SUBSYSTEMS OF SET THEORY AND ANALYSIS
by
Harvey Martin Friedman
SUBMITTED IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF
PHILOSOPHY
at the
MASSACHUSETTS INSTITUTE OF
TECHNOLOGY
August, 1967
Signature of Author.,.......i.-....
Department of Mathematics
August , 1967
Certified by ... rw.
..-...........
Thesis Supervisor
L ;1
by.......
Accepted
-.
ivS
0
.....
- -
- ·
.. -· -
.
-
I.
·.
·0
0
0·
Chairman, Departmental
Committee on Graduate
Students
2
SUBSYSTEMS OF SET THEORY AND ANALYSIS
by
Harvey Martin Friedman
"Submitted to the Department of Mathematics on August , 1967
in partial fulfillment of the requirements for the degree of
Doctor of Philosophy."
Abstract
Usual set theory is formulated in terms of closure
conditions. A typical example is the power set axiom, which
asserts closure under the operation of power set.
In Chapter I we consider set theory based on closure
conditions applied only to definable sets. We formalize this
set theory and call it ZF*. Our principal result of Chapter I
is that, provably in first-order arithmetic, ZF is consistent
if ZF* is. Our proof of this theorem uses a Skolem hull
construction and a syntactic transformation.
In Chapter II, we consider three theories of hyperarithmetic analysis,
A1 -CA, 21 -AC,
and
1 -DC.
:
These are called
theories of hyperarithmetic analysis primarily because the
hyperarithmetic sets form a minimum w-model for each of them.
1
1
We first show that £1 -DC is a conservative extension of
1-CA
for purely
21 sentences. The proof is by means of an inner
model construction. Careful attention has to be paid to limit
the axioms we use to prove relevant sentences about hyperarithmetic sets.
We then show that there are theorems of
which are not theorems of
finding a suitable sentence
1 -AC.
S
1-DC
This is done by first
and considering the auxiliary
theories
:1 -AC + S
and
:-1DC + S.
We then obtain our inde-
pendence result via Gdel's
Theorem, by showing that
Con(Z1 -AC + S)
in
£1-AC
of
is provable
1 -DC
+ S.
Last, we show that
is a conservative extension, for purely
T,
21 sentences,
a natural subsystem of predicative analysis. The proof
uses an inner model construction on certain auxiliary theories.
Thus, a model for each finite subsystem of
Z1 -AC
is obtained
as an inner model of a model of an extension, by the negation
of an instance of induction, of a corresponding finite subsystem of
T.
Thus non-standard models are implicit in the
construction.
Chapter III is concerned with hierarchies (based on the
Jump operator) on recursive linear orderings. Let X be the
set of recursive linear orderings which have no hyperarithmetic
descending chains. Joseph Harrison showed that there are elements of X, which are not well-orderings, on which there are
hierarchies. We first show that under certain weak conditions
on a recursive linear ordering, that if there is a hierarchy
on it, then it must be in X. Finally, we establish the existence of a recursive linear ordering which is in X, yet on
which there are no hierarchies. The proofs of these assertions
use certain Lemmaswhich are proved in the following indirect
way: one assumes the Lemmais false,
and then forms a theory
consisting of the negation of the Lemma together with certain
true
sentences;
then one shows that the resultingtheory
proves its own consistency.
Thesis Supervisor: Gerald E. Sacks
Title:
Professor of Mathematics
TABLE OF CONTENTS
Title Page .........
...................
1
Abstract ................................................. 2
Biographical Note........................................ 6
Acknowledgments.......................................... 7
Bibliography .
............
................................. 8
Introduction...................................... 9
Chapter I
1.
General Situation...............
2.
Remarks on Terminology and Notation............
16
3.
Axioms......................... 0 0
4.
Outline of Proof of Main Theorem a 0 a
5.
Developmentof Ordinals.........
6.
Development of
7.
...............15
,,
0
9 00 00 0
0
&
16
0
.0
00
&
*
0
18
0
a
0
*
0*
00
&
0
0
21
L...............
*
0
0
0
00
0*
*
0
0
25
The System
ZF*' ................
0
0
0
0
00
00
&
&
0
32
8.
The System
ZF' ................
.
0
&
*
00
90
0
0
0
37
9.
The Skolem Argument.............
0
0
0
0
&0
0*
*
&
0
38
ZF, Parameterless ZF, and
0
0
&
*
*6
0.
&
0
0
10.
ZF*.
40
5
Chapter II
48
1.
Definitions of Systems..........
2.
Preliminary Lemmas..............
3.
Conservative Extension Result...
4.
Independence Result.............00000000000.000*00 65
5.
Relation between Predicative and Hyperarithmetic
Analysis.....................
..0..00
.0.
0..000
52
0.....0.000...0
..
59
68
,....0...........0
Chapter III*.............................
.00...0.00..0.000S
77
6
BIOGRAPHICAL NOTE
The author was born in Chicago, Illinois in September,
l948.
The author attended Highland Park High School in
Highland Park, Illinois from September, 1961 to June, 1964,
and Massachusetts Institute of Technology from September, 1964
through August, 1967.
7
ACKNOWLEDGMENTS
The author wishes to express thanks to Professors
G. Sacks, S. Feferman, and G. Kreisel for their valuable
conversations; to Professors Kreisel and Feferman for their
valuable correspondence; and to Professor Sacks for his
remarkable patience and encouragement.
8
BIBLIOGRAPHY
1.
S. Feferman, Systems of Predicative Analysis, Journal of
Symbolic Logic, vol. 29 (1964), pp. 1-27.
2.
Joseph Harrison, Dissertation, Some Applications of
Recursive Pseudo-Well Orderings, August 1966, pp. 4-7,
31-33.
3.
S. C. Kleene, Introduction to Metamathematics,
4.
G. Kreisel, Survey of Proof Theory, to appear.
5.
W. W. Tait, Normal Derivability in Classical Logic, to
appear.
1950.
9
INTRODUCTION
Set theory is usually formulated in terms of closure conditions. A typical example is the power set axiom, which
asserts closure under the operation of power set.
In Chapter I,
we consider set theory based on closure conditions applied only
to definable
sets.
We formalize this set theory and call it
ZF*, and we give a consistency proof of
ZF
relative to
ZF*.
A direct method presents itself for obtaining this relative consistency result; namely, to use a constructible set
construction,
and prove within
ZF*
the relativized to the
constructible sets of each instance of
ZF.
This is, of
course, in analogy with the method of proof for the consistency
of
ZF + AxC relative to
ZF.
However, an examination of the
basic principles needed for such a constructible set construction to go through reveals the need for the least counterexample
principle for ordinals to be provable in
ZF*. By the least
counterexample principle for ordinals, we mean the schema
(3 a)Pa ->
(34a)Pa, where
P
is any formula. It does not
appear that this schema is derivable in
ZF*, even if
restricted to have only one free variable, a.
P
is
Of course, in
ZF, the schema is derivable by means of a closure condition
applied to all sets as follows:
an
a.
Then form
[l0fea & P,
assume
(3 a)Pa,
and fix such
and use Foundation to obtain
("a)Pa. This illustrates the basic difference between
ZF*,
in that the closure condition,
necessarily provable in
ZF*
3
only when
lpea & Pal,
a
ZF
and
is
is given a definition.
10
Such a direct attack seems hopeless.
An outline of our
proof can be found in Section 4 of Chapter I.
Towards the end of Chapter I, we show how to add elements
"on top of" a model of
ZF,
to obtain nonstandard models of
ZF*.
By this means, we obtain results concerning independence
from
ZF*.
In Chapter II, we consider three theories of hyperarith-
metic analysis. These are
CA
A1 -CA, I1 -AC,
and
Z1-DC.
Here
refers to comprehension axiom; AC, to axiom of choice; DC
to dependent choice. The hyperarithmetic sets form a minimum
w-model for each of these theories. We first prove that
£1-Dc
1-cA
is a conservative extension of
for purely
1
sentences. The proof is by means of an inner model construction. We show that given a model of
1-CA,
the submodel of all sets hyperarithmetic
submodel satisfies
1 -DC.
if we then take
in a fixed set, this
Careful attention has to be paid
to the way in which the notion of relative hyperarithmeticity
is
formalized. The formulation in terms of hierarchies seems
to be the correct one here (not
).
The usual proof that the
sets hyperarithmetic in a fixed set always form an
of
2E-DC
is too crude for our purposes.
w-model
It uses the
comparability of all recursive well-orderings, which is a
principle too strong to be provable in
know, in
A 1-CA,
A1 -CA.
However, if we
that given orderings have hierarchies (based
on the Jump operator) on them, we can then conclude, in
AlCA,
their comparability. Thus, the: key point of our proof of the
conservative extension result is the Judicious use of
11
hierarchies on orderings.
Our inner model construction can, in the standard way, be
1_-DC rela-
transformed into a finitary consistency proof of
tive to
A1-CA.
The second result of Chapter II is the independence of
£-DC
from
1-AC.
It would seem to be the case that a Cohen
type argument would be not only useful here, but perhaps necesWe found that quite the contrary is true.
sary.
Our proof
does not use a Cohen type argument, and Cohen type arguments
do not seem to be helpful here, since it seems difficult
to
find Cohen type models (starting with the standard model, the
hyperarithmetic sets, of
following property:
Z1 -AC)
which do not have the
any arithmetical predicate (in x)
satis-
fied to have a solution in the new model has a solution hyper-
arithmetic (in x)
in the new model.
This property can be seen,
from our Theorem 1 of Chapter II, to imply that the new model
satisfies
1 -DC.
Instead, we use G8del's theorem.
and consider the auxiliary theories
We show that
So if
But
+ S
Z-1_DC
1
We choose a sentence
Z1 -AC + S
and
proves the consistency of
1_-DC+ S = Z1 -AC + S,
then
Z1 -DC + S
S is chosen to be a true sentence;
so
S
-1DC + S.
Z11 -AC + S.
is inconsistent.
1_DCf
Notice that the assumption Con(M -DC + S)
1_AC.
is needed for
the independence. We do not know if there is a finitary independence proof (i.e., a finitary proof of consistency of
Z1 -AC + ~F
in
z 11 -oDC).
relative to
-AC,
for some
F
that is provable
12
The key property of
B,
21 -DC
£1 -DC + B,
one can construct, in
key property of the sentence S
is equivalent, in
IT2 sentence. S
is that for any
1
n2
sentence,
an W-model for
is that the theory
B.
Z1-AC + S
21 -DC, to an extension of induction by a
intuitively says that for all sets x,
recursive well-ordering in
x
proves
-DC + S proves
S ->
every
has a hierarchy starting from
More information may be obtained than independence.
see that
The
Con(Il-AC + S);
Con(21-AC + S),
hence
which is a purely
x.
We
£i-DC
21 sentence.
Furthermore, an examination of our proof yields that C:-AC
together with only a finite number of instances of no parameter
£1 -DC
2-DC
is needed to prove
+ S
1
3
proves
S ->
Con(l1 -AC + S).
an w-model for
Z1 -AC + S.
1
we can conclude that there is a purely
provable in
1
Finally,
From all this
sentence which is
ZI-AC together with a finite number of instances
of no parameter
1-DC, but
which is not an w-consequence of
£1-AC.
(An instance
Z -DC,
except the hypothesis,
of no parameter
1 -DC
is the same as
(f)( 3 g)A(f,g),
arithmetical
withno free variables
must have
other than
A
f and g.)
The last result in Chapter II is concerned with the
relation between
1 -AC
ZI
predicative analysis, T.
and a certain natural subsystem of
T
represents the first
of predicative analysis. We show that
extension of
T
for purely
U1
1 -AC
cO levels
is a conservative
sentences. This result,
together with the known characterization of the provable
ordinals of
T,
gives the provable ordinals of
Z1 -AC. Fur-
thermore, our proof of conservative extension uses an inner
13
model construction, which can be transformed into a finitary
1
i-AC
proof of consistency of
relative to
T.
Now
T
known to have a predicative consistency proof; and so,
must
is
then,
21-AC.
The relative consistency result is somewhat surprising,
since the minimum
-model of
the minimum w-model of
T.
ZI-AC
is so much larger than
With this in mind, it is not
surprising that nonstandard models (i.e., non-w-models) must
be essential in our proof.
each finite subsystem of
In the proof, we obtain a model of
£1 -AC
as an inner model of a model
of an extension, by the negation of an instance of induction,
of a corresponding finite subsystem of
T.
In Chapter III, we consider which elements of
hierarchies. We generalize
W*
have
to include recursive linear
orderings whose field is not necessarily
lize hierarchies,
W*
w.
We also genera-
so that at successor we merely have a set in
which the Jump of the set at the predecessor is recursive; at
limits, we have a set in which the effective union of the
previous sets is recursive. Harrison proved that
3 neW*-W
on
which there are hierarchies, in the less general sense W*, W
and hierarchies. He left open whether every
hierarchy.
neW*
has a
We answer it in the negative for our general notion
of hierarchy and the less general notion of
W*.
We also show, under weak conditions, that if
hierarchy, then
n
has a
neW*.
The proofs use certain Lemmas which are proved in the
14
following highly indirect way:
we assume the Lemma is false,
and form a theory by adding true sentences to the negation of
the Lemma; we then show that the resulting theory proves its
own consistency.
15
CHAPTER
1.
General Situation.
axiom
I
Suppose we are given a comprehension
(x1 )...(xn)(3 y)Rxl...xny. We are interested here in
forming the derived schema consisting of the axioms
[(3 tx1 )(3 x2 )...(3 !xn)(FlX
1 &...& Fnxn)
(3x
where the
) (3x2 ) .. .
Fi
(3xn ) (FlX1 & F2 x2 &...& Fn
->
n
& (y)Rx 1 ...
xn),
are formulae with only the free variable xi.
We are purposely vague about the general situation (what
is a comprehension
axiom?), since we have only looked at this
derived schema when the original axiom is drawn from a natural
set of axioms, such as
ZF,
or analysis, or other naturally
occurring systems.
More specifically, we will look at the schema of schema
formed by taking the union of all the schema defined above
corresponding
to each of the comprehension
axioms of
ZF.
We
will not perturb the other (non-comprehension) axioms of
except in minor ways.
We call this derived theory
inessentially modify the Replacement schema in
ZF
ZF,
ZF*.
(We
for
convenience, so that each instance is appropriately placed in
the form
(x1 )...(Xn)(3y)Rxy, so that we may pass to the
derived schema in the manner above.) The axioms of
ZF*
ZF
and
are spelled out in detail, in an elegant form, in the
next section.
We are interested in the relation between
ZF
and
ZF*
as axiomatic theories. Our main result is that
IFET ConZF* ->
system of
2.
ConZF.
It is obvious that
ZF*
is a sub-
ZF.
Remarks on Terminolo
and Notation. The only (standard)
symbols that can occur in a formula of
" e;
2 2-ary relation symbols "=",
the
ZF
(or
ZF*)
are the
2 quantifiers
(3x) and
(x); the propositional connectives; and variables xi,
zi, ui, vi, etc.
Yi,
Everything else is nonstandard; when
nonstandard symbols occur in a formula, they are meant to be
expanded out in such a way that the mere occurrence of a
non-standard symbol implies existence of the corresponding
set.
For example, x =
(3 z) (w)(wez
would be
y
is an abbreviation for
(3 u)(weu & uz))
& x - z].
Also, say,
(3y)[ (z)(zy) & yex].
3.
Axioms.
0.
Axioms for predicate calculus with equality.
1.
Extensionality.
2.
Infinity.
3.
Power set. (xO)(3 l)(X2 )(x2 ex1 - (x3 )(x3ex
2 ->
4.
Sum set.
5.
Replacement
variables
Then
e x
x
For
ZF, we have
(x0
(3 xO)(
3
(x0)(
and
xl)
e x
xl
)
schema.
y
=
2 )(x2
ex
& (x1 )(x1 e x0 ->
(x 2 ) ( x 2
Let
(
ex
l
Axy
-
2
e x 1 ).
x 1 U {x1 le x0 )).
x3 x0 )).
