sn, cn)}n∈ωas in the thesis of Lemma 75. Put N=SnMn,
S0=SnSn. It is straightforward to verify that (N, S0)|=CS−+∀φTh(φ)→S(φ, ε). Let
us check one direction of the quantifier axiom. Assume that (N, S0)|=∃β≥vαS(φ(v), β).
Let kbe big enough so that (Mn, Sn)|=∃β≥vαS(φ(v), β)and φ(v)∈Σ∗
cn−1. It follows
that (Mn, Sn)|=S(∃vφ(v), α). Hence (N, S 0)|=S(∃vφ(v), α). The argument in the rest
of cases is similar.
To check ∆0(LT)–induction, we verify that Sis coded, i.e. for every c∈Nthere exists
ad∈Nsuch that
∀x<cS((x)0,(x)1)≡x∈d
holds (recall that (x)idenotes the projection of xto the i-th coordinate). Take any cand
let nbe big enough so that c∈Mnand each formula smaller than cis of complexity
Σ∗
cn. By using induction in (Mn+1, Sn+1 )we can find the appropriate d∈Mnand since
(Mn, Sn+1Mn)⊆e(N, S 0), this dwill work also for (N, S0).
The proof of Lemma 75 consists in prolonging the given satisfaction class Suntil its
domain catches up with the universes of models constructed along the way. As it turns
out this notion of a satisfaction class being prolongable is well worth isolating.
Definition 76. Suppose (M, S, X )|=CS(X)and for every n,(M, X)|= Σ∗
n⊆X.
1. Sis 0-prolongable if there is Nsuch that M(eN,Nis strongly interpreted in
(M, S)via a satisfaction class SatNand for all φ∈s(X)and α∈Asn(φ)
(M, S)|=S(φ, α)≡ N |=SatNφ[α].
2. Sis n+ 1 prolongable if there is Nsuch that M eN,c∈N\Mand S0⊆N2such
that (N, S0)|=CS(Σ∗
c),S⊆S0and S0is n-prolongable.
4
Let us observe that if Nwitnesses the 0-prolongability of S, then N |=PA. The
following lemma is the key element of our reasoning:
Convention 77. Let GR(X, S, Z )denote the following sentence of L2:
∀φ∈s(X)PrZ(φ)→S(φ, ε).
Moreover, if Sis a satisfaction class on M, then (M, S )|=GR(X, S, S +Y)will ab-
breviate (M, S)|=GR(X, S, {φ∈SentL|S(φ, ε)} ∪ Y).4
Lemma 78. Suppose that (M, S, c)|=CS(Σ∗
c)∧GR(Σ∗
c, S, S +REF(Th)). Then there exists
(N, S0, c0)|=CS(Σ∗
c0)∧GR(Σ∗
c0, S0, S 0+Th)such that
1. (N, S0, c0)is strongly interpretable in (M, S, c).
2. c0∈N\M
34
3. S⊆S0.
Proof. Fix (M, S, c)as in the assumption and work in it. Consider the following LS0∪ {c0}
theory, where c0is a fresh constant and S0is a fresh predicate
GR(Σ∗
c0, S0, S 0+Th) + {φ|S(φ, ε)}+CS(Σ∗
c0, S0) + {c0> a |a∈M}.
We shall show that the theory is consistent. The proof proceeds in two stages. In the
first one, we fix a full model of REF(Th) + {φ|S(φ, ε)}, i.e. a full model (N0,SatN0)
such that N0|=SatN0REF(Th) + {φ|S(φ, ε)}. Such a model exists by the Arithmetized
Completeness Theorem, since by our assumption every consequence of REF(Th)and the
set of true sentences (in the sense of S) is true, in the sense of S(recall that (M, S)|=
PA∗). Let us observe that N0|=SatN0PA, since REF(EA)`PA (and the proof formalizes in
EA) and Th ⊇EA. Now, using N0we formalize the standard argument that any model of
PA admits an elementary extension to a model with a fully inductive, partial nonstandard
satisfaction class. We reason in (M, S, c). We show that
ElDiagSatN0(N0) + CS(Σ∗
c0, S0) + {c>a |a∈M}+GR(Σ∗
c0, S0, S 0+Th)
is consistent. Let Abe a finite (in the sense of M) fragment of this theory and let a−1be
the greatest dsuch that pc>dq∈A. We shall find the interpretation for S0in N0. Con-
sider the arithmetical partial truth predicate Satafor formulae from Σ∗
aand interpret S0as
Sata. Then N0|=CS(Σ∗
a,Sata). Moreover, as N0|=REF(Th),N0|=GR(Σ∗
a,Sata,Sata+Th)
(see e.g. [1]). So Ais consistent and by the formalized compactness theorem, so is the
entire theory. Let (N, S 0, c0)be its full model. Then Nis clearly an end-extension of M
and c0∈N\M. Moreover (N, S0, c0)|=CS(Σ∗
c0, S0)∧GR(Σ∗
c0, S0, S 0+Th)and S⊆S0, by
construction.
