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Then U (x) =
lo,
(
-
PI
if
( )]
p =
*
.
.
.
and
g
s.t.
,
j = n m-I
. m.
+ 1
(xi ) > ) is found from mi by means of explicit defi-
a B l =pd( L a k Vx’f’(p,J,x))
S
lfi6 s
4
Iyl
f o r some
and no element
at stage
A
d
Q
.
%Q
with
x < r(g(z),
z ' e dom g
Uo)
and
. Therefore we have
Q
n
= B:t)
:
Q
A
d (zl+l)
8''
put i n t o
x
.
z' < z
>c d
. We consider two cases was not put i n t o at s t a g e u since x < r ( g ( z * ' ) , w ) n dom g . By o u r induction hypothesis we have
r(g(z"),c)
a
Uo
t h e r e i s some where
x c B(48) n
3
at stage
V
n dom g
Assume f o r a c o n t r a d i c t i o n t h a t some A
Q
x < U
at stage
A
Q
and f o r every stage t
and no element
i s put i n t o
n
i s put i n t o
x n dom g
€
Proof : Induction on i s put i n t o
such
ey
is a stage such t h a t
0 2 Yy'
BPY)
68 n
Then we have f o r every
f o r t h e l e a s t such
v8'
Lemma 4 : Assume t h a t
g(z')=
Ia
i s an i n a c t i v e g(z)-fixpoint at a l l s t a g e s
8n
im2cfoc
dom g
I n the following we w r i t e
4
and the assumption 02p Q
6
r ( g ( z t l ) , u o ) = r ( g q O ( z l t ) , Qo) a t stage
A
such t h a t
2.
0'
>"x
x
w i l l not be
either.
Qo
w a s not put i n t o
and
A
and not
a t stage
U
since there e x i s t s
U'GQ
( z l + l ) n dom g f f c ( z ' + l ) fi dom gQ'
.
:
HIGH a-RECURSIVELY ENUMERABLE DEGREES
Since
( z t + l ) n dom g"
at s t a g e
s ( z ' + l ) n dom guO
3
x i s not put i n t o
r(g(z),r) I r(g(z),o)
f o r all
. Assume t h a t t h e r e a minimal s t a g e > . By t h e preceding no element is
Q
fro
such t h a t
I
r(g(z),co) < r(g(z),w)
w i l l be put i n t o r(g(s),Vo)
A
at some s t a g e
r(g(z),a)
4
where
t
w
5
y < r(g(z),.)
r c
i s d e f i n e d according t o c a s e 2 ) . Since
r(g(z),a) no element 5 wo :
y < r(g(z),uo)
w i l l be put i n t o
Otherwise assume t h a t
r(g(z),rO)
u1
A
> y
and
0' r(g(z),uo)
y
r
whereas
r ( g ( z ) , a o ) < cr at any s t a g e
i s t h e minimal such
i s defined according t o case 1 ) we have
= r(g(z),uo)
Q
i s t h e r e f o r e only p o s s i b l e i f
is defined according t o c a s e 1 ) of t h e d e f i n i t i o n o f
r
A
uo because o f c l a u s e b ) i n t h e c o n s t r u c t i o n .
It remains t o prove t h a t C '
255
c a n ' t be put i n t o
A
r(g
r
. Since
Q
l(z),o,)
at s t a g e
@ ,
as
it w a s shown i n t h e f i r s t p a r t of t h i s proof. Thus we have proved t h a t some X
i s an i n a c t i v e g ( z ) - f i x p o i n t at a l l s t a g e s i n
w0
[@,,OC) whereas t h e r e i s no i n a c t i v e g ( z ) - f i x p o i n t at s t a g e
Since we have
Txt ,C
definition of for all z LU Remark:
If
Q
stage
'L
au
r(g(z),r) a r ( g ( z ) , o )
u s a t i s f i e s t h e assumptions o f Lemma 4
x < sup f r ( g ( z ) , w )
ment
.
t h i s gives a contradiction t o the
and we have proved t h a t
.
