Degrees of Unsolvability. (AM-55), Volume 55 9781400881840

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Annals of Mathematics Studies Number 5 5

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DEGREES OF UNSOLVABILITY BY

Gerald E. Sacks

SECOND E D I T I O N

PRINCETON, NEW JERSEY PRINCETON UNIVERSITY PRESS 1966

Copyright © 1963,1966 by Princeton University Press All Rights Reserved L. C. Card 63-9996 1966 1993

SECOND EDITION REPRINTED

Printed in the United States o f America Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press, Chichester, West Sussex Printed in the United States of America

Princeton University Press books are printed on acid-free paper and meet the guidelines for permanence and durability o f the Committee on Production Guidelines for Book Longevity of the Council on Library Resources

This Book Is For CLIFFORD SPECTOR, 1930-1961 'HS£> 7] i\ov fivr/fir] TtdvrjKoros EPICURUS

PREFACE TO REVISED EDITION It is intended that Section 1 be read before any section that follows it; the same applies to Sections b and 8 . There is a continuous commentary on the priority method which starts in Section b and ends in Section 10. We wish to thank the many persons who helped up prepare and re­ vise this monograph at one stage or another. Among them were R. Friedberg, S. C. KLeene, A. Nerode, A. Robinson, H. Rogers, Jr., J. Rosenstein, J. R. Shoenfield, R. Smullyan, and C. Spector. Special thanks are owed to J. Barkley Rosser. We also thank the National Science Foundation and the Army Research Office (Durham) for financial support. Many results on degrees were obtained after the appearance of the first printing of this monograph. We discuss some of them in Section 1 2 . Nonetheless, we still feel that the subject of degrees is far from fin­ ished.

Many of the weapons developed to attack degrees are now making

inspiring appearances on the other battlefields of mathematical logic, particularly set theory. We hope that what follows will hasten the final victory.

Ithaca, New York April 1, 1 9 6 6

CONTENTS

§1. §2. §3. §4. §5. §6. §7. §8. §9. §10.

P r e lim in a rie s ...................................................................................................... ... a Continuum of M utually Incomparable D egrees....................................... Uncountable Suborderings of D e g re e s ...................................................... The P r io rity Method of Friedberg and Muchnik........................................ , An E xistence Theorem fo r R ecursively Enumerable Degrees . . . The Jump O p e ra to r.............................................................................................. An In te rp o la tio n Theorem fo r R ecursively Enumerable Degrees . Minimal Upper Bounds fo r Sequences of D e g re e s ................................ Minimal D e g re e s................................................................................................... M easure-the ore t i c , Category and D escrip tiv e S e t-th e o re tic A rgum ents...................................................................................................... §11. I n i t i a l segnents of D e g re e s......................................................................... § 1 2 . F u rth er R esults and C o n je c tu r e s ............................................................

55 77 117 123 135

BIBLIOGRAPHY........................................................................................................................

173

ix

1

7 21

*3

153 163 1 69

DEGREES OF IMSOLVABILITY

DEGREES OF UN SOLVABILITY

§1.

Preliminaries

The natural numbers are 0 ,1 ,2 ,...;

let f

and

g

from the natural numbers into the natural numbers. We say have the same degree of recursive unsolvability if f and

g

is recursive in

solvability. f

f

by

f;

g

f

and

Let

and

g

is recursive in

g

is the set of all functions is recursive in

g

U

£ 0.

There are only three cases at stage CASE 1.

such that

t

such that

n > o and i < 2;

let

di(p|+n)

t( j)

It

has not het been

be the least such

t.

We

define di(Pj(J)+n) for all

n > o

and

CASE 2.

-

V

n)

i < 2.

(s)Q = 0

and

(s)3 = 1.

Let

(s)1 = e

and

(s)2 = f.

The purpose of this case it to insure that any function recursive in with Godel number

e

and recursive in

recursive in one of the

bi,s.

d1

with Godel number

i < 2,

For each

partial function whose domain consists of all defined prior to For each

i < 2,the notion

extensions of of and

b

stages

d?.

