Degrees of Unsolvability (North-Holland Mathematics Studies - 2)

DEGREES OF UNSOLVABILITY

Dedicated to

S.C. KLEENE, who made recursive function theory into a theory.

NORTH-HOLLAND MATHEMATICS STUDIES

Degrees of Unsolvability

JOSEPH R. SHOENFIELD Professor of Mathematics Duke University Durham, N.C., USA

1971

NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM-LONDON AMERICAN ELSNIER PUBLISHING COMPANY, INC. - NEW YORK

2

O NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM - 1971 All Rights Reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the.Copyright owner.

ISBN North-Holland 07204 2061 x ISBN American Elsevier00444 10128 4

PUBLISHERS :

NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM NORTH-HOLLAND PUBLISHING COMPANY, LTD. - LONDON SOLE DISTRIBUTORS FOR THE U.S.A. A N D CANADA:

AMERICAN ELSEVIER PUBLISHING COMPANY, INC. 52 VANDERBILT AVENUE NEW YORK, N.Y. 10017

P R I N T E D I N THE N E T H E R L A N D S

CONTENTS DEGREES OF UNSOLVABILITY - J.R. Shoenfield

Introduction 0 - Terminology and Notation 1 - Recursive Functions 2 - Isomorphisms 3 -Algorithms 4 - Relative Recursiveness 5 - Recursive Enumerability 6 - Degrees 7 - Evaluating Degrees 8 - Incomparable Degrees 9 - Upper and Lower Bounds 1 0 - The Jump Operation 11 - Minimal Degrees 12 - Simple Sets 1 3 -The Priority Method 1 4 -The Splitting Theorem 15 - Maximal Sets 16 - Infinite Injury 17 - Index Sets 18 - Branching Degrees

Page

VI I 1 3 8 12 15 20 26

31 39 43 46 49 57

60 66 72 83 93 99

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INTRODUCTION These n o t e s o r i g i n a t e d from a seminar which I gave a t UCLA

i n 1967.

S e v e r a l g r a d u a t e s t u d e n t s i n t h e seminar wrote up

a s e t of n o t e s .

After l e c t u r i n g from these n o t e s a t t h e

C a t h o l i c U n i v e r s i t y of S a n t i a g o i n 1969, I rewrote t h e n o t e s i n t h e p r e s e n t form. A set of n o t e s should have some purpose o t h e r t h e n c o l -

l e c t i n g theorems from t h e l i t e r a t u r e .

My main purpose here

i s t o show t h a t p r o o f s i n degree t h e o r y need not be a s compli-

c a t e d and incomprehensible as t h e y appear t o b e i n t h e l i t e r a ture.

(Some of t h e l a t e r a r t i c l e s of Friedberg and Lachlan

provide pleasant exceptions. )

The main d i f f e r e n c e between

t h e p r o o f s here and t h o s e i n t h e l i t e r a t u r e i s that I have d i -

r e c t l y g i v e n the idea behind t h e p r o o f , i n s t e a d o f t r a n s l a t i n g i t i n t o a n obscure language from which t h e reader must t r a n s l a t e back.

I have a l s o avoided i n t r o d u c i n g t o o much n o t a t i o n and

c a r e f u l l y avoided i n t r o d u c i n g s p e c i f i c enumerations of p a i r s , f i n i t e sets, f i n i t e sequences, e t c .

I a m convinced t h a t i f

t h e e x p e r t s i n degree t h e o r y would f o l l o w these p r e c e p t s i n

w r i t i n g t h e i r a r t i c l e s , o t h e r l o g i c i a n s would read these a r t i -

c l e s i n s t e a d of merely l o o k i n g a t them w i t h awe ( o r d i s g u s t ) .

Mea c u l p a . I have a l s o t r i e d t o show how l i t t l e about r e c u r s i v e func-

t i o n s i s needed.

I have n o t even i n t r o d u c e d what i s u s u a l l y

called a ' r i g o r o u s ' d e f i n i t i o n of a r e c u r s i v e f u n c t i o n .

Of

course, such a d e f i n i t i o n i s needed f o r some a p p l i c a t i o n s of r e c u r s i v e f u n c t i o n t h e o r y , b u t n o t for t h e t h e o r y o f d e g r e e s . Besides s i m p l i f y i n g t h e e x p o s i t i o n , I hope t h i s may i n d i c a t e what s o r t of axioms a r e needed f o r a n a x i o m a t i c t h e o r y of recursive functions. No a t t e m p t has been made t o c o v e r even t h e t o p i c s treated

i n a complete manner.

The main o b j e c t h a s been t o g i v e exam-

p l e s o f t h e t e c h n i q u e s which have proved most u s e f u l .

T h i s book owes much t o d i s c u s s i o n s w i t h s e v e r a l people, p a r t i c u l a r l y Alastair Lachlan, H a r t l e y Rogers, Gerald Sacks, and Mike Yates.

The NSF h a s provided v a l u a b l e f i n a n c i a l

support. J . R. S h o e n f i e l d

Durham, N. C . August 10, 1971

0 . Terminology and Notation

W e use t h e f o l l o w i n g l o g i c a l n o t a t i o n :

or; & We c a l l

and; j f o r 1, v , & ,

for

i m p l i e s ; cs f o r and

f~

vx

a (=

v for

i f and only i f ) .

connectives.

If P ( x ) i s a s t a t e m e n t about x,

h o l d s f o r some x and

7 f o r r&;

3 x P ( x ) means t h a t P ( x )

P ( x ) means t h a t P ( x ) h o l d s f o r a l l x .

T h i s n o t a t i o n w i l l only be used when x i s r e s t r i c t e d t o vary

Sometimes t h i s s e t i s d i r e c t l y i n d i -

through some f i x e d s e t . c a t e d by t h e n o t a t i o n .

Thus

(3xE

y ) P ( x ) means t h a t P ( x )

h o l d s f o r some x i n y; and (‘dx < y ) P ( x ) means t h a t P ( x ) h o l d s

f o r every x such t h a t x


f o r ordered

n-tuples,

i d e n t i f i e d w i t h any c l a s s .

A n a t u r a l number is a non-negative i n t e g e r .

We use N

for t h e c l a s s of n a t u r a l numbers, and u s e i, j , k, m, n, r, s,

and t t o d e s i g n a t e n a t u r a l numbers. An i n f i n i t e sequence i s i d e n t i r i e d w i t h a mapping having N

as i t s domain.

A s usual, we w r i t e

for t h e v a l u e of t h e i n -

TERMINOLOGY AND NOTATION

2

f i n i t e sequence x a t t h e argument n, and d e s i g n a t e t h e i n f i n i t e sequence by

Is).

A f i n i t e sequence o f l e n g t h n i s i d e n t i f i e d w i t h a mapping

w i t h domain 10, 1,

. . . , n-1)

(but

not

with an ordered n - t u p l e ) .

Again xi i s t h e v a l u e of t h e sequence x a t t h e argument i . l e n g t h of a f i n i t e sequence x i s d e s i g n a t e d by l h ( x ) .

The

We use

f o r t h e sequence o f l e n g t h 0 ( a s w e l l a s f o r t h e empty c l a s s ) .

f

1. Recursive Functions

Recursion theory i s t h e a b s t r a c t theory of computations. A l l of t h e computations which we consider w i l l be, a t l e a s t i n

theory, performable i n a f i n i t e length of time.

This means

t h a t t h e o b j e c t s with which we compute m u s t be f i n i t e o b j e c t s ,

i . e . , o b j e c t s which can be s p e c i f i e d by a f i n i t e amount of i n formation. Some examples w i l l c l a r i f y t h e notion of a f i n i t e o b j e c t . A n a t u r a l number i s a f i n i t e o b j e c t , s i n c e i t can be s p e c i f i e d by giving t h e Arabic numeral d e s i g n a t i n g t h a t number.

On t h e

o t h e r hand, a r e a l number i s g e n e r a l l y not a f i n i t e o b j e c t , s i n c e t o s p e c i f y i t w e m u s t , say, give each of i t s i n f i n i t e l y many decimal p l a c e s .