(3 x3 )(x3 ex0 &3 x2 ex )).
be a formula with the free
and possibly more free variables Xl,..,x
n .
17
(X)''' (n) (Y)[(3 Y 1 )(Y2 )(Y2 E Y1 - (3 Y3 )(AY3 Y2 & (Y4)
(Ay3 Y4 ->
(The domain is
is an instance.
Y4 = Y2 ) & Y 3 E
))J
and the axiom asserts the
y
range of the partial function, A'y3 y 2 = AY 3 y2 & (z)
(Ay3 z ->
z = Y2 ), on the domain
Foundation. (xO)(xO
6.
y,
exists.)
(3 x l ) (x l
0 ->
& (x2 )(x 2 ex ->
ex
x2 x O ))).
It is clear that by the usual process of making partial
functions into total functions axiom schema 5 is the same as
the usual formulation in the present context.
Now let
formula
be a formula of 1 free variable. Then the
Ax
(3 y)(x)(x e y - Ax)
For
ZF*
Same as
ZF.
1.
Same as
ZF.
2.
Infinity.
(x1
Fin(y)).
5.
(x1
A
6.
Fin(y)
y
xO - (x2 )(x2
e
x1
e
x2 & Ax2 )),
xe
O ->
CA -- > ( 3 xO)(x
->
Ax
2 )),
A
A
CA -
with 1 free variable, B
ZF.
e xO
1)
( 3 x 0 )(x 1 )(x
1l x
O -
with 1 free variable.
x0 - (3 x2 )(Ax2 & Bx2 x1 & (x3 )(Bx2 x 3 ->
Same as
[X1
with 1 free variable.
Replacement schema. The instances are
E
xU
is a finite ordinal.
Sum set. The instances are
2 ) ( xI
1
will be defined later.
Power set. The instances are
4.
(3x
E xO & (Xl)(X
(3xo)(0
Intuitively it means
3.
CA.
we have
0.
& (y)(y E x 0
is abbreviated as
CA ->
(3 xo)(xl )
x3 = Xl))),
for
with exactly 2 free variables.
)
18
Our formulation of Power set in
Remarks:
ZF*
is seen
to be equivalent to the derived schema of Power set in
ZF
(as given in General Situation) by noticing 1) that if
(3 y)Fy,
if
then
CA,
CA, where
A
These latter are obtained
(3!x)(y)(yEx - Ay).
then
ZF*,
by axiom 1 of
and 2) that
(3y)(Fy & xy)
is
The same remark applies
Extensionality.
to Sum Set.
the same idea yields that the union of the
Essentially
derived schema of the instances of Replacement in
ZF
is equi-
valent, in the present context, to the schema
(CA1 &C
B(x2,x
fxl A 1x,
A) >>-(3xO)(xlExO (3x2 )(Ax
2 &
[x lA2
3,,
x ...
[x l
An 3) & (x3 )(B(x2 ,x 3,[x fA1x,...
{XIAnX})
where
B
free variables,
n + 2
has
-- > X3 =
A
1) ) )
and the
the present context) in Replacement in
ZF*.
A,
{x A2 x3 ,..
ment in
Bxy
(CA1 &...& CAn) ->
as
., x Anx}).
ZF*
is obvious.)
NOTE:
ENTConZF* - ->
ConZF.
ZF*
ZF*,
setting
B(X2 ,xl,fxAXX,
1
"In the present context"
ZF*."
Outline of Proof of Main Theorem.
developing in
(in
(That the above schema contains Replace-
means "using the other axioms of
4.
have 1
But the above is
easily seen to follow from that instance of 5 of
as
Ai
We want to show this schema is contained
free variable.
A
,
The main theorem is
The first step in proving this is
an adequate definition of ordinals, which
19
turns out to be a much more delicate matter than for
ZF, due
to the lack of certain key instances of Replacement in
By an adequate definition of ordinals in
ZF*.
ZF*, we mean a
definition of ordinals such that provably in
ZF*, members of
ordinals are ordinals, and (for a natural definition in
of
w)
w
ZF*
is an ordinal, and ordinals are comparable by
e,
and the e-relation on the ordinals is transitive, and antisymmetric, and antireflexive,
and every ordinal has a (natural)
successor, except possibly the greatest ordinal.
turns out that in
ordinal
ZF*
(It even
we can prove there is no greatest
for our definition of ordinal given later.) The
definition is made and Lemma 1 establishes the above properties
for it in the next section; we even obtain more:
provably in
that,
ZF*, the new definition of ordinal coincides with
the usual definition
given in
ZF,
on definable sets.
(This
is made precise in Lemma 1, f).)
Next we develop an adequate definition of
Among the properties of the predicate x
have, provable in
w
L,
ZF*,
L
every member of an
the new definition of
x
L
L
ZF*.
needed, we must
x E L
is
L,
coincides with the usual
definition of constructibility for definable x,
definable well-ordering of
within
and a
L.
With this machinery, an apparently straightforward
"proof" of our main result comes to mind.
prove the relativized to
L
Namely, Just to
of each instance of
by taking least counterexamples
ZF
of various things in
G6del established the relativization of the axioms of
in
ZF*
ZF*, as
ZF
to
20
L
within
ZF.
But a moment's reflection will reveal that one
can hardly expect that
ZF* will prove any general least
counterexample
principles; i.e., one may well be able to prove
in
(3 x)(xeL & Px),
ZF*
that
(3ix)(xeL & Px)
in
ZF*,
yet not be able to prove
where
the definable well-ordering
of
U
is defined in terms of
In fact, we do not even
L.
see how to prove each instance of the relativized of
L,
within
ZF*
to
ZF*!
In order to get our main result, we form an auxiliary
system
ZF*' C ZF*, whose definition depends on a certain
crucial transformation on sentences, T.
the property that each instance of
This subsystem has
ZF*' semi-relativized
(semi-relativization is a certain modification of relativization) to
L,
is provable in
auxiliary theory
ZF + V = L
is
to
ZF'
ZF*.
ZF*.
We form another
which is related to
It turns out that
ZF*'
about as
ZF'
ZF.
It
also turns out that in the theory obtained by semi-relativizing
each comprehension axiom of
ZF*'
and retaining the other
axioms, one can give a Skolem hull argument for each finite
subsystem of
ZF' that proves the existence of a (suitably
definable) model of this finite subsystem, and hence its
consistency.
Putting all this together we get a finitary proof that (n)
ZF COn(ZF n ) .
So
FENTConZF* ->
ConZF.
21
5.
Development of Ordinals. We define
Ordx = Trans(x) &
e-Conn(x) & (x is semi-closed under succession) & there are
no 3-chains in
&
x,
i.e., Ord(x) = (y)(z)((ycx & zEy) ->
& zx)
(y)(z)((yE-x
-
-
yZ
v y =
V ZE
zcx)
z)) & (y)(yex->
(3z)
[ (w)(WEz -- (wey v w=y)) & (zex v z=x)I) &: (y)(z)(w)([yez &
ZE
& (wey V w=y)]). We define
y Q x) ->
Ord'(x) = Ordx & (y)((Ordy &
(y x v y=x)) - Ord"(x) - Ord'(x) & (y)(Ord'y ->
(xC_y v yC x)).
Ord" will be our notion
It is obvious that
(x) (y)(Ord"x & Ord"y ->
For
Ord(x), we define
of ordinal in
(xc y v yEx v y'=X))
is successor of
y
ZF*.
x
if and only if
y =x U {x.
Lemma 1:
The following are Theorems of
ZF*, where
Ax
has 1
free variable:
a)
If
Ordx & yx,
b)
If
Ord'x
c)
For
x
then
and
with
yEx
Ordy.
then
Ord(x),
Ord'y.
and
not a successor, ltJx
= x.
x
For Ord(x), x a succe ssor, x = y
d)
If
C)rd"(x)
e
If
yr
f)
If yr =
g)
Ord"( (w). (Explained 1below.)
h)
If
= {xAxI,
Proof :
v y=w.
then
Ord'( y),
xlAx}, and
a)
then
wEx
We can't have
So
z EX
WEx,
Ax ->
wey.
aj
nd
To see
and so
Ord"(y).
Ord"(y).
Ord"(x),
yx.
yew v y=w
t.Jx = y.
Ord"(y).
Claim Trains(y).
we have
3-chain.
and
yex
fxiAx}, Ord(y ), then
y r=
Trans (x),
and
yi, we have
Let
zy,
By
then
Ord t" Vly).
wz.
Th,en by
-Conn(x), w Ey V yEW
because we would h;
ave a
-Conn (y), let
zEcw V WEZ v w=z
zey, wey.
by
T'
hen
-Conn(x).
22
zIex
or
z U {zi = x.
z U f{zey
or
z Lt (zz = y
z lJ
yez Ut {z}t, then
3-chain yey, yy,
So
yey.
Ord(y).
for if
z=x, then xC
zex v z=x. But
Hence
since if
yz,
But
By
zex.
So
yey.
and hence
y,
zey v z=y v yez.
e-Conn(x),
a,b,cex,
Towards showing Ord'(y), let
By Trans(x), z C x.
y, Ord(z).
z4x,
U {z.
x.
and we have a 3-chain in
z'
yz
which yields the
yey,
bec, (cea v c=a), a,b,c,ey. Then
Now suppose ab,
By a), we have
But if
The first yields the 3-chain
yez v yz.
the second yields
zey, yez, z=x;
b)
yez U (z}.
or
we have
e-Conn(x),
By
zex, and
zey. Hence
semi-closed, let
y
Towards showing
then
yez,
Y Ey.
Ord(x),
Let
c)
(y)(y U
with
x
every member of a member of
exists.
if Ux
{y) ex. But
yU
member of
x
if Lix exists.
no 3-chains. So Ux C z,
is a member of
a member of
d)
xC
By
z,
then
y _ z.
c),
then
y '.
x.
x.
x,
So
lJx exists and is
yC
z.
If
zex,
Let
If
Ord'(z).
z . x,
then
then
x = z U
{zl,
since x
zx,
But
wz,
If
x.
if it exists.
since if
Ord'(y).
x,
since also every
x =-Ux,
is a member of a member of
x,
So UxC
x.
By semi-closure of
{y}, and so
yeyU
then again Ux C_ x,
z
is a member of
x
yex.
Now let
By Trans(x),
x).
yi
has
But every member of
then
w
is a member of
z.
If
z
z
x.
zex v z=x.
If
z=x,
Then
yez v zy .v
x
z=y.
Hence
x
y v
23
e)
Let
Ord' (y), y = {xJAx .
1)
All elements of
Ord .
For every
members of
y
have
Uy
Since
are
If
z
uez,
yC
y_
y
xC z v z
x.
2)
Ord".
If
y = u
uEz v u=z.
So
y
Some element of
The
Suppose
If
u
then
u C z.
u=z,
then
zC
x.
contains
z.
z.
z C y.
proved that any
be any
Either all
y D z.
Suppose the second holds, i.e., some
Then clearly
z
Then let
is not a successor, by c) we
z.
we have
then
Ord".
or some member of
Then if
y, and so
Ord'(z),
are
we have
xEy
the first holds.
y
Then either
Ord'
(unique)
xEy
is ord".
y
is not
Ord".
We have
such that every member is
e-least element of
is definable, and has every member an
y
Ord"
is
which is not Ord"
Ord".
by suitable use of Replacement schema of
ust
(There is a least
ZF*,
and
Foundation.)
Just apply 1) to obtain a contradiction.
f)
We merely have to show all definable Ords are Ord'.
As in
e) we go down to a definable Ordinal all of whose members are
Ord".
Let
Ordz.
We wish to show
3 wEy
with
If
z
either
w
w U
z C y, z=w,
y
be such an ordinal,
w·z.
then
y =
zEy v z=y.
Suppose
But since Ord"w,
z=w v zw
wc)y v w U
and so
(wU fw]). Hence wU
out, so zC wl
{w],
and so
w] = y.
zey.
So
xfBxl.
Let
zfy & yz.
we have
zcy.
If
wlU
w U {w]Ey.
So
w
z = w U {w) v zEW U (w],
Then
w.
z, w z.
w) = y
Now then
Ord"
we have
zey.
Now
then since
The first is
Ord'(w tl (w})
either one implying
y,
z v z
w
f
{wJ)C_z v z C w U {w).
and since
z
24
g)
We now explain Axiom 3.
and
(z)(zex ->
(3 w)(z=w U
Ordx & (3y) (y=xU I xi).
Fin(x) = Ordx & (3y)(x=y Ordy)
wl)).
We let
x
w
is a successor
be the
x0
in
Ord =
Ax. 2 of
ZF*.
Now clearly
show
is definable,
w
Ord(w).
1)
Trans(w).
Let
successor
Ord
or
0.
successor
Ord
or
0
Ord
SC)
or
0,
and so
new.
Since
Let
n
Then
x
x
is transitive,
and every member of
x
is a
x
is a
is a successor
is a finite Ord (i.e.,
s-Conn (w) .
Suppose 3 new
nn & mn & nm. Tt
ake
There is an
Fin(x)),
k d 0,
any
-least such
k
and
mew,
lm
then
m=l
But if
1
1
1
meW,
k.
and Foundation.
by Foundation
lew . We must have, for some
mel, then mck. If
k=lU Il
me'W, k n &
then
= mvlU
mek. If
l) em,i .e.,
Contradic tion, by the lack of 3-chains in Ords,
and the comparability
directly
and call it
is a sulccessor Ord, and
m k & knm & (lem v me1 v m=l).
3)
n
ZF*
em v 0=m, for any
since
and by Trans(w),
k:=mv kem.
such that for some
-least by Replacement in
and Trans (m). Hence
of
1
and
m.
e under succession. This is insured
Semi-closur
from Axiom 2 of
4)
n ,m, rEW.
have
xen.
XEW.
2)
Now
and so we merely have to
rer,
contradict
No 3-chains
·
By Trans(r),
if
ren.
Ord(r).
ZF*.
Suppose
nem, m er, rn
we have nr.
If
r=n,
v r=n,
By Trans(r)
we also have
rr.
where
again,
These
we
25
Trans ~Jy).
1)
wey;
hence any
Ord".
both
So
3)
z,wE Uy,
If
w'
w
or
Then
zew',
Semi-closure.
Let
Ord"(w),
u&Jy.
then
zez' Iy,ww'EY,z',w'
Without loss of generality,
v w=z.
zew v wz
Hence
zewey, Ord"(w).
Then
ze Uy.
or ZJfzew.
z U {z}Ew
So
z'.
wEw'.
for some
zew
then
uEw, wey.
has
y).
z
z' C w'.
assume
z Jy,
If
uez
s-Conn
2)
So
Ord~(Uy).
We have only to show
h)
z U FzIE Uy.
In the first case,
In the second suppose
some ucy
A)
so
w C u,
but
wu.
has either
w=u
or
has
Hence
weu,
and
z U {z} = we Uy.
ucy
Every
B)
Uy = w = z
u
w.
Then clearly
(z3.
A) and B) are exhaustive.
4)
a,b,c
e Uy.
cec.
If
6.
Then
c=a,
Develo
a,b,c
nt of
are
then
aec,
aeb, bec,
Suppose
No 3-chains.
L.
Ord",
cec.
of course,
aEc.
Contradicts
If
where
then
cea,
Ord(c).
We wish to define a class of sets,
which has a definable well-ordering,
L,
and
(cea v c=a),
and provably so in
L,
ZF*.
We are interested in
will not be an object.
the predicate xL.
We let
n,m,r,p,q
be special variables
We let
a,p,,y,... be special variables
Ord"x.
We write
a+l
for
a 'U {a}.
for elements of
for sets
We let
X
variable for limit (non-successor and non-null)
x
with
be a special
Ord"'s.
w.