If

b

and whose

is a finite extension of

of

d® and

is defined and

g

of

that

d^(n)

d^,

either

proceed directly to stage

s + 1.

n

d^(n)

has been

d^(n) = di(n) .

d^,

n such that

dQ

will be

the unique

of weight well-orders the set of all

CASE 2.1 . There exists an b

such

is not defined and also of the values of

for which b(n)

be

values are given by

is an effective encoding of those

d^(n)

n

let d|

f

finite

then the weight

b(n) b(n)

is defined for those

n

is not. such that for any finite extensions

(e}t>(n)

or

(f)g(n)

isundefined.

We

18

DEGREES OP UNSOLVABILITT CASE 2.2.

Case 2.1 does not hold, and there exist an

finite extensions,

b

and

c,

of

(e)b (n) are defined but are not equal.

n

and

d3 such that and

(e}c(n)

Among all such triads

(n, b, c)

there is

a unique one with the property that 2n . 3weight of b . ^weight of c has

the least possible value; let it be denoted by

Case

2.1 does nothold, there exists afinite extension

that

(e}s (ng) isdefined; let

be that member of

{bs, c3}

gs

We define

dQ(n),

for each d3(n)

n

d.,(n)

for which

ly to stage

d3(n)

hs(n),

gs(n)

.

hs(n),

gs(n)

respectively,

respectively, is defined and

Neither Case 2.1 nor Case 2.2 holds.

s is otherwise.

d3(n),

We proceed direct­

We define

Let

r a be the least n such that s vg be the least n such that d3(n) is un­

= d^ vs^ 55 °*

That completes the construction. fined everywhere.

By Case 1, each

We show that the degrees of

dQ

By Case 3,

dQ and

is recursive in both

and

d1

d1 dQ

and

f

bi

Case 2

d 1.

have no greatest lower bound. dQ

whose degree is greater than the degree of

Let

and w.

Let

be Godel numbers such that do di » = (e) 0 = {f} 1

s

are de­

and

be a function whose degree is a lower bound on the degrees of

d1. We will find a

n

hs

of least weight such that

respectively,to be

is undefined, and let

defined.

Let

such

s + 1.

CASE 3.

an

ofd3

respectively, is not defined. CASE 2.3.

e

g

be the one ofleast weight.Let

(e}hS(ns) * {f)gS(ng)

w

(ng, bs, cs) . Since

be such that

(s)0 * o,

holds at stage

s.

(s)3 = i,

.

(s)., = e

and (s)2 - f.

Thus

If Case 2.1 holds at stage s, then there exists

such that either (e)

is undefined.

do

(n)

or

{f}

di (n)

It follows that Case 2.1 does not hold at stage

s.

If

§2.

19

A CONTINUUM OF MUTUALLY INCOMPARABLE DEGREES

Case 2.2 holds, then there exists an w(n) = {e}

dn

n

(namely,

such that

di (n) ^ Cf) (n) = w(n)

It follows that Case 2.3 holds at stage (i)

ng)

s.

there is a finite extension

Thus for each b

of

n, we have:

d®such that

(e)b (n)

is defined; (ii)

if

b

and

and

c

(e}c(n)

We will use (i) in

bs. Let

Dg

are finite extensions of are both defined, then

and

(ii)

be the domain of

to show that

Dg

|2J
0j-

is a finite extension of

&

where

(e}b (n)

It also follows from (L2) that the values of d® are

effectively computable from j < s.

such that

d®. Then

Dg = F U where

d^

d^

be a function such that for each

Xn|B(i, n) = bi . It is easily checked that xin|A(i, n) . We saw above that the degree of

Xin|B(i, n) Xin|A(i, n)

are at i,

is recursive in was at most

£ T.

20

DEGREES OF UNSOLVABILITY

If we can show that both xin|B(i, n)

dQ

and

d1

are recursive in a composition of

and a predicate of degree

the degrees of

dQ

and

d1

< o",

are at most

then it will be clear that

o".

stage

s

of the simultaneous definition of

then

t(j)

is obtained effectively from

Consider what happens at

dQ

and

d^ and

d1.

d®,

If Case i holds,

and

are finitely extended with the help ofxin|B(i, n).

d^

and

d^

Suppose Case 2 holds.