An n-tuple o f f i n i t e o b j e c t s i s a f i n i t e

o b j e c t , s i n c e i t can be s p e c i f i e d b y s p e c i f y i n g each of t h e n objects i n turn.

A f i n i t e c l a s s of f i n i t e o b j e c t s i s a f i n i t e

o b j e c t ; an i n f i n i t e c l a s s of f i n i t e o b j e c t s i s g e n e r a l l y not a f i n i t e object.

A f i n i t e sequence o f f i n i t e o b j e c t s i s a f i n i t e

o b j e c t ; an i n f i n i t e sequence g e n e r a l l y i s n o t . A space i s an i n f i n i t e c l a s s X of f i n i t e o b j e c t s such t h a t ,

given a f i n i t e o b j e c t x, we can decide whether o r not x belongs t o X.

We g i v e some examples of spaces. (1) The c l a s s N of n a t u r a l numbers i s a space.

( 2 ) If X and Y a r e spaces, then X

xY

is a space.

( 3 ) I f X i s a space, then t h e c l a s s Sub(X) of f i n i t e subc l a s s e s of X i s a space.

3

4

RECURSIVE FUNCTIONS

(4) If X is either a space or a finite non-empty class of finite objects, then the class Sq(X) of finite sequences of elements of X is a space. We use X, Y, and 2 for spaces.

Generally x, y, and z

are elements of X, Y, and Z respectively.

A function from X

Y is a mapping from X to Y; a set in

X is a subclass of X.

A function is always a function from a

space to a space, and a

set is

always a set in a space.

We

use F, G, H, L and M for functions and A, B, C, D and E for sets.

A function F from a Cartesian product of n spaces is called a function F(<xlJ

of

. . .,

n arguments; we write simply F(xlJ

is called a relation

of

for

A set in a Cartesian product of n spaces

xn>).

as 2 function

..., x,)

of

n arguments.

If' we refer to ..x..y..

x,y, we mean the function F defined by

F(x,y) = ..x..y..

.

If we refer to ..x..y.. as 2 relation

of

x,y, we mean the relation A defined by <x,y> E A tt ..x..y.. By the relation

=,

we mean x

=

the relation C , the relation

.

y as a relation of x,y; similarly



E A c 3

@(XI,

Y> E B

G(Y,x),

where F, G , and B a r e r e c u r s i v e , t h e n A i s r e c u r s i v e . The u n i o n , i n t e r s e c t i o n ,

set d i f f e r e n c e of two r e c u r -

or

s i v e s e t s ( i n t h e same s p a c e ) i s r e c u r s i v e . of a set A i n X, d e s i g n a t e d by A',

The complement

i s t h e set X

-

A;

i t i s re-

cursive iff A i s recursive. Any combination of r e c u r s i v e se ts by means of c o n n e c t i v e s i s recursive.

Thus i f A i s d e f i n e d by

x E A t , (x c B g x E C)

&x

E D

where B, C, and D i s r e c u r s i v e , t h e n A i s r e c u r s i v e .

i s n o t t r u e when w e u s e q u a n t i f i e r s .

The same

Thus suppose t h a t w e d e -

fine x E Awhere B i s r e c u r s i v e .

3Y(<X,Y> E B) TO t e s t w h e t h e r x E A o r x

t e s t w h e t h e r <x,y> E B o r <x,y>

p

B f o r each y .

A , w e must

Since there

a r e i n f i n i t e l y many y, t h i s c a n n o t be done i n a f i n i t e l e n g t h of t i m e

.

This problem d o e s n o t a r i s e i f t h e q u a n t i f i e r v a r i e s t h r o u g h

RECURSIVE FUNCTIONS a f i n i t e set.

7

Thus i f A i s defined by

<x,k> E A H (3n




E

where B i s recursive, then A i s recursive.


E A ) f o r a l l

x, then

F i s recursive; f o r w e may compute F ( x ) by examining <x,O>,

<x,l>, t o A.

... i n

t u r n u n t i l we come t o t h e fi'rst one which belongs

2. I s o m o r p h i s q s

An isomorphism of X and Y i s a one-one f u n c t i o n F from X o n t o Y s u c h t h a t F and F - l a r e r e c u r s i v e .

I f such a n iscmor-

phism e x i s t s , w e s a y t h a t X and Y a r e i s o m o r p h i c . We s h a l l c o n s i d e r o n l y p r o p e r t i e s of s p a c e s which a r e i n -

v a r i a n t u n d e r isomorphisms.

F o r example, w e show t h a t r e c u r -

s i v e n e s s of f u n c t i o n i s i n v a r i a n t u n d e r isomorphisms. be a r e c u r s i v e f u n c t i o n from X t o Y .

L e t G be a n isomorphism

of X and XI, and l e t H be a n isomorphism of Y and Y'.

function

FI

Let F

The

from X I t o Y' which c o r r e s p o n d s t o F i s g i v e n b y

t h e commutative diagram

Thus F l

=

HoFoG- 1

.

S i n c e H, F, and G - l a r e r e c u r s i v e , F ' i s

recursive. Isomorphism Theorem. Any two s p a c e s a r e i s o m o r p h i c . A s a first step towards the proof, w e n o t e t h a t t h e i d e n t i t y

mapping f r o m X t o X i s a n isomorphism; t h a t t h e i n v e r s e of a n i s o morphism i s a n isomorphism; and t h a t t h e c o m p o s i t i o n of two i s o morphisms i s a n isomorphism.

It f o l l o w s t h a t t h e r e l a t i o n of

being isomorphic i s a n equivalence r e l a t i o n .

Hence w e need

o n l y show t h a t f o r e v e r y X, N i s i s o m o r p h i c t o X . A l i s t i n g of a set; A i n X i s a one-one r e c u r s i v e f u n c t i o n

from N t o X w i t h r a n g e A .

O t h e r w i s e s t a t e d , a l i s t i n g of A i s

a n i n f i n i t e sequence ix,S

of e l e m e n t s of A i n which e v e r y menber 8

9

ISOMORPHISMS

of A appears exactly once such that given n, we can compute xn.

Lemma

1.If

F is a listing o f X, then F is an isomoqhism

of N and X.

Proof. We need only show that F - l is recursive.

Given

x, we compute F(O), F(1),

... until we find an n such that

F(n) = x.

n.

Then F-l(x)

=

Q.E.D.

Lemma 2. I f A is an infinite set in X, and F is a recursive function from N to X with range A, then A has a listing. Proof. Define a function G from N to X inductively as follows: G(n)

=

F(m), where m is the smallest number such that

...,

F(m) is distinct from each of G(O), G(1), m exists because A is infinite. from N to X with range A. pute G(n) when G(O), G( l), recursive.

G(n-1).

This

Clearly G is a one-one function

Since F is recursive, we can com-

. . .,

G(n-1) are known.

Hence G is

Q.E .D.

Lemma 2. I f X has a listing, then any space Y included in

X has a listing. Proof. Let F be a listing of X. G(n)

=

F(n) if F(n)

E

Y, and G(n)

= yo

Pick yo

otherwise.

E

Y.

Set

Then G is

a recursive function from N to X with range Y;SO Y has a listing by Lemma 2.

Q.E.D.

If x is a finite object, we may write down a complete description of x .

We may suppose that the symbols used in this

description are chosen from a finite class

r independent of x.

ISOMORPHISMS

10

( I t would s u f f i c e t o p u t i n t o

r

a l l symbols used i n E n g l i s h ,

i n c l u d i n g p u n c t u a t i o n marks, and a l l t h e u s u a l mathematical symbols.)

Since

r

i s a f i n i t e c l a s s of f i n i t e o b j e c t s , S q ( r )

i s a space; and a l l of o u r d e s c r i p t i o n s belong t o t h i s s p a c e . Lemma

4.The

s p a c e Sq(

P r o o f . L e t xl, x2,

r ) has ...,

a listing.

xr be t h e symbols i n

r.

Let

F ( 0 ) and F ( 1 ) be t h e empty sequence. I f n > 1, l e t t h e prime pow“2 “k d e c o m p o s i t i o n of n be pln1p2 er . . .pk , where p1 < p 2
E

B)

I n view of t h e Isomorphism Theorem, we may always

N.

