Unformatted text preview: 1 THE INEVITABILITY OF LOGICAL STRENGTH:
strict reverse mathematics
by
Harvey M. Friedman*
Department of Mathematics
The Ohio State University
http://www.math.ohiostate.edu/%7Efriedman/
August 29, 2007
Abstract. An extreme kind of logic skeptic claims that "the
present formal systems used for the foundations of
mathematics are artificially strong, thereby causing
unnecessary headaches such as the Gödel incompleteness
phenomena". The skeptic continues by claiming that
"logician's systems always contain overly general
assertions, and/or assertions about overly general notions,
that are not used in any significant way in normal
mathematics. For example, induction for all statements, or
even all statements of certain restricted forms, is far too
general  mathematicians only use induction for natural
statements that actually arise. If logicians would tailor
their formal systems to conform to the naturalness of
normal mathematics, then various logical difficulties would
disappear, and the story of the foundations of mathematics
would look radically different than it does today. In
particular, it should be possible to give a convincing
model of actual mathematical practice that can be proved to
be free of contradiction using methods that lie within what
Hilbert had in mind in connection with his program”. Here
we present some specific results in the direction of
refuting this point of view, and introduce the Strict
Reverse Mathematics (SRM) program. TABLE OF CONTENTS
1. Many sorted free logic, completeness.
2. Interpretations, conservative extensions, synonymy.
3. PFA(N), EFA(N,exp), logical strength.
4. Five related systems of arithmetic with N.
5. Five related systems of arithmetic with Z.
6. Arithmetic on N and arithmetic on Z.
7. Seven strictly mathematical theories.
8. FSTZ.
9. FSTZ = FSTZD = FSTZS.
10. FSQZ. 2
11. Conservative extensions, interpretability, synonymy,
and logical strength.
12. RM and SRM. 1. Many sorted free logic, completeness.
We present a flexible form of many sorted free logic, which
is essentially the same as the one we found presented in
[Fe95], section 3. In [Fe95], Feferman credits this form of
many sorted free logic to [Pl68], [Fe75], [Fe79], and
[Be85], p. 9799.
We prefer to use many sorted free logic rather than
ordinary logic, because we are particularly interested in
the naturalness of our axioms, and want to avoid any
cumbersome or ad hoc features.
We will not allow empty domains. We allow undefined terms.
In fact, the proper use of undefined terms is the main
point of free logic.
A signature s (in many sorted free logic) consists of
i. A nonempty set SRT(s) called the sorts.
ii. A set CS(s) called constant symbols.
iii. A set RS(s) called relation symbols.
iv. A set FS(s) called function symbols.
v. We require that CS(s), RS(s), FS(s) be pairwise
disjoint, and not contain =.
vi. A function r with domain CS(s) » RS(s) » FS(s), and
with the following properties.
vii. For c Œ CS(s), r(c) Œ SRT(s). This is the sort of c.
viii. For R Œ CS(s), r(R) is a nonempty finite sequence
from SRT(s). This is the sort of R.
ix. For F Œ CS(s), r(F) is a finite sequence from SRT(s) of
length ≥ 2. This is the sort of F.
We make the simplifying assumption that equality is present
in each sort.
The s variables are of the form vna, n ≥ 1, where a Œ
SRT(s).
The s terms of s, and their sorts, are defined inductively
as follows.
i. The s variable vna is a s term of sort a. 3
ii. If c Œ CS(s) then c is a s term of sort r(c).
iii. If t1,...,tk are s terms of sorts a1,...,ak, k ≥ 1, and
F Œ FS(s), F has sort (a1,...,ak+1), then F(t1,...,tk) is a s
term of sort ak+1.
The atomic formulas of s are defined inductively as
follows.
i. If s is a s term then s↑,sØ are atomic formulas of s.
ii. If s,t are terms of the same sort, then s = t, s @ t are
atomic formulas of s.
iii. If s1,...,sk are terms of respective sorts a1,...,ak, k
≥ 1, and R Œ RS(s) of sort (a1,...,ak), then R(s1,...,sk) is
an atomic formula of s.
The s formulas are defined inductively as follows.
i. Every atomic formula of s is a s formula.
ii. If j,y are s formulas, then (ÿj),(j Ÿ y),(j ⁄ y),(j Æ
y),(j ´ y) are s formulas.
iii. If v is a s variable and j is a s formula, then
("v)(j),($v)(j) are s formulas.
The free logic aspect is associated with the use of ↑,Ø,=,@.
As will be clear from the semantics, ↑ indicates
“undefined”, Ø indicates “defined”, = indicates “defined
and equal”, @ indicates “either defined and equal, or both
undefined”. Also variables and constants always denote, and
a term is automatically undefined if any subterm is
undefined.
We now present the semantics for many sorted free logic.
A s structure M consists of the following.
i. A nonempty set DOM(a) associated with every sort a Œ
SRT(s).
ii. For each c Œ CS(s), an element c* Œ DOM(r(c)). This is
the interpretation of the constant symbol c.
iii. For each R Œ RS(s), a relation R* Õ DOM(a1) ¥ ... ¥
DOM(ak), where R has sort (a1,...,ak). This is the
interpretation of the relation symbol R.
iv. For each F Œ FS(s), a partial function F* from DOM(a1) ¥
... ¥ DOM(ak) into DOM(ak+1), where r(F) = (a1,...,ak+1). This
is the interpretation of the function symbol F. 4
A s assignment is a function g which assigns to each s
variable vna, a Œ S, an element g(vna) Œ DOM(a).
We inductively define val(M,t,g), where M is a s structure,
t is a s term, and g is a s assignment. Note that val(M,t,g)
may or may not be defined.
i. Let v be a s variable. val(M,v,g) = g(v).
ii. Let c Œ CS(s). val(M,c,g) = c*.
iii. Let F(s1,...,sk) be a s term. val(M,F(s1,...,sk),g) =
F*(val(M,s1,g),...,val(M,sk,g)) if defined; undefined
otherwise.
Thus in order for val(M,F(s1,...,sk,g) to be defined, we
require that val(M,s1,g),...,val(M,sk,g) be defined.
We inductively define sat(M,j,g), where M is a s structure,
j is a s formula, and g is a s assignment.
i. sat(M,s↑,g) if and only if val(M,s,g) is undefined.
ii. sat(M,sØ,g) if and only if val(M,s,g) is defined.
iii. sat(M,s = t,g) if and only if val(M,s,g) = val(M,t,g).
Here we require that both sides be defined.
iv. sat(M,s @ t,g) if and only if val(M,s,g) = val(M,t,g) or
val(M,s,g),val(M,t,g) are both undefined.
v. sat(M,R(s1,...,sk)) if and only if R*(val(M,s1,g),...,
val(M,sk,g)). Note that condition implies that each
val(M,si,g) is defined.
vi. sat(M,ÿj,g) if and only if not sat(M,j,g).
vii. sat(M,j Ÿ y,g) if and only if sat(M,j,g) and sat
(M,y,g).
viii. sat(M,j ⁄ y,g) if and only if sat(M,j,g) or
sat(M,y,g).
ix. sat(M,j Æ y,g) if and only if either not sat(M,j,g) or
sat(M,y,g).
x. sat(M,j ´ y,g) if and only if either (sat(M,j,g) and
sat(M,y,g)) or (not sat (M,j,g) and not sat (M,y,g)).
xi. sat(M,("vna)(j),g) if and only if for all x Œ DOM(a),
sat(M,j,g[vnax]). Here g[vnax] is the s assignment resulting
from changing the value of g at vna to x.
xii. sat(M,($vna)(j),g) if and only if there exists x Œ
DOM(a) such that sat(M,j,g[vnax]).
We say that a s structure M satisfies a formula j of s if
and only if sat(M,j,g), for all s assignments g. We say that
a s structure M satisfies a set T of s formulas if and only
if M satisfies every element of T. 5 We now give a complete set of axioms and rules of inference
for s. It is required that v is a s variable, c is a s
constant, s,t,r,s1,...,sk,t1,...,tk are s terms, j,y,r are s
formulas, v is not free in j, and t is substitutable for v
in r. It is also required that each line be a s formula.
i. All tautologies.
ii. vØ, cØ.
iii. t↑ ´ ÿtØ, where t is a s term.
iv. tØ ´ t = t.
v. s @ t ´ (s = t ⁄ (s↑ Ÿ t↑)).
vi. F(s1,...,sk)Ø Æ (s1Ø Ÿ ... Ÿ skØ).
vii. R(s1,...,sk) Æ (s1Ø Ÿ ... Ÿ skØ).
viii. s = t ´ t = s.
ix. (s = t Ÿ t = r) Æ s = r.
x. (s1 = t1 Ÿ ... Ÿ sk = tk) Æ F(s1,...,sk) @ F(t1,...,tk).
xi. (s1 = t1 Ÿ ... Ÿ sk = tk) Æ (R(s1,...,sk) Æ
R(t1,...,tk)).
xii. (tØ Ÿ ("v)(r)) Æ j[v/t].
xiii. (tØ Ÿ r[v/t]) Æ ($v)(r).
xiv. From j Æ y derive j Æ ("v)(y).
xv. From y Æ j derive ($v)(y) Æ j.
xvi. From j and j Æ y, derive y.
A theory is a pair T,s, where s is a signature, and T is a
set of s formulas.
Let T be a theory with signature s. A proof from T is a
nonempty finite sequence of s formulas, where each entry
lies in T, falls under ixiii, or follows from previous
entries by xiv, xv, or xvi.
A proof from T of j is a proof from T whose last entry is
j.
We have the following completeness theorem.
THEOREM 1.1. Let T be a theory in many sorted free logic
with signature s. Let j be a s formula. The following are
equivalent.
a. Every s structure satisfying T, also satisfies j.
b. There is a proof from T of j. 2. Interpretations, conservative extensions,
synonymy. 6
Let s,t be signatures in many sorted free logic, and S,T be
theories with signatures s,t, respectively. We want to
define what we mean by an interpretation of S in T.
We will first present a semantic formulation of this
notion. We then discuss syntactic formulations.
It is convenient to first define what we mean by an
interpretation p of s in t. This notion is used for both
semantic and syntactic formulations.
We then define what we mean by an interpretation of S in T.
The notion of interpretation of s in t is quite weak; e.g.,
there is no requirement that the interpretation of function
symbols be partial functions.
p is an interpretation of s in t if and only if p consists
of the following data.
i. For each sort a Œ SRT(s), p assigns a t defined set p(a)
of tuples of objects of various nonzero lengths and various
sorts in SRT(t). Only finitely many lengths are allowed, and
separate formulas are needed for each length. We also need
separate formulas for each sequence of sorts used. Also, p
assigns a t defined binary relation =(a) which is formally
set up to hold only of pairs drawn from p(a). Again,
separate formulas are needed for every pair of lengths. We
allow prospective parameters, so that a finite list of
distinguished free variables is given, for each a, which
are for the prospective parameters.
ii. Since we are allowing parameters, there is no need to
assign data for any c Œ CS(s) of sort a Œ SRT(s). However,
we will be interested in the notion of parameterless
interpretation. So it is best to have p assign data to c Œ
CS(s). p assigns a t defined set p(c), with distinguished
variables for prospective parameters.
iii. For each R Œ RS(s) of sort (a1,...,ak), p assigns a t
defined set p(R) of k tuples (of tuples of various lengths),
with distinguished variables for prospective parameters.
iv. For each F Œ FS(s) of sort (a1,...,ak+1), p assigns a t
defined set p(F) of k+1 tuples (of tuples of various
lengths), with distinguished variables for prospective
parameters. 7 Let S,T be theories in many sorted free logic, with
signatures s,t. We now define the notion of interpretation.
We say that p is an interpretation of S in T if and only if
i. p is an interpretation of s in t.
ii. Let M = T. There exists a choice of parameters from
the various domains of M such that p defines an actual model
of S, with the proviso that equality in each sort be
interpreted as the associated binary relation in p (often
called a weak model of S when giving, say, the Henkin
completeness proof for predicate calculus with equality).
We now give the natural equivalent syntactic notion of
interpretation of S in T, in case S is a finite theory in a
finite signature.
Let p be an interpretation of s in t. Since we are assuming
that S is finite, there are only finitely many
distinguished variables v1,...,vk used for prospective
parameters, in p. Let j be a s sentence. Then pj is the t
formula with free variables v1,...,vk that asserts that
i. p defines a s structure.
ii. j holds in this s structure.
The requirement is that if j is the universal closure of an
axiom of S, then ($v1,...,vk)(pj) is provable in T.
It is obvious by the completeness theorem that this is
equivalent to the original semantic definition, provided S
is finite.
More generally, the semantic notion has a natural syntactic
equivalent if
i. The relational type of S is finite; and
ii. We do not allow parameters.
NOTE: From now on we will only consider finite theories
S,T, and only interpretations without parameters.
Let S,T be theories in many sorted free logic. We say that
T is a conservative extension of S if and only if
i. The signature t of T extends the signature s of S.
ii. S,T prove the same s formulas. 8 We say that T is a definitional extension of S if and only
if
i. S,T have the same sorts.
ii. T is logically equivalent to an extension of S that is
obtained only by adding axioms which explicitly define the
new symbols in T by means of formulas in the signature of S
We say that p is a faithful interpretation of S in T if and
only if p is an interpretation of S in T, where for all
sentences j in the signature of S, S proves j if and only
if T proves pj.
We consider two important conditions on a pair of
interpretations p of S in T, and p’ of T in S.
The first condition, which we call weak synonymy, asserts
that for all models M of S, pp’M ª M, and for all models M
of T, p’pM ª M. Here ª is isomorphism.
THEOREM 2.1. Let p,p’ be a weak synonymy of S,T. Then
i. Let M,M* = S. Then M ª M* ´ p’M ª p’M*.
ii. Let M,M* = T. Then M ª M* ´ pM ª pM*.
iii. p is a faithful interpretation of S in T.
iv. p’ is a faithful interpretation of T in S.
Proof: Let p,p’ be a weak synonymy of S,T. For i, assume
M,M* = S, p’M ª p’M*. Then pp’M ª pp’M* ª M ª M*.
For iii, assume T proves pj. Let M = S. Then p’M = T, and
hence p’M = pj. Hence pp’M = j. Therefore M = j. Hence S
proves j.
Claims ii, iv are by symmetry. QED
The second condition is even stronger, and makes sense when
S,T have the same sorts. We say that p,p’ are a synonymy of
S,T if and only if
i. p,p’ are domain preserving interpretations of S in T and
of T in S, respectively.
ii. For M = S, pp’M = M.
iii. For M = T, p’pM = M. 9
We now show that this notion is the same as another notion
that is commonly used to mean synonymy. The first is also
model theoretic.
We say that S,T are (weakly) synonymous if and only if
there is a (weak) synonymy p,p’ of S,T.
THEOREM 2.2. Let S,T be two theories with the same sorts,
but where the symbols have been renamed, if necessary, so
that S,T have no symbols in common. There is a synonymy of
S,T if and only if S,T have a common definitional
extension.
Proof: Let p,p’ is a synonymy of S,T, where S,T have the
same sorts, and the symbols have been disjointified. Let R
consist of
i. the axioms of S.
ii. the axioms of T.
iii. axioms defining the symbols of T by formulas in the
signature of S, via p’.
iv. axioms defining the symbols of S by formulas in the
signature of T, via p.
We claim that R is a definitional extension of S. To see
this, it suffices to show that i,iii logically imply ii,iv.
We now argue in i,iii. To obtain ii, we need only obtain
the interpretations of the axioms of T by p’. But these
interpretations are provable in S.
We now have to obtain iv. A typical instance of iv would
take the form
1) ("x1,...,xk)(P(x1,...,xk) ´ pP(x1,...,xk)).
From iii, we obtain
2) pP(x1,...,xk) ´ p’pP(x1,...,xk).
Since p,p’ are a synonymy,
3) S proves p’pP(x1,...,xk) ´ P(x1,...,xk)
by the completeness theorem.