We can tell which of the three cases, 2.1, 2.2, or 2.3, holds by composing xinlB(i, n)

and a predicate of degree

£ n.

(We do not exhibit the predi­

cate since it is similar in nature to the one-quantifier form occurring in our argument that

xin|A(i, n)

has degree at most

£ ’.)

With the help of

this predicate we can single out the desired extensions of If Case 3 holds, we extend explicitly define

d^*1

d^

and

and d^ d^+1 in

of a predicate of degree at most

o"

d®

in a trivial fashion. terms of

d®

and

and the function

d^

d^.

Thus we can with the aid

xin|B(i, n) .

follows with the help of the closing remarks of Section 1 that are of degree at most

and

dQ

It

and

d1

o".

Each of the theorems of the present section was proved by means of the diagonal method.

Each of the above constructions amounted to a defini­

tion of a function by induction; the function had to meet countably many requirements, and stage requirement.

s

of the induction was devoted to meeting the

3th

Godel !s construction of an undecidable, arithmetical predi­

cate and Kleene!s construction of a non-recursive, one-quantifier form made similar use of the diagonal method.

In Sections ^-7, 9 and 11 we will

see that the diagonal method lacks the power needed to obtain results about degrees deeper than those of Sections 2 and 3.

§ 3 . UNCOUNTABLE SUBORDERINGS OF DEGREES

in urns sec Lion we sLuay conai Lions wmcn are suiricienL ana, in some cases, necessary for a p a rtia lly ordered set to be imbeddable in the upper sem i-lattice of degrees. Two p a rtia lly ordered sets, M and M1, are called order-isomorphic i f there is a map m-^mT of M onto M’ such th at m < n i f and only i f m' < n*. A p a rtia lly ordered set P is said to be imbeddable in a p a rtia lly ordered set Q if P is order-iscmorphic to some subset of Q. Let T be a p a rtia lly ordered set of cardinality at most th at of the continuum such that each member of T has a t most alephone successors; we show T is imbeddable in the degrees i f and only i f each member of T has a t most countably many predecessors. Let A and B be sets of degrees such that A is countable, B has cardinality less than th at of the continuum, and no member of B is less than or equal to any fin ite union of members of A; we show there exists a degree d such that d is greater than every member of A and incomparable with every member of B. Finally, we show that i f T is a p a rtia lly ordered set of cardinality at most that of the continuum with the property th at each member of T has a t most fin ite ly many predecessors, then T is imbeddable in the degrees. We assume complete fam iliarity with the arguments of Section 2. THEOREM 1 . Let T be ap a rtia lly ordered set, and le t M and Nbe d isjo in t subsets of T such th at M has cardinality less than th at of the continuum, N is countable, and no memberof N is less than any member of M. For each n € N, the set Mn = {m|m e M& m 0,

will denote the unique member of

(B^)

recursive in

v =

if and only if for each

2v(k) - 1 < v(k+i) < 2v(k).

whose index is

(B^)

will have cardinality of the con­

be the set of all sequences of natural numbers such that

(v(k)|k = o, 1, 2, ...}

if

a^.

< n^j

of functions will be extracted from a sequence

of sets of functions.

B^

will constitute

must meet five sets of requirements: if n^ < n^j

the members of

be such

be a function of degree

b^

1 < v(k) < 2k

M

if and only if

2, •••)

at

bj recursive in

The sequence

S

let

and

{b^li = 0, 1, 2 , ...}

fbjji = o>

t c V,

N = {i^li = 0,

A

at < a^

(R1)

(R5)

Bi

V,

We will obtain a sequence of functions

Let

and

and let the given order-isomorphism between

that for each m^
o

and

Bi^m ^

B^(m)

B^

if

r € R^,

*

During the course of the construc­

is defined if and only if for sane

Note that if

' then

will be a member of

such that for each

f^ = % n v ( n ) (r)

tion we will say

23

n,

Qj_nv(n)

is defined by and and hence equal to

ls deflned for a11

v c S

with the property that

w(n) = v(n). Since for each

c N,

is countable, we write

■ 0, ^ For each of

A

and

i

and

j,

let

corresponding to

a^

m^j

2, ...} 55 {tit < ^

.

be a function whose degree is that member under the given order-iscmorphism between

M. The construction of the functions

definition by induction.