26
x= M ( a
We say
->
(3f)(Domf = a+l
)
U f(x) = f(X)) & ()(Bea+l ->
& f(0) =
& (X)(Xca+l
f (+l) = Fodo(f(B)) &
XEX
x (3 xo)(3 n)(x=(zlzey & <y,ey>
Fodo(y) =
x=f(a)), where
n(z)[xo]l & FinSeq(x,y))],
<yey>
n(z)[x
O
satisfies the formula with
the structure <y,ey>
]
which is the sequence starting with
z,
means
G6del
at the sequence of elements of the domain
n
number
where
y, (z,xo) ,
followed by the
sequence xO.
Remark on formalization: We formalize the satisfaction relation
in
ZF*
the same way we do in
Terminoloy
Remarks on
and Notation.
Lemma 2:
For each
Czyl...Yn'
and
Ax
Bzw
with only free variables shown, the
If
and
y ={xjAx. and if
Y1,'''yYn E y,
n(z)[Yl1 ,...,Y
n]}
tzlzey & <Yey>
is the Gdel
n
ZF*:
y},
y U
a well-ordering of
and for each
with 1 free variable,
following is provable in
where
Also see 2.
ZF.
z=
(zlzy & CY
C.
number of
Thus
x,
is
then
1
...yn]
Fodo(x)
set of all sets first-order definable over
Bzw
means the
for definable
sets x.
Proof:
both
Suppose
Czy 1
(zlzey & Czy1 l.. and
is
(3 w)(wez & way 1 ).
{zlzey & <ys
by Replacement on the definable y.
<ysy>
=
is routine.
nlzvyl
]
,
for
zy.
>
We note that
= n(z)[yl]
We want to show
ZF*
certain sets definable in terms of the members of
ZF*,
Czy1
The proof in the case of
What complicates it in the case of
provably exist in
exist
y
ZF
is that
may not
and also that the theory of Gdel
numbering may not be formalizable
in
ZF*.
The latter is not
27
the case, since w
is definable, and hence by Replacement and
Foundation, induction on
we may define +
and
x,
w
provably holds in
ZF*, and also
and prove the relevant properties.
What comprehension axioms are needed to establish our equivalence?
Apparently, what's involved is
ust that the theory
of finitely hereditary sequences of elements of
numbers provably in
Yl,y2 ,y3 Ey, n,mew,
<{yl,n],{n,mll],[y2 ,Y3,fnfl>."
ZF*,
Y1,Y 2,y 3,
and
for sentences
there exists the sequence
Such sentences can clearly be
proved by suitable instances of Replacement in
definable sets
and natural
ZF* have the intended interpretation.
For instance, we must verify provability in
like "for every
y
n,m.
ZF*
for
One assumes in
ZF*
that
such a sentence in false, and goes to definable counterexamples
Y1 ,Y2 ,y3
and
n,m
via the definable well-ordering of
y ! (y,
Bzw.
Lemma 3:
a)
[a = xAx3 & (=a v pea) & (3x)(x=M(B))]
-> 3f
fying
b)
(3x
Each instance of the following is provable in
the conditions given in the definition of
a o)
xJAx
(3 !x0 )(x0 =M(a)).
->
xfAx3 ->
(Xo=M(P)).
Proof:
Assume
have 2 functions
a = {xjAx & (3 x)(x=M(a)). Suppose we
f,g
satisfying conclusion,
take, using Replacement and Foundation in
way,
B
that
3f
f(B)
satis-
x=M(B).
Pea =
Also
ZF*:
to be the
and
g,
-least element of
fg
and
ZF*
a + 1
B
is definable.
use Replacement on
$
to get the
Suppose
fyl (3 y)(ye
g.
We
in the usual
with the property
satisfying definition of
g(P). Then
f
x=M(a),
with
Lim(B). Then
& for all
f
28
satisfying
x=M(a), f(y) = y)l.
definition of
sum set in
ZF*
U.
to get a union,
the definition of
x=M(a),
clearly
contrary to hypothesis, f(B)
Now suppose
when
of
f
= 6+1.
In any
f
f(B) = U.
satisfying
But this is
g(B).
By hypothesis,
f(6)
is fixed
varies over the functions satisfying the definition
x=M(a).
Clearly
f(8+l) = Fodo(f(b))
fying the definition of
independent of
f,
Clearly
x=M(a),
again contradicting
B
f(6+1)
f
satis-
is also
the definition of
i.
exists.
Now suppose for some
B,
and so
for any
0.
So no such
such
We can apply
Bea,
part a) false. Take least
and apply above, since least such
B
is definable.
This concludes part a).
b)
(3!xo)(xo=M(p))
Now suppose
= a.
for all
We conclude the proof of Lemma 3 by
obtaining a contradiction. Suppose
ZF*
write
but not for
We may assume this without loss of generality, by taking
least counterexamples.
in
BEa,
on
a
M(B)
Lim(a). Using Replacement
we can get the set of all M(p)'s, Pea.
for that
with
x
xo=M()
Thus we may take the union by sum set in
(We
if it is unique.)
ZF*
and call this
Now it is not hard to see, under our hypothesis that using
replacement there is an
PEa,
and
consisting of only
(a ,U>, and that this is the required
definition of
from a).
f
U=M(a).
Suppose
So
a = T-l.
<P,M(P)>'s,
f
in the
(3 xO )(xo=M(a)). Uniqueness comes
It is easy to see that
definable and by our hypotheses,
f(y)
is definable.
y
is
It is
U.
29
obvious that elements of
and that
P(f(y))
Fodo(f(y))
P(f(y)).
0
are elements of
exists by power set in
ZF*.
P(f(y))
Furthermore,
exists since it can be gotten by replacement on
Proceed as above to get an appropriate
Fodo(f(y)) = M(a).
a =
Fodo(f(y))
Uniqueness follows from a).
f
to give
The case
is trivial.
We want to insure in
ordering of
L
ZF* there being a definable well-
(among other things). This insurance is easily
obtained by a natural definition of
L
in
ZF,
but not in
We have no choice but to complicate the definition of
L
ZF*.
by
adding on conditions.
We define xEL,
approximately as
(3 a)(3y)(y=M(a) & xy).
But this is not good enough for our purposes.
extra conditions on this
a
(z)(z=M(a) --> z=y). Whenever
say
y:
(a
& xz)] ->
(0)(z)([z=M(3)
1)
and
We define 5
x
and
v
a
and
=a)),
have such a
y,
we
O(x) = a.
2)
and if
yea,
For all
B,y < a,
then
there is a unique corresponding
M(B)
3)
The M(B)'s, Bea,
4)
For every
O(z) < a.
5)
Also if
q
M(y).
and
M(a)
zEM(a), we have
z,w
y=M(B),
M(a),
then
are transitive sets.
(3 p)(O(z) =
)
and
[ (zeM(O(w)) ->
Now, there is a usual definable mapping
O(z)<O(w)].
F
in full
set theory (identity this with the 2-ary relation F(x) = y)
mapping the constructible
ZF*
sets
1-1
may well not be able to prove
into ordinals.
Of course,
(x)(3 y)(F(x) = y).
ition 5) will be that (the 2-ary relation)
F
is a
1-1
Cond-
30
function when restricted to domain
(x)(xM(a) ->
Ord"(F(x))).
M(a),
(This is the
and
F
which, in full
set theory, assigns (GCdel numbers in the form of ordinals)
to each constructible
set a sequence of ordinals, the first
being the rank of the set in the constructible
the rest of the sequence codes in, via
F
hierarchy, a+l;
on the sets in
M(a),
how the constructible set in question is first-order defined
over M(a).)
xeL = (3a)(3 y)(vy=M(a)& XEy & a
We define
and
y
satisfy conditions 1)-5) above).
We define
Lemma 4:
x < y = xEL & yeL & F(x)eF(y), for
The following are provable in
yL).
(y)(yeM(O(x)) ->
Proof:
we have
Let
Also
nition of
yL,
since
the definition of
(3a)(O(z) = a).
yL,
(y)(yex ->
if
yeL),
xeL
if
O(y)
let
•
O(x) .
zM(O (y)). Then
for suppose not.
rest of condition 4) follows for
since
y
x
does, and
To show
zeM(O(x)) .
and that
part of Lemma 4,
Then
Then we get a
xL.
The
M(O(y)) C_ M(O(x)).
O(y), M(O(y))
M(O(y))C_ M(O(x)).
(y)(yex ->
have (assuming yx)
xeL,
because of condition i4)
satisfies5) (i.e., y, together with
y
xcL.
Towards verifying 4 ) in
contradiction via condition 2) in tl
he definition of
y,
then
satisfy 1), 2), and 3) in the defi-
O(z) _ O(x),
being satisfied for
as in 5).
By 4) in the definition of
yM(O(x)).
y, M(O(y)), O(y)
ZF*:
F
yeL),
that
yeL.
notice by
yM(O(x)),
Trans(M(O(x)) we
and so by the first
31
is provable in
(x)(xEM(B) --> xcL),
exists unambiguously by Lemma 3.)
Proof of Lemma 5:
y
Now
is a limit.
=
y.
ZF*.
Then
xM(y).
Let
).
xM(6+l
With this well-ordering
such that there is no
assuming there is an
x.
of
a,
y
Case 1.
y,
There is a definable
and we may use this to definably
M(6)
the finite sequences of elements of
ZF*,
in
Let
+ 1.
in the natural way, proving in
x
does not satisfy
and hence by the definition of
M(5), <,
well ordering on
well-order,
is definable in
M(B)
wEL.
Also
aljac & M(a)
Form
ay,
for some
Case 2.
xEL.
(Note that
ZF*.
Take e-least member, and call it
conclusion}.
xeM(a),
B = {xjAx3 ->
1 free variable,
Ax,
For each
Lemma 5:
that it is a well-ordering.
ZF*
M(6+l),
M(+1),
we take a least, in
M(a)
satisfying condition 1,
This least
{aClaE+l & xEM(a)I
and so we consequently can form
ZF*,
is definable in
x
and take
the e-least member, thereby obtaining a contradiction.
So every
xcM(6+l)
possesses a (unique)
O(x).
Conditions 2)-4) are treated similarly, taking definable
counterexamples and using definable well-orderings. The proof
of 5), after taking least counterexamples,
is much like our
indication of construction of a definable well-ordering
M(+l) )
on the basis of one for
To show
weL,
Lemma 6:
it suffices to prove
EL.
wM(w+l).
The proof
ZF.
HZF (x)(x=uW-_(x=w)), where
to the predicate
above.
M(5),
is like the proof of this fact in
of
A
is
A
relativized
32
Proof:
7.
Left to the reader.
The System
ZF*'. We define a transformation mapping
formula in prenex form in the standard notation (described in
2.
Remarks on Terminology and Notation) into formulae which
contain the
'<( symbol.
If
B
to be the usual prenex form for
is in prenex form define
B.
to be the
T
Take
identity on formulae with no quantifiers, and take
to be
is
B-
T((3 xi)Bxi)
(3 xi)(T(Bxi)& (xj)(xj< x i -> T(B-xj))).T((xi)Bxi)
(xi)(T(Bxi) v (3 xj)(xj < x i & T(B-xj))).
proved by induction that T(B-)
for any prenex
B.
and
XT(B)
It is easily
are equivalent
Recall that the interpretation of
xj ( xi
is xL & xeL & F(xj)eF(x
i ).
We form
ZF*'
as follows: Extensionality & Foundation &
Infinity & Power Set & Sum Set & Modified Replacement. The
latter is the only difference between
other axioms are the same.
follows:
in
ZF*'
ZF*
in
ZF*'
Any instance of Replacement in
provided that the
is obvious that
of
and
Replacement
(z)(z < y --)>T(Bxz)). Bxy
Lemma 7:
ZF*'
Axy
ZF*
ZF*.
The
is as
is an instance
be of the form
T(Bxy) &
having only 2 free variables.
It
ZF*' C ZF*.
Extensionality & Foundation & Infinity are theorems
when relativized to
Proof:
L.
For (Infinity)' take
x0
to be
w.
For
(Extensionality)' and (Foundation)' ust note from Lemma 4 that
& yx) -> yEL.
(xeL
33
The Power set and Sum set axioms (of
Lemma 8:
ZF* when relativized to
theorems of
to
x =
will be equivalent to
L
zeL & wL.
yfAyl.
The relativized
yjyEL & (Ay)'1 & xL.
(z C w)'
observe that the relation
L.
x =
In power set, we have
Proof:
are
ZF*)
is equivalent to
xL
& x =
xO
the set of all
ylyeL &
then there is a set
xoEL
with
subsets
z
such that
zeL.
Now the hypothesis
x
of
is definable, and so
x
IP(x).
We use replacement on
O(y)'s
with
all
yeL
yeL
with
x
y
M(0) C m(a).
with
and
yEL
to get the set of all
yElP(x). This is a definable set of
.J
have
MJtl),
is definable, and so,
U
yeM(J),
The required set
is in
tells us
has a definable power set,
x
IP(x)
and so it has a union,
Ord",
zC w &
So we have to verify that if
(Ay)'),
that
Now
x
1a,
because if
of all subsets y
by Lemma 2; hence
xoEL,
then
of
x
by Lemma 5.
The relativized of Sum Set is checked similarly.
Lemma 9: If
all
(3 y)(y < z & Ayxl...xn ), and
Ea = fxjBxl,
(That is, if
A
then
3
O(y),
with
O(z), O(x1 ),...,
O (xn)
y < z & Ayxl...x n .
any formula with the free variables shown,
any formula with 1 free variable, the above is provable in
Proof:
Z,Xl,...,xn
provably a
M(a)
Lemma 10:
O(y)
to this lemma, and then go to definable
But we obtain a contradiction,
since there is
for these supposed definable counterexamples.
The semi-relativized of each of Replacement in
is a theorem of
ZF*).
ust assumes there are counterexamples
One
counterexamples.
B
ZF*.
ZF*'
34
Proof:
ZF*',
We take a particular
e.g., that one whose
(u)(u < y -->T((
We let
D
Axy
instance of Replacement in
3 z)(w)Czwxu)),
where
be a definable domain,
ZF*
that there is a set
xcL
with
xeD,
y
SL
T(( 3 z)(w)Czwxy) &
is
C
DEL.
of all
is the unique
is quantifier-free.
We wish to show in
yeL
y
such that for some
with
(Axy)'. This is
easily seen to be equivalent to finding a set
SL
of all
yEL
such that for some
1)
xEL & yL & (3 Z))L(w)L(Cwxy V (3 WI)L(W, < w &
& (')L(Z' < z
xD,
(3 w)L(~Cztwxy & (wt)L(wI < W ->
->
Cztwtxy)) &
(u)L{U < y ->(z)LE (3w)L(~Czwxu & (w')L(W'< w ->
(3z')L(zt < z & (w)L(Cz'wxu v (3 w t)L(w
Convenient Notation: If
in 1),
X
then let
X,Y]
and ending with
Let
xeD.
Czw'XY))
X
and
Y
t
Czwxu)) v
< w & 'Cztwtxu))]].
are expressions occurring
be the subformula of 1) beginning with
Y.
U = union of the
O(y)'s such that 1) holds for some
We proceed to place definable bounds on the quantifiers
above in such a way that the new formula is equivalent to 1)
for
y
xD,
yMJ).
(Note that for each
xeD
there is at most 1
satisfying 1).)
for
Let
fl(<x,y>)
zeL
with
of the range of
Let
O(w)
for
E(W)LCz'w'xy]
fl
f2 (<x,y,z>)
wL
be undefined
with
on
if 1 is false:
be
otherwise. Define
O(z)
t!1 = union
D x MJ).
be undefined if
"[Czwxy,
E(w)L,I~Czwtxy]; be
Czw'xy],
otherwise. Define
35
L2'2
=
Let
otherwise bie
,Czw
' xy.