Hence by i,iii, we obtain 1). 10
By the symmetric argument, we also see that R is a
definitional extension of T.
Conversely, let R be a definitional extension of both S,T,
where the symbols of S,T have been disjointified. We have
two axiomatizations of R. The first corresponds to R as a
definitional extension of T, and the second corresponds to
R as a definitional extension of S. Let p be a definition of
the symbols of S by formulas in the signature of T, viewed
as a potential interpretation of S in T. Let p’ be a
definition of the symbols of T by formulas in the signature
of S, viewed as a potential interpretation of T in S.
Let M = T. Then (pM,M) satisfies the first axiomatization
of R. In particular, pM = S. Also (pM,M) satisfies the
second axiomatization of R. Therefore p’pM = M.
Let M = S. Then (M,p’M) satisfies the second axiomatization
of R. In particular, p’M = T. Also (M,p’M) satisfies the
second axiomatization of R. Therefore pp’M = M. QED
THEOREM 2.3. Let S,T be two theories with the same sorts.
There is a synonymy of S,T if and only if there are
interpretations p from S in T and p’ from T in S, such that
the following holds. For all formulas j in the signature of
S and y in the signature of T, S proves j ´ p’pj, and T
proves y ´ pp’y.
Proof: Let p,p’ be a synonymy of S,T. Let M = S, and M = j
[a]. Since p’ is domain preserving, p’M = T, and p’M =
pj[a]. Since p is domain preserving, pp’M = S, and pp’M =
p’pj[a]. Hence M = p’p[a]. Therefore M = j ´ (p’p)[a].
Since a is an arbitrary assignment for the signature of S,
and M is an arbitrary model of S, we have that j ´ p’pj is
provable in S.
Conversely, assume that S,T,p,p’ are as given. By applying
the conditions to atomic formulas j,y, we obtain that for
all M = S, pp’M = M, and for all M = T, p’pM = M. QED
If S,T obey the equivalent conditions in Theorems 2.2 and
2.3, then we say that S,T are synonymous. If there is a
weak synonymy of S,T, then we say that S,T are weakly
synonymous. 11
We will also use the following well known result. A theory
is said to be decidable if its set of consequences (in its
own signature) is recursive.
THEOREM 2.4. Suppose S is interpretable in T, where T is
consistent and decidable. Then S has a consistent decidable
extension with the same signature as S.
Proof: Let p be an interpretation of S in T, where T is
decidable. Let S’ consist of the sentences j in the
signature of S such that T proves pj. Clearly S’ extends S,
and S’ is a recursive set. Deductive closure and
consistency are obvious. QED 3. PFA(N), EFA(N,exp), logical strength.
We now present two very basic and well studied systems of
arithmetic. The most comprehensive current reference to
fragments of arithmetic is [HP98].
PFA(N), EFA(N,exp) are based on the set N of all
nonnegative integers. In the later sections, with the
exception of section 4, we focus on systems based on the
set Z of all integers.
PFA abbreviates “polynomial function arithmetic”, and EFA
abbreviates “exponential function arithmetic”.
PFA(N), EFA(N,exp) build on an earlier system due to R.M.
Robinson, called Q (see [Ro52]). We use the notation Q(N),
to emphasize that Q is based on N and not on Z.
The signature of Q(N) is L(N). L(N) is one sorted, with
0,S,+,⋅,<,=. The standard model for L(N) is the usual
N,0,S,+,⋅,<,=.
The signature of PFA(N) is also L(N). The signature of
EFA(N,exp) is L(N,exp), which is one sorted, with
0,S,+,⋅,exp,<,=, where exp is a binary function symbol. The
standard model for L(N,exp) is the usual N,0,S,=,⋅,<,=,
where we take exp(0,0) = 1.
The nonlogical axioms of Q(N) are as follows.
Q1. ÿSx = 0.
Q2. Sx = Sy Æ x = y.
Q3. (ÿx = 0) Æ ($y)(x = Sy). 12
Q4.
Q5.
Q6.
Q7.
Q8. x
x
x
x
x + 0 = x.
+ Sy = S(x + y).
⋅ 0 = 0.
⋅ Sy = (x ⋅ y) + x.
< y ´ ($z)(z + Sx = y). Note that in free logic, these axioms logically imply that
0Ø, SxØ, x+yØ, x⋅yØ.
The S0(N) (S0(N,exp)) formulas are the formulas of L(N)
(L(N,exp)) defined as follows.
i) every atomic formula of L(N) (L(N,exp)) is in S0(N)
(S0(N,exp));
ii) if j,y are in S0(N) (S0(N,exp)) then so are ÿj, j Ÿ y, y
⁄ y, j Æ y, j ´ y;
iii) if j is S0(N) (S0(N,exp)) and x is a variable not in
the term t of L(N) (L(N,exp)), then ($x)(x < t Ÿ j) and
("x)(x < t Æ j) are in S0(N) (S0(N,exp)).
In [HP98], the terms t in bounded quantification are
required to be variables. This is a minor difference.
PFA(N) is essentially the same as IS0 in [HP98], p. 29. The
nonlogical axioms of PFA(N) are as follows.
1. The axioms of Q(N).
2. (j[x/0] Ÿ ("x)(j Æ j[x/Sx])) Æ j, where j is S0(N).
EFA(N,exp) is essentially the same as IS0(exp) in [HP98], p.
37 (although there only base 2 exponentiation is used). The
nonlogical axioms of EFA(N,exp) are as follows.
1. The axioms of Q.
2. exp(x,0) = S0, exp(x,Sy) = exp(x,y)⋅x.
3. (j[x/0] Ÿ ("x)(j Æ j[x/Sx])) Æ j, where j is in
S0(N,exp).
We introduced the one sorted system EFA = EFA(N,exp) in
[Fr80]. It was also used in the exposition of our work on
Translatability and Relative Consistency, in [Sm82]. See
[HP98], p. 405, second paragraph, regarding some historical
points.
EFA(N,exp) represents the minimum level of formal
arithmetic where standard coding mechanisms in arithmetic
can be done naturally without worry. For example, we do not 13
have to worry about how to code sets of binary relations on
[0,n].
In fact, EFA(N,exp) appears to be quite strong from the
mathematical viewpoint. We conjecture that EFA(N,exp) is
sufficient to prove any normal theorem of number theory
that is adequately formalizable in its language. We can be
liberal about “formalizable” here, using the various
natural codings available in EFA(N,exp).
For example, we conjecture that Fermat’s Last Theorem is
provable in EFA(N,exp). This has never been established.
This conjecture captured the imagination of Jeremy Avigad
who wrote extensively about it, and related issues, in
[Av03].
Accordingly, we now make the following definition.
T has logical strength if and only if
EFA(N,exp) is interpretable in T.
The main point of this paper is the presentation of
strictly mathematical theories with logical strength. See
section 7, and Corollary 11.11. 4. Five related systems of arithmetic with N.
We now introduce six systems of arithmetic on N that are
closely related to PFA(N) and EFA(N,exp).
LEMMA 4.1. There is a S0(N) formula Exp(x,y,z) with only the
distinct free variables shown such that the following is
provable in PFA(N).
i) Exp(x,0,z) ´ z = S0;
ii) Exp(x,Sy,z) ´ ($v)(Exp(x,y,v) Ÿ z = v⋅x);
iii) (Exp(x,y,z) Ÿ Exp(x,y,w)) Æ z = w.
Proof: See [HP98], p. 299. QED
LEMMA 4.2. Suppose Exp(x,y,z) and Exp’(x,y,z) satisfy the
condition in Lemma 4.1. Then PFA(N) proves their
equivalence.
Proof: Let Exp(x,y,z), Exp’(x,y,z) obey the conditions in
Lemma 4.1. Let n,m,r be such that Exp(n,m,r) Ÿ
ÿExp’(n,m,r). Fix n,r, and let m be least such that ($s £
r)(Exp(n,m,s) Ÿ ÿExp’(n,m,s)). Let Exp(n,m,s), 14
ÿExp’(n,m,s), s £ r. Clearly m > 0. Let Exp(n,m1,t), s =
t•n. Then ÿExp’(n,m1,t). Also n = 0 ⁄ t £ s £ r. The
latter is impossible by the choice of m. Hence n = 0, s =
0. Since m > 0, Exp’(n,m,s). This is a contradiction. QED
The sentence EXP(N) is taken to be ("x,y)($z)(Exp(x,y,z)),
where Exp is any formula satisfying the conditions of Lemma
4.1. By Lemma 4.2, this defines EXP(N) up to provable
equivalence in PFA(N).
Let CM(N) = “common multiples” be the following sentence in
L(N).
CM(N). For all n > 0, the integers 1,2,...,n have a
positive common multiple.
The five relevant fragments of arithmetic considered here
are as follows.
Q(N), PFA(N), PFA(N) + EXP(N), PFA(N) + CM(N), EFA(N,exp).
Note that the signature of all of these systems is L(N),
except for the last, which has signature L(N,exp).
The most basic relationships between these theories are
well known, and summarized in the following two theorems.
THEOREM 4.3. Q(N) Õ PFA(N) Õ PFA(N) + CM(N) = PFA(N) +
EXP(N) Õ EFA(N,exp). These Õ are all proper.
Proof: Assume PFA(N) + EXP(N). Write 2x, x ≥ 0, according to
EXP(N). Fix n ≥ 1. We prove by induction on 1 £ m £ n that
1,...,n have a positive common multiple £ 2^m2. This is
obvious for m = 1. Let 1 £ m < n, and x be a positive common
multiple of 1,...,m, x £ 2^m2. Then x(m+1) £ (2^m2)(2m) £
2^(m+1)2. This establishes CM(N).
Now assume PFA(N) + CM(N). Fix n,m ≥ 2. Let x be a positive
common multiple of 1,...,nm. We can assume that x is the
least positive common multiple of 1,...,nm. Show that every
prime factor of x is £ nm. Show that x+1,2x+1,...,(nm)x+1
are pairwise relatively prime. Let y be a positive common
multiple of 1,...,nx+1. Code ntuples as Gödel did, £ y, in
order to develop the geometric progression 1,n,n2,...,nm.
This establishes EXP(N). 15
To see that PFA(N) + EXP(N) Õ EFA(N,exp), write Exp(n,m,r)
for the internal exponentiation relation (in L(N)). Argue
that Exp(n,m,r) ´ exp(n,m) = r using S0(exp) induction,
exactly as in the proof of Lemma 4.2.
To see that Q(N) does not prove PFA(N), consider all of the
polynomials in one variable x with integer coefficients,
which have a positive leading coefficient, or is 0. These
form a model of Q(N) under the usual 0,S,+,⋅,=, with £
defined according to axiom Q8 of Q(N). This model of Q(N)
does not satisfy, for example, ($y)(x = 2y ⁄ x = 2y+1).
To see that PFA(N) does not prove EXP(N), let M be a
nonstandard model of PFA(N). Let x be a positive
nonstandard integer in M, and let M’ be the restriction of
M to the integers of M whose magnitude is at most xn, for
some standard n ≥ 1. Them M’ is a model of PFA(N). It is
easily verified that 2x does not exist in M. QED
THEOREM 4.4. EFA(N,exp) is a definitional extension of
PFA(N) + EXP(N). PFA(N) is interpretable in Q(N). PFA(N) +
EXP(N) is not interpretable in PFA(N).
Proof: For the first claim, note that by the proof of
Theorem 4.3 (that PFA(N) + EXP(N) Õ EFA(N,exp)), EFA(N,exp)
proves that exp(n,m) = r ´ Exp(n,m,r).
It remains to show that every axiom of EFA(N,exp) is
provable in PFA(N) + EXP(N) + ("n,m,r)(exp(n,m) = r ´
Exp(n,m,r)). This is clear by inspection.
The second claim is credited to Wilkie, in the sharp form
on p. 367 of [HP98].
For the third claim, see [Wi86], and [HP98], p. 391. QED 5. Five related systems of arithmetic with Z.
We now introduce six systems of arithmetic on Z that are
closely related to the six systems of section 4. These are
parallel to those systems introduced in section 4, and move
us closer to the strictly mathematical theories of section
7.
We first introduce LOID(Z). LOID abbreviates “linearly
ordered integral domain”. According to Theorem 5.3 below,
LOID is an extremely robust strictly mathematical theory. 16 The signature of LOID(Z) is L(Z). L(Z) is one sorted, with
0,1,+,,⋅,<,=, where – is unary. The standard model for L(Z)
is the usual Z,0,1,+,,⋅,<,=.
The nonlogical axioms of LOID(Z) are
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
o. x+0 = x.
x+y = y+x.
x+(y+z) = (x+y)+z.
x+(x) = 0.
x⋅1 = x.
x⋅y = y⋅x.
x⋅(y⋅z) = (x⋅y)⋅z.
x⋅(y+z) = (x⋅y)+(x⋅z).
x⋅y = 0 Æ (x = 0 ⁄ y = 0).
0 < 1.
ÿx < x.
(x < y Ÿ y < z) Æ x < z.
x < y ⁄ x = y ⁄ y < x.
(0 < x Ÿ 0 < y) Æ (0 < x+y Ÿ 0 < x⋅y).
x < y ´ y < x. Note that in free logic, these axioms imply 0Ø, 1Ø, x+yØ,
xØ, x⋅yØ.
How do we know that we have included all appropriate axioms
in LOID(Z)? We first present some basic development of
LOID(Z). Define x > y ´ x < y, x ≠ 0 ´ ÿx = 0.
LEMMA 5.1. The following are provable in LOID(Z).
i. x = x.
ii. (x+y) = x + y.
iii. x+x = 0 ´ x = 0.
iii. x⋅0 = 0.
iv. x = (1)⋅x.
v. (1)⋅(1) = 1.
vi. x,y < 0 Æ x⋅y > 0.
vii. x < 0 Ÿ y > 0 Æ x⋅y < 0.
Proof:
i. By d, x + x = 0. Hence x = x + (x + x) = (x + x)
+ x = 0 + x = x.
ii. By d, x+y + (x+y) = 0. Hence x + y + x + y + (x+y)
= x + y = (x+y). 17
iii. Use m. If x = 0 then we are done. Assume 0 < x. Then
by n, 0 < x+x, and so x+x ≠ 0, x ≠ 0. Assume x < 0. By o, 0
< x, 0 < x + x = (x+x). Hence (x+x) ≠ 0, x ≠ 0.
iv. (1)⋅x + 1⋅x = 0 = (1)⋅x + x. Hence 0 + x = ((1)⋅x) +
x) + x = (1)⋅x.
v. (1)⋅(1) = 1 = 1.
vi. Assume x,y < 0. By axioms n,o, 0 < x,y, (x)⋅(y) > 0.
Now (x)⋅(y) = (1)⋅x⋅(1)⋅y = (1)⋅(1)⋅x⋅y = 1⋅x⋅y = x⋅y.
vii. Assume x < 0, y > 0. By axioms n,o, 0 < x,y, 0 < (x)⋅y. Since (x)⋅y = (1)⋅x⋅y = (x⋅y), we have 0 < (x⋅y), x⋅y
< 0. QED
The official definition of an ordered field is given in,
say, [Ja85], p. 307:
An ordered field (F,P) is a field F together with a subset
P (the set of positive elements) of F such that
i. 0 œ P.
ii. a Œ F Æ (a Œ P ⁄ a = 0 ⁄ –a Œ P).
iii. a,b Œ P Æ (a+b Œ P Ÿ a⋅b Œ P).
LEMMA 5.2. Let M = (D,0,1,+,,⋅,<) be a model of LOID(Z).
There is an ordered field (F,P) and an isomorphism
j:(D,0,1,+,,⋅) Æ F such that for all x Œ D, jx Œ P ´ x >
0.
Proof: Let M be as given. By axioms ai, M is an integral
domain. Hence the fraction field construction results in a
field F and a canonical isomorphism j:(D,0,1,+,,⋅) Æ F.
Recall that F consists of the equivalence classes of
ordered pairs (x,y), where x,y Œ D, y ≠ 0, under the
equivalence relation (x,y) ª (z,w) ´ xw = yz. Define jx =
[(x,1)]. Obviously j is an isomorphism from M into F.