Q
Iz
6 f
r\
dom g
. Therefore t h e s e s t a g e s
3
0
i s put i n t o
t h e n no e l e A
at any
play a r o l e i n t h i s proof
which i s similar t o t h e r o l e of " t r u e s t a g e s " ( s e e Soare1233) in t h e proof i n ORI. Lemma 5 :
a) b)
l c
S
~
=*
A
For every
e B a we have
and
.
Proof : For convenience we prove a ) and b) simultaneously by
256
MAASS
WOLFGANG
. Assume f o r t h e f o l l o w i n g t h a t
i n d u c t i o n on g"(e)
and t h a t a ) and b) a r e t r u e f o r a l l
el
g-'(e)
such t h a t
= z
g"(el)
.
< z
Observe t h a t t h i s a s s u m p t i o n d o e s i n g e n e r a l n o t imply t h a t
u
(A(e')
a3cfa
I g-'(e')
, which
z
5
u
=*
z1
<
f B ( e t ) I g-'(e')
< z J
from
A(et)
Lemma 2
i s regular :
=*
3
i f we have
i s o f c o u r s e p o s s i b l e ( s e e p o i n t 3 ) of t h e
m o t i v a t i o n ) . But we g e t t h e i n f o r m a t i o n t h a t g"(e')
z
<
B(e')
Since every
t h a t every
I
B(")
i s r e g u l a r we g e t
i s r e g u l a r as w e l l . Then
A(e')
u
implies that
I
U
g-l(e') < z 3
is regular. This
i s t h e o n l y fact which w e u s e from o u r i n d u c t i o n h y p o t h e s i s s o t h a t i n t h e case
elpa =
we d o n ' t need an i n d u c t i o n a t a l l ( t h i s i s
Q
r a t h e r s u r p r i s i n g i f compared w i t h t h e s i t u a t i o n i n ORT For
we write
z+l
J:=
f o r t h e set o f t h o s e s t a g e s
M
u
i s unbounded i n
M
~ A ( ~ g ' -) l (I e ' ) For
-
an <
13 z, n An =
dom g
A(' %
z
by u s i n g t h e r e g u l a r i t y o f
OL
.
I
a define
An
A(")n
pt > hn
:=
gz'(y)
6 i(
.
)
t 1
2 := n
H
sup
An
[An\ n is
B LJ
Z2 L,
3
. It
1,
.
X
some 'c
c
c(
IN
with
c(
We have
1
<
for the
i s constant i n
A
B('X)
An+'
a0
> rr*
a(
since the function
that
2e M
and <
U
define then
.
a - f i n i t e set of a l l
rn i s a n i n a c t i v e g ( z ' ) - f i x p o i n t
(Cr,a) := ft'\ r
dom g h )
f o l l o w s from p r o p e r t y ( 2 ) o f t h e a p p o x i -
m a t i n g f u n c t i o n and Lemma 3 We w r i t e
<
:=
an = B ( v ) ~ an
and t h e f a c t t h a t
)
A(")
B (((z+l) n
s e t s it i s e a s y t o s e e t h a t
are r e g u l a r u - r . e .
e x i s t s . For every g i v en
(V y
B:x)~
A
By u s i n g p r o p e r t y ( 1 ) o f g o (a A(")
0-
are s a t i s f i e d . W e w a r t t o prove
where t h e a s s u m p t i o n s o f Lemma 4 that
, seet231).
s
rt c
Q
3
kr0,a) f o r e v e r y
) z'
.
in
zt L z
Cr,a)
such
f o r some
Then we have t h a t r ( g ( z
r IN
hat
11
*)
a c c o r d i n g t o Lemma 4 1
HIGH &RECURSIVELY
where
ro i s t h e l e a s t element o f
show
sup C r ( g ( z ' ) , e ) I
i n order t o prove t h a t
z' E ( z + l )
dom g l
n
<
M
r e
. Therefore
M
i t i s enough t o
- IN)]
z ' € ((z+l) n dom g
A
sup { r ( g ( z * ) , v ) I
.