(See Lemma 9.)
on
Define
f4
on
f5
union of the range of
Let f 7 (<x,y>)
f6
on
f7
O(z)
on
x M)
O(w)
x M4).
r(w')L,Cz'wxy];
(u)L,Cz 'w'xu];
Cz'w'xu].
be undefined if
for
zL
with
with
f8
wL
U9 = union of the range of
fO(<x,y,u,z,w>)
uO(wt) with
union of the range of
if
=
D x M(U) x MJll) x M(U4) x MJ
).
5
other-
Define U 7 = union
on
f1 0
f9
(Z )L,"CZ 'w'xu];
3w)L,~Cz'w'xu] .
~P[(
Define
D x M(U) x M(U7 ).
f9 (<x,y,u,z>) be undefined if
otherwise be
[(3 w)L,Cz'w'xy];
if
D x M(U).
U 8 = union of the range of
otherwise be
=4
wt < w & .Cz'w'xy. Define U 6 =
,[u < y,
f8 (<x,y,u>)
otherwise be
be
x M(U1).
be undefined
be undefined if
with
of the range of
D x M)
on D x M)
with
PO(w')
pO(u)
[(z')L,Cz'wxyl];
otherwise. Define tJ
be undefined
f 6 (<x,y,z, ',w>)
otherwise be
Let
w' < w &
with [CCz'wxy,Cztw'xyj otherwise. Define U
union of the range of
Let
with
) be undefined if
f5 (<,y,z,2 '>)
be O(w) for wL
Let
w'
U3 = union of the range of
4[z' < z, Cz'w'xy,
with
Let
wise be
D x M ) x Me 1 ).
O(w' ) for
union of the range of
Let
on
D x M(U) x M( l1 ) x M(J2).
Let f4(<x,y,z>]
'O(z')
f2
f3 (<x,y,z,w>) be undefined if (w')L(W' < w ->
Czw'xy);
f3
of
union of the range
.[(3w)L,Czw'xu];
and
[Czwxu,
Czwtxu]. Define
on
D x MJ)
x MU
be undefined if
7
) x MJ
).
[(w')L,Czw'xu];
w' < w & "Czw'xu. Define U
on D x M)) x M
8
7)
xM
8)
=
x Ml
9).
36
Let
fll(<x,y,u,z>)
0O(z')with
otherwise be
union of the range of
Let
be undefined if
[zI < z,CzwtIxu].
fll
on
f 1 2 (<xy,u,z,z'>)
otherwise be
O(w)
for
'[ (3zt,)L,Czw'wxu];
D x M(j) x M J7 ) x M
be undefined
wEL
with
U12 = union of the range of
Define
fl2
=
Ull
8 ).
4 [(w)L,v Cz'w'xu];
if
'[Cz'wxu,,Cz'w'xu . Define
on
D x Ml)
x M(U 7) x MJ
8
)
x MJl)Let
f13(<x,y,u,z,z',w>)
otherwise be
PO(w')
with
union of the range of
w' < w &
fl
,[(3 w,)L,._Cz'wIxu];
be undefined if
on
Cztwtxu. Define U
=
3
D x MVJ) x M(U 7 ) x MPJ 8 ) x
Mu11) X M12)
Note that by suitable instances of Replacement in
all of the above are provably well-defined.
ULi is definable,
that for
xD,
so that each
yMJ(U),
Note that each
It is easily seen
it is the case that
to the predicate Bxy
1 < i < 13
M(Ui) C L.
ZF*,
Axy
is equivalent
obtained by placing the bounds
Mi),
on the appropriate quantifiers in 1).
Now each instance of the following is provable in
If
a =
x < y
xjAx),
iff
and
x < y
a limit,
then for
M(a),
A
of
and
of 1 free variable.
of the schema is like the proof in
well-ordering
x
ZF.
Use the definable
Then relativizing the
quantifiers occurring in the expansions of the "<"ts
for
xED.
Bxy,
The proof in
M(a).
Now let; V = max~J,Ui,O (D)).
occur in
yEM(a),
holds when the quantifiers in the definition
are relativized to
ZF*
a
ZF*:
to
M(V+w),
that
we get the same predicate as
Hence we have shown our
S
we wanted to show
Bxy,
37
originally EL, is first-order definable over
EM(V+w+l), V + w + 1
hence
8.
The System
M(V+w),
and
definable.
ZF'. Making use of the transformation T
defined in the previous section, we form
ZF'
as follows:
first, Extensionality & Foundation & Infinity & Power Set and
Sum Set axioms of
ZF.
x)
(x)(xEL) & (:
In addition, we have
Ord"(x)) & (x)(3 y)(z)(yEx & (zEx ->
(Ordx ->
in
Replacement
ZF'
z
y)).
Let
will be the following:
Bxyyl...;
Yn
be a formula in prenex form with only the free variables s hown.
Then
(Yn)(x) (3 X1) ( 2 ) (x2 eX1
(Yl).
(3 x3)(x3
T (Bx3x2Y1.. 'Yn) & (x4)(x4 < x2 ->
instance.
Lemma 11:
First, we wish to show in
This is trivial for
Suppose
T(A) - A
prenex form with
prenex form.
any
A
quantifiers.
B
B
...Yn)))),
)T(Bx i
ZF'
each instance of
with no quantifiers.
A
having
(3 xi)(Axi). Then
n + 1
quantifiers in
T(A) - A
n + 1
,Axj
))
is provable in
provable for
T(B)
quantifiers.
is
(xj)(xj < x i v T(A-x))). Now FZFT(Axi)
T(Axj) we have
is equivalent to
T(A-xj)
We have to check that
in
ZF t'.
be in prenex f orm with
is
for all A
ZF'
We then wish to show that
Then we will have shown
in prenex form in
Let
Suppose
n
A
is provable in
is provable with
T(A) - A
s an
1
3 x4Y
ZF' D ZF.
Proof:
T(A) - A.
&
(3xi ) (Ax
ZF '.
i)
(3xi ) (Ax i
(3xi)(T(Axi )
Ax i .
&
Since
zFT(A-xj)
& (xj)(xj
< <x1
Xi V
38
Define
Cxy
to be
equivalent with
n
T(Ay & y=x),
(x4 )(x4 ( x2 ->
ZF'.
But
ZFT(Ax
Choose any such
2
&
xi.
= x3 )
2
wff
of
(3 x)(xEx0 & Cxx2 &
Ax 2 & X2 = x 3.
Ax 2 & x2 exO).
Take
O(xi),
and use the above theorem of
M(O(xi))
a transforration on a
~T(Ay & y=x))))), is (equivalent to) an axiom
ZFI
,(x)(3 xl)(x2 )(x2 exl
of
This is, of course,
(x0 )(3 x1 )(x 2 )(x2Exl
quantifiers. Now
of
y = x & T(Ay).
ZF'
Now assume
and set
So
(3 xi)Axi.
xO = M(O(xi)),
to get the set of all elements
having the property A.
(M(O(xi)) is defined and
has required properties since
(x)(xeL) is an axiom of
and
ZF*, previously.) Hence by
xeL
is formalized as in
one of the axioms of
ZF',
ZF', there is a (-least member.
Hence we have shown by induction the equivalence between
and
A,
in
ZF'.
eliminating the
9.
This has the effect of provably in
T's
in the axioms of
ZF',
The Skolem Argument. We wish to show
where
ZFn
is the first
enumeration of them.
suppose
But then
n
axioms of
and so
T(A)
ZF'
ZF' 3 ZF.
(n) ZF* Con(ZFIn)
ZF'
in some natural
If we succeed in showing this, then
uCon(ZF). Then
~Con(ZF'). Then
(3n)ENTCon(ZF).
Since
ENT
(3 n)
Con(ZFA).
is formalizable in
ZF*,
(3 n) IF*Con(ZFn).
Hence
(3 n)(FzF*Con(ZFA) &
FZF*
Con(ZFn)), and so
Con(ZF*). Hence
ConZF* ->
ConZF.
We give, without loss of generality, a Skolem closure
argument within
ZF*
to give, provably in
is a model for 1) Extensionality
3)
Infinity in
ZF, 4)
in
Power Set in
ZF, 2)
ZF*,
a set which
Foundation
ZF, 5)
in
Sum Set in
ZF,
ZF,
39
6)
(x)(xEL), 7)
z
y)),
9)
1) in 7.
(x)(Ordx ->
Let
D(xyT)
The System
free predicate
E(zwxyT).
C,
ZF*I
Ord"x), 8)
(x)(3y)(z)(yEx&
be the formula obtained from taking
and replacing the 4-place quantifier-
with some 5-place quantifier-free predicate
(3 x)(xED & D(x,y,T)).
(D)(T)(3S)(y)(yES
The construction,
in
ZF*,
of the model of these 9
sentences will be much like a Skolem construction in which the
.
initial model is
At each stage
and we take the union as
x:EL,
We simultaneously
U
in
define
n
an
n,
we throw in some sets
ranges over
and
Sn .
w.
We are interested
S
nEw
n
Sn = M(an).
S
= M(/) = 0.
Consider, for each xS ,
n
(zEy ->
aO =
.
the <-least
yx
with
(z)
z x).
Consider, for each
XESn, PL()
Consider, for each
x,yESn,
- set of all
y
x
with
yeL.
element of
x
not in
x
y, the <-least
y.
Consider, for each
that
with
XESn,
UL(x) = set of all
TSn,
the unique
yL
such
(3 z)(zex & yEZ).
Consider, for each
semi-relativized
We continue
of 9) to
L.
SL
satisfying the
(Call this 9)").
"considering" through 9)", closing
Sn
in
effect, under the "Skolem functions" for 9)", in such a way
that, as in
7.
The System ZF*, we have that the Skolem func-
tions produce values definable in terms of the arguments.
40
We take
an+l
= (union of the
considered above) + w.
Take
O(z)'s
Sn+
for the
z's
M(an+l).
We can then use appropriate instances of Replacement in
ZF*
in combination with the definable well-ordering <,
show that if the
n,
Sn
and
an
are not well-defined
to
for each
then there are definable counterexamples to our construction
in the following sense:
for some specifically definable sets,
the sets corresponding to them that we considered above do not
But this is impossible by Lemmas 7,
exist.
our construction is well-defined in
8,
9, and 10.
So
ZF*.
i Sn
is an M(a), a definable, a a
new
In particular, it is transitive.
It also contains
Now our model
limit.
Due to the absoluteness of the definition of
definition of
in
ZF*,
w
in
M(a)'s, a
a limit,
and of the
it is easily seen
putting all this together, that the sentences 1)-9)
are true when the quantifiers range over
since
L
w.
M(a)
Furthermore,
is definable, the definition of satisfaction and
the induction on
to prove, in
M(a).
a
ZF*,
are easily developable in
that
ZF*,
in order
Con( 1)-9) ).
From the remarks at the beginning of this section, we
immediately have
Theorem 1:
10.
F
ConZF* -- > ConZF.
ENT
ZF. Parameterless
partially
ZF, and ZF*.
to further consideration
This section is devoted
of the system
ZF*
Chapter 1, and partially to some other subsystems of
of
ZF.
We define a sentence of set theory to be arithmetical if
it is the relativized of some sentence of set theory to
Corollary 1:
is a conservative extension of
ZF
w.
ZF*
for arithmetical sentences.
Proof:
Let
A
,A) ->
Con(ZF* +
be arithmetical, and
Con(ZF +
IZFA.
by modifying the proof of
A)
.A,
ZF' +
A,
our Lemmas carry over.
Con(ZF* +
A),
Due to Lemma 6,
respectively.
Now since
and so
~Con(ZF +
as
ZF*I, ZF'
Theorem 1 slightly; Just redefine the systems
ZF*' +
We can show
all of
.A), we have
zF*A.
Our next Theorem concerns sentences of the form
(x)(3 !y)Axy,
arbitrary, with only 2 free variables, that
A
ZF*.
are provable in
Now in
which define, provably, in
ZF,
everything and which is 1-1.
Another is
(x)(3 y)(y
Theorem 2:
ZF
Not so in
((Axy & Axz) ->
Then
Proof:
We let
z).
C
we construct a model for
for
X
Q
ZF* + XC,
ZF*.
R
given an arbitrary model
is a 2-ary relation on
as follows:
The domain is to be
is the rationals. The 2-ary relation, Sxy,
defined as
yeX;
is not provable in
X,
Note
is a first-order theory with equality.)
We define
where
C
x) & (x)(y)(z)
(All models are assumed to be equality models.
ZF*
that
=
be a sentence of the above form, and
ZF, G\ = <X,R>, where
pi .
ZF*. Thus,
(x)(3 1y)(Axy & y
C
=
(x)(3 y)(y = IP(x)).
be any formula with only free
variables shown, and let
y
a Skolem function which moves
An example is
{x}).
Let Axy
there are many such sentences
Rxy
true if
if
xX,
x,yeX; x < y
yeQ.
if
X U Q,
is
x,yEQ; false, if
xEQ,
U
We claim
ZF* +C.
satisfies
First, we show that the elements of
indistinguishable
in
for
G
in the sense that if
xicQ, yjEX,
then so does
Q
Q,
same order relations in
are
Axl...xnYl...ym
holds
AZl...znyl...y
for
if the two sequences of rationals,
zieQ
3
in
xi,
m
have the
i
(i.e., there is a 1-1 order
To see this, it suffices to show that, given
preserving map).
such a pair of similar sequences of rationals, there is an
X
which keeps the elements of
automorphism of
fixed,
and which maps, in an order-preserving way, the sequence xi
onto
zi.
And such an automorphism is easily given by any map
which fixes the elements of
itself, which maps the
X
into the
xi
yj.
j (x)(3,y)(Axy). Then by indistinguish-
Now supposeB
it; is clear that for
ability,
and maps the rationals 1-1 onto
some yX,
for otherwise we would have
hence
for
Axz
So
8
Q3
x 2 , xl
x1 , x0 E Q.
a(k
satisfies axiom
in the domain.
of
0
~
=
S
satisfies
ZF*.
(x2 )(x2 x 0 - x 2 ex 1),
Then either
- x 2 cx 1 ),
and so since
In the second case, we have
1/2(x0 + x1 )
and
x,yEQ,
xI
and
xO EX,
or
In the first case, we can conclude that
(x2 )(x2 EXO
xO = x 1.
for
C.
To verify 1 of ZF*, suppose
for
for
Axy
by indistinguishability, assuming xQ.
does not satisfy
Clearly
Axy
But now I claim that
z = y + 1.
since yX,
A(x+l,y),
C
we have
xEQ,
x1 .
But then
x0 = x 1 .
1=ZF,
we have
1/2(x 0 + x 1 ) < xO
iff
43
To verify 2 of ZF*, set
1d3 =0
see that
x U
Also,
x)
al
from
EX0 ,
since
to
.
(3
in
6
remains unchanged for
x
axiom 2 of
easy
0
xeX,
f.
in
when we pass
a.
of
x0
are the
. Also, the subsets in at
in
ZF*
. Putting this together, we see that
is satisfied in
"in the same way" as it
E
4 ,
To see that 3 of ZF* is satisfied by
satisfies
CA.
suppose
Then the unique element defined in
CA
by indistinguishability. We let this element be
we are looking for is a power set of
claim that
y = P(x)
in the model
have to show that (2
,
in
subsets of
x
x.
. We
does the trick.
We
But this is
in
x
What
are the members
in
3
are the
.
in
ZF*
is checked similarly.
To see that axiom 6 of
If
y
must be
in the model (
satisfies y = P(x).
and the only subsets of
Axiom 4 of
xeX U Q.
x
aL
obvious, since the only members of
y
to
.
is in
of
is
or any of its members, are identical with the corres-
X0 ,
ponding elements in (
eX,
It
is same as
. Also, the members in
same as the members of
of
x0 = w of
xeX,
take
ZF*
xO
is satisfied in
g,
let
in foundation as an R-least
memberof x in ® . If xeQ, take x0 to be , in aL (or
2). Obviously I2j p e x.
Axiom 5 of
CA
defines in
ZF*
63
is the most complicated.
is again
interested in the range in
6
e X.
Call it
D.