Define P = {[(x,y)]: x,y > 0 ⁄ x,y < 0}. I.e., P =
{[(x,y)]: x,y have the same nonzero sign}. It is obvious
that 0F = [(0,1)] œ P.
We claim independence of representatives, in the sense that
for all x,y,z,w, if [x,y] ª [z,w], then
*) x,y have the same nonzero sign ´
z,w have the same nonzero sign.
To see this, assume [x,y] ª [z,w]. Then 18
x⋅w = y⋅z, y,w ≠ 0.
case 1. x ≠ 0. Then x⋅w ≠ 0, y⋅z ≠ 0, z ≠ 0. By inspection,
using Lemma 5.1, vi),vii), we see that *) holds.
case 2. x = 0. Then y⋅z = 0, z = 0. By inspection using
Lemma 5.1, vi),vii), we see that *) holds.
Now let [(x,y)] Œ F. If x,y have the same nonzero sign then
[(x,y)] Œ P. If x,y have opposite nonzero signs, then –x,y
have the same nonzero sign, and hence –[(x,y)} = [(x,y)] Œ
P. Finally, if x = 0 then [(x,y)] = 0F.
Now let [(x,y)], [(z,w)] Œ P. By the independence of
representatives (claim above), we can assume that x,y,z,w >
0. Note that
[(x,y)] + [(z,w)] = [(x⋅w + y⋅z,y⋅w)] Œ P.
[(x,y)] ⋅ [(z,w)] = [(x⋅z,y⋅w)] Œ P.
This establishes that (F,P) is an ordered field.
Now let x Œ D. If jx = [x,1] Œ P then obviously x > 0. If
jx = [x,1] œ P then ÿx > 0, using the independence of
representatives. QED
THEOREM 5.3. A purely universal sentence in L(Z) is true in
the ordered field of real numbers if and only if it is
provable in LOID(Z). LOID(Z) can be axiomatized as the set
of all quantifier free formulas in L(Z) which are
universally true in the ordered field of real numbers.
Proof: It suffices to show that every purely existential
sentence in L(Z) that is true in some model of LOID is true
in the ordered field of real numbers.
Let M be a model of LOID satisfying the purely existential
sentence j in L[Z]. By Lemma 5.2, let (F,P) be an ordered
field extending M, with an isomorphism j:(D,0,1,+,,⋅) Æ F,
and for all x Œ D, jx Œ P ´ x > 0.
We define < on F by
x < y ´ yx Œ P.
We claim that 19
j:(D,0,1,+,,⋅,<) Æ (F,<)
is an isomorphism. To see this, let x < y in M. Then yx >
0, and so j(yx) Œ P. Hence j(y)–j(x) Œ P, and so x < y in
(F,<). Hence j is preserved, and so j holds in (F,<).
Now every ordered field (F,<), where < is defined as above
from P, extends to an ordered real closed field, whose <
agrees with the < of (F,<). Hence j holds in some ordered
real closed field. Since j is a first order sentence, j
holds in all ordered real closed fields. Hence j holds in
the ordered field of real numbers.
The final claim follows from the previous claim using the
observation that the axioms of LOID(Z) are quantifier free
formulas in L(Z) which are universally true in the ordered
field of real numbers. QED
The S0(Z) (S0(Z,exp)) formulas are the formulas of L(Z)
(L(Z,exp)) defined as follows.
i) every atomic formula of L(Z) (L(Z,exp)) is in S0(Z)
(S0(Z,exp));
ii) if j,y are in S0(Z) (S0(Z,exp)) then so are ÿj, j Ÿ y, y
⁄ y, j Æ y, j ´ y;
iii) if j is S0(Z) (S0(Z,exp)) and x is a variable not in
the term t of L(Z) (L(Z,exp)), then ($x)(t < x < t Ÿ j) and
("x)(t < x < t Æ j) are in S0(Z) (S0(Z,exp)).
Henceforth, we will use x £ y as an abbreviation for x < y ⁄
x = y, and x ≥ y as an abbreviation for y < x ⁄ y = x.
The signature of PFA(Z) is L(Z). The nonlogical axioms of
PFA(Z) are
1. LOID(Z).
2. (j[x/0] Ÿ ("x)(j Æ j[x/Sx])) Æ (x ≥ 0 Æ j), where j is
in S0(Z).
Note that PFA(Z) proves the axiom of discreteness: there is
nothing in (0,1). To see this, let j = x ≥ 0 Ÿ (x < 1 Æ x £
0). Use 2 for j, to obtain (x ≥ 0 Ÿ x < 1) Æ x £ 0. Now
suppose x < 1. If x > 0 then x £ 0. Hence x £ 0.
L(Z,exp) is one sorted, with 0,1,+,,⋅,exp,<,=, where – is
unary and exp is binary. The standard model for L(Z,exp) is 20
the usual Z,0,1,+,,⋅,exp,<,=, where exp(x,y) is the usual
xy, which is defined if and only if y ≥ 0, and where x0 = 1.
The signature of EFA(Z,exp) is L(Z,exp). The nonlogical
axioms of EFA(Z,exp) are
1.
2.
3.
4.
is LOID(Z).
exp(x,0) = 1.
y ≥ 0 Æ (exp(x,y+1) = exp(x,y)⋅x Ÿ exp(x,y1)↑).
(j[x/0] Ÿ ("x)(j Æ j[x/x+1])) Æ (x ≥ 0 Æ j), where j
in S0(Z,exp). LEMMA 5.4. There is a S0(Z) formula Exp(x,y,z) with only the
distinct free variables shown such that the following is
provable in PFA(Z).
i) Exp(x,0,z) ´ z = S0;
ii) y ≥ 0 Æ (Exp(x,y+1,z) ´ ($v)(Exp(x,y,v) Ÿ z = v⋅x));
iii) (Exp(x,y,z) Ÿ Exp(x,y,w)) Æ z = w;
iv) Exp(x,y,z) Æ y ≥ 0.
Proof: This is a straightforward adaptation of Lemma 4.1 to
PFA(Z). QED
LEMMA 5.5. Suppose Exp(x,y,z) and Exp’(x,y,z) satisfies the
condition in Lemma 5.4. Then PFA(Z) proves their
equivalence.
Proof: This is a straightforward adaptation of Lemma 4.2 to
PFA(Z). QED
The sentence EXP(Z) is taken to be ("x,y)($z)(Exp(x,y,z)),
where Exp is any formula satisfying the conditions of Lemma
5.4. By Lemma 5.5, this defines EXP(Z) up to provable
equivalence in PFA(Z).
Let CM(Z) = “common multiples” be the following sentence in
L(Z).
CM(Z). For all n > 0, the integers 1,2,...,n have a
positive common multiple.
Note that CM(Z) is formally the same as CM(N), but it is
still convenient to use the notation CM(Z), CM(N).
The five relevant fragments of arithmetic considered here
are as follows. 21
LOID(Z), PFA(Z), PFA(Z) + EXP(Z), PFA(Z) + CM(Z),
EFA(Z,exp).
The most basic relationships between these theories are
summarized in the following three theorems.
THEOREM 5.6. LOID(Z) Õ PFA(Z) Õ PFA(Z) + CM(Z) = PFA(Z) +
EXP(Z) Õ EFA(Z,exp). These Õ are all proper.
THEROEM 5.7. EFA(Z,exp) is a definitional extension of
PFA(Z) + EXP(Z). PFA(Z) + EXP(Z) is not interpretable in
PFA(Z). PFA(Z) is not interpretable in LOID(Z).
Proof: Theorem 5.6, 5.7, with the exception of the final
claim of Theorem 5.7, can be proved by an adaptation of the
corresponding proofs of Theorems 4.3, 4.4.
For the final claim of Theorem 5.7, the essence of the
matter is that Q(N) is not interpretable in the theory of
ordered real closed fields, ORCF. If this were not the
case, then, since the theory of ordered real closed fields
is decidable, by Theorem 2.4 we have Q(N) has a consistent
decidable extension with signature L(N). This contradicts
the well known essential undecidability of Q(N); see
[Ro52]. Obviously PFA(Z) is not interpretable in LOID(Z),
since Q(N) is trivially interpretable in PFA(Z). QED 6. Arithmetic on N and arithmetic on Z.
We establish some relationships between the five systems of
section 4,
Q(N), PFA(N), PFA(N) + EXP(N), PFA(N) + CM(N), EFA(N,exp).
and the five systems of section 5,
LOID(Z), PFA(Z), PFA(Z) + EXP(Z), PFA(Z) + CM(Z),
EFA(Z,exp).
In sections 4,5, we have discussed the relationships
between the theories in each of the two groups separately.
We can interpret PFA(N) in PFA(Z), by taking the domain to
be the nonnegative elements, and defining 0,S,+,⋅,<,= in the
obvious way. We call this interpretation p(N,Z). 22
We can interpret PFA(Z) in PFA(N) by taking the domain to
be the pairs (n,0), n > 0, and (n,1). Here (n,0) represents
the negative integer –n, and (n,1) represents the
nonnegative integer n. We define 0,1,+,,⋅,<,= in the
obvious way. We call this interpretation p(Z,N).
THEOREM 6.1. p(N,Z), p(Z,N) is a weak synonymy of PFA(N),
PFA(Z), and is also a weak synonymy of PFA(N) + EXP(N),
PFA(Z) + EXP(Z).
Proof: It is obvious that p(N,Z) is an interpretation of
PFA(N) in PFA(Z), and p(Z,N) is an interpretation of PFA(Z)
in PFA(N). It is also obvious that p(N,Z) is an
interpretation of PFA(N) + EXP(N) in PFA(Z) + EXP(Z), and
p(Z,N) is an interpretation of PFA(Z) + EXP(Z) in PFA(N +
EXP(N).
For weak synonymy, let M = PFA(N), and within M, form the
(n,0),(n,1) interpretation of PFA(Z), according to p’,
obtaining p’M = PFA(Z). Within p’M, form the nonnegative
element interpretation of PFA(N), according to p, obtaining
pp’M. The nonnegative element interpretation just uses the
(n,1). Clearly we have an isomorphism from pp’M onto M by
sending each (n,1) to n.
Let M = PFA(Z), and within M, form the nonnegative element
interpretation of PFA(N), according to p, obtaining pM =
PFA(N). Within pM, form the (n,0),(n,1) interpretation of
PFA(Z), according to p’, obtaining p’pM. Clearly we have an
isomorphism from p’pM onto M by sending each negative n to
(n,0), and each nonnegative n to (n,1). QED
We extend p(N,Z) and p(Z,N) in the obvious way to
p(N,Z;exp), p(Z,N;exp).
THEOREM 6.2. p(N,Z,exp), p(Z,N,exp) is a weak synonymy of
EFA(N,exp), EFA(Z,exp).
Proof: Argue as for Theorem 6.1. QED
Note that p(N,Z) and p(Z,N) are not domain preserving, and
so we cannot use them to establish synonymy. We give new
interpretations for this purpose.
We can interpret PFA(N) in PFA(Z), by taking the N to be
0,1,1,2,2,... 23 with the obvious corresponding definition of 0,S,+,⋅,<,=.
Specifically, we first define, in PFA(Z), the function f:Z
Æ Z by f(x) = the position in the above sequence = 0 if x =
0; 2x1 if x > 0; 2x if x < 0. Then we define 0’,S’,+’, ⋅’,
<’,=, uniquely, in such a way that f is an isomorphism from
Z,0’,S’,+’,⋅’,<’,= onto {x: x ≥ 0},0,+1,+,⋅,<,=. Call this
p’(N,Z).
We can interpret PFA(Z) in PFA(N), by taking the Z to be
...6,4,2,0,1,3,5,...
with the obvious corresponding definition of 0,1,+,,⋅,<,=.
Specifically, we first define, in PFA(N), the function g:N
Æ N ¥ {0,1} by g(2n+1) = (1,n+1), g(2n+2) = (0,n+1), g(0) =
0. Then we define 0’,1’,+’,’,⋅’,<’,=, uniquely, in such a
way that g is an isomorphism from N,0’,1’,+’,’,⋅’,<,= onto
{(x,0): x > 0} » {(x,1): x ≥ 0} with its usual 0*,1*,+*,*,⋅*,<*,=, that makes it look like the arithmetic of Z. Call
this p’(Z,N).
LEMMA 6.3. p’(N,Z), p’(Z,N) is a synonymy of PFA(N), PFA(Z),
and also of PFA(N) + EXP(N), PFA(Z) + EXP(Z).
Proof: It is obvious that p’(N,Z) is an interpretation of
PFA(N) in PFA(Z), and p’(Z,N) is an interpretation of PFA(Z)
in PFA(N). This is also obvious with EXP(N) and EXP(Z).
For synonymy, let M = PFA(N), and within M, form the
...6,4,2,0,1,3,5,... interpretation of PFA(Z), according to
p’, obtaining p’M = PFA(Z). Within p’M, form the 0,1,1,2,2,... interpretation of PFA(N), according to p, obtaining
pp’M. Note that in p’M, 0,1,1,2,2,... is the 0,1,2,3,...
of M. Hence pp’M = M.
Let M = PFA(Z), and within M, form the 0,1,1,2,2,...
interpretation of PFA(N), according to p, obtaining pM =
PFA(N). Within pM, form the ...6,4,2,0,1,3,5,...
interpretation of PFA(Z), according to p’, obtaining p’pM.
Note that in pM, ...6,4,2,0,1,3,5,... is the
...,3,2,1,0,1,2,3,... of M. Hence p’pM = M. QED
We extend p(N,Z) and p(Z,N) in the obvious way to
p(N,Z;exp), p(Z,N;exp). 24
LEMMA 6.4. p’(N,Z,exp), p’(Z,N,exp) is a synonymy of
EFA(N,exp), EFA(Z,exp).
Proof: Argue as for Lemma 6.3. QED
We now construct a certain model M of Q(N). The domain will
consist of certain polynomials in variables xa, a < w1, with
integer coefficients. We will not be using the ordering of
variables.
Let P be such a polynomial. The maximal monomials of P are
the monomials of P that are maximal with respect to the
divides relation. Note that if P is not the trivial
polynomial, 0, then P has at least one maximal monomial.
We take dom(M) to be these polynomials which are either 0,
or whose maximal monomials all have positive coefficients.
For M, we use the ordinary 0,S,+,⋅. We define < as in axiom
Q8 of Q(N).
LEMMA 6.5. M is a model of Q(N).
Proof: We first need to verify that dom(M) is closed under
+,⋅. Let P,Q Œ dom(M). We can assume that P,Q are not the 0
polynomial. Let a be a maximal monomial of P+Q. If a occurs
in P and not in Q, or in Q but not in P, then it retains
its coefficient, which must be positive. If a occurs in P
and Q, then its coefficient in P+Q is positive. Hence P+Q Œ
dom(M).
It is trickier to establish that PQ Œ dom(M). Let a1,...,an
be the monomials of P, and b1,...,bm be the monomials of Q.
Since we are assuming that neither P nor Q is 0, we have
n,m ≥ 1. Let S be the set of all aibj that are maximal among
all of the aibj (even if the coefficient of aibj in PQ is £
0). Suppose aibj Œ S. Then obviously ai is maximal in P and
bj is maximal in Q. Now any apbq = aibj, where ap is a
monomial in P and bq is a monomial in Q, must have that ap
is maximal in P and bq is maximal in Q. Hence the
coefficient of aibj in PQ must be positive (since it is the
sum of the coefficients contributed by each of these apbq =
aibj). Therefore aibj is a monomial of PQ with positive
coefficient.
We now claim that every maximal monomial aibj of PQ lies in
S. To see this, let aibj be a maximal monomial of PQ, where 25
aibj œ S.
previous
positive
monomial Let apbq Œ S be a proper multiple of aibj. By the
paragraph, apbq is a monomial of PQ (in fact, with
coefficient), contradicting that aibj is a maximal
of PQ. We have now shown that every maximal monomial aibj of PQ has
a positive coefficient. Thus PQ Œ dom(M).