A s an example, t h e equivalence

(1)

x

E

wIe*

3n(x E

wIJn),

t o g e t h e r with t h e f a c t t h a t x

f

W I , ~ i s a recursive r e l a t i o n

of x, I , n, shows t h a t x E WI

i s an F Z r e l a t i o n of x , I .

Si-

milarly, (2)

[11(x)

shows t h a t [ I ] ( x )

=

=

Y

* 34 [IIn(x)

=

Y)

y i s an RF, r e l a t i o n of I,

A r e c u r s i v e s e t A i s RE; for x

i s a r e c u r s i v e r e l a t i o n of x,y.

E

x, y .

At)3y(x

E A),

and x

E

A

The converse i s f a l s e , as t h e

above examples show. A s e l e c t o r for a r e l a t i o n A i n X

x

Y i s a p a r t i a l function

F from X t o Y such t h a t for a l l x, F ( x ) i s defined i f f t h e r e i s a y such t h a t <x,y> E A , and, i n t h i s case, ( x , F ( x ) >

E

A.

Thus

F s e l e c t s a y = F ( x ) such t h a t <x,y> E A, p r o v i d e d t h a t such a y exists.

S e l e c t i o n Theorem. If A i s an RE r e l a t i o n i n X X Y , then t h e r e i s a recursive s e l e c t o r f o r A . Proof'. There i s a r e c u r s i v e B such t h a t <X,Y>

E A-

~ Z ( < X , Y J Z >E B ) .

20

RECURSIVE ENUMERABILITY L e t {) be a l i s t i n g of y

x

2.

21

We d e s c r i b e a n a l g o r i t h m

I from X t o Y by d e s c r i b i n g t h e p r o c e s s of computing a c c o r d i n g

t o I w i t h t h e i n p u t x.

The n t h s t e p i n t h i s p r o c e s s c o n s i s t s

of computing B ( x , yn, z n ) .

If t h e r e s u l t i s T r , w e g i v e t h e

o u t p u t ynj o t h e r w i s e , w e go on t o t h e n e x t s t e p . that

111 i s a Corollary

selector f o r A.

1.A

It i s c l e a r

Q.E.D.

set A i s RE i f f i t i s t h e domain of a r e c u r -

sive p a r t i a l function. P r o o f . Such a domain i s RE by ( 1 ) .

x

E A c* 3 y ( < x , y > E

If F i s a recursive

B) w i t h B r e c u r s i v e .

s e l e c t o r f o r B, t h e n A i s t h e domain of F . Corollary

2. I f

If A i s RE, t h e n

Q.E.D.

B i s RE and A i s d e f i n e d by

x

E A-~Y(<x,Y>

E

B),

t h e n A i s FiE. P r o o f . S i n c e A i s t h e domain of any s e l e c t o r f o r B, t h i s f o l l o w s from t h e theorem and C o r o l l a r y 1. As

F i s RE.

Q.E.D.

an example, t h e range A of a r e c u r s i v e p a r t i a l f u n c t i o n F o r if I i s a n a l g o r i t h m f o r F, t h e n

x

E BCJjy([I](y) =

x)

and [ I ] ( y ) = x i s a n RF, r e l a t i o n of x,y by ( 2 ) . By C o r o l l a r y 1 and t h e Isomorphism Theorem, a s e t A i n X

i s RE iff A = WI f o r some I e A l g ( X , N ) .

Any such I i s t h e n

c a l l e d an i n d e x of A . Parameter Theorem. If A i s a n RE r e l a t i o n i n X X Y, t h e n

22

FZCURSIVE ENUMERABILITY

t h e r e i s a r e c u r s i v e f u n c t i o n F from Y t o Alg(X,N) such t h a t X € WF(y)@

A

<x,Y>

f o r a l l x and y. P r o o f . L e t I be a n i n d e x of A , and l e t F(y) be t h e a l g o r i t h m J such t h a t t h e p r o c e s s of computing a c c o r d i n g t o J w i t h input

x i s t h e same as t h e p r o c e s s 3f computing a c c o r d -

i n g t o I w i t h i n p u t <x,y>.

X E WF(y)'d CJ

Then F is r e c u r s i v e and

CXJy> 1' E A-

<x,Y>

Q.E.D.

L i s t i n g Theorem. An i n f i n i t e set A i s RB i f f i t h a s a l i s t i n g P r o o f . I f A has a l i s t i n g , i t i s t h e r a n g e of a r e c u r s i v e f u n c t i o n and hence i s RE.

x

E A

t-)

3y(<x,y>

S e t F(<x,y>)

=

E

Suppose t h a t A i s RE.

B) w i t h B r e c u r s i v e , and choose xo

E A.

x i f <x,y> E B and F(<x,y>) = xo o t h e r w i s e .

Then F i s r e c u r s i v e and h a s r a n g e A . and Lemma 2 of

Let

f 2 , A has a l i s t i n g .

By t h e Isomorphism Theorem Q.E.D.

The raph of a p a r t f a l f u n c t i o n F f r o m X t o Y i s t h e r e l a t i o n A i n X % Y d e f i n e d by <x,y> E A * F ( x )

G r a m Theorem.

=

y.

A p a r t i a l function i s recursive iff i t s

g r a p h i s RE. P r o o f . The g r a p h of [ I ] i s RE b y ( 2 ) .

Since t h e only

s e l e c t o r for t h e g r a p h of F i s F i t s e l f , t h e c o n v e r s e follows f r o m t h e S e l e c t i o n Theorem.

Q.E.D

Complementation Theorem. A set A i s r e c u r s i v e i f f b o t h A

RECURSIVE ENUMERABILITY

and A'

23

a r e RE. P r o o f . I f A i s r e c u r s i v e , then A C i s r e c u r s i v e ; s o A

and A'

a r e RE.

Suppose t h a t A and A C a r e RE with i n d i c e s I

and J r e s p e c t i v e l y .

For each x, x

E

s o we m y

WI o r x E WJ;

d e f i n e a r e c u r s i v e f u n c t i o n F by F ( x ) = p n ( x E WI,n Then x

E

AC+x E W I , F ( x ) ;

SO

v x

E

1.

W

J,n A i s recursive. Q.E.D.

We now e s t a b l i s h some connections between r e c u r s i v e n e s s and limits.

I f lxn]

i s an i n f i n i t e sequence i n X, we say t h a t

x i s t h e l i m i t of fxn\ and w r i t e l i m xn s u f f i c i e n t l y l a r g e n.

If

=

x i f xn

[ ~ i s~ anf i n f i n i t e

=

x for all

sequence of func-

t i o n s from X t o Y, we say t h a t F i s t h e l i m i t of {Fn$ and w r i t e l i m Fn = F if lim F,(x)

= F(x)

for a l l x .

I f l i m Fn

=

F, a

modulus f o r .IFn) i s a f u n c t i o n H from X t o N such t h a t n 2 H(x) 3 F n ( x ) = F ( x ) f o r all n and x . A sequence { F ~ )i s r e c u r s i v e i f F J X )

i s a r e c u r s i v e func-

t i o n of n , x . Modulus Lemma. If A i s RE and F i s r e c u r s i v e i n A, then t h e r e i s a r e c u r s i v e sequence [Fn) such t h a t l i m F,

= F and

a modulus of IFn) which i s r e c u r s l v e i n A . Proof. W e suppose t h a t a l l spaces involved a r e N. I be an index of A and l e t A,

and l i m An

=

A.

= WI,~.

Then {An\

We o b t a i n a modulus G for {An\

Let

i s recursive

by s e t t i n g

24

RECURSIVE ENUMERABILITY

G(k)

= pn ( k

E An)

i f k E A and G ( k ) = 0 i f k

A; and G i s

recursive i n A . F o r any r, r

L e t J be a n a l g o r i t h m f o r F i n A .

L e t H ( r ) be t h e smallest number rn such t h a t r E rn

I8J,m

A WJ.

