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is
(chain.
(E 2)
is o b v i o u s .
and ~yy, (y) = y'. N o w d e f i n e
~ y
y 6 ~yT, (X)
=
is n o r m a l
and < ~ y Normality
~ y = id
yy'
X 6 U7 (4) Uy
f~ ~,,
, Uy >
is a m e n a b l e
To p r o v e
y < y'
.
amenability
set U n = U
•
T
It s u f f i c e s
to s h o w t h a t U n 6 ~ But this T Y" O Z  d e f i n a b l e from ~yy, and U nY 6 0 ~ y ÷ .
Since
that z yy'
~
7,
: < ~
N O W set
< < o Z y , Uy >
Y
is i m m e d i a t e
since U n is Y
n = U ny, of U n is u n i f o r m we get ~yy, (Uy) Y is c o f i n a l we have:
the d e f i n i t i o n
So o b s e r v i n g
(5)
N ~n Y
, Uy>
,
T* = sup I ,
, < ~yy.
, ,
U
, >
I* = I U {y*}. C l e a r l y , O~y, ~ O ~
[ y £ I >
I Y 6 I > , < Zyy,
since ef(IIl)
~ and:
62
X 6Uy~
Our
next
.
aim
~ X 6 ~y
is to s h o w
(7) ? (y) D
and
that
M Y { ~n
3 y< y*
I yc_X
(gZy 6 M Y
for
. We n e e d
first:
y 6 I ,n
for (~.
NY .
the p r o o f
such
~yy, (~y)
= ~y, .
fl j U y ~ J~Y ~+i Hence
Set
there
methods
is a
Z l(NY)
since
it is
'
N Y is a m o u s e ,
be a r e c u r s i v e
By s t a n d a r d
that
of L e m m a 1.3. U O~y 6 J~+l ¥ "
enumeration (cf.
the
of the proof
of
first
63
Lemma and C~y
1.2),
if ~
By
one
v 6X i Y
, ~
~i(~'~)
the
for
Now
I'
set
f o r C Tt,
~
=
' (]rAy b
< ~>
X i 6 N Y such Y
that
Xi £ U Y Y
, < ~ >
6 [ X i  ~ in Y
, then
~ i ( ~' T%) "
choice
i i~< ~ X Y*
is
O~y~
we
"
have
Then
~
OZ.
a mouse.
So
Before is
and
choose
Zyy,(X$)
=
X y, i
In p a r t i c u l a r ,
y < y'
since
N Y~
canonically
< v
canonical
y 6 x y, i
since
may
But
I'
and was
I' 6 K is
proceeding
a natural
I ~I'
further
prewellordering
is
defined
the
let
of
I'
the
set
us
good
for
OZy~
from
N Y*
and
we
were
remind
mice
is
hence
N Y*
£ K,
looking
for.
reader
that
the
which
,
defined
there
as
follows
M
It
is
~ N
shown
iff
in
M 8 6 N8
[4
] that
where
~
@ >M,N
restricted
is
to
regular.
the
core
mice
is
a well
ordering. We
often
use
Let ~
(¥)
M
D
Our = K.
(cf.
the
HM
= KM
~,T
is
1
Then be
regular.
of
if an
which
It
M
is
is
satisfy
easily
seen
critical,
iterative
of
M
then at
M,
. Hence
model
models
H M< 6 M .
that,
if M. '
W
this
iff
M.
HM < H i <
Let
all
[ 4])
w.
a structure
w1Erd~s
that
and
application
conjecture
show
immediate.
following
there
proved
is
< is
Then
a model
r be
in w h i c h
proposed
Let
where
1
1
hand
and
~ = w2, ~'
= ~.
starting Kunen
from
showed
existence
of O #.
We
strengthen
in S i l v e r ' s
result
cannot
be
weakened.
Theorem
1.7:
Assume
~Erd~s
in K.
Proof:
Let
A6~(~)
a good
set
of
fice,
Chang's
N K
and
conjecture.
set
indiscernibles
since
there
will
then
.