& t € m|

At stage

s

{Q^ ^ }

A

takes the form of a

of the construction, either or both

of two kinds of activity may take place: (1) for

some

i

and

their common set of arguments (2)

n the functions

R^

and

irrevocable commitments are made concerning the values to be

taken by some of the functions them.

CfynV l1 < k < 2n}

are defined;

when it becomes time to define

All commitments will be capable of expression by one of two utter­

ances: (a)

for all

v,

as soon as one of

Bj(m)

and

B^(pjI+I^ 1^ )

is assigned the value it must take when it is defined, the other must be assigned the same value and must take that value when it is defined; (b)

if

then for all

6 • p“+t(iJ) iS put in k

such that

assigned the value

a ^ (m)

R^

1 < k < 2n ,

when

is defined,

Qink(6*p“+t^i^ )must be

and must take that value when it is

defined. All commitments will be honored at the earliest possible stage. describe the status of

p“+r(13),

6 • p1^ ^ )

We will

respectively in utterance

DEGREES OP UNSOLVABILITY

2k

(a), (b) respectively, by saying that

p^l+r^i^ ,

6•

respectively,

has been committedto a reserved, closed respectively,classification in the ith partition.

We will

say that B^(m)

has received a

value if it has been

defined or if it has been assigned a value it must take when it is defined. If a commitment of type (a) has been made at stage the first one of

Bj(m)

value of the other.

and

B^(p“+r^i^ )

s,

then we will say

to receive a value induces the

Before we proceed with stage

s

of the construction,

we must state an induction hypothesis concerning what has happened prior to stage

s: (Hi)

Only a finite number of the sets

been defined. has been

For each

defined for all

such that

and k o

and

and

j : if

j

p1^,

positive integer ber of the form

p^,

6 • p1^

respectively,

is restricted to a finite range of values. 6 • p®

r(ij), p“

For each

respectively, has been committed to a reserved,

closed respectively, classification in the t(ij)

i^*1 partition, then there is a

respectively, such that every natural num­

(m > r(ij)),

6 • p1^

(m > t(ij))

respectively, has

been committed similarly and such that no natural number of the form (m < r(ij)),

6 • p: “

furthermore, for each

(m < t(ij)) v € S

and natural number

m,

Bj(m)

of type (a) and if

been put in

then

R ^ prior to stage

s,

= for all

k

such that

(H3)

For each

p“

respectively, has been committed similarly:

are joined together by a commitment

6 •

g

was committed to a reserved classification in

classification in some partition are of the form where

Rik

(i, j, m)

has been committed to a reserved or closed classification in the tition.

have

has been defined, then

1 < g < 2n . There are only finitely many triples

such that for some the

i

{R^^li > 0 , n > o)

and

B^(pjl+r^i^

6 • p^+b(ij)

a1;)(m)

1 < k < 2n . i

and

j,

a natural number of the form

p®,

respectively, has been committed to a reserved, closed respectively,

§3.

classification in the 22 •3J • 51

< s

25

UNCOUNTABLE SUBORDERINGS OP DEGREES ith partition only if

2 • 3^ •

< s

and

j ^ i,

respectively.

At stage

s

there are five possible cases.

Cases 1-3 correspond

to (R1 ) - (R3). CASE 1.

s= 2 • 3J* •

only finitely many m prior to stage finitely many stage

s.

s. m

and

such that

such that for some

Suppose

(p?)

there must be a

B^(p*?)

has received a value

k

v,

B^(p^)

was defined prior to

had its value induced prior to stage

closed classification in the

s

for sane v,

The first part of (Hi) tells us that there are only

follows from (H2) and (H3) that

stage

n^ < ni . We claim that there are

p^

It

was not committed to a reserved or

ittl partition prior to stage

such that

s.

B^(p^)

(t = p^1)

and thereby induced the value of

s.