E

and

2 G(k) f o r every k used i n t h e computation o f [ J I A ( r ) . An

If n > H ( r ) , t h e n [J],

A

A

( r ) = [ J I n ( r ) = [ J ] ( r ) = F ( r ) by t h e

[J]2(r)

when r E WAn J,n and F n ( r ) = 0 o t h e r w i s e , t h e n l i m Fn = F and H i s a modulus for Thus i f w e set F n ( r )

Use Principle.

=

F i n a l l y , H i s r e c u r s i v e i n A ; f o r a n o r a c l e f o r A eniFn\. A a b l e s u s t o f i n d t h e z used i n t h e Computation of [ J ] (r) and t o compute G ( z ) f o r each such z .

Q.E.D.

A set A i n X i s RE ir; H if t h e r e i s a

space Y and a s e t

B i n X X Y r e c u r s i v e i n H such t h a t

x f o r all x. t o H.

E A~+~Y(<x,Y>

E

B)

We can t h e n r e l a t i v i z e a l l of t h e above r e s u l t s

H i s a n I E Alg(X,N) such t h a t A =

An index of A

E H*,

proving t h a t

A i s r e c u r s i v e i n H*.

S i n c e H i s r e c u r s i v e i n H and hence RE i n H a H i s r e c u r s i v e i n H*.

However, H* i s n o t r e c u r s i v e i n H.

If i t were, t h e n

every set RE i n H would be r e c u r s i v e i n H, c o n t r a d i c t l n g t h e r e l a t i v i z a t i o n t o H of t h e p r e v i o u s l y proved r e s u l t t h a t not every RE set i s r e c u r s i v e .

6 . Degrees We w r i t e F

sRG

if F

i s recursive i n G .

(1)

F

A

nn+li f

A set A i s

x where B i s

E:

E A-

n:

B)

it has a d e f i n i t i o n

't/y(<x,y> E B )

( I n the literature,

1; and

E

En

and

ana r e g e n e r -

t o d i s t i n g u i s h them from o t h e r k i n d s

of s et s which w e do n o t c o n s i d e r . )

Note t h a t t h e

sets

a r e j u s t t h e RF. s e t s .

We now g i v e some r u l e s f o r showing t h a t se ts a r e

nn.

xn or

I n each c a s e , t h e proof i s by i n d u c t i o n on n .

The

c a s e n = 0 i s omitted i n t h e p r o o f , s i n c e i t f o l l o w s from previous r e s u l t s . ( A l ) If A i s

xn

(Tn) and

B i s d e f i n e d b y x E B-

F ( x ) E A where F i s r e c u r s i v e , t h e n B i s

En

(n,).

Proof. Trivial. ( A 2 ) If A i s

and

Zrnor

rm f o r some m < n,

of x,y.

E A);

Hence A i s

nn.

If n

E

B ) where B i s

31

>

is

Now

relation

= 1, A i s a l s o

zn. If n vnv2.By t h e

nm case

znnl.

and b y ( A l ) , x E A i s a

s i m i l a r proof shows t h a t A i s 3y(<x,y>

Em; t h e

By t h e i n d u c t i o n h y p o t h e s i s , A i s

x E Ac;,vy(x

En

nn.

P r o o f . We suppose t h a t A i s

similar.

then A i s

nn-l;t h e n

1, x E A

a

d

induction hypothesis,

32

B is

ITVALUATING DEGREES

lTnm1; so

In.

A is

c In (qn), then A

(A3)-If A i s

En.

P r o o f . Say A 5.s where B i s

nn-l.

is

1,

and

(A4) I f A i s

(En).

Then x E A *

3y(<x,y> E B ) By t h e

rn-l;s o A C i s vn.

€3

i s d e f i n e d by x E B c - )

~ Y ( < x , Y >E A), t h e n B i s

B i s d e f i n e d by x

nn.

E

En.

B e \dy(<x,y>

<x,y> E A - 3 z ( < x , y , z >

fln-l.

If A i s

q nand

B), then B

E

zn, s o t h a t

P r o o f . Say A i s

where C i s

nn

Hence x E ACC-).\dy(<x,y> E B e ) .

i n d u c t i o n h y p o t h e s i s , BC

is

is

E C)

Then

x

E

Bt-, 3y3z(<x,y,z> +7

E C )

3w(<x,F(w),G(w)> E C )

where w v a r i e s through XXY, F ( x , y ) = x, and G ( x , y ) = y .

(Al), < x , F ( w ) , G ( w ) >E C i s a

(A5) If are

A and B a r e

Tn-, r e l a t i o n

of x,w;

zn (v,),t h e n A

U

By

B and A

n

In,so

that x E A

3 Y ( < x , y > E C ) and x E B C ) ~ Z ( < X , Z > E D ) where C and D a r e Then X E

A u €3 * ~ Y ~ z ( < x , Y E> C v <x,z> E D ) .

BY (Al) and t h e i n d u c t i o n h y p o t h e s i s , t h e p a r t f o l l o w i n g is a

lln-l

B

En ('ITn).

P r o o f . Suppose A and B a r e

mn-,.

2,.

so B i s

r e l a t i o n of x, y, z.

Then t h e p a r t f o l l o w i n g

e

EVALUATING DEGREES

3y

z n relation

is a

33

so A u B i s

of x,y;

by (A4).

We

t r e a t A n B similarly.

( A 6 ) If A i s

En

(v,),and

E

B-

<XJk>

E

c t t (3 n

'in

t h e n B and C a r e Proof. If A i s

<x,k>

5

k ) ( < x , n , k > c A), k)(<x,n,k>

A),

E

(nn).

TnJwe

B O vn(k

E

<


B and C are d e f i n e d by

have

n v <x,n,k> E A);

u s i n g (A2), ( A 5 ) , and (A4), we conclude t h a t B i s

n,.

If

A i s En, t h e n <X,nJk> where D i s

Vn-,. <x,k>

E

E

A H 3y(<x,n,k,y>

E

D)

Then B


0' i s

46, 0'

2 0'.

5

a ' for a l l a ; s o e v e r y d e g r e e i n t h i s

We now show t h a t , c o n v e r s e l y , e v e r y d e g r e e

t h e jump of some d e g r e e .

Theorem ( F r i e d b e r g ) . I f 0' such t h a t b 1 = b u 0'

5

a , then t h e r e i s a degree b

a.

=

While t h i s theorem g i v e s a n i c e p r o p e r t y of t h e jump o p e r a t i o n , i t can be used t o show t h a t t h i s o p e r a t i o n h a s p r o p e r F o r example, i t i s n o t one-one.

t i e s which a r e n o t s o n i c e .

For t h i s , u s e t h e theorem t o o b t a i n a d e g r e e b u 0' = 0".

Then b 1 = O",

but b

b such t h a t b ' =

+ 0'; for t h i s

would i m -

p l y t h a t b u 0' = 0' .f 0'1. I n f a c t , we can have a ' = b' w i t h any o f t h e f o u r a l t e r n a t i v e s a = b, a l b , a sible.


= k Since the s e t o f

3B(dc6 & 8

on T

&

B

[ I ] (m'l = k : .

on T i s t h e r a n g e cf a r e c u r s i v e p a r t i a l

f u n c t i o n and hence i s €03, t h i s e q u i v a l e n c e i m p l i e s t h a t t h e g r a p h of [ I ]

A

is RE; so [IIA i s r e c u r s i v e by t h e Graph Theorem.

A

L e t [ I ] (m' = k .

Choose

fi C

6 A so that [ I ] (m) = k.

For a l l b v t f i n i t e l y many & C A , we have a c d Hence t h e r e i s a B 8 Then [ I ] ( m ) = k .

Now l e t

o (

cb

above, t h e r e i s a Since

8

and y

on T such t h a t

with

I

8

c6

O(

C A and 6 C 6

on T and [ I ] b ( m

on T such t h a t

C%

C

y

=

6 .

and 6 C

k.

.

By t h e

and [ I ] I( m ' = \ I ] * ( m ) .

A

do n o t I - s p l i t d , [ I ] ( m ) = k .

A t r e e T i s I - s p l i t t i n g i f whenever T ( & ( ' ) :

Q.E.D.

and T ( d (I), ,

a r e defined, they I - s p l i t T ( a ) . Lemma

2 . I f T i s I - s p l i t t i n g and A i s on T , t h e n

A is

I -minimal.