The verification of Q2, Q4  Q7 is by the ring laws for
polynomials. Q8 is by definition. Q1 follows from the fact
that 1 œ dom(M). For Q3, let P Œ dom(M), P not 0. If P is
nonconstant then P1 Œ dom(M). If P is constant then P ≥ 1,
and so P1 Œ dom(M). QED
LEMMA 6.6. Q(N) and PFA(N) are not weakly synonymous.
Proof: We use the model M of Lemma 6.5. Let M’ be a model
of PFA(N) defined in M without parameters. We show that M’
is countable.
Recall that in interpretations, we allow the domain to
consist of tuples of varying lengths. We also allow the
equality relation to be interpreted by an equivalence
relation. This equivalence relation must be definable in M
without parameters.
We call two polynomials isomorphic if and only if they are
identical up to a permutation of variables. We call two
finite sequences of polynomials isomorphic if and only if
they are coordinatewise isomorphic via a single
permutation. We call the equivalence classes under this
equivalence relation on the finite sequences of
polynomials, shapes.
Note that by the symmetry of M, for any two tuples of
polynomials that are isomorphic, one lies in dom(M’) if and
only if the other lies in dom(M’).
case 1. Any two tuples of polynomials in dom(M’) that are
isomorphic, and lie in dom(M’), are interpreted to be equal
in M’. Since the number of shapes is countable, we see that
dom(M’) is countable.
case 2. Let (P1,...,Pn), (Q1,...,Qn) be isomorphic elements
of dom(M’), which are not satisfied to be equal in M’. Let
a be an automorphism of the variables {xa: a < w1}, that
interchanges (P1,...,Pn) and (Q1,...,Qn). Then a extends 26
uniquely to an automorphism a* of M of finite order, which
in turn induces an automorphism b of M’ of finite order.
Since b interchanges the distinct elements [(P1,...,Pn)] and
[(Q1,...,Qn)] of dom(M’), we see that b has finite order.
But no model of PFA(N) can have an automorphism of finite
order because of the definable linear ordering <, with
parameters. So this case is impossible.
Since M’ is countable, and M is uncountable, it is clear
that we cannot define, in M’, an isomorphic copy of M.
Hence Q(N), PFA(N) are not weakly synonymous. QED
We summarize the synonymy and mutual interpretability
results.
THEOREM 6.7. PFA(N), PFA(Z) are synonymous. PFA(N) +
EXP(N), EFA(N,exp), PFA(Z) + EXP(Z), EFA(Z,exp) are
synonymous. There are no other synonymy, or even weak
synonymy, relations between the 10 systems. Q(N), PFA(N),
PFA(Z) are mutually interpretable. PFA(N) + EXP(N),
EFA(N,exp), PFA(Z) + EXP(Z), EFA(Z,exp) are mutually
interpretable. There are no other mutual interpretability
relations between the 10 systems. LOID(Z) is interpretable
in Q(N), but not vice versa.
Proof: The first claim is by Lemma 6.3. For the second
claim, PFA(N) + EXP(N), PFA(Z) + EXP(Z) are synonymous by
Lemma 6.3. PFA(N) + EXP(N), EFA(N,exp) are synonymous by
Theorems 4.4 and 2.2. PFA(Z) + EXP(Z), EFA(Z,exp) are
synonymous by Theorems 5.7 and 2.2. EFA(N,exp), EFA(Z,exp)
are synonymous by Lemma 6.4.
For the third claim, PFA(N) + EXP(N) is not interpretable
in PFA(N) by Theorem 4.4. That Q(N), PFA(N), PFA(Z) are not
interpretable in LOID(Z) comes from (the proof of) Theorem
5.7. That Q(N), PFA(N) are not weakly synonymous comes from
Lemma 6.6.
For the fourth claim, use the first claim together with the
interpretability of PFA(N) in Q(N), from Theorem 4.4.
The fifth claim follows from the second claim.
The sixth claim follows from PFA(N) + EXP(N) not
interpretable in PFA(N), and Q(N) not interpretable in
LOID(Z). The former is by Theorem 4.4, and the latter is by
the proof of Theorem 5.7. 27 The seventh claim is by the proof of Theorem 5.7, and the
interpretability of ORCF in Q(N) in [FF02]. 7. Seven strictly mathematical theories.
Among the twelve theories considered in section 6, only
Q(N) and LOID(Z) are strictly mathematical. The rest rely
on induction stated for all bounded formulas. However, Q(N)
and LOID(Z) do not have logical strength, in the sense used
in this paper (see the end of section 3).
We now present six strictly mathematical theories. We will
extend the one sorted signatures from sections 35,
L(N), L(Z), L(N,exp), L(Z,exp),
with the new many sorted signatures
L(Z,fst), L(Z,fsq), L(Z,fst,fsq), L(Z,exp,fst),
L(Z,bexp,fst), L(Z,exp,fsq).
Here fst abbreviates “finite sets of integers”, and fsq
abbreviates “finite sequences of integers”. Also bexp
abbreviates “binary exponentiation”.
L(Z,exp,fst) is two sorted. We use Z for sort 1, and fst
for sort 2. Here fst abbreviates “finite sets of integers”.
We use 0,1,+,,⋅,exp,<,= on the Z sort. We use Œ between
sort Z and sort fst.
L(Z,exp,fsq) is two sorted. We use Z for sort 1 and fsq for
sort 2. Here fsq abbreviates “finite sequences of integers
”. We use 0,1,+,,⋅,exp,<,= on the Z sort. We use the unary
function symbol lth from sort fsq into sort Z. We use the
binary function symbol val from sort fsq cross sort Z, into
sort Z.
The standard model for L(Z,exp,fst) has first sort Z, with
0,1,+,,⋅,<,= as usual, and exp(n,m) = r if and only if nm =
r Ÿ m ≥ 0, where n0 = 1. Thus exp(n,m) is defined if and
only if m ≥ 0. The second sort, fst, consists of the finite
subsets of Z, where Œ is as usual.
The standard model for L(Z,exp,fsq) has domain Z, wi†h
0,1,+,,⋅,<,= as usual, and exp(n,m) = r if and only if nm =
r Ÿ m ≥ 0, where n0 = 1. Thus exp(n,m) is defined if and 28
only if m ≥ 0. The second sort, fsq, consists of the finite
sequences from Z, where lth is the length function, which
takes values in the nonnegative elements of Z. Also
val(x,n) is the nth term of x, counting from 1, and so is
defined if and only if 1 £ n £ lth(x).
We also work with the elimination of exp in L(Z,exp,fst),
L(Z,exp,fsq).
The signature of FSTZ is L(Z,fst). The nonlogical axioms of
FSTZ are stated informally as follows.
1.
2.
3.
4.
5.
6. Linearly ordered integral domain axioms.
Finite interval. [x,y] exists.
Boolean difference. A\B = {x Œ A: x œ B} exists.
Set addition. A+B = {x: x+y: x Œ A Ÿ x Œ B} exists.
Set multiplication. A⋅B = {x: x⋅y: x Œ A Ÿ x Œ B} exists.
Least element. Every nonempty set has a least element. The signature of FSQZ is L(Z,fsq). The nonlogical axioms of
FSQZ are stated informally as follows.
1. Linearly ordered integral domain axioms.
2. lth(a) ≥ 0.
3. val(a,n)Ø ´ 1 £ n £ lth(a).
4. The finite sequence (0,...,n) exists.
5. lth(a) = lth(b) Æ a,a+b,a⋅b exist.
6. The concatenation of a,b exists.
7. For all n ≥ 1, the concatenation of a, n times, exists.
8. There is a finite sequence enumerating the terms of a
that are not terms of b.
9. Every nonempty finite sequence has a least term.
Before giving formal versions of these axioms, we make some
remarks about the nonlogical axioms of FSQZ.
a. Ø indicates “is defined”. See section 1.
b. Axioms 48 are presented in terms of the length and
values of the finite sequence that is asserted to exist. In
the case of axiom 8, this involves the ring operations.
c. Axiom 7 uses n as a variable (not a standard integer).
We now give formal presentations of FSTZ and FSQZ. 29
The nonlogical axioms of FSTZ are given formally as
follows.
1. Linearly ordered commutative ring axioms.
2. Finite interval.
($A)("x)(x Œ A ´ (y £ x Ÿ x £ z)).
3. Boolean difference.
($C)("x)(x Œ C ´ (x Œ A Ÿ ÿ(x Œ B))).
4. Set addition.
($C)("x)(x Œ C ´ ($y)($z)(y Œ A Ÿ z Œ B Ÿ x = y+z)).
5. Set multiplication.
($C)("x)(x Œ C ´ ($y)($z)(y Œ A Ÿ z Œ B Ÿ x = y⋅z)).
6. Least element.
($x)(x Œ A) Æ ($x)(x Œ A Ÿ ("y)(y Œ A Æ y £ x)).
The nonlogical axioms of FSQZ are given formally as
follows.
1.
Z.
2.
3.
4. The above linearly ordered commutative ring axioms for 0 £ lth(a).
val(a,n)Ø ´ 1 £ n £ lth(a).
The finite sequence (0,...,n) exists.
($a)(lth(a) = n+1 Ÿ ("k)(1 £ k £ n+1 Æ val(a,k) =
k+1)).
5. lth(a) = lth(b) Æ a,a+b,a⋅b exist.
lth(a) = lth(b) Æ ($g)("n)(val(g,n) @ val(a,n)) Ÿ
($g)("n)(val(g,n) @ val(a,n)+val(b,n)) Ÿ ($g)("n)(val(g,n) @
val(a,n)⋅val(b,n)).
6. The concatenation of a,b exists.
($g)("k,n)((1 £ k £ lth(a) Æ val(g,k) = val(a,k)) Ÿ (1 £
n £ lth(b) Æ val(g,lth(a)+n) = val(b,n))).
7. For all n ≥ 1, the concatenation of a, n times, exists.
lth(a) = n Æ ($b)(lth(b) = n⋅m Ÿ ("q,r)(0 £ q < m Ÿ 1 £
r £ n Æ val(b,n⋅q+r) = val(a,r))).
8. There is a finite sequence enumerating the terms of a
that are not terms of b.
($g)(("k)(($n)(val(g,n) = k) ´ (($n)(val(a,n) = k) Ÿ
ÿ($n)(val(b,n) = k))).
9. Every nonempty finite sequence has a least term.
1 £ lth(a) Æ ($k)("i)(1 £ i £ lth(a) Æ val(a,i) £
val(a,k)).
In axiom 5 above, we use the symbol @ from free logic, which
means “either both undefined, or equal”. See section 1.
The signature of FSTZEXP is L(Z,exp,fst). FSTZEXP extends 30
FSTZ by
i. exp(n,0) = 1.
ii. m ≥ 0 Æ (exp(n,m+1) = exp(n,m)⋅n Ÿ exp(n,m1)↑).
iii. The finite set {exp(n,0),...,exp(n,m)} exists.
We will find that FSTZEXP is quite weak. We let FSTZEXP’
extend FSTZ by
i. exp(n,0) = 1.
ii. m ≥ 0 Æ (exp(n,m+1) = exp(n,m)⋅n Ÿ exp(n,m1)↑).
iii. n ≥ 2 Ÿ 0 £ m < r Æ exp(n,m) < exp(n,r).
iv. The finite set {exp(n,0)+0,exp(n,1)+1,...,exp(n,m)+m}
exists.
The signature of FSQZEXP is L(Z,exp,fsq). FSQZEXP extends
FSQZ by
i. exp(n,0) = 1.
ii. m ≥ 0 Æ (exp(n,m+1) = exp(n,m)⋅n Ÿ exp(n,m1)↑).
iii. The finite sequence (exp(n,0),...,exp(n,m)) exists.
Recall CM(Z) from section 3, stated in L(Z).
Thus the seven strictly mathematical theories considered
here are
FSTZ, FSQZ, FSTZ + CM(Z), FSQZ + CM(Z), FSTZEXP, FSTZEXP’,
FSQZEXP
in the respective signatures
L(Z,fst), L(Z,fsq), L(Z,fst), L(Z,fsq), L(Z,exp,fst),
L(Z,exp,fst), L(Z,exp,fsq).
We offer the following remarks in comparing the strictly
mathematical nature of FSQZ and FSTZ.
i. Finite sequences of integers are more commonplace in
mathematics than finite sets of integers.
ii. The pointwise ring operations on finite sequences of
integers, and the concatenation of finite sequences of
integers (including indefinite concatenation), is more
commonplace in mathematics than the Boolean ring operations
on finite sets of integers, and the addition and
multiplication of finite sets of integers. 31 8. FSTZ.
In this section we show that FSTZ is a conservative
extension of PFA(Z). This follows from the particularly
convenient axiomatization of FSTZ given in Theorem 8.28:
THEOREM 8.28. FSTZ can be axiomatized as follows.
1. LOID(Z).
2. ($A)("x)(x Œ A ´ (y < x < z Ÿ j)), where j Œ S0(Z,fst)
and A is not free in j.
3. Every nonempty set has a least element.
Recall the axioms of FSTZ.
1.
2.
3.
4.
5.
6. Linearly ordered integral domain axioms (LOID(Z)).
Finite interval.
Boolean difference.
Set addition.
Set multiplication.
Least element. We will often use scalar addition and scalar
multiplication. We write A+x = x+A = A+{x}, and A⋅x = x⋅A =
A⋅{x}.
In Lemmas 8.1  8.27, it is understood that we are
asserting provability within FSTZ.
LEMMA 8.1.
i) ÿ(x < y Ÿ y < x+1);
ii) (a,b),[a,b),(a,b] exist;
iii) ∅,{x} exists;
iv) x⋅A = {x⋅y: y Œ A} exists;
v) every nonempty set has a greatest element;
vi) every set is included in some interval [a,b];
vii) sets are closed under pairwise union and pairwise
intersection;
viii) for standard n ≥ 0, {x1,...,xn} exists;
ix) the set of all positive (negative, nonnegative,
nonpositive) elements of any set exists.
Proof: For i), assume 0 < x < 1. By LOID(Z), 0 = 0⋅x < x⋅x <
1⋅x = x. Hence there is no least y such that 0 < y < 1. By
finite interval, (0,1) exists. By least element, there is a
least y such that 0 < y < 1. This is a contradiction. So
ÿ(0 < x < 1). Now suppose x < y < x+1. Then 0 < yx < 1,
which is a contradiction. 32 For ii), note that by i), (a,b) = [a+1,b1], [a,b) = [a,b1), (a,b] = [a+1,b].
For iii), note that ∅ is the interval (x,x), and by i), {x}
is the interval [x,x].
For iv), note that x⋅A = {x}⋅A, and apply set multiplication.
For v), Let A be nonempty. Then A = {1}⋅A has a least
element x. Clearly x is the greatest element of A.
For vi), let A be given. Then A Õ [min(A),max(A)].
For vii), note that A « B = A\(A\B). Also note that A » B =
C\(C\A « C\B), where A,B Õ [min(min(A),min(B)),
max(max(A),max(B))].
For viii), note that {x1,...,xn} = {x1} » ... » {xn}.
For ix), let A be given. By vi), let A Õ [a,b]. Then the
set of positive elements of A is A « [1,b]. The other cases
are handled similarly. QED
We write A for (1)⋅A, and AB for A+(B).
LEMMA 8.2. Let d ≥ 1 and x be an integer. There exists
unique q,r such that x = dq + r and 0 £ r < d.
Proof: For uniqueness, let x = dq + r = dq’ + r’, 0 £ r,r’ <
d. Then d(qq’) + rr’ = 0, d(qq’) = r’r. Hence dqq’ =
r’r < d. So qq’ < 1, and hence q = q’. Therefore 0 =
r’r, and so r = r’.
For existence, fix d,x as given, and first assume x > 0.