K < K ~+" P first By
two
un
Chang's
P e n~'
Then
is c o u n t a b l e
and
con
6B
6 ~'.
L e t b':
~~
~
where ~
a n d b' (~) = ~ .
Our
choice
b' (e) = i s
U is
rug T ~ and g ~
a regularity
(y,~)regular.
(~) < gT,(~)}.
sequence
Contradiction!
for
U and ~
< ~
for
~ 6 Y.
70
We model T s
return
to
of
+ V = K and
ZF
: KT s
Case
the
main
Hence ~
I:
is
the
T 6 E
such
of
our
notion
KT c K S T
~ K +"
There
line
proof. of
We
mouse
note
is
that
absolute
KT
is
a
in
KT t
since
s
that
{s 6 C
IK T
J KT
} 6 U.
T Let
~
Set
W
be
be
tion
L e m m a
s Y i > T
such
KT S
{aE
7s
and
Let
T 6 E,
. Then
~(~)
let
:+ 7~ < ~
2.4:
> T
Suppose 
M.. 1 that
KT
M
= M ~,
2.5: WN C
for
L,
a
since
are
IT 6 r n g
s
since
mouse
KT N s
7~ < ~
There T
K ~
AM, cy.
at
C
set
since
iterate
WN
and
otherwise
ME K { , ~
, s6
all
N
since
for
for
have
a6
the
Hence
and
Bi
7
S
T £ rng
KT ~
at
a
Let
M s with
if
T > T.
a c~ ,
itera
M is
= 7~}
, such ~
T: S
KT ~
6 > ~ , y .
Let
N =
the
large
Let
~ < T
Since
since
contradicts
S
~T
a N f M i . Hence
and
T
of
.
= y s.
M 6K T
arbitrarily
~T S
and
~ < ~
a 6~(~)
 This
iterate
M exists.
N6
K
ith
:+ i < s
~ l
not
We
M~ be 1
n K TmM ~  
then
Lemma
at
Then
N ~ M
~th
that
7~
AM, 6 Nc But
}. T h e n for each ~ £ W there is a core mouse s the ~th iterate of M s is a mouse at a 7 < T . Let s
point
follows Then
such.
J K~
a mouse
Proof: a6
least
{~IK [ ~
Ms ~ K~ Ms
the
M ~ N.
KT
is
a
unique our
T 6 E
But
choice
such
of
K < +t
core
it
(N).
then
model
core
that
mouse
of
ZF. with
an
M.
that
E U.
+ Proof: fine
Let T ~ 6 E,
~ < K f
. We
6 sT
show ~+
by
that
there
induction
is on
T > ~ ~ < K
with as

T~
=
the
least
{a6 WN CTI
T
such
that
T > T,~,
f~ (s) < T s } 6 U;
SUP ~ sup
{T
least
I~ < { a n d
i such
~6 C
that
. T
Let
T = sup
T ~ and
= {e6
C I
sup
T h e n C is c l o s e d Z
= {~6
T~ < f f
unbounded
= sum
{~16
(~) h a s
the
On>
NOW
is
Lemma
Set
2.6: that
Let
we
Z 6 U such
(a) T
~ E Z set
the y~
are
X6~(y~) This by
Lemma
We may
Set
Now
;f
let
This
some
shows
T = sum ~
_ ~a} 6 U. set
Xn
:
Then
Z =
{~ 6 Y N
and
,
fn* = ~ ( f
) . Then
~6
Xn
X* = z n
*
=
hence
pick
~
WTIT 6 rna~ ~
{~If ~ (~) 6 f ~ ( v ) } , n+1 n
N ~ ~
where
N6
K=+.
T ~ T such

and
~(Xn )
T~_> @~} 6 U.
Let
and
*
*
{~I fn+1 (~) 6 f n ( V ) } ,
But
.
~
{~[T
that
~:
hence
*
fn+1(~)6
fn(~)
Contradiction!