Proof. We assume t h a t F A

=

[ I I A i s t o t a l and prove t h a t

ip, F. Let B be t h e s e t o f d such t h a t T(o l h ( T ( G ( n ) ) ) ; s o l h ( T ( G ( n ) ) ) 2 n.

It follows t h a t A(n) = T(G(n+l)),.

Q.E.D.

Next w e show how t o c o n s t r u c t I - s p l i t t i n g t r e e s . i s an a l g o r i t h m f o r T, t h e n d and 8 I - s p l i t

If J

y and a r e on T

iff

y c

A

o (

A. 7 C 8 &

36([Jl(6) =

~ n 3 i 3 j ( [ 1 1 " < ( n )= i

@

4

& [I]

T h i s i s a n RE r e l a t i o n of I , J , d ,6 ,

&

(n;

3/.

36([Jl(6) = & )

= j

k:

i

+ j).

Hence by t h e S e l e c -

MINMAL DEGREES

52

t i o n Theorem, t h e r e a r e r e c u r s i v e p a r t i a l f u n c t i o n s Lo and L1 such t h a t i f

7

I - s p l i t s on T, t h e n L o ( I , J , 7 j

a r e on T and I - s p l i t L1(I,J,

y)

7;

and L 1 ( I , J , - y / )

while otherwise, L o ( I , J ,

7 ) and

a r e undefined.

Now l e t 6 be on T, and d e f i n e T I (

y

) b y i n d u c t i o n on l h (

y)

a s follows: T I ( 0 ) = 6,

TI(^(^)) It i s e a s y t o v e r i f y t h a t

= TI

L ~ ( I , J , T I ( 1~) . has t h e following p r o p e r t i e s :

( a ) TI i s a t r e e ; ( b ) T I i s a s u b t r e e of T;

( c ) 6 i s on T I ;

( d ) e v e r y s t r i n g on TI i s a n e x t e n s i o n of 6; ( e ) TI i s I - s p l i t ting;

( f ) i f T 1 ( 7 ) i s d e f i n e d and I - s p l i t s on T,

and T 1 ( Y ( ’ ) j a r e d e f i n e d . of T -

then

T1(7(*))

We c a l l TI t h e I - s p l i t t i n g s u b t r e e

6.

To p r o v e Theorem 1, we c o n s t r u c t a s e t A i n N such t h a t dg A i s minimal.

L e t (Is] be a l i s t i n g of Alg(N,N).

w e must i n s u r e t h a t A f [ I , ] and t h a t A i s I,-minimal

Then

f o r each s .

A t s t e p s , w e d e f i n e a t o t a l t r e e T, and a s t r i n g 6, on T,.

Our i n t e n t i o n i s t o t a k e A on T, s o t h a t b S C A . t h a t A i[I,] and t h a t A i s I,-minimal words, we w i l l choose Ts+l

and

€jsi1

We i n s u r e

a t s t e p s+l.

I n other

s o t h a t i f A i s on Ts+l

and

%+1 C A , t h e n A f [I,] and A i s I s - m i n i m a l . We must a l s o be s u r e t h a t we w i l l be a b l e t o choose such an A .

For t h i s , we choose Tsfl

p r o p e r e x t e n s i o n of 6 , and Ts+l

and 6,+1

so that

i s a s u b t r e e of T,.

bSi1

is a

The f o r m e r

MINIMAL DEGFiEES

53

guarantees t h a t t h e r e i s a unique A such t h a t b S

cA

f o r a l l s.

Since 6 s i s on T, f o r a l l s, t h e l a t t e r guarantees t h a t 6,,

.. . a r e

6s+1,

a l l on T,;

s o A i s on Ts.

We now d e s c r i b e s t e p s .

A t s t e p 0, we s e t TO = I d and

Now suppose t h a t s t e p s i s completed; we s h a l l do

6o = 0 .

s t e p s+l.

Since 6, i s on Ts and Ts i s a t o t a l t r e e , 6 , has

two incompatible extensions on T,.

Hence t h e r e i s a proper

extension 6 of 6, on T, such t h a t 6 i s incompatible with [I,]. We t a k e 6,+1

t o be an extension of 6; t h i s guarantees t h a t bS+1

i s a proper extension of 6, and t h a t A

There a r e now two c a s e s .

[I,].

Suppose f i r s t t h a t some exten-

s i o n of 6 on T, does not I s - s p l i t on T s . be such a n extension, and t a k e Ts+l = T s .

We then l e t 6,+1 Then A i s Is-minimal

by Lemma 1.

Now suppose t h a t t h e r e i s no such e x t e n s i o n .

l e t 6,+1 = 6 and l e t T,+l f o r 6.

be t h e I , - s p l i t t i n g

Then w e

s u b t r e e of Ts

Using ( d ) and ( f ) i n t h e p r o p e r t i e s of s p l i t t i n g sub-

t r e e s and t h e f a c t t h a t every extension of 6 on T, I s - s p l i t s on Ts, we e a s i l y prove b y i n d u c t i o n on l h ( 7 ) t h a t T s + 1 ( 7 )

i s defined.

Thus Ts+l i s a t o t a l t r e e ; and A i s Is-minimal

by Lemma 2.

Q.E.D.

One can show t h a t t h e s e t we have j u s t constructed has degree 5

Theorem a

5

0'.

I n s t e a d , we prove a b e t t e r r e s u l t .

011.

2

( S a c k s ) . There i s a minimal degree a such t h a t

MINIMAL DEGREES

54

The c o n s t r u c t i o n f o r Theorem 2 i s a m o d i f i c a t i o n of t h a t Again we d e f i n e 6, a t s t e p s s o t h a t 6,+1

for Theorem 1.

a p r o p e r e x t e n s i o n o f 6,, t h a t 6s C A for a l l s .

and t a k e A to b e t h e unique s e t such However, t h e t r e e s a r e t r e a t e d d i f f e r -

A t s t e p s , we d e f i n e t r e e s T

ently.

For i

n o t be t o t a l ) .




so,

2

i

since

55

MINIMAL DEGREES We may t h e r e f o r e choose s 2 so

otherwise our r e s u l t i s clear.

s o t h a t kscl ks

2

i

>

Then t h e k a t s t e p s+l i s i-1.

= i.

k, Case 2 o c c u r s a t s t e p s+l; s o T:-"

Since

Ti-l, s+l

=

If our r e s u l t i s f a l s e , t h e n t h e r e .is a smallest t such t h a t k t + l = i .

TS+l = Tit and Ti-l s+l = Ti-l; t i t h e k a t s t e p t+l i s n o t

If ks to T ':

>

. . ., kt

S i n c e kS+*,

i for a l l s

for a l l s 2 s o .

s o T! j-1;

2

1

=

are all

Ti-l. t

>

>

s

i, w e have

This implies t h a t

s o kt+l =f i, a c o n t r a d i c t i o n .

s o , t h e n T:

i s d e f i n e d and e q u a l

We l e t Ti b e TYo.

S i n c e 6 , i s on

TS whenever T! i s d e f i n e d , 6 s i s on Ti; so A i s on Ti. i S We s h a l l show t h a t i f Ti+l i s d e f i n e d , t h e n e i t h e r i t i s an I i - s p l i t t i n g not I i - s p l i t

s u b t r e e of T,!

or TS i+l

=

TY and some 6n d o e s

Assuming t h i s , we s e e t h a t t h e same r e -

on TT.

s u l t h o l d s when a l l of t h e s u p e r s c r i p t s s a r e dropped; and i t t h e n follows by Lemmas 1 and 2 t h a t A i s Ii-minimal.

If s

W e p r o v e t h i s r e s u l t by i n d u c t i o n on s .

=

S

0, Ti-I-1

Assume t h a t t h e r e s u l t h o l d s for some s . s+l so t h i s c a s e i s i+l, t h e n Ti+l = TY+l and T i f 1 = T;!

cannot b e d e f i n e d ,

If ks+l trivial.

>

If ks+l




s , w e say t h a t

0; o t h e r w i s e ,

x is active

x i s inactive a t step t .

x i s permanent; o t h e r w i s e , x i s temporary.