Let A = {xdq: q Œ [0,x]} = {x}  d⋅[0,x]. By Lemma 8.1 ix),
let A’ be the set of all nonnegative elements of A. Then A’
is nonempty since xdq > 0 for q = x. Choose q such that
min(A’) = xdq. Obviously 0 £ xdq and q Œ [0,x].
If q = x then d = 1 and xdq = 0, in which case we are
done. So we can assume that q < x.
Suppose xdq ≥ d. Then xd(q+1) ≥ 0 and q+1 Œ [0,x],
contradicting the choice of q. Hence 0 £ xdq < d. Set r =
xdq. Then x = dq + r and 0 £ r < d. 33
We still have to handle the case x £ 0. The
trivial, and so we assume x < 0. Write x =
d. Then x = d(q)  r. If r = 0 then we are
assume 0 < r < d. Then x = d(q1) + dr, 0 case x = 0 is
dq + r, 0 £ r <
done, and so we
£ dr < d. QED LEMMA 8.3. Let k ≥ 0. The following is provable in FSTZ. For
all r ≥ 2, the elements of [0,rk+1) have unique
representations of the form n0r0 + ... + nkrk, where each ni
lies in [0,r). If n0r0 + ... + nkrk = m0r0 + ... + mkrk and
each ni lies in (r/2,r/2), then each ni = mi.
Proof: It is important to note that k is treated as a
standard integer.
For uniqueness, suppose n0r0 + ... + nkrk = m0r0 + ... + mkrk,
where each ni,mi Œ [0,r). Let i be greatest such that ni ≠
mi. We can assume that ni < mi. Here we think of i as a
standard integer defined by a large number of cases.
Now subtract the second representation from the first. Then
we obtain an inequality of the form
p0r0 + ... + pi1ri1 ≥ ri,
where p0,...,pi1 Œ (r,r).
Note that p0r0 + ... + pi1ri1 £ (r1)(r0 + ... + ri1) = ri1.
This is the desired contradiction.
The second claim can be established in the same way by
subtraction, since any two elements of (r/2,r/2) must
differ by < r, and hence at most r1.
For existence, we proceed by external induction on k. The
case k = 0 is trivial. Suppose existence for all r ≥ 2 and x
Œ [0,rk+1), has been proved for a given k, where k ≥ 0. Let
r ≥ 2 and x Œ [0,rk+2). Write x = rk+1nk+1 + y, 0 £ y < rk+1.
Note that 0 £ nk+1 < r. By induction hypothesis, write y =
n0r0 + ... + nkrk, n0,...,nk Œ [0,r). Then x = n0r0 + ... +
nkrk + rk+1nk+1, n0,...,nk+1 Œ [0,r). QED
Until the end of the proof of Lemma 8.12, we fix a standard
integer k > 0.
LEMMA 8.4. For all r > 1, S[r] = {n0r0 + n1r2 + ... + nir2i +
... + nkr2k: n0,...,nk Œ [0,r)} exists. Every element of S[r]
is uniquely written in the displayed form. 34 Proof: S[r] = [0,r)⋅r0 + [0,r)⋅r2 + ... + [0,r)⋅r2k. The second
claim follows immediately from Lemma 8.3. QED
For x Œ S[r], we write x[i] for ni in this unique
representation.
LEMMA 8.5. For all r > 1 and i Œ [0,k], {x Œ S[r]: x[i] =
0} and {x Œ S[r]: x[i] = 1} exist.
Proof: The first set is
[0,r)⋅r0 + ... + [0,r)⋅r2i2 + [0,r)⋅r2i+2 + ... + [0,r)⋅r2k.
The second set is
[0,r)⋅r0 + ... + [0,r)⋅r2i2 + r2i + [0,r)⋅r2i+2 + ... +
[0,r)⋅r2k. QED
LEMMA 8.6. Let r > 1 and i,j,p Œ [0,k]. Then {x Œ S[r]:
x[i] + x[j] = x[p]} exists.
Proof: Let r,i,j,p be as given. If i = p then {x Œ S[r]:
x[i] + x[j] = x[p]} = {x Œ S[r]: x[j] = 0}. If j = p then
{x Œ S[r]: x[i] + x[j] = x[p]} = {x Œ S[r]: x[i] = 0}. Both
of these cases are covered by Lemma 8.5.
We now handle the case i = j ≠ p. We wish to show that A =
{x Œ S[r]: 2x[i] = x[p]} exists, where i ≠ p.
Now D = {x: ($a Œ [0,r)(x = ar2i + 2ar2p} exists, since D =
[0,r)⋅(r2i + 2r2p).
Let T be the sum of the sets [0,r)⋅r2q, where q Œ [0,k]\{i}.
We claim that A = (D + T) « S[r].
To see this, obviously every element of A lies in (D + T) «
S[r]. On the other hand, let x Œ (D + T) « S[r]. Write
x = ar2i + 2ar2p + y
where a,b Œ [0,r) and y Œ T. Since 2a < r2, this is the
representation of x Œ S[r]. Evidently, x Œ A.
We now handle the case i ≠ j ≠ p. We wish to show that B =
{x Œ S[r]: x[i] + x[j] = x[p]} exists. 35
Now E = {x: ($a,b Œ [0,r))(x = ar2i + br2j + (a+b)r2p}
exists, since E = {x: ($a,b Œ [0,r))(x = a(r2i + r2p) + b(r2j
+ r2p))} = ([0,r)⋅(r2i + r2p) + [0,r)⋅(r2j + r2p)).
Let V be the sum of the sets [0,r)⋅r2q, where q Œ
[0,k]\{i,j,p}. We claim that B = (E + V) « S[r].
To see this, obviously every element of B lies in (E + V) «
S[r]. On the other hand, let x Œ (E + V) « S[r]. Write
x = ar2i + br2j + (a+b)r2p + y
where a,b Œ [0,r) and y Œ V. Since a+b < r2, this is the
representation of x Œ S[r]. Evidently, x Œ B. QED
We define ab ´ ($c)(b = a⋅c).
LEMMA 8.7. For all r > 1 and i,j Œ [0,k], {x Œ S[r]:
x[i]x[j]} exists.
Proof: If i = j then {x Œ S[r]: x[i]x[j]} = S[r], which is
handled by Lemma 8.4. Assume i ≠ j. We want to prove that A
= {x Œ S[r]: x[i]x[j]} exists.
Now E = {x: ($a,b Œ [0,r))(x = ar2i + abr2j)} exists, since E
= {x: ($a,b Œ [0,r))(x = a(r2i + br2j}} = [0,r)⋅(r2i +
[0,r)⋅r2j).
Let D be the sum of the sets [0,r)⋅r2q, where q Œ
[0,k]\{i,j}. We claim that A = (E + D) « S[r].
To see this, obviously every element of A lies in (E + D) «
S[r]. On the other hand, let x Œ (E + D) « S[r]. Write
x = ar2i + abr2j + y
where a,b Œ [0,r) and y Œ D. Since ab < r2, this is the
representation of x Œ S[r]. Evidently x Œ A. QED
LEMMA 8.8. For all r > 1, i Œ [0,k], and A Õ [0,r), {x Œ
S[r]: x[i] Œ A} exists.
Proof: Note that {x Œ S[r]: x[i] Œ A} is [0,r)⋅r0 + ... +
[0,r)⋅r2i2 + A⋅r2i + [0,r)⋅r2i+2 + ... + [0,r]⋅r2k. QED
LEMMA 8.9. Let j be a propositional combination of formulas
xi = 0, xi = 1, xi+xj = xp, xixj, xi Œ Aj, where i,j,p Œ 36
[0,k]. The following is provable in FSTZ. For all r > 1 and
A0,...,Ak Õ [0,r), {x0r0 + ... + xkr2k: j Ÿ x0,...,xk Œ [0,r)}
exists.
Proof: For atomic j, this follows from Lemmas 8.4  8.8.
The propositional combinations are handled by the fact that
the subsets of S[r] form a Boolean algebra. QED
LEMMA 8.10. For all r > 1, i Œ [0,k], and E Õ S[r], {x Œ
S[r]: ($y Œ E)("j Œ [0,k]\{i})(x[j] = y[j])} exists.
Proof: We first claim that A
[0,k]\{i})(x[j] = y[j])} Õ E
suppose x Œ S[r], y Œ E, and
Since the coefficients of r2i
we see that xy Œ (r,r)⋅r2i.
E + (r,r)⋅r2i. = {x Œ S[r]: ($y Œ E)("j Œ
+ (r,r)⋅r2i. To see this,
"j Œ [0,k]\{i}, x[j] = y[j].
in x and y both lie in [0,r),
Hence x Œ (r,r)⋅r2i + y Õ We claim that A = (E + (r,r)⋅r2i) « S[r]. To see this, let
x Œ (E + (r,r)⋅r2i ) « S[r]. Write
x = y + a⋅r2i
where y Œ E and a Œ (r,r). In this equation, the
coefficient of r2i is the coefficient of r2i in y plus a,
which must lie in (r,2r). Hence this must be the
representation of x Œ S[r]. Evidently, x agrees with an
element of E (namely y) at all positions other than at r2i.
QED
LEMMA 8.11. Let j be a propositional combination of
formulas xi = 0, xi = 1, xi+xj = xp, xixj, xi Œ Aj, where
i,j,p Œ [0,k]. Let m Œ [1,k]. Let y = (Qmxm Œ [0,r))...(Qkxk
Œ [0,r))(j). The following is provable in FSTZ. For all
A0,...,Ak Õ [0,r), {x0r0 + ... + xm1r2m2: y Ÿ x0,...,xm1 Œ
[0,r)} exists.
Proof: Here Qi is " or $. Lemma 8.9 handles j. Lemma 8.10
handles existential quantifiers. Universal quantifiers are
taken care of by relative complementation. QED
LEMMA 8.12. Let r > 1, E Õ S[r], i1 < ... < ip Œ [0,k], and
x1,...,xp Œ [0,r). Then {y Œ S[r]: y[i1] = x1 Ÿ ... Ÿ y[ip] =
xp} exists.
Proof: Note that this set is A « B1 « ... « Bp, where for
all j Œ [1,p], Bj = {y Œ S[r]: y[ij] = xj} = [0,r)⋅ r0 + ... 37
+ [0,r)⋅r2k where the term with exponent 2j is replaced by
xjr2j. QED
We now release the fixed standard integer k. For formulas j
without bound set variables, and integer variables z not in
j, Let jz be the result of relativizing all quantifiers in j
to [z,z].
LEMMA 8.13. Let j be a formula without bound set variables
whose atomic subformulas are of the form xi = 0, xi = 1,
xi+xj = xp, xixj, xi Œ Aj. Let y,z be distinct integer
variables, where z does not appear in j. Then FSTZ proves
that {y Œ [0,z]: jz} exists. Also FSTZ proves that {y Œ [z,z]: jz} exists.
Proof: Note that the conclusion should be viewed as a
separation principle with parameters (represented by the
free variables of j other than y.
By changing variables, we can assume that y is x0, the free
variables of j are among x0,...,xm1, and the quantified
variables are among xm,...,xk. Also replace the
relativizations to [z,z] with relativizations to [0,z], by
appropriately modifying the formula.
Now apply Lemma 8.11 with r = z+1. We obtain {x0r0 + ... +
xm1r2m2: j’ Ÿ x0,...,xm1 Œ [0,z]}. Now apply Lemma 8.12 with
p = m1, i1,...,ip = 1,...,m1, and r = z+1. We obtain {x0 Œ
[0,z]: j’} = {y Œ [0,z]: jz}.
The second claim follows from the first. QED
LEMMA 8.14. Let j be a formula without bound set variables
whose atomic subformulas are of the form s = t, s < t, st,
or t Œ Aj, where s,t are terms without •. Let y,z be
distinct integer variables, where z does not appear in j.
Then FSTZ proves that {y Œ [z,z]: jz} exists.
Proof: By inductively introducing existential quantifiers
needed to unravel the terms. A bound can be placed on the
existential quantifiers introduced which depends only on j
and the value of the bound z. Since the terms do not use •,
the expansion stays within the form in Lemma 8.13. QED
Formulas of the form in Lemma 8.14 are called special
formulas. 38
Note that we do not allow ⋅ in special formulas. We first
need to use Lemma 8.14 to obtain some basic number theory
before we can handle ⋅ appropriately.
LEMMA 8.15. x,y ≠ 0 Æ gcd(x,y),lcm(x,y) exist. x > 1 Æ x
is divisible by a prime.
Proof: For the first claim, let x,y ≠ 0. By Lemma 8.14, {a
Œ [1,xy]: ax Ÿ ay} exists. Then gcd(x,y) is its
greatest element. By Lemma 8.14, {a Œ [1,xy]: xa Ÿ ya}
exists. Then lcm(x,y) is its least element.
For the second claim, let x > 1. By Lemma 8.14, {p Œ [2,x]:
px} exists. Let p be the least element. Then p is a prime
divisor of x. QED
LEMMA 8.16. Suppose x,y > 1 and ax + by = 1. Then there
exists cx + dy = 1, where c Œ (0,y), d Œ (x,0). Suppose
x,y > 0 and ax + by = 1. Then there exists cx + dy = 1,
where c Œ [0,y], d Œ [x,0].
Proof: Let x,y,a,b be as given. By symmetry we can assume
that a ≥ 0.
Let A = {s Œ [0,ax]: ($t Œ [1ax,0])(xs Ÿ yt Ÿ s + t =
1)} = {s Œ [0,ax]: ($t)(xs Ÿ yt Ÿ s + t = 1)} = {s Œ
[0,ax]: there is a multiple of x and a multiple of y, which
add up to 1}. Note that A exists by Lemma 8.14, and A is
nonempty since it includes s = ax, with t = by. Let cx be
the least element of A.
Write cx + dy = 1. Note that (cy)x + (d+x)y = 1. By the
choice of c, ÿ(0 £ cy < c), and so cy < 0 or cy ≥ c.
Hence c Œ [0,y).
Note that 1 = cx + dy £ xy + dy = (x+d)y. Hence x+d > 0, and
so d > x. Hence d Œ (x,0].
Note that c ≠ 0 and d ≠ 0 because of cx + dy = 1, x,y > 1.
For the second claim, we need only consider the case (x = 1
⁄ y = 1). By symmetry, assume x = 1. Then take c = 1 and d
= 0. QED
We say that x,y are relatively prime if and only if x,y ≠ 0
and the only common divisors of x,y are 1 and 1. 39
LEMMA 8.17. Let x,y be relatively prime. Then there exists
a solution to ax + by = 1.
Proof: We fix a positive integer t. We wish to show by
induction (equivalently, least element principle) that the
following holds for every 0 < s £ t. For all 0 < x,y £ s, if
x,y are relatively prime then ax + by = 1 has a solution.
We need to express this condition by a special formula.
("x,y Œ [t,t])((0 < x,y £ s Ÿ x,y relatively prime) Æ
ax + by = 1 has a solution).
("x,y Œ [t,t])((0 < x,y £ s Ÿ nothing in [1,x] divides
both x,y) Æ ax + by = 1 has a solution in [s,s]).
Here we have used Lemma 8.16, which provides a bound on
solutions to ax + by = 1.
The basis case s = 1 is trivial. Suppose true for a fixed s
≥ 1. Let x,y £ s+1 be relatively prime. We can assume 1 < y
< x = s+1. Write x = qy + r, 0 £ r < y. Since x,y are
relatively prime, we have 0 < r < y.
Note that y,r are relatively prime and positive. Hence by
induction hypothesis write cy + dr = 1. Now dx +(cdq)y =
1.
We still have to consider the case where x or y is
negative. But then we can merely change the sign or signs
of one or more of a,b. QED
LEMMA 8.18. Let p be a prime and suppose pxy. Then px or
py.