But then

received a value prior to

B^(p^). Thus

p^

was committed

to a reserved classification in the

ktJl partition and

received a value prior to stage

The last part of (H1) says that there

are only finitely many triples

s.

(k, i, t)

duced prior to stage

s.

have it induced or defined. least positive

r

m,

p^

tion in the

Bj(m)

can receive a value is to

m > 0

v,

s.

B^(p^)

had its value in­

Let

such that for all

was

and all

r(ij)

be the

B^(p^+r)

has not

It follows from (H2) and (H3) that for

has not yet been committed to a reserved or closed classifica­ ith partition.

fication in the of

B^(p^)

B^(p^)

p£

Consequently, there are

That proves our claim.

received a value prior to stage all

s.

for some v,

The only way

v,

kth partition and

subsequently received a value prior to stage such that

subsequently

such that for some

committed to a reserved classification in the

only finitely many m

B^(p^)

and

We now commit

i^*1 partition for all B^(p^+r^ ^ )

to a reserved classi­

m > 0:

for all

v,

as soon as one

receives a value, the other must receive the

same value; if the first has received a value prior to stage

s,

the other

must receive that value now. CASE 2. many stage

m

s= 22 • 3^ • 51 .We claim that there are only finitely

such that for some s.

The only way

duced or defined. in Case 1 . Let

v,

B^(6 • p3^)

B^(6 • pj1)

has received a value prior to

can receive a value is to have it in­

Our claim is easily proved by means of the argument given t(ij)

be the least, positive

t

such that for

m > 0

and

26

DEGREES OF UNSOLVABILITY

all

v,

B^(6 • p^+ 0:

such that

Before we proceed with Case 3, we fix what is meant by a finite extension of

B^

if

6*

arguments of

B^

B^(m^)

m

Let

B^(m0)

some value of

B^

v s.

B^

at stage

For example, the assignment Bj

which might induce

B^fm^.

B^n^)

and

make matters still worse, the assignment of a value to for infinitely many

Suppose we assign a value to values in two different ways.

may be induced.

has not be the

(j < z). Unfortunately, the situation is com­

which might induce a value for

m

B^(m)

s might consist simply of

be possible to assign values independently to

induced; second,

It follows from

such that

might induce some value of

duce the value of B^(m)

and consider

< m 1 < ... < m z-1

plicated by the possibility of induced values. of a value to

and

which have not received a value prior to stage

Then a finite extension of

assigning values to

s.

i

1 < k < 2n .

at stage

(H1) and (H2) that there are infinitely many received a value prior to stage

s.

6•

at some future stage, then at that stage,

must be set equal to

z

Of

has not yet been committed to a reserved or closed classifi­

closed classification in the

first

s.

Thus it may not B^n^).

B^Cn^)

To

might in­

m.

B^(m). This assignment may induce

First, the value of

may be of the form

Bj(p“+

may be

p£+ r (ik) and the value of

B^(n)

It follows from (H2) that the sets ■jj lvalue of

Bj(p“+r^ i^)

induced by

B^(m)j-

and j^(k, n) lvalue of B^(n) induced by B^(m) & m = p£+ r ^iic^ jare finite.

However, each value thus induced may itself induce a value in

two different ways. such that value of ..., at

ak+1

and

aQ, a1, ..., a^

be a finite sequence of values

such that for each

ak+1

k < t,

in one of the two ways described above.

a chain of induced values of length

Suppose that

aQ = B^(m)

Let

t.

a^

induces the

We call

We claim that

aQ, a1, t < 2s.

is an induced value of the first kind; then it is easily seen cannot be an induced value of the second kind.

Thus to prove

§3.

UNCOUNTABLE SUBORDERINGS OF DEGREES

27

our claim, we need only show that if all the induced values in a chain are of the same kind, then the chain has length at most lows the fact that there are only finitely many n

s.

k

But this last fol­

such that for some

has been committed to a reserved classification in the

(H2).