The following lemma isolates the relation between prolongability and global reflec-
tion.
Lemma 79. Suppose that (M, S, c)|=CS(Σ∗
c)where cis nonstandard. Then the following
conditions are equivalent
1. Sis n-prolongable.
2. (M, S, c)|=GR(Σ∗
c, S, S +REFn(EA)).
Proof. We prove by induction on nthat
∀M∀S∀c∈M\ω(M, S, c)|=CS(Σ∗
c) =⇒
Sis k-prolongable ⇐⇒ (M, S, c)|=GR(Σ∗
c, S, S +REFn(EA)).
For the base step fix M,S,Xas above and assume first that Sis 0-prolongable. Fix N
as in the definition of 0-prolongability. Working in (M, S)take any proof pof a sentence
φ∈s(Σ∗
c)from EA. Since EA is a finite theory N |=SatNEA. Then, since pis a proof from
true axioms in the sense of SatN, then SatN(φ, ε). Hence, since φis a formula from s(Σ∗
c),
S(φ, ε)holds by the definition of 0-prolongability.
Suppose now that (M, S, c)|=∀φ∈s(Σ∗
c)PrS
EA(φ)→S(φ, ε). Consider the follow-
ing (M, S, c)-definable theory Th := {φ∈Σ∗
c|S(φ, ε)}. By our assumption, (M, S, c)|=
35
ConTh. Work in (M, S, c). By the Arithmetized Completeness Theorem (we use the as-
sumption that (M, S, c)|=PA∗), we have a full model N |=SatNTh. Hence, obviously
SatNand Scoincide on s(Σ∗
c)(observe that they need not coincide on Σ∗
c).
Now assume that the equivalence holds for n. Fix M,S,Σ∗
cas above and assume first
that Sis (n+ 1)-prolongable and pick (N, S0, c0)|=CS(Σ∗
c0)such that S⊆S0and S0is
n-prolongable. By the induction assumption (N, S0, c0)|=∀φ∈s(Σ∗
c0)PrS
REFn(EA)(φ)→
S(φ, ε). By compositional clauses, it follows that for every φ(v)∈M
(N, S0, c0)|=S(p∀vPrREFn(EA)(φ[v]) →φ(v)q, ε).
Hence in particular, if ψ∈Mis any axiom of EA +REFn+1(EA), then (N, S0)|=S(ψ, ε).
So suppose p∈Mis a proof of a sentence φ∈s(Σ∗
c)from the axioms of REFn+1(EA).
Work in (N, S0). By the previous argument, if θis a premise of p, then S(θ, ε)holds. Since
all the formulae occurring in pbelong to Σ∗
c0, then we have (N, S0)|=S(φ, ε). However,
S0and Scoincide on s(Σ∗
c).
Now, working in (M, S, c), assume ∀φ∈s(Σ∗
c)PrS
REFn+1(EA)(φ)→S(φ, ε). We apply
Lemma 78 to Th := REFn(EA)and conclude that there is (N, S 0, c0)as in the thesis of
the lemma. By our inductive assumption applied to (N, S0, c0), it follows that S0is n-
prolongable. Hence Sis n+ 1-prolongable, which ends the proof.
Corollary 80. The following conditions are equivalent for a model (M, S, c)|=CS(Σ∗
c, S)
1. (M, S, c)is n-prolongable.
2. There exists a (M, S, c)-definable sequence of models {(Mk, Sk, ck)}k≤n∈ωsuch that (M0, S0, c0) =
(M, S, c)and for each k < n
(a) MkeMk+1 and Sk⊆Sk+1.