Proof: Let p,x,y be as given. Suppose the contrary. Then
x,y ≠ 0, and p,x are relatively prime, and p,y are
relatively prime. By Lemma 8.17, write ap + bx = 1, cp + dy
= 1. Then apcp + apdy + bxcp + bxdy = 1. Note that p
divides every summand, and so p divides 1, which is a
contradiction. QED
LEMMA 8.19. Let x,y be relatively prime and let x,z be
relatively prime. Suppose xyz. Then x = 1 or 1. 40
Proof: Let x,y,z be as given. Write ax + by = 1 and cx + dz
= 1. Then axcx + axdz + bycx + bydz = 1. Since x divides
every summand, x divides 1. Hence x = 1 or 1. QED
LEMMA 8.20. Let x,y be relatively prime and xyz. Then xz.
Proof: Let x,y,z be as given. We can assume that z ≠ 0. It
suffices to prove this for x,y,z > 0.
Now x/gcd(x,z) divides y(z/gcd(x,z)) via the integer factor
yz/x. Also note that x/gcd(x,z) and y are relatively prime.
We claim that x/gcd(x,z) and z/gcd(x,z) are relatively
prime. To see this, suppose they have a common factor u >
1. Then gcd(x,z)u is a factor of x and also a factor of z,
contradicting that gcd(x,z) is the greatest common factor
of x,z.
By Lemma 8.19, x/gcd(x,z) = 1. I.e., gcd(x,z) = x. So xz.
QED
LEMMA 8.21. Let a,b be relatively prime. Then the least
positive common multiple of a,b is ab.
Proof: Let a,b be as given, and let x be a positive common
multiple of a,b. Write x = ay.
Since bay, by see by Lemma 8.20 that by. Hence b £ y.
Therefore x = ay ≥ ab as required. QED
Lemmas 8.22, 8.23 finally tell us how to handle •
appropriately.
LEMMA 8.22. There is a special formula j with free
variables among x,y such that the following is provable in
FSTZ. For all z there exists z’ > z such that ("x,y Œ [z,z])(x = y2 ´ jz’).
Proof: Let j express x+y = lcm(y,y+1). Let z be given. If y
œ [1,0] then gcd(y,y+1) = 1, and hence by Lemma 8.10,
lcm(y,y+1) = y(y+1). Therefore ("x,y Œ [z,z]\[1,0])(j ´
x+y = y(y+1)). Hence ("x,y Œ [z,z]\[1,0])(j ´ x = y2).
The quantifiers in j can be bounded to an integer z’ that
depends only on z. 41
We still have to modify j in order to handle [1,0]. Take
j’ to be (j Ÿ x,y œ [1,0]) ⁄ x = y = 0 ⁄ (x = 1 Ÿ y = 1).
QED
LEMMA 8.23. There is a special formula y with free
variables among u,v,w, such that the following is provable
in FSTZ. For all z there exists z’ > z such that ("u,v,w Œ
[z,z])(u⋅v = w ´ yz’).
Proof: Let y = ($x,y,a,b)(x = y2 Ÿ y = u+v Ÿ a = u2 Ÿ b = v2
Ÿ 2w = xab). Let z be given. Then ("u,v,w Œ [z,z])(u⋅v =
w ´ y). Use the j from Lemma 8.22 to remove the first,
third, and fourth displayed equations, to make y special.
The quantifiers can be bounded to z’ > z, where z’ depends
only on z. QED
We now extend Lemma 8.14.
LEMMA 8.24. Let j be a formula without bound set variables
whose atomic subformulas are of the form xi = 0, xi = 1,
xi+xj = z, xi⋅xj = xp, xi Œ Aj. Let y,z be distinct integer
variables, where z does not appear in j. Then FSTZ proves
that {y Œ [z,z]: jz} exists.
Proof: Let j be as given. Replace each atomic subformula of
the form x⋅y = z by the y of Lemma 8.23, with an appropriate
change of variables. Call this expansion r. Let z be given.
Then there exists z’ > z depending only on z such that for
all y Œ [z,z], jz ´ rz’. By Lemma 8.14, {y Œ [z’,z’]: rz’}
exists. Hence {y Œ [z’,z’]: jz} exists. Hence {y Œ [z,z]:
jz} exists. QED
LEMMA 8.25. Let j be a formula without bound set variables.
Let y,z be distinct integer variables, where z does not
appear in j. Then FSTZ proves that {y Œ [z,z]: jz} exists.
Proof: Let j be as given, and let z be given. Expand the
terms appearing in j using existential quantifiers. Apply
Lemma 8.24 with appropriately chosen z’, where z’ depends
only on z and the terms that appear. QED
We now define the class of formulas of FSTZ, S0(Z,fst).
i) every atomic formula of FSTZ is in S0(Z,fst);
ii) if j,y are in S0(Z,fst), then so are ÿj, j Ÿ y, y ⁄ y,
j Æ y, j ´ y; 42
iii) if j is in S0(Z,fst), x is an integer variable, s,t are
integer terms, x not in s,t, then ($x Œ [s,t])(j) and ("x Œ
[s,t])(j) are in S0(Z,fst).
LEMAM 8.26. Let j be in S0(Z,fst). Let x1,...,xk be an
enumeration without repetition of at least the free
variables of j. The following is provable in FSTZ. Let r >
1. Then {x1r1 + ... + xkrk: x1,...,xk Œ [0,r) Ÿ j} exists.
Proof: By induction on j. Let j be atomic. Then this
follows from Lemma 8.25. Suppose this is true for j,y in
S0(Z,fst). Let r be among ÿj, j ⁄ y, j Ÿ y, j Æ y, j ´ y.
Then obviously this holds for r.
Now suppose this holds for j in S0(Z,fst). Let y = ($x Œ
[s,t])(j). Let x1,...,xk be an enumeration without
repetition of at least the free variables of y. Then
x1,...,xk,x is an enumeration without repetition of at least
the free variables of j.
We want to show that
A = {x1r1 + ... + xkrk: x1,...,xk Œ [0,r) Ÿ ($x Œ [s,t])(j)}
provably exists for all r > 1. We know that
B = {x1r1 + ... + xkrk + xrk+1: x1,...,xk,x Œ [0,r) Ÿ j}
provably exists for all r > 1. We can define A from B
appropriately so that we can simply apply Lemma 8.25. QED
LEMMA 8.27. Let j lie in S0(Z,fst). Let z be an integer
variable that does not appear in j. Then FSTZ proves that
{y Œ [z,z]: j} exists.
Proof: From Lemmas 8.25 and 8.26. QED
THEOREM 8.28. FSTZ can be axiomatized as follows.
1. LOID(Z).
2. ($A)("x)(x Œ A ´ (y < x < z Ÿ j)), where j Œ S0(Z,fst)
and A is not free in j.
3. Every nonempty set has a least element.
Proof: Axiom scheme 2 is derivable from FSTZ by Lemma 8.27.
For the other direction, first note that we can derive
("A)(A exists). Hence every set has a greatest element.
Then it is easy to see that finite interval, Boolean 43
difference, set addition, and set multiplication are
special cases of axiom scheme 2 above. QED
THEOREM 8.29. FSTZ is a conservative extension of PFA(Z).
Proof: By Theorem 8.28, FSTZ proves PFA(Z). It now suffices
to show that any model M of PFA(Z) can be expanded by
attaching sets to form a model of FSTZ. Take the sets of
integers to be those sets of the form
{x Œ [n,n]: j}
where j is a formula in S0(Z) with parameters allowed,
interpreted in the model M. The verification of FSTZ in the
expansion is straightforward. QED 9. FSTZ = FSTZD = FSTZS.
We show that FSTZ is equivalent to two interesting
weakenings of FSTZ. These results are of independent
interest, and not central to the paper.
The signature of FSTZD is the same as that of FSTZ, which
is L(Z,fst). FSTZD = “finite sets of integers with
duplication”. The nonlogical axioms of FSTZD are as
follows.
1.
2.
3.
4. Linearly ordered commutative ring axioms.
Finite interval.
Boolean difference.
Duplicate set addition.
($B)("x)(x Œ B ´ ($y)($z)(y Œ A Ÿ z Œ A Ÿ x = y+z)).
5. Duplicate set multiplication.
($B)("x)(x Œ B ´ ($y)($z)(y Œ A Ÿ z Œ A Ÿ x = y⋅z)).
6. Every set has a least and greatest element.
Axiom 4 asserts the existence of A+A, and axiom 5 asserts
the existence of A⋅A.
Lemmas 9.1  9.8 refer to provability in FSTZD.
LEMMA 9.1. i)iii),v)ix) of Lemma 8.1.
Proof: Straightforward. QED
LEMMA 9.2. Let A Õ [x,x], x ≥ 0, and y > 3x. Then A+y
exists. 44 Proof: Let A,x be as given. By Lemma 9.1, let B = A » {y}.
Then B+B is composed of three parts: A+A, A+y, {2y}. We
don’t know yet that the second part is a set. Note that A+A
Õ
[2x,2x] and A+y Õ [x+y,x+y].
First assume y > 0. Note that 2x < x+y and x+y < 2y. Hence
these three parts are pairwise disjoint. Since B+B and the
first and third parts exist, clearly the second part
exists.
Now assume y < 0. Note that 2x > x+y and x+y > 2y. Hence
these three parts are pairwise disjoint. So the second part
exists. QED
LEMMA 9.3. Let A Õ [7z/8,9z/8], z > 0, w < z/2. Then A+w
exists.
Proof: Let A,z,w be as given. Let B = A » {w}. Then B+B is
composed of three parts: A+A, A+w, {2w}. Note that A+A Õ
[7z/4,9z/4] and A+w Õ [7z/8 + w,9z/8 + w].
Note that 9z/8 + w < 7z/4. Hence the first two parts are
disjoint. Therefore A+w is among B+B\A+A, B+B\A+A » {2w},
B+B\A+A\{2w}. Hence A+w exists. QED
LEMMA 9.4. A+y exists.
Proof: Let A Õ [x,x], x > 0, and y be given. We can assume
that y is nonzero. Write y = z+w, where z > 3x, A+z Õ
[7z/8,9z/8], w < z/2. By Lemma 9.2, A+z exists. By Lemma
9.3, A+z+w exists. But A+z+w = A+y.
It remains to show how z,w can be chosen. Set z =
9max(x,y). Set w = yz = y9max(x,y). Note that A+z Õ
[x+z,x+z] Õ [7z/8,9z/8].
We have only to verify that w < z/2. I.e., y  9max(x,y)
< 9max(x,y)/2, which is y < 9max(x,y)/x. This follows
from x > 0 and y ≠ 0. QED
LEMMA 9.5. A+B exists.
Proof: Let A,B be given. Let A,B Õ [x,x], x ≥ 0. By Lemma
9.2, let C = B+4x. Consider A»C + A»C. This is composed of 45
three parts: A+A, A+C, C+C. We don’t know yet that the
second part is a set.
Note that A+A Õ [2x,2x], A+C Õ [3x,5x], C+C Õ [6x,10x].
Hence these three parts are pairwise disjoint. Since A»C +
A»C and the first and third parts exist, clearly the second
part exists. I.e., A+C exists.
Observe that A+C = A+B+4x, and so A+B = A+C+4x, which
exists by Lemma 3.4. QED
LEMMA 9.6. A exists.
Proof: Let A be given. First assume that A Õ [1,x], x > 1.
Let B = A » {1}. Note that B⋅B = A⋅A » {1} » A, where we
don’t know yet that A exists. However, A is disjoint from
A⋅A » {1}. Hence A exists.
Now assume that A Õ [x,1], x > 1. Using Lemmas 9.1 and
9.4, let B = A+x3. Consider B»{1} ⋅ B»{1}. This is
composed of three parts: B⋅B, B, {1}, where we don’t know
yet that B exists. Note that B⋅B Õ [x3+1)2,(x3+x)2], B Õ
[1+x3,x+x3]. Hence the three parts are pairwise disjoint.
Therefore B exists.
Finally, let A be arbitrary. Write A = A+ » A » A0, where
A+ is the positive part of A, A is the negative part of A,
and A0 is the 0 part of A, which is {0} if 0 Œ A and ∅ if 0
œ A.
Note that A = (A+) » (A) » A0, and so A exists. QED
LEMMA 9.7. A⋅x exists.
Proof: First assume A Õ [y2,y3), y > x > 1. Consider A»{x} •
A»{x}. This is composed of three parts: A⋅A, A⋅x, {x2}, where
we don’t know yet that A•x exists. Note that A⋅A Õ [y4,y6],
A⋅x Õ [xy2,xy3]. Hence the three parts are pairwise disjoint.
Therefore A⋅x exists.
Now assume A is arbitrary and x > 1. We can choose y > x
such that B Õ [y2,y3], where B is a translation of A. Then
B⋅x exists.
Let A = B+c. Then A⋅x = (B+c)⋅x = B⋅x + {cx}. Therefore A⋅x
exists. 46
The case x = 0 is trivial. Finally suppose A is arbitrary
and x < 1. Then A⋅x = (A⋅x), and x > 1. Therefore A⋅x
exists. QED
LEMMA 9.8. A⋅B exists.
Proof: Let A,B be given. We first assume that A,B Õ [1,x],
x > 1. Let C = A » B. Then C⋅C exists. Its negative part is
obviously A⋅B, which therefore exists. Note that A⋅B = (A⋅B), and therefore A⋅B exists.
For the general case, write A = A+ » A » A0, B = B+ » B »
B0. Then A⋅B is the union of the nine obvious cross
products. There are only three of them that we have to
check exist, the other six obviously existing. These are
A+⋅B, A⋅B+, A⋅B. However, it is easy to see that these are,
respectively, (A⋅B), (A⋅B), A⋅B, and therefore exist. QED
THEOREM 9.9. FSTZ and FSTZD are equivalent.
Proof: By Lemmas 9.5 and 9.8. QED
We now present another variant of FSTZ which we call FSTZS
= “finite sets of integers with scalars and squares”. Here
we replace A•B in favor of scalar multiplication and
squares.
The signature of FSTZS is L(Z,fst). The nonlogical axioms
of FSTZS are as follows.
1.
2.
3.
4.
5. Linearly ordered ring axioms.
Finite interval.
Boolean difference.
Duplicate set addition.
Scalar multiplication.
($B)("x)(x Œ B ´ ($y)(y Œ A Ÿ x = y⋅z)).
6. Squares.
($A)("x)(x Œ A ´ ($y)(0 < y Ÿ y < z Ÿ x = y2)).
7. Least element.
Axiom 5 asserts that each c⋅A exists. Axiom 6 asserts that
each {12,22,...,n2}, n ≥ 0, exists.
Lemmas 9.10  9.27 refer to provability in FSTZS.
LEMMA 9.10. i)ix) of Lemma 3.1. A+B exists. 47
Proof: For the first claim, we need only observe that by
scalar multiplication, A exists. From this we obtain that
every nonempty set has a greatest element. For the second
claim, we can repeat the proof of Lemma 9.5. QED
To show that FSTZS is equivalent to FSTZ, it suffices to
prove that A⋅B exists in FSTZS. We do not know a clean way
of doing this. Instead, we recast the proof of Lemma 8.23
for FSTZS in order to derive that A•B exists. Much of the
proof will be the same. The key point is to avoid use of 
in the auxiliary languages, and instead use a monadic
predicate for “being a square”.
LEMMA 9.11. Let d ≥ 1 and x be an integer. There exists
unique q,r such that x = dq + r and 0 £ r < d.
Proof: See Lemma 9.2. QED
LEMMA 9.12. Let k ≥ 0. The following is provable in T2. For
all r ≥ 2, the elements of [0,rk+1) have unique
representations of the form n0r0 + ... + nkrk, where each ni
lies in [0,r). If n0r0 + ... + nkrk = m0r0 + ... + mkrk and
each ni lies in (r/2,r/2), then each ni = mi.
Proof: See Lemma 9.3. QED
Until the end of the proof of Lemma 8.21, we fix a standard
integer k > 0.
LEMMA 9.13. For all r > 1, S[r] = {n0r0 + n1r2 + ... + nir2i +
... + nkr2k: n0,...,nk Œ [0,r)} exists. Every element of S[r]
is uniquely written in the displayed form.
Proof: See Lemma 9.4. QED
LEMMA 9.14. For all r > 1 and i Œ [0,k], {x Œ S[r]: x[i] =
0} and {x Œ S[r]: x[i] = 1} exist.