The set that
Then we are
of the partial function in
,
x3 =x 1 ), on the domain
Bx2 x 1 & (x3 )(Bx2 x3 ->
X2
with
if
S(x2 ,D),
S(x2 ,D)
X
Bx2 x & (x3 )(Bx2 x 3 -> x3 =x1 ), then x1 EX.
and
, Cx2xl
Tl
holds in
Replacement in the model
Replacement
,
in
6
,
such that, for
Cx2xl
holds in
.
x2 EX,
Then by
we would have (this instance of)
It remains to show
with the free variables shown,
Axl...x
,n
al
which holds in
Bx1 ...x n
there is a formula
,
iff
Cx2 xl
and we would be done.
that for each formula
holds in
Now every
x2 EX, and so, by indistinguishability,
has
Now suppose there is a formula
xl
D.
iff
Ax1
.. . x n
xi X.
when
It suffices to prove by induction that for any formula
AXl...n,
into Q, there is a formula Bxil ... xik
with xiEX iff
A 1x...n
To see
this
XiExj,
Axl...x
iDom(f),
instances of
by
n
(3v)(v=v)
Put
quantifier-free, take
JEDom(f), by
or
by
if
Axl...xn
f(i)
if
to
2) replacing all
xi = xi,
or
xi=xj,
(3v)(v=v)
B
by 1) replacing all instances
x =xi, iEDom(f), J
3) replacing all instances of
xjxj,
i,JcDom(f),
or
xiExj,
n
=Bxil...Xik.
iff
be the formula obtained from A
of
k
iEDom(f), we have
if
f(i)
i Dom(f), xi
for
i
til<i<n)
such that for any sequence x...x
Dom(f)1,
fill< i < n & i
3
from
and for any partial function f
xi
ex ,
j
(3v)(v=v)
xi=xJ,
(3v)(v=v)
f(i) < f(J),
f(J),
or
Dom(f),
if not;
if not, respectively.
in prenex form, and suppose our claim is
true for all formulae with less quantifiers.
We may assume that
Axlx. . . n
is
(xO)CxOxl .,
.xn
since the existential case follows from this case by taking
45
negations.
into
Q.
Let
f
fill
be a partial function from
Now 2 finite partial functions g,h
i < n)
w ->
from
Q
are said to be of the same type if 1) they have the same
D, 2) g(x) < g(y)
domain
iff h(x) < h(y).
Consider the set of partial functions on
which are identical to
a finite #
f
on
ill
i
n.
f.
=
of types represented in this set. Pick a repre-
Let
Di, 1
i
inductive hypothesis for
k,
X
1
n,
= fi (l)
and for any
for
Cx0 xl...xn
xO...xn
{PlP
Then we take B
indistinguishability,
q) =
to be
fi;
for
I
for
k},
{r
o
Dom(fi),
(13
<
i.e., each Di
xeX
with
1eDom(fi), we have
-Xp
DiXpl.
ffl,f2,,f.
be the formula given by the
has exactly the free variables x
O .• <
i < n)
There are only
sentative from each type, and call this set
fl
filO
for
CxOxl...xn
r
n & r
(xo)D1 &
lDom(fi),
~
iff
Dom(fi ) } .
Di. By
it is easily seen that
B
and
A
satisfies the conclusion of our claim for the function f.
This concludes the proof of Theorem 2.
We now define some new subsystems of
the same as
ZF
except for the Replacement
instead, only allows the
to have
n + 2
other ones.
ZF.
Axy
ZF n
schema.
is to be
ZF n,
in the Replacement schema at most
free variables; x
It is easy to see that
and
ZF*
y
and possibly n
is a subsystem of
ZF© . We also have
Corollary 2:
Proof:
ZF*
ZF 0
.
Consider the model of
ZF*
constructed in the
proof of Theorem 2.
(ly & (z)(z0 ->
->
ZFO
provable in
.
(x)(3y)(0ex
Consider the sentence A
(zEx <->
zy)))).
But our model of
We believe strongly that
stronger conjecture that
ZFO
the details of a proof of
ZF*
does not satisfy A.
ZF,
and in fact in the
ZFn+ l,
ZFi
ZFO
ZF
A is obviously
for all
n.
Although
have not yet been carried
out, we can give the definition of a very promising model of
ZFO+ (3 x)(y)(3 z)(--(zy <-We let
M(a)
z=x)).
be the minimal model of
ZF.
any Cohen generic set of natural numbers over
MS (a) be the corresponding Cohen model for
We let
FS
We let
M(a).
ZF + V
S
be
We let
L.
be the set of all sets of natural numbers
which are finitely different from
difference from
S
Recall that
S
(i.e., whose symmetric
is finite).
R(I+1)
Consider the sets
=
U
IP(R(1)), R(X) =
xeMS(a)
R(1).
such that for any formula
Ayzx1 ,
we have, in
let
X
be the set of all such x.
xeX
such that any finite combination of union and power set on
x
in
1MS(a),[zl(3y)(yex& A(x,y,FS))) f
MS (a) gives a set in
For each
,
let
lative hierarchy up to
Define, for each
R'(a) n Y,
fg (z)
R' ()
5
We let
We define
fX)=
be the set of all
be the unique rank in the cumu-
a function f
and by the equations f+l(x)
y),
MS (a).
whose domain is
{YfyeY & (3 z)(zex &
y
U fx
xcZ as
We
X.
in the model
5,
Y
xl.
[(3B3)(<a d x
Range(fg
)) &
xEY].
47
We conjecture that
<Z,e>
is the desired model.
We feel that a detailed analysis of the relations between
the theories
Z
distinguish each
n,
hopefully by finding natural sentences to
ZFn
from
ZF n+ ,
would involve non-trivial
applications of the notion of forcing. In particular, careful
attention seems to be required as to the model-theoretic
properties of models of theories obtained by forcing; e.g.,
the definability or indistinguishability of various elements
of the models constructed.
CHAPTER II
1.
Definitions of Systems. We will have for our language,
number variables,
n,m,p,q,r;
set variables,
x,y,z,w,u,v,...;
the relation nex
between numbers and sets; and the constant
number "O" which can hold only between variables of the same
type.
We introduce function variables in the usual way by
defining them in terms of sets of natural numbers in the usual
We will have full number theory at our disposal, since
way.
all systems considered here will have the unrestricted induction axiom schema (called I)
where
A
[AO & (n)(An ->
An')] -->
(n)An,
is any formula with possibly free variables of both
kinds, and can have both number and set quantification.
Also, all systems considered here will have the axiom of
extensionality
(that any two sets with the same members are
equal), and we will tacitly assume that the system we will call
I
includes this axiom.
Thus
I
will be the unrestricted
axiom schema of induction plus the axiom of extensionality.
In addition, all systems considered will have the recursively enumerable comprehension axiom schema (called ReCA),
3 y)(n)(ney- (3k)T-e,n,k,xl,
.. . ,) ) 1
(x1)... (xl)(e)(
The predicates Ti
are understood to be written out in
the usual way with bounded quantifiers.
(Or, we may have
instead introduced them as primitive, and defined them by
adding axioms of primitive recursion.) The predicates TA
above are what Kleene would call
the characteristic
T
above
are
what
Kleene
would
call
T,f
where
,
function for the set,
x i.
where
fi
See Kleene
is
[3],
p. 291.
Before we get into the mathematics of the systems we will
be considering, we will state, informally, some propositions
concerning what can be done in the system
Proposition 1: We can, in
I + ReCA,
I + ReCA.
Justify all uses of
coding normally found in the development of hierarchy theory.
Thus, we may Justify the use of such symbols as
(respectively (2n + l *
(X)n
where
pn
3
m+llnex & mey],
x,y>
and
and {mlpm+lex],
is the nth prime).
We define the relation between functions and sets alluded
to above, as
l
2 n+
f(n) = m
iff
n,m>
f,
where
n,m> =
3m+l.
Proposition 2:
In
ReCA + I,
we may provably perform
"collapse of like quantifiers". In other words, the usual way
of collapsing 2 successive universal set quantifiers (or function
quantifiers) into one can be completely Justified on the basis
of
ReCA + I.
Def. 1:
essentially
A predicate A(n,xl,
. .,.
xp )
of
n
is said to be
1TI if it is in prenex form followed by a matrix
T(e,n,m,xl,...,P,fl,...,f
,nl,...,nk),
where there are no set
quantifiers in A, and the fi
occur as universal function
quantifiers, in any order, mixed together with possibly number
quantifiers
(ni)
and
function quantifiers in
(3 nj)
A.)
and
(3 m).
(No existential
An essentially
4
formula is
Just the prenex form of the negation of an essentially
T11
5o
formula.
Proposition 3:
For every arithmetical predicate (with
(f)(3 m)T(e,n,m,xl,...,x
2,
parameters) there is a predicate
nl,
. . .,n p )
which is, provably in
I + ReCA,
equivalent.
)
There is also a predicate (3f)(m),T(i,n,m,xl...,x,·nl,...,np
which is, provably in
Proposition 4:
y = (x)(
xi+l
)
-
3
I + ReCA, equivalent.
as the
(x0 ,...,x,)
a sequence
= Jump of xi .
( )
(x)
Define
such that
(x)(1)
(Thus, y
-th Jump of
x
x.
= x
and
is a predicate of 3
Then in
variables, defined in the usual way.)
I + ReCA
we
= (x)()).
may prove (3!y)(u
This is proved by induction on
2.
Consider
L) (x)(p)(e) (n) (f ) (3m)T2(e,n,m,f,x) (3g)(r)~-T2(p,n,r,g,x)I
-- > (3y)(n)(ney (f)(3m)T2(e,n,m,f,x))}.
2)
The schema,
where
A
(x)((n)[Anx
(x)(e)((n)( 3 f)(m)-T
3)
1,
is essentially
2
(3 y)(n)(ny-
Bnx] ->
B
Anx)},
is essentially I1
l(e,O,m,n,f,x)
->
(3 y)(n)(y)n
is a
function& (m)~T
2 1 (e,0O,m,n,y,
x )))
(x)((n)( 3 f)A(n,f,x) ->
4)
A(n,yn,x))),
where
A
3 g)(m)T
(x)(e)((f)(
5)
(m)-T
3 (e,,m,y n
is essentially Z 1.
3
(e,0,m,f,g,x) ->
where
A
(f)(3 y)(n)(yo = f &
yn+ l ,x) ) ).
(x)((f)(3g)A(f,g,x)->
6)
(3y)(n)((Yn) is a function &
is essentially 2 1
(f)(3y)(n)(yo = f & A(Yn·Yn+lX)))
having only the free variables
f,g,x.
We
call I + ReCA
+ 1), the pure
We call I + ReCA + 1), the pure
A; II +
ReCA
+ 2),
A1-CA;
+ ReCA + 2),
I
51
essentially
E1 -AC; I + ReCA + 4),
I + ReCA + 3), pure
A-CA;
essentially 41-AC; I + ReCA + 5), pure :1 -DC; I + ReCA + 6),
essentially 2 -DC.
NOTE: "CA" is supposed to mean "comprehension axiom"; "AC",
axiom of choice; "DC", dependent choices.
Proposition 5:
In pure
predicate, there is a pure
21-AC, for every essentially
1
predicate
11
(f)(3n)T(,m,n,...)
which is provably equivalent.
To see this, first use Proposition 2 to collapse adjacent
like quantifiers. Then use pure
£1 -AC
to interchange a
number and function quantifier, and then use Proposition 2 and
then pure
Z1-AC,
etc.
Proposition 6:
Pure
21 -AC D pure
The idea is that, in pure
m
Al-CA,
there is a solution to one of two n1
uses pure
Z1 -AC
set in pure A-CA
1
in the
1-CA.
one has that for each
predicates, and one
to form a Skolem function. Then the required
is obtained, in pure
Z 1 -AC,
recursively
ump of the Skolem function.
Proposition 7:
Pure
£1 -AC = essentially £1 -AC.
By Proposition 5, one has only to consider the case
(n)(3 f)(3 g)(m)fT--> 3
the
1
£1 -AC
f
and
g
Skolem function. But we may collapse
quantifiers in the usual way, and apply pure
to get a Skolem function, which will have recursive in
it the Skolem function wanted in the implication above.
A similar argument shows
Proposition 8:
Pure
2 1-AC D essentially i-CA.
52
4-CA
Conjecture: We do not know whether pure
A-CA
or whether pure
essentially A-CA,
Z1 -AC - essentially
=
-
Z-AC, or whether
AI-CA, and we conjecture that none of the
three statements is correct.
Proposition 9:
call
essentially
Pure
21 -DC ->
1_AC, and pure
2Z-AC.
(By Prop. 5, we
2:-AC, just
1 -AC.)
£1 -DC - pure
1 -DC.
Hence also
Proposition 10: Essentially
Proposition 11: We can prove in
ReCA + I
the recursion
theorem (with parameters).
2.
Preliminary Letmmas. We will eventually show that Z1 -DC is
a conservative extension for purely
21
sentences of pure hi-CA.
We define, in I + ReCA, several notions.
Def. 1:
P(n,m) is defined as the CGdel number (in the usual 1-1
onto G8del numbering of pairs of natural numbers) of the pair
<n,m>.
Def.
2:
epX is defined in the usual way as the nth partial re-
cursive function in x.
Note that in I + ReCA we can prove
(n)(x)(3 c)
Def. 3:
C {0,l' & (m)
RLOX(n) is defined as "Range (cpX)
(meDom(ep)) & Cpx defines a linear ordering," where
o
defines
a
linear ordering means that 1) (m)(cpX(P(m,m))= 0), 2) (p)(q)(r)
(Cppx(P(p,q)) . 1 & PX(P(q,r)) - 1) ->
3) (p)(q)(qnx(P(p,q))
- 1 ->
- 1)).
,x
(P(p,r))
cpX(P(q,p)) - 0. 4) (p)(q)(p
= 1 v cp(P(q,p))= 1). We define
Xcp(P(p,q))
= q v
RLO(n) =- RLO0(n).
53
Def. 4:
define
We define
n
q = RLOX(n) & (cpn(P(pq))
p <
n
x =
,
q = RLOX(n) & n(P(p,q))
<
we use the subscript n
When
1 v p = q).
n .
instead of
We
1.
denotes
the empty set.)
We define
WX(n)
Suc (pq) =
n
RLOX(n) & q < p
onx(P)
.SUx (p))-
-
n
n
HY n
n
n
condition
ye
(P(r,s) s < xP & re(z)s
=
)
n
Y=
). Thus H x(Z)
n
z
n
xq )·
&
xP
n
((Z)p)(l)
n
(x)p
r
n
(q)(p
RLOX(n) &
4 variables asserting that
starting from
(r)(r < xP ->
RLO(nX) & (p) (m)(Suc nx (m,)
(Z)
(P) ->
Def. 8:
v (mey &
Lim x( ) = RLOX(n) & (3q)(q <
(z) m ) & (p) (lim x(P) -> (x)p
n
(p)(o
&
n
Suc x(p) = (3 q)Suc n(p,q).
De . 7:
& (y)(3 m)(y=
m < xr))).
n
(r) (rey ->
Def. 6:
RLOX(n)
as
=
&
is the predicate of
is a hierarchy on the
RLO, n,
It will be useful later on to include the
(3p) (kE(z)p)), in the definition above.
(k)(kez ->
is the predicate of 5 variables asserting that
HYx( )
"I
z
is a hierarchy on the
starting from
Def. 9:
x y
y.
RLO, nx
Thus if
p
is defined as
n
x,
uR to but not including
then
(z)p=
0.
"x is recursive in
y"
in the
usual way.