(b) (Mk, Sk, ck)|=CS(Σ∗
ck).
(c) ck+1 ∈Mk+1 \Mk.
(d) (Mk+1, Sk+1, ck+1)is strongly interpretable in (Mk, Sk, ck).
Corollary 81. The following conditions are equivalent for a model (M, S, c)|=CS(Σ∗
c, S)and
for every n∈ω
1. M |=REFn+1(EA)
2. There exists b∈M\ωsuch that Sbis n-prolongable.
Proof. (2) ⇒(1) follows immediately from Lemma 79. We prove (1) ⇒(2). Fix (M, S, c)
as in the assumptions. Let Satkbe an arithmetically definable truth predicate for Σ∗
k(as
in the proof of Lemma 78). By induction in (M, S, c), we have
∀φ∈Σ∗
kSatk(φ, ε)≡S(φ, ε)
for every k. Since M |=REFn+1(EA), then, for every k, we have
M |=∀φ∈Σ∗
kPrSatk
REFn(EA)(φ)→Satk(φ, ε).
Hence for every l∈ω,(M, S, c)|=∀x<l∀φ∈Σ∗
xPrSx
REFn(EA)(φ)→Sx(φ, ε).By the
overspill principle we can find a b>ω,b≤c, such that
(M, S, c)|=∀x<b∀φ∈Σ∗
xPrSx
REFn(EA)(φ)→Sx(φ, ε).
Hence Sb−1is n-prolongable.
36
Since REF(EA) = PA, the following is the most memorizable version of our main
lemma:
Theorem 82. The following conditions are equivalent for a model (M, S, c)|=CS(Σ∗
c)and for
every n∈ω
1. M |=REFn(PA)
2. There exists b∈M\ωsuch that Sbis n-prolongable.
Now, we have all that is needed to finish our conservativity proof for CT0:
Proof of Lemma 75. Suppose (M, S, c)|=CS(Σ∗
c)and M |=REFω(Th). Let Satnbe the
arithmetical partial truth predicate for formulae in Σ∗
n. By assumption on Mit follows
that for every n
M |=∀φ∈Σ∗
nPrSatn
REFn(Th)(φ)→Satn(φ, ε).
By induction, for every n∈ω, Satnand Sncoincide. Hence for every n∈ωwe have
(M, S)|=∀x<nGR(Σ∗
x, Sx, Sx+REFx(Th)).
By overspill it follows that for some d > ω
(M, S)|=GR(Σ∗
d, Sd, Sd+REFd(Th))
We define the chain (Mn, Sn, cn)by induction. Assume that (Mk, Sk, ck)has been de-
fined and it satisfies GR(Σ∗
ck, Sck, Sck+REFd−k(Th)). To get (Mk+1, Sk+1, ck+1)we apply
Lemma 78 to Th0=REFd−(k+1)(Th).
Corollary 83. The arithmetical consequences of CT0+∀φTh(φ)→S(φ, ε)and REFω(Th)
coincide.
Proof. Conservativity part follows from Theorem 74. That CT0+∀φTh(φ)→S(φ, ε)`
REFω(Th)was established in Corollary 51.
It might seem that in the above proof the limit model is a very specific model of CS0.
Quite surprisingly every model of CS0is of this form. One of the crucial steps in the proof
is worth isolating as a lemma
Lemma 84. Suppose that (M, S)|=CS0,(M0, SM0)⊆(M, S),d1∈M0and (M0, Sd1M0)|=
CS−(Σ∗
d1). Then for every d0∈M0such that d1−d0is nonstandard, (M0, Sd0M0)|=CS(Σ∗
d0).
Proof. Let us fix M, S, M0, d1, d1as above. Since (M0, Sd1M0)|=CS−(Σ∗
d1), it follows
that
(M0, Sd0M0)|=CS−(Σ∗
d0).
We prove that induction axioms hold in (M0, Sd0M0)as well. By the assumptions, it
follows that (M, Sd1)|=∀φ∈Σ∗
d1PrPA(φ)→S(φ, ε), hence also
(M0, Sd1M0)|=∀φ∈Σ∗
d1PrPA(φ)→S(φ, ε).
Let SatΣd0be as in the proof of Theorem 59 and let us abbreviate S(φ, ε)with T(φ)and
SatΣd0(x, ε)with T rd0(x). Observe that T rd0∈Σ∗
d0. It follows that
(M0, Sd1M0)|=∀φ∈Σ∗
d0T(T rd0(φΣ)) ≡T(φΣ).