Case In
2.2:
this
case
we
theorem
of
Un
normal
K
is
+
(K+) K =
K
actually
[ 3 ] it in
show
suffices K.
This
that
to is
U N L[U]
show
that
equivalent
is
is a m e n a b l e K the following
a
and
+ Claim:
There
are
arbitrarily
large
T <
.
+ Let For W
T
y < K
the =
moment,
{~ 6 C vT :
We
. We
first
~
shall let
PK T = K } 6 U. ~ T~
{X6~(a)nK
see
that
T 6 E be
T
there
arbitrary.
For
~ 6 W
is
T > y
Recall
satisfying
the
claim.
that
define:
I~(X)}.
show
WT = { ~ 6 WTIVT6 6~K} To
show
this
pick
T 6 E,
U. T ~ T.
Then
Y =
{6 6 W  I T 6 r n g
T}
6 U.
Let
a6
T
Let A
A
= o ~ U u0 ~M U u o
= ~
0 = ~ ~
z ~eq~
:]~s
= l~
~ON
£IIeU~ 7 ~M ~ M =
~
z0
pu~
:%~s
'eAT~ISUel]
ST
i0
alaq~ ,D9 ~
"~IOa
IeTOnXo
~q~
£eId
~
s~an~onx~s
~q~
'/ooxd
l~ ~
:1~
1o
'Z ~ {0} 9 i ao 7
~u~s~xd
~q%
uI
o~ 9 x
"g% 3 ]eq~
qons
co> I
sT
~a~N%
pu~
o~9
u~qm
~ ~sooq0"
ST
qons
~)q
pu~
x ~q~ 96A
1
n3 ql~>
A ~]!uT7 ~)
dns
1
( nuo)dns
e
~eq~
n3
X)
o>
" {~n 3 (~) ~0
•~ 9 { ~ =
I D3
~]
{g ~ [g > @ ( i ~ ) q } l ~no~
e UT
~IqeuTl~p
= ~ u~q& = X 6"g
(g~)q}
ox~q~ %~S
'~>
oS
M
x u~q~
l~ U
~eq~
"~ uI
"s~Inu~xo7
{(x) T~
70
~ou~ H
= x qons x ~q
II~qs sT
~ ~
~AIS
I<X'T>}
:q
H
~.eq
= g~
:7oo~d
:OL'~
,D
~M
0 uaq&
uoI~ex~mnu~
[V]@q3x
0 ~ eq
H9
~oqs
papunoqun
~q;
~0 p u e O
~Iq~uIT~P
]eq~
• ~D0
0 s!
" ~1 D ~X
"X3 ~ ~q
eurmaq £ q
= 0 ~S ~ ~o7
"uo!%o~fIq
l
"F/3
'~umi~q
ZL
78
Lemma
2.11:
{sIT
Let
W 6 U,
f 6 ~6
cf(~) +.
Then
there
is T 6 E s u c h
that
>  f(e) } 6 U .
Proof:
Suppose
set
= {~ 6 W f l C ' I T ~ ~ < f ( ~ ) }
YT
not.
(b) (c)
above,
the
K many
predecessors
For
gT
~ 6W
are
let h and
define
pairwise
(mod U)
g
inject
f(a)
into
gT 6 6 ~ T
distinct
mod
(6 < T~,T 6 and v Then
= v
We
now
prove
C
T
we
(?B~) < B) •
then
,
get:
6 [C] n such
that
~ .
Let
f. 6 W, 1
f• :
~ , B , O n R M such iteration
~
Let
some
n. For
= ~o,~+B
(f)
Now define
(x 6 H)
that
only and
an
.
iterable
L~[M]
premouse
is a d m i s s i b l e .
set C = {Kili < ~}.
~ = ~M,a,B
n ~] , < ~ > and
f,
6 [C n 7] <w,
6 W such
1
that
f.
1
:
K
n
~
6 K such
that M,a,6
are
countable
in
is c o n s i s t e n t .