Proof: See Lemma 9.5. QED
LEMMA 9.15. For all r > 1 and i,j,p Œ [0,k], {x Œ S[r]:
x[i] + x[j] = x[p]} exists.
Proof: See Lemma 9.6. QED 48
Note that we cannot use Lemma 9.7 here since it involves
multiplication of sets, as opposed to just scalar
multiplication of sets.
LEMMA 9.16. For all r > 1, i Œ [0,k], and A Õ [0,r), {x Œ
S[r]: x[i] Œ A} exists.
Proof: See Lemma 9.8. QED
LEMMA 9.17. For all r > 1 and i Œ [0,k], {x Œ S[r]: x[i] is
a square} exists.
Proof: Use Lemma 9.16 with A = {12,...,r2}. QED
LEMMA 9.18. Let j be a propositional combination of
formulas xi = 0, xi = 1, xi+xj = xp, Sq(xi), xi Œ Aj, where
i,j,p Œ [0,k]. The following is provable in T4. For all
A0,...,Ak Õ [0,r), {x0r0 + ... + xkr2k: j Ÿ x0,...,xk Œ [0,r)}
exists.
Proof: See Lemma 9.9. Sq(xi) means “xi is a square”. QED
LEMMA 9.19. For all r > 1 and i Œ [0,k] and E Õ S[r], {x Œ
S[r]: ($y Œ E)("j Œ [0,k]\{i})(x[j] = y[j])} exists.
Proof: See Lemma 9.10. QED
LEMMA 9.20. Let j be a propositional combination of
formulas xi = 0, xi = 1, xi+xj = xp, Sq(xi), xi Œ Aj, where
i,j,p Œ [0,k]. Let m Œ [1,k]. Let y = (Qmxm Œ [0,r))...(Qkxk
Œ [0,r))(j). The following is provable in T4. For all
A0,...,Ak Õ [0,r), {x0r0 + ... + xm1r2m2: y Ÿ x0,...,xm1 Œ
[0,r)} exists.
Proof: See Lemma 9.11. QED
LEMMA 9.21. Let r > 1, E Õ S[r], i1 < ... < ip Œ [0,k], and
x1,...,xp Œ [0,r). Then {y Œ S[r]: y[i1] = x1 Ÿ ... Ÿ y[ip] =
xp} exists.
Proof: See Lemma 9.12. QED
We now release the fixed standard integer k.
LEMMA 9.22. Let j be a formula without bound set variables
whose atomic subformulas are of the form xi = 0, xi = 1,
xi+xj = xp, Sq(xi), xi Œ Aj. Let y,z be distinct integer 49
variables, where z does not appear in j. Then FSTZS proves
that {y Œ [0,z]: jz} exists. Also T4 proves that {y Œ [z,z]: jz} exists.
Proof: See Lemma 9.13. QED
LEMMA 9.23. Let j be a formula without bound set variables
whose atomic subformulas are of the form s = t, s < t,
Sq(t), or t Œ Aj, where s,t are terms without •. Let y,z be
distinct integer variables, where z does not appear in j.
Then FSTZS proves that {y Œ [z,z]: jz} exists.
Proof: See Lemma 9.14. QED
We call the formulas given in Lemma 9.23, good formulas.
LEMMA 9.24. Let x = y2, y ≥ 0. Then the next square after x
is (y+1)2, and this is at most 3x+1.
Proof: Suppose y2 < z2 < (y+1)2. We can assume z ≥
y < z < y+1 since squaring is strictly increasing
nonnegative integers. For the second claim, first
y £ x. Then observe that (y+1)2 = y2+2y+1 = x+2y+1
QED 0. Clearly
on the
note that
£ 3x+1. LEMMA 9.25. x = y2 if and only if x is a square and the next
square after x is x+2y+1. The next square after x is at
most 2x+1.
Proof: The forward direction is by Lemma 9.24. For the
reverse direction, let x be a square and the next square
after x is x+2y+1. Let x = z2. Then the next square after x
is (z+1)2. So (z+1)2 = z2+2y+1 = z2+2z+1. Hence y = z. QED
LEMMA 9.26. There is a good formula j with at most the free
variables among x,y, such that the following is provable in
FSTZS. For all z there exists z’ > z such that ("x,y Œ [z,z])(x = y2 ´ jz’).
Proof: Let z be given. We can assume that z ≥ 0. Let j(x,y)
be (y ≥ 0 Ÿ Sq(x) Ÿ Sq(x+2y+1) Ÿ ("w)(Sq(w) Æ ÿ(x < w <
x+2y+1)))). Note that j expresses that x is a square, y ≥ 0,
and x+2y+1 is the next square after x. Note also that when
bounded to [3z+1,3z+1], the meaning remains unchanged.
This works for x,y Œ [0,z], and can be easily modified to
work for x,y Œ [z,z]. QED 50
LEMMA 9.27. There is a good formula y with at most the free
variables u,v,w, such that the following is provable in
FSTZS. For all z there exists z’ > z such that ("x,y,z Œ [z,z])(u⋅v = w ´ yz’).
Proof: Let z be given. As in the proof of Lemma 3.23, use y
= ($x,y,a,b)(x = y2 Ÿ y = u+v Ÿ a = u2 Ÿ b = v2 Ÿ 2w = xab)
and Lemma 5.26. We can easily bound the quantifiers to an
appropriately chosen [z’,z’]. QED
THEOREM 9.28. FSTZ, FSTZD, FSTZS are equivalent.
Proof: It suffices to show that A⋅B exists within FSTZS. Let
A,B be given, where A,B Õ [z,z]. Then A⋅B Õ [z2,z2], but
we don’t know yet that A⋅B exists.
Let z’ be according to Lemma 9.27 for z2. Then A⋅B = {y Œ [z2,z2]: ($u,v,w)(u Œ A Ÿ v Œ B Ÿ yz’)} = {y Œ [z2,z2]:
($u,v,w)(u Œ A Ÿ v Œ B Ÿ y)z’} which exists by Lemma 9.23.
The second claim follows immediately from the first. QED 10. FSQZ.
We now give a very simple interpretation of FSTZ in FSQZ,
which is the identity on the Z sort. It follows immediately
that FSQZ proves PFA(Z). We then show that FSQZ is a
conservative extension of PFA(Z).
Recall the axioms of FSQZ.
1. Linearly ordered integral domain axioms.
2. lth(a) ≥ 0.
3. val(a,n)Ø ´ 1 £ n £ lth(a).
4. The finite sequence (0,...,n) exists.
5. lth(a) = lth(b) Æ a,a+b,a⋅b exist.
6. The concatenation of a,b exists.
7. For all n ≥ 1, the concatenation of a, n times, exists.
8. There is a finite sequence enumerating the terms of a
that are not terms of b.
9. Every nonempty finite sequence has a least term.
The interpretation of the integer part is the identity. The
interpretation of the sets of integers in T0 are the
sequences of integers in FSZ. The Œ relation is interpreted
as 51
n Œ x if and only if n is a term of a.
We write (n upthru m) for the finite sequence a, if it
exists, such that
i. lth(a) = max(0,mn+1).
ii. For all 1 £ i £ lth(a), val(a,i) = n+i1.
LEMMA 10.1. The empty sequence exists. For all n, (n)
exists.
Proof: The empty sequence is from axiom 9 of FSQZ.
Clearly (0) exists by axiom 5. Let n > 0. Form (0,...,n1),(0,...,n) by axiom 5, and delete the latter from the
former by axiom 9, to obtain (n). If n < 0 then form (n),
and then form (n) by axiom 6. QED
LEMMA 10.2. For all n there is a sequence consisting of all
n’s of any length ≥ 0.
Proof: Let n be given. Form (n) by Lemma 10.1. Let k ≥ 1.
The sequence consisting of all n’s of length k is obtained
by axiom 8. QED
LEMMA 10.3. For all n,m, (n upthru m) exists. The
interpretation of Finite Interval holds.
Proof: Let n,m be given. We can assume that n £ m. By axiom
5, form (0,...,mn). By Lemma 10.2, form (n,...,n) of
length mn+1. By axiom 6, form (0,...,mn) + (n,...,n) = (n
upthru m). QED
LEMMA 10.4. The interpretation of Boolean Difference holds.
Proof: Let x,y be given. By axiom 9, we obtain the required
sequence. QED
LEMMA 10.5. The interpretation of Least Element holds.
Proof: Let a be nonempty. Apply axiom 10. QED
We now come to the most substantial part of the
verification  Set Addition and Set Multiplication.
We first need to derive QRT = quotient remainder theorem.
This asserts that 52 for all d ≥ 1 and n, there exists unique q,r
such that n = dq+r Ÿ 0 £ r < d.
LEMMA 10.6. For all d ≥ 1 and n, there is at most one q,r
such that n = dq+r.
Proof: Let d ≥ 1, dq+r = dq’+r’, and 0 £ r,r’ < d. Then d(qq’) = r’r. Suppose q ≠ q’. By discreteness, qq’ ≥ 1, and
so d(qq’) ≥ d. However, r’r < d. This is a
contradiction. Hence q = q’. Therefore r’r = 0, r = r’.
QED
LEMMA 10.7. Let d ≥ 1 and n ≥ d. There is a greatest
multiple of d that is at most n.
Proof: By Lemma 10.3, form (d,d,...,d) of length n. By
Lemma 10.3 and axiom 6, form (1,2,...,n)⋅(d,...,d) =
(d,2d,...,dn) of length n. By Lemma 10.3 and axiom 5, form
(n,...,n) and (d,2d,...,dn). By axiom 5 form (nd,n2d,...,ndn).
We now wish to delete the negative terms from (nd,n2d,...,ndn). If ndn ≥ 0 then there is nothing to delete.
Assume ndn < 0. Obviously, the negative terms of (nd,n2d,...,ndn) are in [ndn,1]. Form (1,...,dnn) by Lemma
10.3 and axiom 6. By axiom 9, delete (0,...,dnn) from (nd,n2d,...,ndn). The result lists the nonnegative nid, 1 £
i £ n. By axiom 10, let nid be least among the nonnegative
nid with 1 £ i £ n. Then id is the greatest multiple of d
that is at most n. QED
LEMMA 10.8. The Quotient Remainder Theorem holds.
Proof: Let d ≥ 1 and n be given.
case 1. n ≥ d. By Lemma 10.7, let dq be the greatest
multiple of d that is at most n. Then ndq ≥ 0. If ndq ≥ d,
then d(q+1) = dq+d is a greater multiple of d that is at
most n. This is a contradiction. Hence 0 £ ndq < d, and set
r = ndq.
case 2. 0 £ n < d. Set q = 1, r = n.
case 3. –d £ n < 0. Set q = 1, r = n+d.
case 4. n < d. Then –n > d. By case 1, write –n = dq+r, 53
where 0 £ r < d. Then n = d(q)r = d(q1)+dr, and 0 < dr
£ d. If r = 0 write n = d(q)+0. Otherwise write n = d(q1)+dr, 0 < dr < d.
QED
LEMMA 10.9. Let lth(a) = n and m be given. The sequence a(m)
given by axiom 8 is unique and has the same terms as a.
Proof: Recall from axiom 8 that a(m) has length nm, and for
all q,r with 0 £ q < m Ÿ 1 £ r £ n, val(a(n),n⋅q+r) =
val(x,r))).
According to the QRT, this defines all values at all
positions of a(m). Thus a(m) is unique and obviously has the
same terms as x. QED
LEMMA 10.10. Let lth(a) = n+1 and lth(b) = n, where n ≥ 1.
Let a(n) and b(n+1) be given by axiom 8. For all 1 £ i £ j £
n, val(a(n),jnin+j) = val(a,i) and val(b(n+1),jnin+j) =
val(b,j).
Proof: Let a,n,i,j be as given. Note that jnin+j = (ji)(n+1)+i = (ji)(n)+j. Since 0 £ ji < n,
val(a(n),jnin+j) = val(a(n),(ji)(n+1)+i) = val(a,i).
val(b(n+1),jnin+j) = val(b(n+1),(ji)(n)+j) = val(b,j).
QED
LEMMA 10.11. Let a,b be nonempty. There exists g,d such that
i. a,g have the same terms.
ii. b,d have the same terms.
iii. lth(g) = lth(d)+1.
iv. Let a be a term of a and b be a term of b. Then there
exists 1 £ i £ j < lth(d) such that val(g,i) = a and
val(d,j) = b.
Proof: Let a,b be nonempty, lth(a) = n, lth(b) = m. Let u be
the last term of a and v be the last term of b. Let a’ be
aum, and b’ = bvn, where here multiplication is
concatenation. Then lth(a’) = lth(b’), a’ has the same
terms as a, and b’ has the same terms as b. Finally, let g =
a’a’u, and d = b’b’. Obviously lth(g) = lth(d+1). Clearly
every term in a is a term of the first a’ in a’a’u, and
every term in b is a term in the second b’ in b’b’. We never
have to use the last term of d since the last two terms of 54
b’ are the same. QED
LEMMA 10.12. The interpretations of Set Addition and Set
Multiplication hold.
Proof: Let a,b be given. We can assume that a,b are
nonempty. Let g,d be as given by Lemma 10.11, say with
lengths n+1,n. Let a be a term of a and b be a term of b. By
Lemma 10.11, there exists 1 £ i £ j £ n such that val(g,i) =
a and val(d,j) = b. By Lemma 10.10, there exists k such that
val(g(n),k) = a and val(d(n+1),k) = b. Hence
a+b is a term of g(n)+d(n+1).
a⋅b is a term of g(n)⋅d(n+1).
On the other hand, by Lemma 10.9, g(n) has the same terms as
a and d(n+1) has the same terms as b. Hence
i. the terms of g(n)+d(n+1) are exactly the result of summing a
term of a and a term of b.
ii. the terms of g(n)⋅d(n+1) are exactly the result of
multiplying a term of a and a term of b.
Thus
iii. g(n)+d(n+1) witnesses Set Addition for a,b.
iv. g(n)⋅d(n+1) witnesses Set Multiplication for a,b.
QED
THEOREM 10.14. The interpretation of every axiom of FSTZ is
a theorem of FSQZ.
Proof: By Lemmas 10.3, 10.4, 10.7, and 5.12. QED
THEROEM 10.15. FSQZ proves PFA(Z).
Proof: Since the interpretation of FSTZ in FSQZ used here
is the identity on the Z sort, the result follows
immediately from Theorem 5.29 and Lemma 10.14. QED
THEROEM 10.16. FSQZ is a conservative extension of PFA(Z).
Proof: By Theorem 10.15, it suffices to expand any model M
of PFA(Z) to a model of FSQZ. Use the bounded S0(Z) binary
relations of M which are univalent, with domain some
{1,...,n}, as the finite sequences. QED 55 11. Conservative extensions, interpretability,
synonymy, and logical strength.
The ten systems of arithmetic considered here are
Q(N), PFA(N), PFA(N) + EXP(N), PFA(N) + CM(N), EFA(N,exp).
LOID(Z), PFA(Z), PFA(Z) + EXP(Z), PFA(Z) + CM(Z),
EFA(Z,exp).
These were presented in sections 4,5, and relationships
between these twelve systems were established in sections
4,5,6  especially see Theorem 6.7.
The seven strictly mathematical theories considered here
were presented in section 7:
FSTZ, FSQZ, FSTZ + CM(Z), FSQZ + CM(Z), FSTZEXP, FSTZBEXP,
FSQZEXP.
Recall that FSTEXP extends FSTZ by
i. exp(n,0) = 1.
ii. m ≥ 0 Æ (exp(n,m+1) = exp(n,m)⋅n Ÿ exp(n,m1)↑).
iii. The finite set {exp(n,0),...,exp(n,m)} exists.
FSTZEXP’ extends FSTZ by
i. exp(n,0) = 1.
ii. m ≥ 0 Æ (exp(n,m+1) = exp(n,m)⋅n Ÿ exp(n,m1)↑).
iii. n ≥ 2 Ÿ 0 £ m < r Æ exp(n,m) < exp(n,r).
iv. The finite set {exp(n,0)+0,exp(n,1)+1,...,exp(n,m)+m}
exists.