37
Since Sd0coincides with Sd1on (Σ∗
d0)M0, then
(M0, Sd0M0)|=∀φ∈Σ∗
d0T(T rd0(φΣ)) ≡T(φΣ).
Since T(T rd0(xΣ)) defines and Sd0M0in (M0, Sd1M0)it is sufficient to show that
(M0, Sd1M0)|=η[T(T rd0(xΣ))/P ],(∗)
where ηis an arbitrary instance of an induction axiom for a fresh predicate letter P, in a
semi-relational form. Fix η. Since M0|=PrPA (η[T rd0(xΣ)/P ]) and η[T rd0(xΣ)/P ]∈Σ∗
d1,
we have
(M0, Sd1M0)|=Tη[T rd0(xΣ)/P ].
(∗) follows as in [19] and Theorem 59.
Theorem 85. Suppose (M, S)|=CS0has cofinality κ. Then there is a chain {(Mα, Sα, cα)}α∈κ
such that Sα∈κMα=M,Sα∈κSαand for every α < β < κ
1. MαeMβand Sα⊆Sβ.
2. (Mβ, Sβ)|=CS(Σ∗
cβ).
3. cβ∈Mβ\Mα.
We note that the above chain need not be continuous.
Proof. Fix (M, S)|=CS0and a cofinal sequence {dα}α∈κ. In the base step we build
(M0, S0, d0)|=CS(Σ∗
d0). Consider the formula θ0(x, y, z)
∃cSeq(c)∧len(c) = x∧c0=z∧ ∀i<x(2ci< ci+1)∧θ(x, y, c)
where θ(x, y, c)is
∀φ∈Σ∗
y∀i<x∀α < ciφ<ci∧α∈Asn(∃vφ)∧S(∃vφ, α)→ ∃β < ci+1 β∼vα∧S(φ, βφ).
Thus θ0(x, y, z)expresses that there is a witness-bounding sequence cof length xand
starting from z, which works for those formulae of Σ∗
ycomplexity which are below some
of the elements of the sequence. Let ebe such that d0−eis nonstandard. We reason in
(M, Se)|=CS(Σ∗
e)and by a straighforward induction conclude that
∀xθ0(x, e, e).
We let abe an arbitrary nonstandard number and we fix cwitnessing θ0(a, e, e). We define
M0:= sup{ci|i∈ω}.
Clearly if b1, b2< ci, for some i, then b1+b2, b1·b2< ci+1 by the assumption on c.
Hence M0⊆ M. We check that (M0, SeM0)|=CS−(Σ∗
e). The unique non-trival step
is the one for quantifiers: fix any formula φ(v, ¯w)∈Σ∗
e∩M0,α∈M0and assume that
(M0, SeM0)|=S(∃vφ(v, ¯w), α)and α, φ < ci. Then
(M, S)|=S(∃vφ(v, ¯w), α)
38
Hence by the properties of cthere is β < ci+1 ,β∼vαsuch that
(M0, Se)|=S(φ, β).
To conclude that S0:= Sd0is fully inductive, we apply Corollary 84 for M0=M0,d0=d0
and d1=e.
In the successor step assume that (Mα, Sα, cα)has been constructed and w.l.o.g. as-
sume that dα+1 /∈Mα. We pick e∈Msuch that e−dα+1 is nonstandard and repeat the
reasoning from the base step with dα+1 replacing d0. In the limit step, we assume that
for every α < β,(Mα, Sα, cα)has been constructed. We take (M0
β, S0
β) = Sα<β(Mα, Sα)
and assume (by the regularity of κ) that γis the least such that dγ/∈M0
β. We put cβ=dγ
and repeat the reasoning from the base step with cβreplacing d0.
The construction from Lemma 75 enables us to extend the result from Section 4.
Theorem 86. CT0+ Σ1(LT)-REF(UTB +T)is Π1(LT)conservative over CT0.