<M,~,B>
= bl(e),
there
foregoing
is < M , e , 8 > 6 K + such
There
M , ~ , B 6 X. L e t b:
But M,~,8
of the
formula
be as in the
H~~X
6 = bl(6).
are
countable
is n o t h i n g
<M,~,8> 6 H L[Vo]= K0
where
lemma.
Let
H is t r a n s i t i v e .
Then
H = H~,~,~
in K +. We
claim
to be proved.
Otherwise
K
argument
. The
X ~ H,
same
~ = w in K + such
Set M = b  l ( M ) ,
and~,~,~ that
that
is c o n s i s t e n t .
< M , ~ , B > 6 K.
If K = K +,
K + = L [ V o] and shows
that M , e , 8
are
countable
K0
in K.
We
are n o w
and ~ = ~ M , ~ , 8 theorem part L~[a]
ready be as
to f i n i s h
in C o r o l l a r y
let 0~6 K be a m o d e l
of O i i s b A(a).
the p r o o f
transitive. It s u f f i c e s
Let
of ~ . a be
to show:
3.12. We m a y
of T h e o r e m By
the
assume
3.1.
Let M,~,8
Barwise compactness that
the ( ~  i n t e r p r e t a t i o n
the w e l l
founded
of ~. T h e n
87
Claim:
L[a] ~ A(a)
Let ~' be the Z F  l a n g u a g e w i t h the c o n s t a n t ~ and ordinal c o n s t a n t s (~ £ On). Define a class S of ~ '  s e n t e n c e s • Let 6 [C] n, fi 6 W, fl: Kn
~
6 [C N ~ ] n ~ < ~ > ~ < T > and L~[a] b ~(~(~)))
I n d i s c e r n i b i l i t y a r g u m e n t s show that this is a c o r r e c t d e f i n i t i o n and that (i) S is a consistent, (2)
r~(~)q 6
(3)
rBx ~(x) I 6 S
S
iff
d e d u c t i v e l y closed class of sentences
L~[a] b ~(~) iff
Bt £ T
where T is the class of (4)
rBx 6 On ~(x) I 6 S
Now l e t ~ b e
iff
for ~
)
of
algebra
(F(~) ,e,
otherwise,
have
show
T (b i
smallest
if
B has
to
~
to ( 2 a ° s + 1 )
< ~
with
, and
result
such
. Now by
choice
of
prin
(P%0,z(~) l d >_ r =
. Thus,
below
all
~ ( x , y I .... ,yt ) 6 ~o set
By
existence
with
bounded M =
~
definable will
subsets
remain
can
now
be
so
that:
established:
2.3
Theorem:
(i)
For
all
finite
If
M ~
PA
(eiJi
sets
, M ~
[a,b]
e. 2 l
n
a < e 1• < b
< max(a,c)),
(I)
F
'
i+I
< max(a
¥x
, then
is
n 6 IN
there
is
a sequence
, satisfying
'
~0(x,y I . .• . . y t ) 6 M ~
there
c)
'
? , i°
n ~
thus
f(m)
its
~ g(m)) . We
S c
T
(the
so
graph
is
to
show
of
3.1),
< g(a)
that < ci(
c,
> g(m)) .
want
T
, a 6 IN , b = f(a)
total
~ g(c)
e i the
definition
of
ci, the
~ x)
i < k}
i K k, graph
"g(y)
of
g
, ~ x"
.
system
of
~ PA.
is
Now
~
function,
S U {c ~ c o } U { V x
the
~ T')
Vm
> n
a ~ d
arithmetical
M'
I'
dicting
total.
~ c O } U { Vx
a model
IN
B m
PA) :
g:IN
a recursive
6 IN
in
in
If
e o , . . . , e k 6 [a,b]
(IN , a , e ° .... ,e k)
abbreviating
:
B n £
be
Z1definable g
(provability
< ci(~
this, If w e
M'
then had
~ g(c)
Itformula
g(c)
~ x) ]i 6 IN}
the
c
I' ~ d
by
= g(c) ^ d
and
M' l ~
< c
I ' ~Z
d
M'
consistent.