Si-

milar d e f i n i t i o n s hold w i t h B i n place of A . We now d e s c r i b e s t e p s .

W e do n o t h i n g u n l e s s k

suppose t h a t F$., i s ( l k ) . k

AS

C,

BS.

L e t s be i n row n, and f i r s t

Suppose t h a t t h i s i s t h e c a s e .

no a c t i v e requirement, p u t k i n t o A .

E

Cs

and

If k i s i n

Otherwise, choose n

minimal such t h a t k i s i n a n a c t i v e n-requirement.

If n i s

a n A-number, p u t k i n t o B; i f n i s a B-number, p u t k i n t o A . (Thus t h e c o n d i t i o n

51 w i t h

the smallest value of n i s taking

p r i o r i t y over the others. ) Now suppose t h a t R, i s (21). a number k ment i s




n)-

5 G(n) for all n, then YF SR G; for v,(n) = p s ( \d t 5 G(n))(t 2 s 3 F(t) > n).

If vF(nj

The construction of an RE set A in N is F-restricted if whenever n is put into A at step s, then n

2F(s).

If F is

restricting and the construction of A is F-restricted, then

A

SR

so s

For if n is put into A at step Thus n E A C-, n E A * F b ) vF(n).

'vF.




n.

At step

Since s > G(n), k LS(F(s)) 5 LA(F(s)). LA(F(s))

>

s,

Then

which is in A is in AS.

s

s

2 G(n),

then

we put into A a number k 5 LS(F(s)).

AG(n); so k

>

LA(").

Since A S C A,

Combining these three inequalities,

LA(n); so F(s)

>

n.

Example. Let F be a listing of an RE set A in N. F is one-one, it is restricting. ting F ( s ) into A at step s.

Since

We can construct A by put-

Since F(s)

< Ls(F(s)),

struction is F-restricted and F-supported; so dg A

=

this condg

VF.

77

MAXIMAL SETS

Lemma

3.

If a i s RE and a ' = O f t , t h e n t h e r e i s a r e s t r i c t -

i n g f u n c t i o n F such t h a t V F i s dominant and dg YF = a . P r o o f . Let A be an RE s e t i n N o f d e g r e e a .

Since

A i s n o n - r e c u r s i v e , i t i s i n f i n i t e ; s o i t h a s a l i s t i n g G by

By Lemma 1, t h e r e i s a dominant f u n c -

t h e L i s t i n g Theorem.

By t h e Modulus Lemma, t h e r e i s a re-

t i o n H recursive i n A . c u r s i v e sequence

iHnI

w i t h t h e l i m i t H and a modulus M of {Hn]

which i s r e c u r s i v e i n A . L e t F ( s ) be t h e s m a l l e s t number n

5

G ( s ) such t h a t H s ( n ) .f

Hs+l(n), i f t h e r e i s such a n n; o t h e r w i s e , l e t F ( s )

>

Clearly F i s recursive. V G ( n ) , then F ( s )

>

If s - M(m)

n.

= G(s).

for a l l m 2 n and s 2

It f o l l o w s t h a t F i s r e s t r i c t i n g and

that vF(n)

5

max, s, S:(j)

and i, j y! AS. =

(Fa,

. ..,

5

This implies t h a t j

i




(6).

proving

s+l

If L s + l ( n ) = L s ( n ) , t h e n Sn

( L s + l ( n ) ) 2 S i ( L s ( n ) ) by

From t h i s and (61, w e s e e t h a t for s with s.

2

S:(Ls(n))

so,

(4)

increases

it

S i n c e t h e r e a r e o n l y f i n i t e l y many n - s t r i n g s ,

From t h i s and ( 6 \ , w e s e e t h a t

e v e n t u a l l y becomes c o n s t a n t .

L s ( n ) e v e n t u a l l y becomes c o n s t a n t ; s o l i m L s ( n ) e x i s t s . For a l l s u f f i c i e n t l y l a r g e s, L ( n ) = Ls(n)

so

A";

I f m f n, L s ( m ) f L s ( n ) for a l l s; s o L(m) =# L ( n ]

L(n) $A.

These f a c t s show t h a t A i s c o i n f i n i t e .

It remains t o show t h a t Sn i s a l m o s t c o n s t a n t on A'. 6 be t h e s m a l l e s t n - s t r i n g

many i

E

such t h a t S n ( i ) = U

for infinitely

It w i l l s u f f i c e t o assume t h a t S n ( i )

A'.

Let

>QC

for

i n f i n i t e l y many i E A C and d e r i v e a c o n t r a d i c t i o n .

i.

S:(i)

= a and L A ( " ) 5

L e t B be t h e s e t of i

E

Then B i s i n f i n i t e .

L e t Cs be t h e s e t of i such t h a t

=o(

A C such t h a t S n ( i )

We show t h a t

and i i s n - h e l d a t s t e p s .

(8)

s

2 t 3 B

n Cs

It w i l l s u f f i c e t o show t h a t B

6

C Ct.

Cs C C s f l .

and l e t j be t h e h o l d e r of i a t s t e p s .

i s a h o l d e r a t s t e p s, i , j f As+1.

Let i

E

B f~ Cs,

S i n c e i E A C and j

We have LS+1(n) 5 LA(") 2

MAXIMAL SETS

82




i.

Then S n ( i )




S;+l(j)

S n ( i )

j with j

j)

We c l a i m t h a t

(5); and t h e l a s t two a r e f i n i t e .

clear i f i

(4) and ( 5 ) .

The f i r s t two a r e RE, s i n c e

LA(”)].

by ( 4 ) and

by

r e c u r s i v e r e l a t i o n of i,s, i t

is a

s u f f i c e s t o v e r i f y t h a t BC i s RE. f o u r sets A ,

o(

i E Cs.

L/

i $T‘ B

=

can choose such a

(5’, S:(i)




E

y with m

T h i s f o l l o w s from t h e f a c t t h a t i f y n As+1

=




with m




E

C, then <m,r>

belongs t o o n l y f i n i t e l y

many r e q u i r e m e n t s .

For i f < m , r >

E Cs,

then <m,r>

i s not put i n t o a require-

ment a f t e r s t e p s e x c e p t , p o s s i b l y , a t t h e s t e p a t which i t i s put i n t o A .

87

INFINITE I N J U R Y ( C ) If < m , r >

E

C, t h e n <m,r>

permanent n-requirement w i t h n

i f f <m,r>

E A

belongs t o no

5 m.

A member of A can belong t o no permanent r e q u i r e m e n t .

Suppose < m , r > <m,r>

E

A.

By ( B ) , we can choose s so l a r g e t h a t

Cs and e v e r y temporary requirement c o n t a i n i n g

Since <m,r>

i s inactive a t step s .

E Cs

- AS and i s not

p u t i n t o A a t s t e p s , t h e r e i s a n n-requirement with n which i s a c t i v e a t s t e p s and c o n t a i n s < m , r > .

5

m

This require-

ment must b e permanent. We l e t En be t h e s e t o f arguments of permanent n - r e q u i r e rnent s. ( D ) En and C ( n )

-

A(n) are finite.

We prove t h i s by i n d u c t i o n on n .

show t h a t En i s RE.

Our f i r s t s t e p i s t o

Let E i be t h e set of arguments of perma-

nent n-requirements c r e a t e d a t s t e p s . I s ( k E E E ) , i t s u f f i c e s t o show t h a t k

r e l a t i o n of k, s .

Since k E

E

En

EE i s a r e c u r s i v e

Given k and s, we can d e c i d e i f a n n - r e -

quirement with argument k i s c r e a t e d a t s t e p s; and, i f s o , we W e must now d e c i d e whether y i s

can f i n d t h i s requirement y . permanent o r temporary. f o r each < m , r >

E y with

By ( A ) ,

m




n-requirement w i t h a n argument t h i s i m p l i e s t h a t En

k i s created a t step s .

But

is finite.

From t h e r e s u l t j u s t proved and t h e i n d u c t i o n h y p o t h e s i s , w e conclude t h a t Em i s f i n i t e for m

5

n.