Proof. Fix (M, T )|=CT0. We shall find (M, T )⊆(N, T 0)|=CT0+Σ1(LT)-REF(UTB+T),
which suffices to end the proof. Firstly, turn (M, T )into a model (M, S)|=CS0in the
canonical way. Secondly, assume that there exists {(Mi, Si, ci)}i∈ωsuch that (M0, S0) =
(M, S), the rest of the chain is as in the thesis of Lemma 75 and for each i > 0,(Mi+1, Si+1, ci+1)
is strongly interpreted in its predecessor (Mi, Si, ci). Let Sati+1 denote the satisfaction re-
lation witnessing the interpretability of (Mi+1, Si+1, ci+1 )in (Mi, Si, ci). Put (M∞, S∞) =
Si∈ω(Mi, Si)and define T∞:= {φ∈M∞|S(φ, ε)}. By the proof of Theorem 74,
(M∞, S∞)|=CS0, hence (M∞, T∞)|=CT0, so it is sufficient to show that
(M∞, T∞)|= Σ1(LT)-REF(UTB +T).
Suppose that for some φ(x)∈Σ1(LT)and c∈M,(M∞, T∞)|=PrT
UTB(φ(c)). Let pbe the
witnessing proof and fix any i∈ωsuch that p<ci. Hence (Mi, Si)|=PrS
UTB(φ(c)) . We
reason in (Mi, Si). Since every next model in the chain is strongly interpretable in the
previous one, we have (in (Mi, Si))
(Mi+1, Si+1)|=Sati+1 ∀v(Mi+2, Si+2 , ci+2)|=Sati+2 CS(Σ∗
ci+2 )∧ci+2 > v.
Consider (in (Mi, Si)) the model (Mi+1, Si+2 Mi+1 ). This model is interpretable in (Mi+1, Si+1),
hence by Proposition 20 it is a full model (in the sense of (Mi, Si); i.e. it is strongly inter-
pretable in (Mi, Si)). Let Sat0
i+1 denote the respective satisfaction class. We shall show
that
(Mi+1, Si+2Mi+1 )|=φ(c).
This suffices to end the proof, since φ(c)is a Σ1(LT)formula and (Mi+1, Si+2 Mi+1 )⊆e
(M∞, T∞). Reasoning in (Mi, Si)we see that since Si+2Mi+1 is definable in (Mi+1, Si+1 ),
which satisfies full induction, also (Mi+1, Si+2 Mi+1 )satisfies full induction. Moreover,
since Si+1 ⊆Si+2Mi+1 we know that for every e(∈Mi)(Mi+1, Si+2Mi+1 )|=Sat0
i+1
CS(Σ∗
e). Hence,
(Mi+1, Si+2Mi+1 )|=Sat0
i+1 UTB.
Moreover, if ψis an assumption of proof pand an arithmetical sentence, then ψ∈Tω,
hence, by the choice of i,ψ∈Si. It follows that (Mi+1, Si+2Mi+1 )|=Sat0
i+1 ψ, since Sat0
i+1
coincides with Sati+1 on sentences from Mi. We conclude, still working in (Mi, Si), that
Sat0
i+1 makes every premise of ptrue in (Mi+1, Si+2Mi+1 ). So p’s conclusion, φ(c), must
39
be deemed true in (Mi+1, Si+2 Mi+1 )by Sat0
i+1. Since φ(c)is a standard formula with a
parameter, we can conclude that (Mi+1, Si+2Mi+1 )|=φ(c).
Now, we show how to justify the existence of the chain {(Mi, Si, ci)}i∈ω. Let BΣ1(LT)
denote the extension of CT0with Σ1collection scheme for the language with the truth
predicate. As shown in [20], BΣ1(LT)is Π2conservative over CT0. So we can assume that
the above model (M, T )is a countable recursively saturated (in the extended language)
model of CT−+BΣ1(LT). By the classical result of Wilkie–Paris [26] there exists a proper
end-extension (M0, T 0)|=CT0of (M, T ).15 Hence it is sufficient to start the construction
of the chain from (M0, T 0
a), where a∈M0\Mand then proceed as in the proof of Lemma
75.
6 Two Open Problems
We conclude our paper with two open problems:
Question 1 Does the statement of Theorem 85 remain true if we require for for every
α < β,Mβis strongly interpretable in Mα?
Question 2 Can we strengthen Theorem 86 by showing that Σ1(LT)-REF(UTB +T)is
in fact provable in CT0+EA?
Let us stress that, by the proof of Theorem 86, the positive answer to Question 1 im-
plies that the answer to Question 2 is positive as well.
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