2.3,(ii),(3) , then
for
1
is
we
some
As
we
determine
had
(contra
i 6 IN might
an
, since
have
added
o Th(IN ) t o 3.4
T'
, we
Corollary
obtain
(Paris,
Harrington)
: PA
U Th
(IN)
~
~
,
nI noting the
that
recursive
total
proof
there
is
2.3(ii) I ~
just
of
3.4:
M ~
for
, a 6 I < b
had
(Xlformulas
I ~e
contradicts
M
Let b
£ M
models . If
~
of
were end
the minimality
of
f
(a+1) a
M ~
b'
(a+1) a a
there
is
with
. Hence
remark
any
~
extensions) b
the
IN . S i n c e
[a,b]
6 I
. By exceeds
, a 6 M ~
arithmetic)
to
I ~ Th 1 ( I N ) .
up
that
true
there
going
Th(IN)
so
of
totality
~b[O,b]
too.
PA
which
a ~
the
function,
a smallest (only
states
function
recursive
Second
we
a
I =
to
1.4
provably
N ~
a
. Thus
an
I c
[a,b']
M ~
[a,b']
I ~
~ a
~
, by
e ~
M
,
a (a+1) a,
(a+1) a
, and
by
'
167
Remark:
In
finable iff
Paris
function
there
graph
[2]
of
Y:MxM ~ M
is a n
I c M ~e
Y
a
countable
has M ~ PA
Y(a,b)
, then
indicator
M ~ PA
the
in PA,
and
for a l l
, which
which
sentence
the We
can
just
countable
be proved
MI,M 2 ~ PA
one has
MI ~
M3 ~e M2
. This
of
PA
system
formula
~
initial
segment
account,
in
in
by
of
a turns and D.
3.5 C o r o l l a r y : segments
{X c ~
o u t to b e Jensen
There
a nice
asked
are models
(up to i s o m o r p h i s m )
Let
determined
by
language
Define
M 2'
M I' {c i li 6 ~ of P A
S S y ( M I) = SSy(M2) , h e n c e
if
this
MI
part
of t h e
Note
that
for
any in
[5]
3 b 6 M I a z
, cf.
of o n e
. Since
Y(x,y)
for:
T' b e as in the p r o o f
countable.
in a n y
satisfies
in M } , t h e n
S S y ( M I) = S S y ( M 2)
an i n d i c a t o r
is an i n d i c a t o r
JX =
M 2 . Furthermore,
if t h e
for unprovability.
shows
>
that
but
by a r e s u l t
[4] t h a t
M I , is
parameters
, then
Ehrenfeucht
the
of
with
MI,M 2 ~ PA
M I'
also
: Y(a,b)
. He shows
, then with
c a n be r e p l a c e d
showed
a,b 6 M
, Thz1 (M I) ~ Th(M2) , S S y ( M I) = S S y ( M 2 ) ,
standard
Proof:
stated
M ~ PA
M I ~ M2,
H. F r i e d m a n
all
requiring
this
P A to be a d e 
V z H y
(c+I) c} c
If
I c M e
M ~
defines Vx
latter
exploited
easily
for
for
, a 6 I < b
reading:
indicator
that
= max{cl[a,b]~
indicatorproperty any
such
, I ~ PA
indicatorproperty.
function
an i n d i c a t o r
Z1definition
true but unprovable of the
defines
are mutually that
Ce MI' 2
U T'
substructure of
ThzI(MI)
are m u t u a l l y
M
M I ~ a , M 2 ~ ~ s.
~ Th(~)
the r e d u c t s
initial
of
, M I'
M I' , M 2' = ThzI(M2)
initial
to and
segments,
168
and
by
M I' l= T'
As theorem
a second that
PA
i~
~
3.6
Lemma:
provable
The
PA
is n o t
All
sentences
provable,
Let
us
but
for
now
this for
these.