Since each chain has length at most

2s

n,

k*'*1 partition

and since each induced value

can induce only finitely more values in each of the two ways, it follows that the set {(k, n) |B^(n)

receives a value determined by the value assigned to

B^(m) is finite.

and by commitments made prior to stage

Thus the assignment of a value to

B^(m)

sj-

has only finitely

many repercussions despite the commitments we are forced to honor immedi­ ately.

Note that it is perfectly possible for the assignment of a value to

B^(mQ)

to determine a value for

B^Cm^.

By a finite extension of

B^

We cannot rule out such an event.

of length

not only the receiving of values by the first have not received values prior to stage

s,

z

z

(j < z)

CASE 3.

which

but also the assignment of all tB^I;) =

1> 2> •••)

course the values assigned

B^(nij)

defined prior to stage

n s.

0;

Let

to

B^(mj)

let

f

Bl[(m)

B^

for which

and if the act of putting

u2,

tension of

B^

B^(m)

f(m)

induces the value of

Let

q(2), q(l)

respectively, be the greatest mem­

ug, u 1

q(2), q(l)

respectively,

respectively,

be if

-1.At stage f(m) »

has received a value prior to stage equal to s

f(m)

for all

m

for which

B^(n),

We require

then

s bJ

would constitute a finite ex­ f

to be complete in the

is defined and if assigning the value

equal to the induced value of

re­

respectively, has not been

is a finite extension of B^

as described above. if

nk |

R ^ , Rin

has not received a value prior to stage

following sense:

independently.

respectively, if

otherwise,

s we say a partial function m

and

such that

Rk#q(2)_i> Ri,q(i)-i

is greater than

for all

(j < z)

s = 23 • 3k • 5e • 71

spectively, be the least

B^(m)

we mean

must be consistent with our commitments, since it may not be pos­

sible to assign values to

ber of

Of

s

argument of

values thereby determined in any of the functions as a result of commitments.

at stage

f(n)

f(m)

to

is also defined and is

B^(n). Thus the partial function

f

assigns

28

DEGREES OF UNSOLVABILITY

consistent values to the first ceived values prior to stage thereby induced in

B^.

z

arguments of

s,

which have not re­

and in addition, assigns all values

We define the weight of

of Theorem 1 of Section 2.

f

The partial function

as we did in the proof

(e) , where

f

is a

partial function, was defined in Section 1. CASE 3a.

q(l) = q(2) = q.

we will define

^

^iqg

the property that

v(q) = g,

Godel number

e

1 < g < 2q,

let w(g)

g

such that

such a way

we will have

B^

be the unique member of

(k)(k > q r bethe least

n

for a11

For each

S

with

B^

g

with

such that

defined by

w(g)(k+l) = 2w(g)(k)).

such that

f

v € S

;

received a value prior to stage is a partial function

1 < g < 2q ,

not recursive in

when the construction is complete.

v(g)(q) = q

Let

For each

p£ > u2

s;

r

and such that

B^(p£)

is well-defined by (H3).

which is a finite extension of B ^ g^

has not If there

and which

has the property that (1)

fe)f(p£)

is defined, we say

g

is not null.

weight) such finite extension of let

If

g

B ^ g^

is not null, let the least (by

be the needed one at stage

Zg be the resulting value of (1), and let

ment of B ^ g^

If

g

is

null,

let t

be

t = max |t 11 < g < 2qj- + p£

R iq " {n,U1 < n

(2)

g

be the greatest argu­

which receives a value as a result of making the needed

finiteextension.

For each

tg

s,

and h

< 4 J V

such that

=

1 < g < 2q

u 1 + 1.

We define:

;

{n,U2 < n < and

4

•

h € Riq, we define:

value received at some stage prior to stage s = i value received above at stage s (g not null) o otherwise.

It follows from (H3) that for any

v € S,

B^(p£)

could not have had its

value induced as a consequence of the finite extensions described by (2). Thus we are at liberty to define:

§3.

1 - z if g is notnull and h = p£ = -j value received at stage sor prior to ^ 0 otherwise and g such that h € R ^ and 1 < g < 2q

(h) for

each h

CASE 3b.

q(l) ^ q(2).

Rk, q(2)+h = ^u 2 +

1 q(2).

.