Def. 10:
Inx
< ImXI
proper imbedding of
means
nx
into
RLOX
(n) & RLOx (m ) &
mX, i.e.,
<
(p) (q) { (p nxq
is a
-- >
I
54
f(p)
m
xf(q)) & ([p<xq
=
In X I
< ImxI = InXl =
ImXI
Def. 11:
(f)] -> p
E Range
means that the range of the
InX
y
& q
m
mXI v
InX
W.
is
(HX (z)) &
& HZ (x) &
(x)(y)(z)([WZ(n)
n
I + ReCA. Also,
provable in
y = x
HZ
nz
(y)] -> y=x)
is
-
(y)]
[W'
Z(n) & HZ_(x) & HZz
np
np
is also provable.
Proof:
element
n,x,y,z
with
I + ReCA.
By
Given
(P I W p74(Y)P
q.
then, since
()q
above
TZ)
Lemma 1:
(Y)q
Range (f))1.
< ImX .
(3n)(3z)(RLOX(n) &
HypX (y)
f
E
in
WZ(n), HZz(y).
n
WX(n),
Form
we have a < -least
n
Clearly we must not have
(s)(s <
If Suc x(qs),
n
n
0 x(q).
n
lim x(q),
If
n
xq -> (X) s = (Y)s), we have
then since
(X)S = (Y)S
(X)q
we have
= (Y)q.
The following is provable in pure
Lemma 2:
wZ(n) & wZ(m) &
three holds:
a)
A1-CA:
if
(3x)
(4
x)).* then exactly one of the following
there is a unique imbedding of the ordering
nz onto a proper initial segment of the ordering
mz
and
55
there is no imbedding of the ording
ment of the ordering
mZ
onto an initial seg-
.
nz
there is a unique imbedding of the ordering n z
b)
whole ordering m,
onto the
and vice versa, and there are no imbeddings
onto proper initial segments in either direction.
c)
the interchanging of
n
with
Furthermore, denote
x(1)
m
and
( ).
•fx
for some
I
In a), if there is an imbedding,
a proper initial segment of
g
Then all the
depending on
n
(p), Suc z(p)
n
and
of
n
onto
I + ReCA, (klf(k)
g(k)),
and
Running through the 3 cases,
p.
in a straightforward way,
lim z(p)
n
f,
then it must be unique, since
m
in
is another, form,
take an n-least member,
O
e w.
x,
m.
Proof:
if
a).
for the result of applying
iterations of the Jump operator to
above maps may be found
in
n
using the linearity of
n
and
m,
we get a contradiction.
Same with c).
of
If there is an imbedding, g,
segment of
with
f
n,
take
-1
g 1,
m
and note that
onto an initial
-l
g 1
must disagree
somewhere. Then follow the procedure above.
If there is an imbedding of
then there is one from
m
to
that we can form inverses by
n
onto the whole ordering m,
n
by taking inverse.
ReCA.)
(Note
And there are no proper
imbeddings, using least counterexample argument above.
So the main thing is the existence of these imbeddings.
We define a function
f z(p)
greatest k
such that
56
(3q)(Suck z(p,q)), where
Suck
(This is well defined, since
is
k iterations of
(p)(q)(r)(Suc
n
->
r = q)).
There is always a greatest
Suc.
z(p,r) & Suc z(p,q) ]
n
k,
since otherwise
we would have an arithmetically defined chain through n z , and
finitely many iterations of ReCA (+ I) would realize this claim
as an object, and would contradict WZ(n).
Fix
n,m, with WZ(n), WZ(m), (x) (Zz (x)). Let f (p) be
n
n
abbreviated f(p). We claim that, for each p, either there is
an imbedding of the ordering
the ordering
n,
so that
m
onto an initial segment S
keS---> k <
p,
of
or there is an
n
imbedding of
kJk < zp) of
n, onto an initial
segment of m,
and these imbeddings are to be found < T(x)p(l(f(P)+l)).
(This is the result of applying 10(f(p)+l) iterations of the
Jumpoperator to (x)p.)
Weassumethis is false,and, as usual, form plclaim is
false}. We take an n-least
clearly claim is true for
f(r) + 1 = f(q).
member,
q.
If
q
Suc
If
(q), then
(q,r), then certainly
So there exists a comparison mapping,
< T()r(lO(f(r)+l)), between the segment of
m.
0
n
up to
r,
and
It is then easy to see that there is a (unique) comparison
map between the segment of
n
up to q, and m, IT(x(lO(f(r)+l))
The limit case is similarly easy; the uniqueness
of these compar-
ison maps is used heavily, and that the property of being a
comparison map is low arithmetical, and that the function f
is low arithmetical.
57
Lemma :
az nd
WY(n)
The following is provable
WY (m),
InY I < lmy I,
and
(3z) (z < T(x)p (o (f(f)+l)) & HY (z)),
mY
in n,
is
in
I + ReCA:
an d
If
(3x) (HY (x)),
nY
wh ere
p
of the range of the imbedding of
then
is the l.u.b.
in
and f
n,
defined in the proof of Lemma 2.
f y
Proof:
Similar to the proof of Lemma 2.
Hy
that if
One notes,
e.g.,
has a solution, it must be unique.
np
If
Lemma 4:
of the image of the
nZ
and p is the l.u.b. in
then
imbedding, and (3y)(HZz(y)),
m
< InZI,
m
where f is f z
(3x)(H z (x) & y < T() (l(f(P)+l))),
n
np
is provable in
Proof:
I + ReCA.
One can prove first, as
in Lemma 2, that
(3x) (H (x)).
Then, again like Le mna
with
one can prove that
HZZ(x),
This
2, starting with this
y < T(x)(lO(f()+l))
x
(0:
np
course, it is provable in
ReCA +
(3x)(HZz(x) ->
[ that
np
(3!x) HZ (x)).)
np
Reas (nx )
Def. 12:
(3q) Suck
n.
(q,P)->
X (n) & (p) (Onx(P) v Lim x(p) v (3k)
RLO
x(P,q) & limn (q) ) &
(q)(q < xp )] .
(3r
-)(O (r)) &C(P )
(Reas (n) reads
Lemma 5:(z)(n)(x)(y)(m)([Reas(mZ)
"n
& Hzz
n
W Z (m)
->
(3 p)(x < T( y
is provable in
)p
I + ReCA.
(3q) (Suc
is reasonable.")
(x) &HZz(y)&
m
& there are infinitely many q >mzp)),
58
Assume Reas(mz), ffZ (x), H z (y), WZ (m). Then
n
zleast member
there is a set, w, which has no mZ-least member.It is
Proof:
then easy to see, using
I + ReCA, that
(3g)(k)(g(k+l) (
g(k)).
n
(p) (k)((x)p
We will show that
T (Y)g(k~ BY Reas(mZ), this
is sufficient.
For, we take, in I + ReCA, pl(3k)(~((x)p < T(Y)g(k)))•.
Wecan do this, since the predicate we are taking the extension
of is arithmetical in
call it
0
1)
q,
by
x,y.
WZ(n).
And we take an
clearly
We then obtain a contradiction.
z
T(Y)g(k))
(k)(z
T( Y
)g (k ) '
If
lim
If
(g(k)), then by Reas(mz), we have
r,s
with
() s(r)
Suc
lim(s),
and so
and so80 z
T()g(k)
(q,r).
(x)(l)
T(Y)g(k1),
z
T(Y)s
(X)r < T (
Then
all k > 0,
)g
Sucz
K T(Y)q
(q )
,
use
The case when
If
(k)'
(q)
some
Sucrz(g(k),s),
all
and hence
lime
then
So clearly
k for the following general reason:
(y)(l)
z(g(k)),
0m (g(k)), also easy.
Suc
all
such p,
z (q). To get a contradiction for this case, it suffices
to show that
2)
n-least
if
k.
(Y)g(k)
But then
(X)q T(y)g(k),
p < zq, then
is easy. When
Reas(mz) as in Case 1) above.
m
3)
lim
> (X)1
a set is
(q).
Let
T(Y)g(k+10)
(x)1 ,
k
be arbitrary. We know ()
)
-
( <
zq
Furthermore, the question of whether
is a question low arithmetical in
the uniqueness of Hierarchies on
RLO's,
,
due to
and even hierarchies
59
on initial segments of
in
(Y)g(k+lO)'
RLO's.
and hence
Hence
(x)q is low arithmetical
(x)q will re recursive in
mZ,
g(k+l10)' Hence, by reasonableness of
we have
(,X)q < T(Y)g(k).
Lemma 6:
"The predicate WX(n)
Proof:
We have, in
I1
for predicates
in
ReCA + I,
x,
is
the Kleene normal form
(3 f)(m)4T(e,n,m,f,x)). Now
there is an explicit recursive function
3
WX(g(k,n))]. Hence
g,
for which we can
(k)(n)E(f)(3m)T(k,n,m,f,x)-
ReCA + I, that B)
such that (n)(f)(3m)T(n,O,m,f,x) -
(3 f)(m)~'.T(e,n,f,x)), is provable in
(f)(3 m)T(m,O,m,f,x) of
predicate
x"
and we use that to formalize the
So assume A) (n)(WX(n)
prove in
in
I + ReCA.
provable in
Lemma.
E1
is not
I + ReCA,
n
is
1
since the
in
x,
by A)
and B), and Kleene's normal form Theorem is provable. Now
substitute
3.
e
for
to get a contradiction.
Conservative Extension Result.
Theorem 1:
S
n
is any purely
E 1 -DC + S
where
Proof:
pretation
J
Given any model
k
S
of pure
A1 -CA + S,
sentence, there exists a model M'
is (3 y)(g)(3n)T(g,O,n,y,z),
A model of pure
A1 -CA + S
X,
J,+,x,O,'; 2) having same
+,x,O,,
and a binary relation
We define a new model
HypY(M) = M',
R(J,x)
where
of
for some
e.
consists of an inter-
of the natural numbers, and
with a set of objects,
XEX.
M
together
R(j,x), JEJ,
as 1) having same
but restricted to those
60
JEJ, xX
that
with
Hyp(x).
M
Now since M
S, choose y
(g)(3 n)T(i,O,n,y,g). Then clearly M'
M
(e,O,n,y,g).
Mt
So
We claim
M'
(g)(3 n)T
S.
=
E 1-DC.
It is sufficient to show that the
relativized of each axiom of
a theorem of pure
such
A 1 -CA.
X-1DC to the predicate Hyp
is
(We define the relativized, T(A),
of a formula A, to the predicate HypY(x)
by
T(A) v T(B), T(A & B) = T(A) & T(B), T(A)
-= T(A),T(3xA)=
T(A v B) =
(3 x)(T(A) & ypY(x)), T((x)A) = (x)(HypY(x) ->
Q, Q quantifier
free,
T(A)), T(Q) =
T(3nA) = (3n)T(A), T((n)A) = (n)(T(A))).
In other words, the universal closure, obtained by inserting
the universal quantifier
axiom of
1)
(y), of the relativized of each
-1_DC to HypY
is a Theoremof pure
Induction. It is
instance of induction to
clear
HypY
ion, and so is provable in pure
2)
ReCA.
To see this, let
ny
first
10'k
l .
l1 -CA.
integers
it suffices
(x)(HypY (x) -> Hypy (x ( l
i,
So
set
Define a new ordering
x()
) )).
< T(1)
k = f y(i) + 2.
If
mx by adding on the
on top of n, and make,in the trivial
way, the ordering my
Imy
relativized of each
is again an instance of induct-
Hy (z), W Y(n), x < T
has a greatest element,
not, set k = 2.
the
To prove the relativized of ReCA,
to prove, in pure A1-CA, that
If
that
A1-CA.
total, so that
RLOY (m)
and
Iny
l
<
Now clearly by Lemma 4, (3w)(Hmy(w)), and so clearly
(3w)(HY y(w)),
m
by
I + ReCA. But, also by Lemma 4,
z(1)
< Tw
.
Hence
3)
HypY(z(l)).
Suppose (f)(HpY(f) ->
(e,O,m,f,g,x))), where
the predicate
nO
. . . ,n k
n l
Pn
n
(3g)(HypY(g)& (m)~T
3
HypY(x).
Let
HypY(f). We consider
is a member of a finite sequence
k 2 0,
such that I) (i)WY(ni). II)
sequence of sets xO ,xl,...,x
j
, J
k,
For any
such that
(i)(HY (xi ) ) ,
ny
i
3
we have that
and
z0 = f
another sequence
and
(i < J)(m)T
3
such that each
(e,O,m,z
,X). i+ 1
J < k,
sequence xo,xl,...,x,
zi
with
zirx
i
,
III) For any
(i < J)(RHY(xi)), it is
(z) & p has infinitely
(3 z)(3p)[Hy
(nj+1 )p Zi ~~T~~i, ¶1+1• TZmany
q >
many q > y p & 3Zozl,S, zz j j+ l
TXi' Zj+1
TZ'
i
n.+1
not the case that
)(m)T3(eO
(s <
with
,m,ZsZs+l,1x)].
Also consider Qn
n O ,.. . ,
n
such that
3
sequence of sets
with I) (i)HYy(xi). II)
3
a sequence z i
that each
zi < Txi
III)
-1
Let
is a member of a finite sequence
O, Reas(ny ),
, k
xOXl,...,Xk
n
zo = f
and
and (i)(m)-T3 (e,O,m,zi,zi+
1 ,x
• j < k. It is not the case that
Et • TXJ+l & HY
(z) & p
such
(3 z)(3 p)
has infinitely many
q >
(n3 , )y
& 3
i_ TXi'
,Zj+l,
,Zl,...,z
j
ZJ+1
TZ'
with
(S < )(m)
~,T
3 (e ,O,m,z s ,zs+1 ,x) ] .
We first note that
that
Qn
is a
21
Pn
is a
predicate in
We wish to show,
in
TI predicate in
f,x,y
f,x,y.
I + ReCA, that
(n)(Pn - Qn).
and
)
·
62
Suppose
Pn.
Assume that
nO,nl,...,nk have the
properties mentioned in the definition of
(i)(3 xi)HYy (xi).
take
i
For, if for some
i, '(3xi)HYy (xi),
then
to be least such (induction), and then by (i)WY(ni)
(q)([WY(q) & (3 z)(Hy (z))] ->
we have that
qY
and, in fact,
qY
is imbedded into
infinitely many points
must be
Pn. We claim that
•
T
some w
p >
with
itely many points
s > yr.
condition III) of
Pn
negative.)
ni
yS.
niy
H
J = i-l.
when
Pn,
z
having
with
(w), where
r
(We allow
For Just apply condition 2) of
II) of
i
with 1.u.b.
Hence the
y
< InYl),
Hy (z)
has infin-
We claim that this violates
(f)(3 g)(m)-uT relativized to
condition
ny
qY
HypY ,
use that
Pn,
and that
J
to be
and that
Hypy (f). In
(s)(s < i ->
(3 xi)HY(xi)),
n½
and use the
x i.
So we want to show
with
of
HYy(xi).
II) in
i
Pn.
III) in
Qn
Qn,
Qn
and we may choose XO,xl,...,xk
follows from 2) in the definition
follows from III) in
<T
plus the
have unique solutions
observation that the predicates HY
(n
Pn
p
in (X)p.
So
Pn --
Suppose
and taking
i
> Qn.
Qn. We first to show
least with
WY (ni),
(i)WY(ni).
we contradict
in the same way as the argument above for
one uses II)
of
Qn,
Pn ->
Using
III)
Qn.
Lemma 5,
of
Qn
For,
and the relativization of (f)(3g)(m)~T.
63
(i)WY(ni).
So
Pn
Now II) of
ni
of hierarchies on
Pn
III) of
follows from II) of
because of
Qn
by uniqueness
WY(ni).
follows from III) of
for the same reason.
Qn
Next, we show that we can eliminate the parameters
x
Pn
in
and
Qn,
1
respectively,
let
el, e2
have
f
and
and have the same meaning and still be,
and
For, since Hypy
E1.
WY(el)
(x )
Hypy
and
,
TZ
WY(e2 ) & (3z)(HYy(z) & x
and
(f )
e1
with Gdel
number
k) & (3w)(HYy(w) & f
T
with
G8del
2
number 1).