Cf.
one
be
so
Theorem that
a
even
V x3
y[x,y]
(~+I)[
remark the
to
v xV
RyllNardzewski's First
we
n
~
(n+1) n
being
provided
give
to
, n 6
IN , a r e
one
given
in
Wilkie)
, even
axiomatizable.
Remark:
By
using
3.6
is
[2].
fact,
3.7
:
For
S U Th
all
(~)
restricted
the
"from
i~ a
provable,
do
far
is
Ramsey
outside"
on
.
theorem
by
finite
are
Infinite
will
which
by
Paris
and
than
the
simpler
a corollamyof
S ~
PA
there
. In p a r t i c u l a r ,
2.3,(i)
is PA
n 6 is
not
n
truthpredicates,
induction
of
a proof In
Elsentences,
(Z)y
following
~I
finitely
~
induction
here
the
true
.
version
But
proved,
with
1.4
using
topic.
, being
y H z[x,y]
a definable
this
once
S I~ s
n
~
superscripts,
(Paris,
PA with
reprove
axiomatizable.
=
n
n
to:
can
eincompleteness:
the
needs
for
gametheoretic
3.7
use
fixed [6]
can
an
H y[O,y]
that
2 . 3 , (i) Wilkie
we
.
mention
Theorem
of
sentences
are
s .
finitely
form
PA
~
application,
the
in
Proof:
, M2 b
the to
result
can
E formulas
easily
for
be
fixed
extended r 6
r
does
not
Proof:
imply
Let
a finite
meters)
S
set
induction
all
be £
axioms in
such
i_~f (M, (ei) i E ~
the
a
n
a finite of
EofOrmulas
contained a way ) b
subset
in
of
PA.
By
the
~(x,Yl,...,y S
(which
may
be
proof
of
t) m a t c h e s
2.3,(ii) with
assumed
without
(where
TF
the para
that
PA U {c~
< ci+1 Hi 6 ~ }
U TF
is
,(3)
the
.
169
last
group
of
then
the
initial
Set
M b
Th(~
according [a,b]
~
to
% ~
the
. By
Remark:
the
As
Paris
model
models than
~
PA
(the
T
of
Proof: are
for
is
a model
Let
S
(eili
6 ~
of
so
(M,eo,...,ek) ~
the
in
number
in
F )
) models
M
S
for
Theorem:
There
not
even
with
such
are
.
S
, so
that
, hence (~)
nl [a,b']
, but
not
s
, n
~
(n+1) n n
would
means there
that are
whereas
not
3.8
segments
in models
of
initial
will
segments
imply
modelling
that
PA
in
, other
3.9, from
PA
nonexistence
paragraph
M ~ PRA
defined
, so in
model by
of
, as
same
~
ce M
argument
as
recursive
that
arithmetic)
(M' (ei)i encoded
of
an
PRA of
. By
( ~ , e o , . . . , e k)
recursive
a sequence
a set
T
of
2
(primitive
, ei 6 M
subset
S
on
~
has
b T
by
an
initial
seg
Eoindiscernibles.
3.2
and
S
. Hence,
. Using for
6 ~
PRA
~
~
b
a
there
for tro(~,x)
2 . 3 , (ii) ,(I) ,(2)
~ ~(x) yields
) as d e s i r e d .
3.9
,
with
, a ~ ei < b
U Th
I b
) contained
that ~
6 ~
TF
3.7
nonstandard
a finite
6 ~
element
S
6 ~
a natural
ei,i
initial
6 ~ )
6 ~
any
(eili
formulas
.
element,
extract
(eili
the
all.