S i n c e two m-require-

ments w i t h t h e same argument cannot be a c t i v e a t t h e same s t e p , t h e r e cannot be two permanent m-requirements w i t h t h e same a r gument.

Hence t h e r e a r e o n l y f i n i t e l y many permanent m-re-

quirements w i t h m

5

n.

It f o l l o w s from ( C j t h a t C ( n )

-

A(n)

is finite. ( E ) I f t h e r e i s a permanent n-requirement with argument A k, t h e n [ I n ] ( k ) = F a . We suppose t h a t t h e requirement y i s c r e a t e d a t s t e p s , and l e t x be a s i n t h e d e s c r i p t i o n of s t e p s . s u f f i c e t o show t h a t x A A = t h a t Qs(x) n A f S i n c e x n A~ =

9,

pernianent, < m , r > A a t step s.

p.

=

8;

y; s o < m , r >

&,(XI

< m.

A

A f

A A with

so <m,r>

E C s and

so

m minimal.

Since y i s

A'.

<m,r>

$,

i s not put i n t o

Since <m,r>

belongs t o a n z A A,

E

z

S i n c e z i s a c t i v e a t some s t e p , i t f o l l o w s

from ( A ) t h a t t h e r e i s a < m l , r t > Then < m 1 , r 1 >

E

It f o l l o w s t h a t a t s t e p s , < m , r >

a c t i v e n'-requirement z w i t h n' i s temporary.

Suppose t h a t x

Choose < m , r >

Q,(X) n 'A

9

$.

It w i l l c l e a r l y

E Ps(m,r);

E z

n A with m1

so by ( 2 ) , < m l , r l >

E


]

F o r each Onl,rl> E P ( m , r ) ,

choose a

c l o s e d s e t c o n t a i n i n g u l , r f >by t h e i n d u c t i o n h y p o t h e s i s .

The

union of t h e s e sets and {O,r>]i s c l o s e d .

( G ) If [I,]

A

(k) i s d e f i n e d , t h e n t h e r e are only f i n i t e l y

many n-requirements w i t h argument k . F o r l a r g e s . t h e n computation of [I,]A

t h e computation of [I,]

A

(k).

S

( k ) i s t h e same a s

Hence t h e r e i s a f i n i t e s e t x

such t h a t f o r e v e r y s f o r which

S

( k ) is d e f i n e d , each p a i r S

used n e g a t i v e l y i n t h e computation o f [ I n ] A ( k ) i s i n x . ( F ) , t h e r e i s a closed s e t y i n c l u d i n g x. w i t h argument k is i n c l u d e d i n y .

By

Every n-requirement

Each time t h a t such a re-

quirement becomes i n a c t i v e , some m e m b e r of y is p u t i n t o A . Thus t h e r e are only f i n i t e l y many temporary n-requirements w i t h

INFINITE INJURY

90

argument k; and there i s a t most one permanent n-requirement w i t h argument k . A

(HI [In]

If D.

We assume that [ I n I A = D and derive a contradiction. Since En i s f i n i t e by ( D ) and D i s simple, we can choose a A Then [ I n ] (k) = D(k) = F a . BY (D) and ( G ) , k E ( D LIE,)'. we can choose s i n row n so l a r g e t h a t : ( a ) every permanent n-requirement i s a c t i v e a t s t e p s; quirement w i t h an argument AS

( c ) [InIs ( k ) = Fa; ( d ) k

5 5

( b ) every temporary n-re-

k i s i n a c t i v e a t s t e p s;

s.

I f a t s t e p s t h e r e i s an ac-

5

t i v e n-requirement w i t h a n argument m nent; so D(m) = [I,] since k

9

En.

A

k, then i t i s perma-

(m) = Fa by ( E ) ; so m

D".

Also m

+ k,

These f a c t s show t h a t an n-requirement with

an argument 5 k i s created a t s t e p

But t h i s i s impossible

s.

by ( a ) and ( b ) .

It follows from (D) and (H) t h a t a l l of the conditions

are satisfied.

Q.E.D.

W e s h a l l need some f u r t h e r f a c t s about t h e construction

j u s t made i n the next s e c t i o n . cursive i n C .

F i r s t , we show t h a t A i s r e -

W e suppose t h a t an o r a c l e f o r C i s given, and

show how t o compute whether o r not <m,r>E A by induction on m.

I f <m,r>

C,

then <m,r>f A; s o we suppose t h a t <m,r>

We f i n d a n s such t h a t a , r > s t e p s can contain <m,r>

E Cs.

A

only i f < m , r >

E

C.

requirement created a f t e r E A;

so a l l t h e perma-

91

INFINITE I N J U R Y

nent requirements containing <m,r> a r e a c t i v e a t s t e p s. <m,r>

by ( C ) ,

E

A i f f no n-requirement x with n

5 m which

a c t i v e a t s t e p s and contains < m , r > i s permanent.

Thus is

I n view

of ( A ) and t h e induction hypothesis, we can t e s t whether o r

not t h i s i s t h e case. Now we consider what can be proved i f we do not assume t h a t C i s piecewise recursive.

t h a t A i s recursive i n C.

We can prove j u s t a s above

I n proving ( D ) , piecewise recur-

siveness was only used t o show t h a t C ( m ) i s recursive f o r m < . n . Hence i f we assume t h a t C ( m ) i s recursive f o r rn

-

conclude t h a t C ( n )

< n,

we can

A(n) i s finite.

Now suppose t h a t C ( m ) i s recursive f o r every m

>

0.

We

can then s t i l l prove t h a t A i s a t h i c k subset of C and t h a t D

&

and D

A, provided t h a t we assume t h a t D i s strongly simple

& C'O).

The only change required i n t h e proof i s i n

(D).

Since C ( O ) may not be recursive, we can only conclude

that k

E

EE i s recursive i n C ' O )

and hence t h a t En i s RE i n

Since D i s strongly simple and D

C'O).

&

C ( O ) , t h i s i s enough

t o i n s u r e t h a t En i s not a n i n f i n i t e subset of DC; and Chis Again, i f we assume t h a t C ( m ) i s r e n, we can conclude t h a t C ( n ) - , ( n ) is

i s a l l t h a t i s needed.

cursive f o r 0


i s p u t i n t o A and t h e n

If w e assume t h e index K of D i s f i x e d ,

giving t h e output 0.

t h e n t h e index I of A i s a r e c u r s i v e f u n c t i o n of t h e index J of

c. We summarize t h e s e r e s u l t s as f o l l o w s .

p l e set i n N.

Then there i s a r e c u r s i v e f u n c t i o n F from

A l g ( N X I?, I?) t o Alg(N X N, WF(J),

for rn

N) such t h a t i f C

t h e n : ( a ) A i s r e c u r s i v e i n C;


0, t h e n D g, A .

17. Index S e t s We s h a l l now use t h e r e s u l t s of t h e l a s t s e c t i o n t o e v a l u a t e t h e d e g r e e s of c e r t a i n s e t s . The i n d e x s e t o f a , d e s i g n a t e d by I x ( a ' , i s t h e s e t o f I

E

Alg(N, N ) such t h a t dg WI

i s not RE.

=

a.

O f cc)urse Ix(a'1 =

$

if a

We s h a l l show t h a t d g ( I x ( a ) ) = a 3 i f a i s RE.

If dg G = d g H, t h e n t h e same s e t s a r e r e c u r s i v e i n G

a s a r e r e c u r s i v e i n H; s o t h e

Zn[G]

(rn[G]

same a s t h e Zn[H]

sets.

This j u s t i f i e s t h e f o l -

(rn[H])

lowing d e f i n i t i o n : a set i s x n [ a l

(I7Jal)

( TTn[H] ) where H i s a f u n c t i o n of degree a.

sets a r e the

if i t i s

zn[H]

The d e s i r e d re-

s u l t on d e g r e e s of index sets i s t h e n i m p l i e d by t h e f o l l o w i n g theorem. Index S e t -complete Z,[a]

Theorem ( Y a t e s ) . If a i s RE, t h e n I x ( a ) i s a

set.