P'n
Then take
HY (w) & x < T
Gdel
with
(x)(f)(w)(z)(HYy(z) &
to be
k & f
number
with G'del
Tw
ey
P(n,f,x)). Take
->
number 1
(Q(n,x,f) & Hy (w) & x
number
with Gdel
).
Tz
Q'n
Gcdel number
with
el,e2, k ,
Here,
(3x)(3f)(3w)(3z)
to be
and
k & f
T
are constants,
2
not variables.
Now,
Q'n
Q'n - P'n.
clearly
So define
(m)(Rm ->
(3p)(WY(p) &
InYlJ
of
pY
1.
is
IpYl).
up to
Rp).
Also,
Rm -
(3n)(Q'n & ImYl
WY (x)), and
Fix such a
We wish to form
R
p.
So
InYl).
By
So
l. So by Lemma 6,
WY(p) & (n)(Q'n ->
A -CA,
Call this set
W Y(p), let
pO
segment
since this is
[<r,s>P's & ~.(thesegment of
rl(Bs)(<r,s> e S!.
since
[<r,s>lQ's & Isl ithe
rj<jls)1, form this (these) set(s).
T =
is
r ). We can, using pure
also equivalent to
WY(n)),
(n)('n->
pY
S.
be the
up to
Let
< y-least
py
upper bound of this set. Let
pY
segment of
bound for
Inyl <
T,
qY|).
ment of
q
q
have
WV(q),
up to and including po1.
take
q = p.
I1). Hence, clearly
q
the
If there is no upper
In any case, clearly
Also, clearly ()(3n)(Q'n
up to
with
(n)(Q'n ->
InYl 2
->
the seg-
(j)(3 z)(Ky(z)).
(3 w)(Hy (w)). The only hard case is when
We wish to show
q
qY
has no greatest point.
But we can form, in pure
{<,J>l(3z)(Hq (z) & Jez)),
ition, has the
1
1
I 1 defin-
since it, besides this
definition
A1-CA,
(<,J>j(z)(H
(z) -> jez)),
I
and we may, arithmetically in this set, get a
w
with
Hy (w).
qy
Finally, we claim that the sequence (not necessarily
unique) needed to verify the relativized z1 -DC
Hyp y, can
to
HYy (W), and hence would
qY
have what we wanted all along, namely satisfying HypY . To see
be defined arithmetically in
w
with
this, it suffices to show that any finite sequence n,...,nk
satisfying the conditions in the definition of
extended to
nl,nl,...,nk+l, satisfying Pn,
Pn
and also that
there are sequences satisfying the definition of
with.
Take
To see the latter, let
(k|f
a(Z)k,
and let k
ey
ISYI = Ithe segment of
f < Tz
with
be a l.u.b.
up to
k
can be
Pn
to begin
H Yy(z), W(e).
Take
s
with
with an appropriate
finite number of points added on top so as to make, by the
Lemmas, (3 w)(H y(w) & z
s = e.
Take
nO = s.
Tw ) . If there is no l.u.b., take
Then the sequence
<no> satisfies the
65
the definition of
Pn.
The same trick allows one to see how to
extend a sequence nO, n l' , .*,.
nO
n
l ,..
nk
l
nk +
(f)(3 g)(m)~T
satisfying Pn
satisfying Pn,
2-AC.
<nO,nl,...,nk> satis-
Qn.
Independence Result.
independent of
to an
for one uses the fact that
relativizes, and use that
fies (the definition of)
4.
nk
21-DC
We wish to show here that
1
i-AC
It suffices to show that
is
I £-DC.
We do this by showing that, for a suitably chosen sentence S
with
Con(S +
-DC),
we can prove in
Con(S + 2:-AC).
For then, if
proves
-DC),
S +
Con(S +
1-_DC
S + 4i-DC
1 -DC
that
then
and hence by Gdel's
S+
DC
theorem,
would be inconsistent.
Lemma 1:
If
S
is any sentence in prenex form starting
with universal number and set quantifiers from left, followed
by existential number and set quantifiers, followed by a matrix
Pr(X1-DC + S
containing only number quantifiers, then
(coding into natural numbers)
Proof:
w-model satisfying S").
Collapse all the universal number quantifiers in
the left part of
quantifiers
S,
and collapse all the existential number
on the right part of S (before the matrix)
that
S becomes
(3 m)
into the matrix, to get
S' = (n)(x)(
3 m)( 3 y)A(n,x,m,y).
can talk of (
(z)(3 w)C(z,w),
being a closure of
i.e.,
LC
Dom(4B) & C(z,w))).
so
Then push
(n)(x)( 3 y)B(n,x,y).
Now in set theory, given two w-models
(w
a
3
at
ML
and
,
we
under the sentence
and (z)(z
Dom(fl) ->
(3w)
66
Instead of talkingof real w-models, talk of codings of
them into a set of natural numbersin a natural (arithmetical)
Then we can prove in
way.
the
n
(x)(3 y)B(,x,y), k < n.
sentences
This depends
(n)(x)(3 y)B(n,x,y) is provable in
heavily on that
£1-DC + S.
Now, consider the arithmetical predicate D(z,w)
where
y
is a (coding of) pair
(k,x), w
a (coding of) pair
(x)(3 y)B(,x,y),
D,
p < k+l."
"z
is
(k+l,y)
x
under the
ow applying
k
Z1-DC
sentences
to this predi-
we obtain a sequence of dependent choices in which
the desired w-model of
S
can be obtained recursively.
S
Now suppose we found a true sentence
Lemma 1,
=
is a (coding of) finite w-model which is a closure of
the (coding of) finite w-model
cate
and for
n
a finite closure of this w-model under
-model, 3
any finite
for any
2z-DC + S that
such
I + S
that
£1 -AC.
(remember all w-models provably satisfy
consistent,
since
Then we would be
would be provably consistent in
I + S
done, since
proves
in the form for
S
I)
and
ZI-DC + S
i1 -DC + S
is
is true.
By well-known techniques, ReCA
is finitely axiomatizable,
since we need only consider the case of two parameters. We
take
S
to be the conjunction of this finite axiomatization
with the sentence
(3y)(X (y))). Then
(x)(n)(Wx(n) ->
S
clearly of the proper form.
We define
(y) ->
(W*)X(n) = RLOX(n)& (y)(HypX
nX-least element of
We know that
y,
S + I
provided
y
ReCA + I,
3 an
0).
r
so we can use Lemmas of
the previous chapter about provability in
ReCA + I.
is
67
Lemma 2:
n
with
The following is provable in
WX(n),
nx
Proof:
in
(W*)X(m) & ~WX (m ),
onto an initial segment of
that if
WX(p)
is an
mX.
S
guarantees that
WX(n).
(n)(3 f)(m)4T(e,O,m,f,n,x). The predicate of
(3z)I x(z)) & ReasX(p) is
with
But
ust on the basis of
Now suppose
p,
f(m)
(W*)X(n) & (3z)(Hxx(z)), then the
conclusion of the Lemma is true.
n
we have
P(w)
We can prove, like Lemma 2 of the previous section,
I + ReCA,
(3z)(Hxx(z))
for every
there is a coding of a function f:w ->
such that whenever
imbedding of
S + I:
satisfies it,
3
£1,
k
and so, since every
such that
(3z)(HxX(z)) &
Reasx(k) & Wx (k.). Then by Lemma5, (z)(Hx',(z) ->
Clearly we have (n)( 3 f)( 3 z )[l
x (z)
p
Hyp (z)).
& (m)"hT(e,O,m,f,n,x)3.
Collapsing the quantifiers in the usual way, and putting the
(n)(3 h)(m)
result in Kleene Normal Form, we end up with
~T(q,O,h,n,x). Furthermore, any Skolem function for this
sentence would have, recursive in it, a Skolem function for
(n)(3 f)(m)'T(e,O,m,f,n,x).
that
So it remains to show, in
S + I,
(n)(3h)(m)~T(q,O,m,h,n,x) has a Skolem function.
Now,there is a standard recursive function F
(n)(RLOX(F(n)) & any set for which there is no
element, has recursive in it a solution
& any solution h
is no
h
such that
F(n)X-least
to (m)~.T(q,O,m,h,n,x)
has recursive in it a set for which there
F(n)X-least element). So
Now consider the
(n)((W*)X(F(n)) & ~wX(F(n))).
El-predicate Pk
a coding of a function from
w into
=
RLOX(k) & (3f)(n)(f
P(w)
such that
f(n) is
68
an imbedding of
this
into F(n)X).
predicate holds of all
1
for some
f
kx
r
.
Let
X
g(n)
g
n,
f(n)
WX(k),
Then we have a function
is an imbedding of
Then recursively in
X, f,
---> P(w),
of a function
rx
into
we can find a
such that, for each n,
is a non-empty set for which there is no
element.
and so holds
be any non-empty set for which there is no
rX-least member.
coding
with
WX(r), by Lemma 6.
with
such that for each
F(n)X
k
By Lemma 2 of this section,
F(n)X-least
Hence, by the special property of the recursive
function F,
we may obtain the desired Skolem function for
(n)(3 h)(m)T(q,O,m,h,n,x)
recursively in
Thus we have verified that
I + S
g.
proves
E1 -AC,
and we
immediately have
Theorem 2:
5.
1i'-DC
is independent of
£1 -AC.
Relation Between Predicative and Ivyperarithmetic
Analysis.
£1-A
C
(or i-CA
1
or E-DC)
are considered reasonable
formulations of so-called hyperarithmetic analysis, in view of
the fact that they are natural systems whose minimum w-model
consists of exactly the hyperarithmetic sets of natural numbers.
In this section we compare
1 -AC
with a system T
which
represents the formalization of a small part of predicative
analysis (see Feferman, E
The system
(y)(3x)(
is the Gdel
T
is
(x)), where
).
I + ReCA + the infinite list of axioms
n
varies, and
k
is fixed, and
number of a natural well-ordering of type
k
cO.
69
(Thus the infinite list of sentences is obtained by changing n).
We will show that
T
£1 -AC
with respect to all purely
is a conservative extension of
N2 sentences.
(ReCA is formalized as one sentence, in the standard way,
i.e., using only 2 parameters xO x1 .)
It is easily seen by well-known techniques that the proof
of conservative extension can be made finitary. So we obtain a
finitary proof of
Con(ZE1-AC) relative to
Con(T).
Finitary
generally means here, in PRA (primitive recursive arithmetic).
A widely used index of complexity of an axiomatic theory
is how large is the least upper bound of its provable ordinals.
a
In the present context, an ordinal
is said to be a provable
ordinal of a given fixed theory, if there is an
n
with
RLO(n)
and n has order type a and the theory proves W(n). In
T1 that we will prove,
view of the conservative extension for
it is clear that the provable ordinals of
the provable ordinals of
1 -AC
are exactly
T.
It follows from work of Feferman [l) and Tait
least upper bound on the provable ordinals of
T
represented by the so-th critical function at
0.
l], pp. 14-16).
5] that the
is the ordinal
(See Feferman
Furthermore, it also follows from their work
that in PRA + rule of primitive recursive induction on the
natural RLO corresponding to the
£0 -th critical function at O,
a consistency proof may be given for
T.
So, by our results,
such a consistency proof may be given for
By an instance
of induction
on
kn
1-AC.
we mean a statementof
70
the form
[A(q) & (p)(tp <
> (p)(p < i n->
n & (m)(m<
Ap), where q
p ->
Am)] ->
has Ok(q),
and
Ap)P
is any
A
formula (in the language of Analysis, described at the beginning of this chapter), with possibly free variables, and
A
is
in prenex form.
By the complexity of a formula in prenex form, we will
mean here the total number of quantifiers occurring. Let
Comp(A) be the complexity of
For each integer m,
on
A.
there is a natural predicate T_(n)
GCdel numbers of formulae of complexity
T_(n)
says that the formula with GCdel number
n,
such that
n
is true.
m
The formalization of these truth predicates involve placing
formulae in a weak Kleene normal form (i.e., no attention is
paid to the form of the quantifiers, but only that the matrix
be the T-predicate of the appropriate variables).
By the reflection principle, for a theory
language) of complexity
(n)([n
Pr(S,n)] ->
m,
S,
(of this
we mean the single sentence
Gdel number of a sentence of complexity (i &
T_(n)). We call this
Rm (s).
m
We need some facts from proof theory which follow from
work of Tait (see [5]),
Feferman (see [1], p. 23) and Kreisel
(see [43 ).
Fact 1: 3
primitive recursive functions F, G, and
H
such
that the following are provable in PRA:
1)
For each m,n > O, F(m,n)
an instance of induction on
complexity < H(m),
kG(n)
gives the Gdel
applied
with no parameters.
number of
to a formula
of
71
2)
For each
the theory
m,n
we have that
Rm(In)
In + the formula represented by
defined as the subsystem of
I
J
Pn
F(m,n).
is
In
consisting of induction
< n.
applied only to formulae of complexity
Let
is provable in
be the sentence "for every purely
Y1
(possibly with parameters), (3 2)(q)((q < _-
predicate
> Pq) &
k
< _p & (r)(r >
(2 = p v(
k
->
-Pr))))."
We let
T' = T +
k
Jp.
the infinite list
In the theory
T,
the axioms
(y)(3x)H (x), as
n
kn
varies, have bounded complexity.
Choose
(y)(3 x)(HY
conjunction of any sentence
c'
such that the
(x)) with
ReCA
and
kn
any purely
Z
some fixed constant
m = c'.
c'.
(x) + ReCA + J.
kp
has complexity < c!,
We are interested in Fact 1, when
c = H(c').
We let
(y)(3x)Hy
Jp,
sentence and any
TP = In +
We define new theories
for m = c
Applying Fact 1
P
we get
Fact 2:
3
primitive recursive functions F
the following
1)
instance
and
G
such that
are provable in PRA:
For each
n > 0,
of induction
on
F(n)
kG(n)
gives the Gdel
number of an
applied to a formula with no
parameters of complexity < c.
2)
For each
in the theory
n > O,
we have that
Rc,(In) is provable
In + the formula represented by
F(n).
Now, since Comp(A) = Comp(-A), we see that
formally implies
Con(TP
n + B)
within
TP
n + B,
R c,(In)
where
p
is any
72
B
integer, and
is any purely
formula represented by
if
TP + B
4
sentence. So hence the
F(n) must not be provable in
is consistent, by Gdel's
matter what
n
and
p
T
+ B
Theorem. This is so, no
are.
hew n
Pew
Fact
:3
F, G,
primitive recursive functions
PRA, under the assumption
Con(T + B),
B
sentence, we can derive, for some constant
such that, in
some fixed
4
c,
(n)(p)Con(T + B + .(the sentence "F(n)")) & (n)(F(n) represents an instance of induction on
kG(n)
formula with no parameters of complexity
by finitary
Con(Zl-AC + B).
means, that
It suffices to show (n)Con(z1-AC with only
Proof:
induction
c).
we can conclude,
From Con(T' + B)
Lemma :
•
applied to a
In + B).
For each
n, we get a model for
"B +
1-AC
with only I"
by applying the inner model technique to the theory
TG(nc3) + B + .(thesentence"F(n+c
)"
).
n+c3
Now
on
k
F(n+c3 )
,
G(n+c 3 )
parameters.
represents a certain instance of induction
and let this induction be applied to
Now
B
says
property); we fix such an
ivized of each instance of
Qs,
no
(3f)(f satisfies some specific
f.
1
Then we wish to show the relat-
"B +
1-AC with only induction In
1n
73
Rx = (3s)(3y)(Hf
to the predicate,
(y) & x < T
& (r)(r <
s
k
ks
TG(n + c 3 ) + B + (the sentence
n+c 3
"F(n+c3)"). This would give a consistency proof of "B + El-AC
->
Qr)),
with only
"F(n+c3)") .
is provable in
relative to
I n
Con(TG(n+c) + B +
n+c3
,(the sentence
This in turn is immediately generalizable to a
consistency proof of
Z1 -AC + B
relative to
Con(T' + B).