(eili
which
be
6 ~}U
[2],
nonstandard
. Thus
EofOrmulas
b
application
M
eo,...,e k
of
at
of
M ~ PRA,
< Ci+lii
any
we
3.1),
be
are
to
by
smallest
there
Itinduction
any
a sequence
n
in g e n e r a l
models,
is
the
,
} satisfies i b' 6 I with
above
there
ment
6 M
~
induction
third
For
determined
in
present
Lemma:
M
restricted
pointSout
, are
the
3.1
< e
of
of
nonstandard
element
d
# ~
For
b
minimality
prefixrestricted which
of
2 P A U {c i
) ~
in
2.3, (i)
existence
contradict
T
, a 6 M ~
2.3, (i),
{d 6 Mi 3 i 6 ~
since
3.8
of
segment
), M
(n+1)~
(M' (ei)i 6 ~ I =
axioms
no
recursive
recursive addition.
nonstandard
models
of
PRA
,
170
Proof:
Let
M b PRA,
(M'((e)i)i I 6 ~
6~
, the
) b
T
proof
of
(*)
H x I Vx2...
Set
A =
of
Using
Hence PRA be
we
an
2.3,(ii)
G~del
take
initial ,(3)
numbers
one
e 6 M
segment
shows
a 6 M
as
the
but
on
n)
of
realizes
containing
induction
get
taken
the
3.8,
H x n ~(Xl,...,x
tr °
p(z) by
. By
so
I
that
that
given
for
any
by
(e) i
open
~
,
in
the
PA
set
formula
# ~
. For
of
language I b
M
iff
9(y)
that
the
A =
formula
{n
6 EO
there
3 x1 0
fixed
~ a
, tr
n
the
variables,
(3 e tr
E truthpredicate n
gn:
(c,d,e)
~
~
~
g i v e n by (n)
[x,y]
~
(z+1) z
gn(a) be
tr

g b
the
with
3 e < b
n
relation
~
t h e n becomes
formula
saying:
gn(a)
(c,d,e)))
Hndefinable. ~
required
gn(min(X)) formula
a
n =
< Card(X)" ~
=
Let plus
Z
homogeneous
Ix,y]
E n
. The
(z+1) z
Z
"the
for
n
[x,y]
sets
X
(instead
of
z
(z+1) z)
can the
be
chosen
special
such
case
gn
that = id
in
the
. Set
(n) VxV
z 3y[x,y]
~
(z+1) z z
i
n
(n) is
then
a
Hn+2sentence
, [x,y]
have
(z+1) z is z
a
H formula. n
172
The paragraph
second
proof
3 generalizes
4.2 L e m m a : language
For
each
is an i n i t i a l
then
PA U Th
the
to
an
for
independence . First,
n 6 ~ :
of P A so t h a t
there
for
If
all
~
there
~(x,y,z)
a
given
is n o t h i n g
is a
M b PA:(if
substructure
(~)
of
new
in
En+1formula
M k ~(a,b,c), I ~ PA,
I ~e M with
in
in the
c > ~ , then
I _ c M, n
a 6 I < b),
V x V z ~ y ~(x,y,z) .
~n+] Remark:
Such
indicator,
a
~
this
has
the m a i n
time for
property
of
~ substructures n
a definition
which
of an
are m o d e l s
of
PA
.
(n) NOW we
show
that
[x,y]
~
(z+1)zz
indeed
defines
such
a
Z indicator: n
(n) 4.3 L e m m a :
If
then
is an
there
M ~ PA,
a,b
I ~ PA,
£
M,
c 6 M \ ~
I eC M,
I ~
, n > O, M ~
M
with
[a,b]
~
a 6 I
[a,b]
< b
~
c
(c+I) e,
.
n
(n) Proof:
[a,b]
sequence
c (c+I) c
~
Xoindiscernibles
of
in 2.3, (ii).
Since
a homogeneous
set
[a,b] as
M
for
being
any
<eo,...,ei>
infinite
segment
in
of
6 ~ ),
partition g~(ei)
primitive
recursive
by
6 ~)
< b
, as
[[a,b]]
d
function,
condition
ei+1,
the
. By 2.3, (ii)
from
all
than
be
arises
In p a r t i c u l a r ,
for
I
a
described
, the
if w e w i s h .