FSQZEXP extends FSQZ by
i. exp(n,0) = 1.
ii. m ≥ 0 Æ (exp(n,m+1) = exp(n,m)⋅n Ÿ exp(n,m1)↑).
iii. The finite sequence (exp(n,0),...,exp(n,m)) exists.
LEMMA 11.1. FSTZEXP’ proves PFA(Z) + ("n)({exp(n,0),...,
exp(n,m)} exists).
Proof: Recall from Theorem 8.28 that FSTZ proves bounded
S0(Z,fst) separation: 56
*) ($A)("x)(x Œ A ´ (y < x Ÿ x < z Ÿ j)),
where j Œ S0(Z,fst) and A is not free in j.
We argue in FSTZEXP’. Fix n ≥ 0. Let A = {exp(n,0)+0,...,
exp(n,m+1)+m+1}. Note that for all 0 £ r £ m, the next
element of A after exp(n,r)+r is exp(n,r+1)+r+1. Let B be
the set of successive differences of elements of A. Then B
= {exp(n,r+1)+r+1(exp(n,r)+r): 0 £ r £ m} = {exp(n,r)+1: 0
£ r £ m}. Hence B {1} = {exp(n,r): 0 £ r £ m}. QED
We use nm for the partial exponential function, according to
PFA(Z). By [HP98], p. 299, the relation r = nm is given by a
bounded formula in PFA(Z). Note that the relation “r is of
the form nm” is also given by a bounded formula in PFA(Z).
LEMMA 11.2. FSTZEXP’ proves that for n,m ≥ 0, every exp(n,m)
is of the form nr.
Proof: Fix n,m ≥ 0. By Lemma 11.1, let A = {exp(n,0),...,
exp(n,m)}. Let B be the set of all elements of A that are
not of the form nr. Let exp(n,t) be the least element of B.
Then t > 0 and exp(n,t1) is of the form nr. Hence exp(n,t)
is of the form nr. This is a contradiction. QED
LEMMA 11.3. FSTZEXP’ proves n,m ≥ 0 Æ exp(n,m) = nm.
Proof: We can assume that n ≥ 2 and m ≥ 0. Let A =
{exp(n,0)+0,...,exp(n,m)+m}. We first show the following.
Let exp(n,r2)+r2, exp(n,r1)+r1 both be of the form ns+s,
where r ≥ 2. Then exp(n,r) = nr.
ultimately change m to s and s to t.
By Lemma 11.2, write
exp(n,r2) = np.
exp(n,r1) = np+1.
np+r2 = ns+s.
np+1+r1 = nt+t.
s < t.
Hence
np+1+r1(np+r2) = np+1np+1 = nt+t(ns+s) = ntns+ts.
Also by np+r2 = ns+s, we have p £ s. 57 case 1. p+1 < t. Then np+1 £ ntns < ntns+ts = np+1np+1 £
np+1, which is a contradiction.
case 2. t £ p. Then ntns+ts £ npnp+pp = 0 < np+1np+1,
which is a contradiction.
case 3. t = p+1. The only possible case.
So t = p+1, r1 = t, p = r2, exp(n,r2) = 2r2. Hence
exp(2,r) = 2r.
Next we claim that every element of A is of the form ns+s.
Suppose exp(n,r)+r is the least element of A that is not of
the form ns+s. Clearly r ≥ 2 and exp(n,r2)+r2, exp(n,r2)+r1 are both of the form ns+s. By the claim, we have
exp(n,r) = 2r, and so exp(n,r)+r = nr+r. This is a
contradiction.
In particular, exp(n,m2) and exp(n,m1) are of the form
ns+s, and so by the claim, exp(n,m) = nm. QED
LEMMA 11.4. FSTZEXP’ is a definitional extension of FSTZ +
CM(Z).
Proof: By Lemma 11.3, FSTZEXP’ proves FSTZ +
“exponentiation is total”. Hence FSTZEXP’ proves FSTZ +
CM(Z). Also, by Lemma 11.3, FSTZEXP’ proves exp(n,m) = nm,
defining exp. QED
LEMMA 11.5. FSQZEXP proves n ≥ 0 Æ exp(n,m) = nm.
Proof: By Theorem 10.15, FSQZEXP proves PFA(Z). We now
argue in FSQZEXP. Fix n ≥ 0. Let a be the sequence
(exp(n,0),...,exp(n,m)). By using the ring operation axioms
of FSQZ, we obtain the sequence b =
(<0,exp(n,0)>,<1,exp(n,1)>,...,<n,exp(n,m)>)
where <x,y> = (x+y)2+x.
By the separation in Theorem 6.28, and Theorem 10.13, we
obtain a sequence g whose terms comprise the terms of b
which are not of the form <t,nt>. (Only bounded quantifiers
are involved in this construction). Let <k,exp(n,k)> be the
least term of g. Then k > 0, and <k1,exp(n,k1)> is of the
form <t,nt>. I.e., exp(n,k1) = nk1. Therefore exp(n,k) = 58
nk, and <k,exp(n,k)> is a term of b of the form <k,nk>. This
is a contradiction. Hence g is empty. Therefore every term
of b is of the form <t,nt>. In particular, <m,exp(n,m)> is
of the form <t,nt>. Therefore exp(n,m) = nm. QED
LEMMA 11.6. FSQZEXP is a definitional extension of FSQZ +
CM(Z). FSQZEXP is a conservative extension of EFA(Z,exp).
Proof: The first claim is immediate from Lemma 11.5 and
Theorem 10.15. For the second claim, first note that
FSQZEXP proves EFA(Z,exp). This is because given any
formula in S0(Z,exp), we can replace all occurrences of exp
in favor of internal exponentiation, using Lemma 11.5,
thereby obtaining a S0(Z) formula, to which we can apply
induction in PFA(Z) Õ FSQZEXP.
Now let M be a model of EFA(Z,exp). We can expand M to M’
by adding the finite sequences, and associated apparatus,
that is internal to M. Then M’ = FSQZEXP. QED
LEMMA 11.7. FSTZEXP is a conservative extension of FSTZ.
Proof: Let M be a model of FSTZ. For n ≥ 0, define exp(0,n)
= 1 if n = 0; 0 otherwise, and exp(1,n) = 1. Now let n ≥ 2.
Clearly {nm: m ≥ 0} is unbounded. If nm exists, define
exp(n,m) = nm. If nm does not exist, m ≥ 0, then define
exp(n,m) = 0. Note that the sets {exp(n,0),...,exp(n,m)}
are already present in M. Hence (M,exp) is a model of
FSTZEXP. QED
LEMMA 11.8. Q(N), FSTZ, FSQZ, FSTZEXP are mutually
interpretable.
Proof: Since Q(N) and PFA(Z) are mutually interpretable, it
suffices to show that PFA(Z), FSTZ, FSQZ, FSTZEXP are
mutually interpretable. Since PFA(Z) is provable in FSTZ,
FSQZ, it suffices to show that FSTZ, FSQZ are interpretable
in PFA(Z). It therefore suffices to show that FSTZ, FSQZ
are interpretable in PFA(N).
Let M be a model of PFA(N). In M, look at the cut I of all
n such that the internal 2n exists. If I is a proper cut,
then by cut shortening, we can assume that I forms a model
of PFA(N). We can then expand I with all of the internal
subsets of I bounded by an element of I, and all of the
internal sequences from I whose length is an element of I,
and whose terms are bounded by an element of I, to form the 59
required models of FSTZ, FSQZ, FSTZEXP (using the proof of
Lemma 11.7) QED
LEMMA 11.9. EFA(N,exp), FSTZ + CM(Z), FSQZ + CM(Z),
FSTZEXP’, FSQZEXP are mutually interpretable.
Proof: EFA(N,exp) is interpretable in PFA(Z) + CM(Z) by
Theorems 5.6 and 6.7, which is provable in FSTZ + CM(Z) and
FSQZ + CM(Z) by Theorems 8.28 and 10.15.
FSTZ + CM(Z) is provable in FSTZBEXP by Lemma 11.4. FSQZ +
CM(Z) is provable in FSQZEXP by Lemma 11.6.
So it suffices to interpret FSTZBEXP, FSQZEXP in
EFA(N,exp). Interpret the finite sets and finite sequences
by finite coding in EFA(N,exp). QED
LEMMA 11.10. EFA(N,exp), FSTZ + CM(Z) are synonymous.
Proof: By Theorem 6.7, EFA(N,exp) and EFA(Z,exp) are
synonymous. It now suffices to show that EFA(Z,exp), FSTZ +
CM(Z) are synonymous.
We interpret EFA(Z,exp) in FSTZ + CM(Z) by preserving the Z
sort, and interpreting exp as internal exponentiation in
PFA(Z) + CM(Z). We interpret FSTZ + CM(Z) in EFA(Z,exp) by
preserving the Z sort and interpreting the finite sets by
codes in EFA(Z,exp). Let M be a model of EFA(Z,exp). M will
be sent to a model M’ of FSTZ + CM(Z) with the same ordered
ring, but where exp is gone. Since CM(Z) must still hold,
we have an internal exponentiation in M’, and so when going
back, we recover the old exp. For the other compositional
identity, let M be a model of FSTZ + CM(Z). Then the Z part
of M is a model of PFA(Z) + CM(Z), and therefore has an
internal exponentiation. M is sent to a model M’ of
EFA(Z,exp), where exp agrees with the internal
exponentiation in M. When we go back, we must have the same
ordered ring structure, and the sets are those given
internally from the ordered ring structure of M.
Thus it suffices to verify that in M, the sets are exactly
those sets coded internally in the ordered ring structure
of M. By Theorem 8.28, FSTZ proves separation for formulas
in S0(Z,fst). We can use this to prove in FSTZ + CM(Z) =
FSTZ + EXP(Z) that every set is coded internally in the
ordered ring structure, as in the proof of Lemma 11.2.
Recall that internal exponentiation is used in that 60
argument to make sure that the induction or separation
needed has only bounded quantifiers. QED
We now summarize these results.
THEOREM 11.11. FSTZEXP’, FSQZEXP are definitional
extensions of FSTZ + CM(Z), FSQZ + CM(Z), respectively.
FSTZEXP is a definitional extension of FSTZ. FSQZEXP is a
conservative extension of EFA(Z,exp). EFA(N,exp), FSTZ +
CM(Z), FSQZ + CM(Z), FSTZEXP’, FSQZEXP are mutually
interpretable. EFA(N,exp), FSTZ + CM(Z) are synonymous.
FSTZ, FSQZ, FSTZEXP are conservative extensions of PFA(Z).
Q(N), FSQZ, FSTZ, FSTZEXP are mutually interpretable.
COROLLARY 11.12. FSTZ + CM(Z), FSQZ + CM(Z), FSTZEXP,
FSTZEXP’, FSQZEXP are strictly mathematical theories with
logical strength. I.e., they interpret EFA(N,exp). 12. RM and SRM.
The Reverse Mathematics program originated with [Fr75],
[Fr76], and the widely distributed manuscripts [Fr75,76],
which refer to some of our earlier insights from 1969 and
1974. Also see [FS00].
RM is the main focus of the highly recommended [Si99]. This
book has unfortunately been out of print soon after it
appeared, but there are ongoing efforts to have it
reprinted.
In RM, the standard base theory, RCA0, introduced in [Fr76],
is not strictly mathematical. However, in RM, we add
strictly mathematical statements to RM and classify the
resulting theories according to implications, equivalences,
and logical strengths.
Often, the formulations of the mathematical statements
investigated in RM involve coding. Usually these codings
are rather robust, but nevertheless constitute another
place where elements that are not of a strictly
mathematical nature appear.
Fortunately, there are substantial areas of mathematics and
a substantial variety of mathematical statements whose
formulations are sufficiently robust to support the
vigorously active development of RM. RM has continued to
grow very substantially since its inception in the 1970s. 61
We fully expect an accelerating development of RM for the
foreseeable future.
However, there is a much greater body of mathematical
activity which is currently not in any kind of sufficiently
robust logical form to support an RM treatment. The bulk of
the relevant mathematical statements are probably too weak,
in terms of logical strength, for an RM development, since
RM starts at the logical strength level of PRA (primitive
recursive arithmetic). PRA is far stronger, logically, than
the nonzero level of logical strength on which this paper
is based  that of EFA(N,exp), or equivalently, IS0(exp),
and lower.
We view this paper as an introduction to what we call
Strict Reverse Mathematics, or SRM.
In SRM, the focus is on theories where all statements are
strictly mathematical  including all axioms in any base
theory. In a sense, SRM is RM with no base theories at all!
Here we have shown that one can achieve logical strength
using only strictly mathematical statements. Without this
fundamental fact, there cannot be any SRM.
The major goal of SRM is to rework and extend RM using only
strictly mathematical statements. SRM should strive to
create sensible logical structure out of a vastly increased
range of mathematics, going far beyond what can be analyzed
with conventional RM.
An integral part of SRM is to take the standard natural
formal systems developed in the foundations of mathematics
 whose axioms are very far from being strictly
mathematical  and reaxiomatize them with strictly
mathematical statements. Such axiomatizations may take the
form of conservative extensions or mutually interpretable
or synonymous systems, as we have done here for PFA(N)
(i.e., IS0), and for EFA(N,exp) (i.e., IS0(exp)).
In this vein, we mention some SRM challenges.
1. Find a strictly mathematical axiomatization of PFA(Z),
in its signature L(Z).
2. Find a strictly mathematical axiomatization of PFA(Z) +
EXP(Z), in its signature L(Z). 62 Some work in the direction of 1,2 is contained in [Fr00].
3. Find a strictly mathematical axiomatization of
EFA(Z,exp), in its signature L(Z,exp).
4. FSQZ appears to be too weak to be naturally synonymous,
or even synonymous, with FSTZ. The same can be said of FSQZ
+ CM(Z), FSTZ + CM(Z), and also FSQZEXP.
However, we can extend FSQZ to FSQZ# and obtain synonymy.
FSQZ# is axiomatized by
1.
2.
3.
4.
5. LOID(Z).
Discreteness.
lth(a) ≥ 0.
val(a,n)Ø ´ 1 £ n £ lth(a).
S0(Z,fsq) comprehension for finite sequences.
(n ≥ 0 Ÿ ("i)($!j)(j)) Æ ($a)(lth(a) = n Ÿ ("i)(1 £ i £
n Æ j[j/val(a,i)])), where j is a S0(Z,fsq) formula in
which a is not free.
6. Every sequence of length ≥ 1 has a least term.
The challenge is to give a strictly mathematical
axiomatizations of FSQZ#, FSQZ# + EXP(Z), FSQZ#EXP, in
their respective signatures L(Z,fsq), L(Z,fsq),
L(Z,exp,fsq).
The notion “strictly mathematical” is sufficiently clear to
support the SRM enterprise. However, there are still fine
distinctions that can be profitably drawn among the
strictly mathematical. We have drawn such distinctions in
our discussion of the relative merits of FSTZ and FSQZ at
the end of section 7.
It is clear from the founding papers of RM, [Fr75,76],
[Fr75], [Fr76], that we envisioned a development like SRM.
We spoke of raw text, and our original axiomatizations of
main base theory RCA0 and our other principal systems WKL0,
ACA0, ATR0, and P11CA0, of RM, were considerably more
mathematical than the formally convenient ones that are
mostly used today. However, any major development of SRM
before that of RM would have been highly premature.
[Fr01] and [Fr05a] are technical precursors of this paper,
dealing with FSTZ and FSQZ, respectively. [Fr05] is a
preliminary report on SRM, attempting to develop SRM at 63
higher levels of strength, and in many ways goes beyond
what we have done very carefully here. However, this
earlier work will undergo substantial revisions and
upgrading in light of this initial publication.
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*This research was partially supported by NSF Grant DMS0245349. ...
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 Math, Logic, Model theory, Finite set

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