Actually, it suffices to consider the case when
n > 5.
This lower bound will be convenient later.
since the predicate
and
R
has small complexity compared to
is available in
I
provably relative to
In
So, certainly all instances of
R
c3 ,
TG(n+c 3 )
n+c3
n+c 3
Certainly, B
provably relativizes, since the numbers, and
hence all the arithmetical relations, remain unchanged by taking
this inner model, by fiat.
For similar reasons, clearly ReCA
model.
holds in this inner
The inner model is non-empty, since Qq,
since we assumed that induction on
Q
where
is false.
To verify the last, and most important axiom of
with only
in
In",
ReCA + In
Hence
unique
y
"B +
1-AC
first note that since n 2 5, we have that,
we can prove
(a)(A < _G(n+c3) ->
k
TG(n+c).
n+c 3
O.(q),
k
W(i)
(3ty)H! (y)) is provable in
ki
From now on, whenever
above by
by well-known techniques.
Q < _G(n+c3 ),
k
H f . Then also
we denote the
(2)(p)(t: ap
< G(n+c3 ) & p
&
k
_(n+c
k
3 )]
->
Hf < TH)
k
k
k
TG (n+ c )
is provable in
n+c
(p)(3 g)A(p,g,x) holds relativized to
Now suppose that
the predicate R,
where we also have
Rx.
We wish to conclude,
TG(n+c3 ) + B + ,"F(n+c3)" that 3
in the theory
Skolem
n+c
function for the above sentence, h,
(p)(3 g)(A(p,g,x) & Rg),
So we have
consider an integer
nition of
see that
Rg.
such that
s
A
Rh.
arithmetical. But
satisfying the properties
Since induction fails on
s < _ G(n+c 3 ).
in the defi-
c3
k(
Q,
for
we
It is straightforward to see, using
U
AU
W(k)
and
ReCA, that
(p) (3s)(s
1)
Hf
(A(p,g,x)) & (t)(t <_s ->
k
A(p,g,x)).
(3ts)
t
kt
wit]h
kt
Hf
p
H
<
kt Ps
s).
G(n+c3 ) & (r)(r <
(3 p)(stuff to the right of (3!s)
Qr)]).
(s)(Ps ->
Applying the axiom
k
we get a l.u.b. in
[(s < _
J
G(n+c)
k,
call it
ko < _G(n+c3). Then clearly
k
ko = G(n+c3 ), then
11
Hence consider the
It is clear that
>
s
)
T
which holds iff
and
11
can be written in a natural way in
s ->
form (since t <
holds of
< TH f
does not have a
It is also easy to see that the part of this sentence
to the right of
predicate
3 ) & (3g< TH )
< _G(n+c
k
ks
ko,
on the
(r)(r <
kt
0 ->
with Ps.
Qr).
Now
So if
k
(r)(r < _G(n+c3 ) ->
k
s
Qr),
contradicting
75
that induction fails on
k 0 < _G(n+c 3 ).
So
G(n+c3 )
with
Q.
(r)(r < _k0 ->
Hence, since
Qr),
we
have (r)(r < kk 0 -> ar). A similar argument shows that G(n+c3)
cannot be reached by a finite number of iterations of successor
from
ko, in the ordering k.
But by sentence 1), it is clear
that a Skolem function for our original sentence can be found
recursive in at most a few jumps of
H
.
Hence a Skolem
function can be found satisfying the predicate
R.
In retrospect what we have shown is that for each sufficiently large n, "
TG(n+c3
n+c 3
)
with only
I
+ B"
is consistent if
(F(n+c3 ))" is. Consequently any
+ B +
provable in
1 -AC
£1-AC
must also be provable in
1T'sentence
T'.
We now
observe
Lemma 2:
Proof:
in
T.
even
T'
T.
It suffices to show that each
Jp
is provable
But this is clear, since it is well known that
T
ust I + ReCA) proves induction for any formula on
(or
k,
p
for each
p.
TC- Z11-AC.
Lemma3:
Proof:
(y)(3x)H
y
We have to show that for each
(x) is provable in
Z1 -AC.
Let
p,
the sentence
A(q)
be the predi-
kp
cate
(y)(3 x)H y
k
ordering
kp.
is provable in
(x), and we apply induction to
A
on the
q
(Remember, the full schema of induction on
E21-AC). If
Suc_(r,q), then clearly
Ik
kp
A(q) ->
76
A(r),
by taking J-umps, using
ReCA.
1 im_ (q)
Suppose
&
(r)
k
(r < _q ->
k
A(r)).
Then
(r)(r < _q ->
k
( 3 x)H Y
kr
(x)).
form a Skolem function for this sentence, within
W e can
1
1 -AC.
When
we do, we can find, recursive in at most a few Jumps of the
Skolem function, a
z
with
Hy (z).
This is true of all
kq
and so we have
A(q ).
(q)(q < _p ->
k
A(q)).
Hence by induction on
kp,
we have
By a similar argument to the above, we
obtain (y)(3x)Hy (x).
kp
Wehave immediately
Theorem 3:
sentences.
£1 -AC
is a conservative extension of
T
for
1
2
77
CHAPTER III
In this chapter, we consider the question of
ust which
recursive linear orderings can have certain structures placed
It is convenient to consider
on them; namely, hierarchies.
more general notions of recursive linear orderings and
hierarchies, than in Chapter II.
We say, in this chapter, RLO+(n)
iff
ape defines a
recursive linear ordering (as in Chapter II) whose field is
p
Field(n), we let
the name for the subordering of
n,
whose field is all
q < p.
RLO+.
recursively enumerable. For any
Thus
np
H+(x) as 1) (k)(kex ->
We define
Field(n)).
itself is an
2)
(q)(q
->
with
(3 m)(ke(x)m & m E
[(On(q) & 9 < T(x)q) v
(x)q)]).
we can define the satisfaction
Note that in ReCA+ I,
We can do this, since we
relation for (codings of) w-models.
can prove in
q
be
T(X)q) v (limn(q) & [P(r,s) s < nq &
(SueC(q,r) & (X)(1)
r e (x)S} <
Field(n)
np
ReCA + I
that (x)(J)(3y)(y
H e (x), we define the corresponding Mx
as (a coding of) the sets < T
define
Reas+(e)
RLO(e)
by
in some
x ( j) ) .
Given
(a coding of an w-model)
(x)m, m
Field(n).
We
the same as in Chapter II, except replace
RLO+(e).
We define lim(e)
as
(p)(3 q)(p E Field(e)
-> P < eq).
We define
except replace
W+(e)
RLO(e)
and
by
W*+(e)
the same as in Chapter II,
RLO+(e).
78
We say that
NW(e,X)
founded with respect to
2)
RLO+(e) & e
iff
i.e.,
X,
Lemma 1:
Let
have
x
for corresponding M x,
(3n)(nEX
1)
there is no e-least member of
is not well-
& n
E
Field(e))
X.
He+(x), Reas+(e), lim(e).
we have
Mx
~ W*+(e).
Then
In fact this
I + ReCA.
Lemma is provable in
Proof: Weconsider the theory T = I + ReCA+
(3x) (3e)
(they satisfy hypotheses of the Lemma but not the conclusion).
We will show that
T,
In the theory
ing
proves its own consistency.
T
fix such an
Mx . We will show, in
M
It is clear that
)
W*(e),
= W(r) & < TX
Hr(X) & M
Fix such an
r.
z <
(Y)t)
(y)s
with
s
and hence
rW(r).
So
Mz
J
Y
with
Lim e
such that 1)
Consider
and with
M
and
holds.
with
NW(e,Y).
(s
).
(This
NW(Y, e).
(y)s.
(t)(t < es
There is a
->
For, since M
M l= Hr (X) & W(r),
must be in the model
Y,
(n) &
(z)t
=
We cla im that for these values, our
claim about satisfaction in
X,
X
and an
s E Field(e)). 2)
H+ (z),
& Reas+ (e) & (ReCA + I),
r
there is a set
Reas(e).
n = e .
We set
g= ReCA.
there is an
is taken to imply implicitly
set
M
-~w,(n)).
Now choose
We can do this, using
T.
(3z)(3n)(%H(z) & Reas+
lim(n) & corresponding M z
~
M f
I.
It remains to show M
M
M = correspond-
and
that this
T,
Also, since lim(e), we have
Since
x, e
M z,
s ). So certainly
W*+(e
and
M
~
W(e )
we see that
Mz
/= (Mz
certainly
l
W*+
79
Mz e M,
(es )). Note that
since
lim(e).
This completes the proof that
Hence
T
proves
Con(T),
Theorem,
T.
since the Soundness Theorem
can be formalized and proved in
So, by GCdels
M
T
ReCA + I.
is inconsistent. So the Lemma
has been established. In fact, in view of the inconsistency, the
ReCA + I.
Lemma is provable in
Theorem 1:
of
If
Reas+(e), W*
Reas+(e) and W*+(e)
and
W*+(e)
RLO + (e)), then there is no
Proof:
If
lim(e),
corresponding M,
are identical notions, since
x
H(x).
with
then if there was such an
by Lemma 1, has
W*+(e).
Mx
since every hyperarithmetic set is in
must have
If
Mx
(of course, in view
M
x,
then
However,
(because W+(e)) we
.W*+(e), which is a contradiction.
.
-lim(e), then by
Reas(e),
there is an
lime(m) & there are only finitely many
n > em.
m
So
with
W*
+ (e
m)
.
Then argue as above.
Let NTWO(n) be RLO+(n)& no tail
i.e., (p)(p
Field(n) ->
of
n
the subordering on
is well-ordered,
qlq > np)
is
not well-founded).
Lemma 2:
NTWO(n),
If
Reas+(e) & H(x) & corresponding M x
Mx
then
I + ReCA + S,
where
S
is the sentence
of Chapter II.
Proof:
(n)(WZ(n) ->
NTWO(n).
Let
zeM.
We wish to show Mz
i ReCA + I +
(3w)(HZ(w))). Clearly.lim(e), since Mz
Hence
Mz
I + ReCA.
=
8o
Clearly
(3s)(s e Field(e) & z
n e Field(e)
{qls
eq
model
Mx
such that the subordering of
en)
Let
e
defined on
is satisfied not to be well-founded in the
satisfies that there is a hierarchy y
Mx
Then
.
T(x)s).
(in the generalized sense of this chapter) on a non-well-founded
RLO+, k, the subordering of
T'
where
Mx
w
M
in view of
m
r e Field(n)
Choose
subordering of
n
NW(k,X).
Now consider the
n
t
to
If
(3q)(q > np & NTWO(nq)).
r > np
with
and the
.W+ (n).
N1 predicate, Pt
from
I + ReCA + S:
{s p • ns < nrj is not well-
determined by
We can do this since
subordering of
(y)p
(3w)( (w) &
I + ReCA + (3x)
Field(n), then
NTWO(n) & W*+(n) & p
founded.
(n)(Wz(n) ->
The following is provable in
Lemma 2:
Proof:
defined above, such that
Then the obvioas generalization of Lemma
Ok (p).
5 in Chapter II gives
Ty)),
e
r
t < nr & the
If
is well-founded.
P
has
no solutions, we are done, for then NTWO(nr).
[2])that
has shown
Harrison (see
which has a solution in
every
Field(n), where
Nl predicate
W*+(n),
has an
n-least solution. His proof uses only principles provable in
I + ReCA,
orderings.
excepting the comparability of recursive wellIn Chapter II, we showed this comparability Lemma
is provable in
I + ReCA + S, by Lemma 2 of Section 2.
So there is an
call it
way
r
q,
n-least solution to the predicate Pt,
and we can prove this in
was chosen, it is clear that
I + ReCA + S.
q > np.
By the
And by the way
q
is defined, it is clear that
Lemma 4:
ing
|
Mx
Let
NTWO(q).
Reas+(e), H(x),
lim(e). Then correspond-
~NTWO(e).
Proof:
T = I + ReCA + (3e)(3x) (they
Consider the theory
satisfy hypotheses, but not conclusion). As in Lemma 1, we
wish to show that
T
proves its own consistency.
So, we argue in
T,
that if
e
and
T,
that
M
MI
+ ReCA.
f (3z)(3n)( (z) & Reas+(n) &
It remains to show M
lim(n) & corresponding Mz
.
NTWO(n)).
By Lemma 2 of this chapter, M
1
-AC.
we have
M
< eP
(k)(k
Now for each p
f (k)(k < ep ->
- > NWM(ep
.(Y)k)
M
with
M
M
NTWO(ep),
(3Y)
) '
So again,
(p)(p
with
Hence
(3X)NW(ep,X)). Hence M
(q > ep & (3Y)(k)(k < eq ->
Z
S + I + ReCA.
Field(e)
By Lemma 3 of this chapter,
ReCA + S + W*+(e).
and we will
T.
Clearly, as in Lemma 1,
M C
are chosen so
M = Mx;
that they violate this Lemma, then let
show, in
x
M
i
(p)(p E Field(e) ->
NW(eq, (Y)k))),
by
M
Field(e) ->
since M
1
(3q)
I +
1-AC, we obtain a
(Z)p
is
(Y,q) with
q > ep & (k)( < eq -- > NW(eq,(Y)k))).
Fix such a
2)
Z
T(X)r.
Z,
Let
and let
s > er
(x)s. There is a natural
(t)(t < e s ->
r
with
have 1)
M
y K T()s
(Y)t = (x)t).
claim about satisfaction in
We set
M.
r
Field(e),
= NTWO(s).
such that
Consider
H
z = y, n = e s,
(y),
and
in our
82
Clearly
lim(es) & Reas+ (es). It remains to show corres-
My
ponding
NTWO(es).
We have
is in
TWO(s),
M
Hence My -
M.
Z e My.
and
So every set
AZ
TWO(eS).
This completes the proof of the self-consistency proving
of
T,
and hence the inconsistency
of
T.
Hence Lemma4 must
be true.
Lemma 5:
There is an
e
with
W*+(e) & Field(e) = [nln
is even) such that 1) (3x)H(x), 2)
(y)(H(y) ->
(3X)(Xi,
y(10) & NW(e,X))).
Proof:
We define a total recursive function F
on indices
of the partial recursive functions. We define G(n)
to be the
Gadel number of the RLO (field w) associated with the
sentence "RLO(n) v (x)(~.H(x))." So for every
the Gdel
domain
number of some
{nln
is even}
RLO.
Let
defined by
F(n)
be the
P < G(n)q
By the recursion theorem, there is an
Fix such an
field is
e.
Then
(nin
e
is even).
is the Gdel
y(10))).
Hence
W*+(e).
(3x)(NW(e,X) & Hyp(X)),
e
G(n)
RLO+
is
with
iff 2p <F(n)2q.
tpe = PF(e)
with
number of an
It is clear that
(H(y) & y < T X(10))) & (y)(H+(y) ->
n,
T1
RLO+
whose
(X)(NW(e,X) ->
(3y)
(3x)(NW(e,X) & X
For, if not, then
W+(e),
and
contradicting the 1st conjunct of the
above conjunction. We claim
~W(e). For, if not, then
(3y)
(H+(y)), contradicting the 2nd conJunct of the above conjunction.
So by the 1st conjunct, we have
the properties
(3y)(H+(y)). So
stated in this Lemma.
e
has all
83
Theorem 2:
,)
Take
e
W*(n)
as in Lemma 5.
e
RLO with 1) the ordering
ural
Field(n)
=
w copies of
(hence Field(n)
=
n
to be the nat-
is an initial segment, 2)
(i.e.,
e x
e).
'H(x). Then
(3 y)(NW(e,y) & y E Mx).
found recursively in such a
n
Take
corresponds to
n
w, 3) the ordering
Now suppose
of
with
n
such that (x) (H n (x)).
Proof:
But
There are
Mx
Hence there is a
y,
Mx
I
NTWO(n).
by Lemma 4.
z
which can be
with the property that no tail
is well-founded with respect to
this contradicts
NTWO(n)
z.
And
z E Mx
.
But