I n 4'3, w e showed t h a t i f F i s r e c u r s i v e i n G and G i s re c u r s i v e i n H, t h e n F i s r e c u r s i v e i n H.

The p r o o f showed how

t o o b t a i n a n a l g o r i t h m f o r F i n H from a n a l g o r i t h m f o r F I n G and a n a l g o r i t h m f o r G i n H.

Thus t h e r e i s a r e c u r s i v e func-

t i o n L such t h a t i f [JIH and [L(I,J)I

H

H

a r e t o t a l , then

).

Now we show t h a t i f a i s RE, t h e n I x ( a ) i s 2 3 [ a ] . A be a n RE set i n N of degree a .

I

E

Ix(a)

t,

c,

Let

With L a s above:

dg WI = dg A 3 J 3 K ( W I = [J]

t+3J3K(WI

93

= [J]

A

&

A = [KIWI)

A

&

A = [L(K,J)]

A

1.

=

INDEX SETS

94

A A Thus w e need only show t h a t WI = [ J ] and A = [ L ( K , J ) ] a r e

n2[A].

Now

WI

= [Jl

A

e \dn((WI(n)

A [Jl ( n ) = T r )

&

(lWI(n)

[ J I A ( n )= F a ) ) .

X!

A S i n c e W,(n) i s RE and [ J ] ( n ) = k i s RE i n A, both a r e Z1[A]. A It f o l l o w s t h a t each of WI(n), [J] ( n ) = Tr, l W I ( n ) , and

n2[A]. We

[ J I A ( n ) = Fa i s n 2 [ a ] ; s o WI = [ J I A i s A A = [L(K,J)] similarly.

2. I f

Lemma vy(<x,y>

E

then x

A i s RE and B i s n l [ a ] ,

E

E D ) where D

i s r e c u r s i v e and h a s a modulus H r e c u r s i v e i n A .

x

f B

++ V z ( 1 i m

( s i n c e l i m Dn(x,z) always e x i s t s ) .

<x,z,s>

++ 3 n ( n >

C1

E

Then

Dn(x,z) = T r )

~ ~ z ~ s 3 > ns (6( n<x,z>

E

D ~ )

E

Dn).

Let s

6r

<x,z>

I n t h e d e f i n i t i o n o f C f , we may r e p l a c e 3 n

Then C f i s RE.

H(x,z).

T h i s shows t h a t C ' i s r e c u r s i v e i n H and

Now l e t Y = Z A N, F ( z , n ) = z, G ( z , n ) = n, and

<x,Y>

G

c

.c-,

<x,F(Y),G(Y)>

Then C h a s a l l t h e d e s i r e d p r o p e r t i e s .

Lemma

i s re-

By t h e Modulus Lemma, D = l i m Dn where I D n )

cursive i n A .

hence I n A .

B ts

C ) where C i s r e c u r s i v e i n A and RE.

P r o o f . We have x E B # v z ( < x , z >

by 3 n

treat

2. L e t

E

c'.

Q.E.D.

A be RE and l e t B be Z 2 [ A ] .

Then t h e r e i s

a C, r e c u r s i v e i n A and RE, such t h a t C ( x ) i s r e c u r s i v e i f x

E

B

95

INDEX SETS

and A i s recursive i n C ( x ) if x $ B. Proof. We suppose that A i s a s e t i n N. x

t, 3

B

E

t Vs(<x,t,s>

where D i s recursive i n A and RE. <x,m,n>

E C

c)

E

By Lemma 1

D)

Let

( 3t 5 m ) ( t / s 5 n)(<x,t,s>

E

D) v m

Then C i s recursive i n A and RE. Fixing x, l e t E = C ( x ) .




m 2 t, then E ( m ) = N . If

If <m,n> E E, then <m,nl> E E

5 m)(V s 5

Thus E ( m ) = N c t m

l a r g e n.

D).

If

Hence i n t h i s case E i s recursive.

x $ B, then f o r each t t h e r e i s an

Hence, given m, ( 3t

E

n)(<x,t,s>

so m

E A;

T h i s shows t h a t A C i s RE i n E.

s such t h a t <x,t,s> f D.

E

AC

E D) C)

i s false f o r

gr(<m,r>

E).

But A i s RE, hence FE i n E;

s o A i s recursive i n E by t h e Complementation Theorem.

Q.E.D.

We can now complete t h e proof of t h e Index Set Theorem. Let a be RE; we must show t h a t every 1 3 [ a ] s e t i s reducible F i r s t we l e t a = 0.

to Ix(a).

( T h i s case i s due t o Rogers.)

L e t A be an RE s e t i n N of degree 0 ' .

If B i s

z3,

then B i s

r 2 [ A l ] f o r some A ' which i s El and hence recursive i n A; s o B Is x 2 [ A ] .

sive, x

E

Choose C as i n Lemma 2.

B i f f C ( x ) i s recursive.

t h e r e i s a recursive F such t h a t

F(x)

E

Since A i s not recur-

By t h e Parameter Theorem, = WFcX).

Then x

Ix(O), proving t h a t B i s reducible t o I x ( 0 ) .

E

B c3

E A

96

INDEX SETS

>

Now l e t a

s e t D of d e g r e e a .

x

512, t h e r e i s a simple

By Theorem 1 o f

0.

Let B b e z 3 [ D ] . E

BC

Then

vk(<x,k> E C )

t+

By Lemma 2, t h e r e i s a n E, r e c u r s i v e i n

where C i s x 2 [ D ] . D and RE, such t h a t

4E ( x , k )

i s recursive,

(1)

<x,k> E C

(2)

<x,k> f C -+ D i s r e c u r s i v e i n E ( & k )

Choose a r e c u r s i v e G by t h e Parameter Theorem so t h a t E(,)

=

S e t H(x) = F(G(x)). ‘G(X); and l e t F be as a t t h e end of $16. We complete t h e proof by showing t h a t x E B f, H(x) E I x ( a ) . S e t t i n g A, ( b ) if E(x’m)

=

WHcX), we have: ( a ) Ax i s r e c u r s i v e i n E (XI.,

i s r e c u r s i v e f o r rn

finite; ( c ) if Since E

SR D, ( a ) i m p l i e s t h a t


f C .

- Ax (‘1

E

Ix(a).

Now l e t x

i s p i e c e w i s e r e c u r s i v e ; s o by ( c ) , D

and H ( x ) $ I x ( a ) .

flB.

SRA x .

E

so

By (l), Thus dg A,




0

Since b


E A a l l J and x.

f o r a l l I and x; and s e t I = F ( J ) .

x

E WI

w <J,x>

Then E WJ

++

E A

e+ E A.

Q.E.D.

C o r o l l a r y ( F i x e d P o i n t Theorem). I f F i s a r e c u r s i v e f u n c t i o n from Alg(X,N) t o Alg(X,N), t h e n t h e r e i s an I such t h a t 1 '

=

'F(1)' P r o o f . Define t h e A o f t h e Recursion Theorem by Q.E.D. W F(I)' D e n s i t y Theorem ( S a c k s ) . If a and b a r e RE d e g r e e s such

s, x i s a c t i v e t a t step t i f x A = 4; o t h e r w i s e , x i s i n a c t i v e a t s t e p t . We say x i s e f f e c t i v e a t s t e p t i f x i s a c t i v e a t s t e p t and no n-requirement f o r B w i t h argument k and v a l u e i i s a c t i v e a t s t e p t ; o t h e r w i s e , w e say x i s i n e f f e c t i v e a t s t e p t . element r of x i s a key element i f r i s i n row m and m

>

An k

+

n.

BRANCHING DEGREES

101

A

A

We now d e f i n e f i n i t e s e t s Ps(n,k) and Q s ( n ) by induction A on s. The idea i s t h a t Ps(n,k) i s the s e t o f numbers which we do not allow an n-requirement f o r A with argument k t o keep o u t of A a t s t e p s; and QAs ( n ) i s the s e t of numbers kept out of A a t s t e p s by m-requirements f o r A with rn

r

E

5 n.

Precisely,

A

Ps(n,k) if r belongs t o an n-requirement f o r A a c t i v e a t

s t e p s and r

E

A

i n the same row a s r; and r

E