Let
(eili
" So w e h a v e
sequence
is s m a l l e r
Zoindiscernibility.
determined
this
on s o m e
< ei+1,
(c+I) c c
a ~ ei
2.3,(ii)
(sequencenumber)
by
M
some
yields
dominates
i 6 ~
(eili
b y the p r o o ~
(~)* (c+1)Cc
gn
implies
all
ei
initial only
I ~
M n
remains
to b e p r o v e d .
We
show
I ~
M, m
~ n, b y
induction
on
m
.
~ 6 ~m+1
"
Zm By
I eC M w e h a v e
already
I ~ Z M.
Let
m
< n,
I ~
O
~ =
H x ~ ( y I ..... Y t , X ) ,
I ~NmM .
So
let
Equivalently replacing 0
we only
sets
some
and b y i n d u c t i o n
to r e n d e r
For
for
entails
3x the
this
<e i + 2 t r n ( . . . ) , t same
b
holds
in
I .
we need
n in the p r o o f
from
of
an i n f i n i t e
< Card(X)
for
any
~
~ s
set,
the
so it is n o t
fixed
function
f:~.
obtain
n > O
an
Moreover,
if
is a t r u e M ~
PA,
Nn+2sentence
M e
~
,
there
is
independent an
~n+1 I eC M,
I
N PA,
so
that
I
~
M
and
I
~
~
~
n
n
Remark: case
i6
For
the f i r s t
M ~ Th(~
Corollary:
For
of i n i t i a l
complete
extensions
stronger
all
Ehrenfeucht result
completions that
for
with
regard
of
n = O
of
4.5,4.3
is o n l y
needed
in the
special
)
infinity
Remark:
part
that PA 4.6
n 6 ~
any nonstandard
ZnSUbStructures of
PA
and D.
satisfying
of
Th(~)
pairwise
has
an
different
.
Jensen
proved
a nonstandard satisfied
by
immediately
to the r e s u l t
model
model
in
[3] f o r
of P A h a s
n = 0 2
~o
its i n i t i a l
substructures.
generalizes
to m o d e l s
by Gaifman
quoted
in the
of
remark
the distinct Note PA
U Th
following
(~), 3.4.
174
References
[i]
J. Paris,
L. Harrington:
Ar i t h m e t i c ,
A Mathematical
in: H a n d b o o k
Incompleteness
of M a t h e m a t i c a l
Logic,
ed.
in Peano
Jon Barwise,
11331142. [2]
J. Paris:
Some
The Journal
[3]
Proc.
1971
of the
Lecture
H. Gaifman: Proc.
Lecture
J. F. Knight: The Journal
[7]
A Note
J.C.
of Number S6r.
Types
Astr.,
Bull.
vol.
43, No.
1976,
Summer
in Math.,
School vol.
vol.
40,
Model
XII,
of A r i t h m e t i c , London
255,
1975,
1970,
128144.
Models
of A r i t h m e t i c ,
317320.
Polonaise
No.
2,
Logic,
539573.
for a Free V a r i a b l e
l'acad&mie vol.
in:
Logic,
in U n c o u n t a b l e vol.
arithmetics,
in M a t h e m a t i c a l
337,
in M a t h e m a t i c s ,
de
in e l e m e n t a r y
and Submodels
Logic,
725731.
223245.
in M a t h e m a t i c a l
Phys.,
4, 1978,
of Set Theories,
A Nonstandard
Theory,
Math.,
Models
Omitted
of S y m b o l i c
Sheperdson:
XCII,
on Models
Notes
for Peano A r i t h m e t i c ,
Some p r o b l e m s
Cambridge Notes,
of the C o n f e r e n c e
Sp r i n g e r [6]
Mathematical, Countable
Results
Logic,
D. Jensen:
H. Friedman:
Sp r i n g e r
[5]
of S y m b o l i c
A. E h r e n f e u c h t , Fundamenta
[4]
Independence
1964,
Fragment
des Sciences, 7986.