a
25 7
ENUMERABLE DEGREES
Q C
M
<
A
Thus assume f o r a c o n t r a d i c t i o n t h a t
Vc
-
< a 3 u e M ~ Z E' ( ( z + I ) n dom g
This implies t h a t f o r every
-
K P L,
C
3
c-)
K
Q
The p a r t
-
u c M 3 z ' e ( ( z + l ) n dom g
(sup K < r(g(zt),w)
IN)
we assume t h a t
It+''
s a t i s f y t h e r i g h t s i d e . By Lemma 4
. Therefore t h e r e i s at
r(g(zt),v) f i x p o i n t at
such t h a t
we have t h a t
which means t h a t
Q
t h e r e i s no stage a t stage
A
that
y
L
'c
Cc+l
and
2
X
C, n 1 = C A n
. Therefore t h e r e i s no
- C,
we have proved t h a t
y
s i n c e otherwise some
active g(e')-fixpoint i n
1
%
= C n A
.
1 y
k
and
At
A
By Lemma 4
t %
since
c
MIt
can be expressed
would be an in'12'
6
which shows t h a t
IN
a;
a;+l
:= p z > X;((the
C,
1; = C n A;
n
c OL
we define
A
A;+,
4
u
B(<') . But . Thus we have proved g 3 .
A
zt
Q
( z + l ) n dom
'c
.
< a
C S: A
by
same as i n t h e d e f i n i t i o n o f ( a t stage
-C
C sa B ( * I )
I n order t o prove a ) assume f o r a c o n t r a d i c t i o n t h a t
For
. Thus
K c L,
a-recursively i n B(
S := s u p f r ( g ( z ' ) , a ) I r e M
that
auch
e
The equivalence which w a s j u s t proved implies t h a t t h i s i s absurd because we have
I
1 i s put i n t o
c
X
, contradicting
[%,=)
Cw n
since
r
i s not an i n a c t i v e g ( z t ) -
such t h a t an element
au
is
r(g(zt),e)
some g ( z ' ) - f i x p o i n t
Q
sup K <
do
Q,z',K
defined according t o case 2 ) of t h e d e f i n i t i o n of
c IN
.
C,)
K c La-
A
of t h i s equivalence i s obvious from our
it+tt
assumptions. For a proof of
72'
.
IN)('E < r ( g ( z t ) , v ) )
L,
Q
An+,)
A
t h e r e e x i s t s a computation of
.
258
WOLFGANG M A S S
"C
-C
for
A"
6:
such t h a t
H
s
5
26 c
- C"
A'
:= sup
- A)) .
L,
We have again t h a t given
L,
1n
since the function
ci
n c r A;
. Now it
i s a g ( z ) - f i x p o i n t at
i s defined f o r some
definition of
c o
is
This implies t h a t
Q 6
f o r every
u
z2 d e f i n a b l e
F o r t h e p r o o f o f b) we choose
have
A(e)
-
-
r1 =
g e t h e r shows t h a t
A(e)
D = B(O) =4 A(')
S QUA'
C,
( u s e Lemma 4 )
A
Sz
hyperregular and
0
E
S n
v
x < rcf A
i s now very easy. We g e t
from Lemma 5
x < rcf A have
K
36 >
A
S n
x
rcf A
f ( x ) (7<&6> p e
OL
x
K
c A*
Q
La-
UZLa &A1 $2.)
this
S"
can be w r i t t e n a s a
i C
cqA
i s non-
D
Q
A) 0
6.
.
"K s L a -
n2 formula.
S1I
. CJ .
S )
such t h a t f o r a l l
Concerning t h e computation f o r
"K
and
x
Then we
we observe t h a t
Since we have
( s e e t h e first p a r t of t h e proof o f Theorem 2 b) i n
r2f a c t
It
which i s weakly
o(
f ( x ) (7C&d> E. A ) 6