We first define

otherwise whenever

q(2) < q(2) +

h 0,

be

-1 . Define

(s)1 = k

such that

let

u

R^

(s)2 =e1

let

and

(s)3 = e2 . Let

has not been defined prior to stage Rk ;otherwise,

be the greatest member of

w(g) as in Case 3a.

eachd e D,

For

Suppose

stage s

q(2)+h g^u 2 + h) = value received at some

s~7 0

stage prior to

29

UNCOUNTABLE SUBORDERINGS OP DEGREES

d(o) = (d)Q andd(i) = (dj^. For each

will finitely extend

)

and

s.

let

D = {2a •3^11 < a < b< 2%

Let

B ^ ^ 1))

q

u .

d € D

we

j_n SUch a way that when the

construction is complete, either (3)

u(uyr](§£(d(0))(y), e,, m, y))

will be false for some some

m

m

=

u(nyT] (§£(d(1 ^ (y), e2, m, y))

or at least one side of (3) will be undefined for

or the left side of (3) will be a function recursive in

a^, a ^ ,

..., akg. The order in which these pairs of finite extensions will occur is the natural order of since of

1

2 . 3

2

B a n d

suffer

2^-1

D.

Thus we begin by extending

is the least member of extend

,

Thus for each

b£^

n € {i|i < d) n D,

g,

at stage

s.

Let

will

d € D 3^3

we have extended

We now dispose of

we may have already extended

n < d;

,

successive, finite extensions, and the union of these will be

according to the above plan. and

and

Then we make a further extension

and so on.

the needed finite extension of that for each

D.

1^

d.

and suppose gj*(n(l))

Thus we extend for the sake of some

in that event, we will further extend it.

We regard

B ^ ^ 1^

similarly. CASE ^a. (U)

There is an

m

such that the left side of

u(nyr](^(y), e,, m, y))

=

u(nyT] (?2 (y), e2, m, y))

is not defined for any partial function of

B ^ d^0^ ,

or there is an

m

defined for any partial function

f1

which is a finite extension

such that the right side of (U) is not f2

which is a finite extension of

30

DEGREES OP UNSOLVABILITY

3^(d(l))^

The needed finite extensions are both trivial (empty). CASE hb.

gW(d (°))

There is an

and an

m,

an

f1 which is a finite extension of

which is a finite extension of

both sides of (4) are defined but are not equal. (m, fv f1

and

f2) v

we assign the weight is the weight of

such that

2m •3M • 5V ,

To each such triple where

n

is the weight of

fg . Then there is a unique triple of least

weight from which we extract the needed finite extensions of gwCdO)) ^ an

m

and

follows that when the construction is complete, there will be

(namely, the

m

associated with the triple of least weight) such

that both sides of (3) are defined but are not equal. (ha.) and (^b) are false.

CASE ^c.

side of (U) is defined for some and for each

m

Then for each

m

the left

f1 which is a finite extension of

all finite extensions of

side of (h) give it the same value.

b^

^ 0^ which define the left

This means that when the construction

is complete, the left side of (3) will have the property that if it is a function defined for all that portion of Fix

m

m,

B ^ d ^0^

then its values are completely determined by

which has been determined prior to stage

s.

and suppose the left side of (3) is defined when the construction

is complete.

To determine the value of the left side of (3), we merely

take any finite extension of

b £ ^ 0^

at stage

s

which defines the left

side of (fc) and insert that finite extension in the left side of ( h ) ; result will be the value of the left side of (3).

the

The hypothesis of Case

(he) tells us that the finite extension we need does exist and that it does not matter what finite extension we take.

Suppose that when the construc­

tion is complete, the left side of (3) is

a function defined for all

let

g

be that function.

number

Then

g

isrecursive in

e1 . But as we have just seen

portion of

b J ^ ^ 0^

g

b£

^°^

m;

with GSdel

is actually recursive in that

determined prior to stage

s.

We now find ourselves

in the same situation we encountered in Case 2.3 of the second half of the proof of Theorem 3 of Section 2. whose domain consists of all value prior to stage Then

g

s

m

Let

B s be the unique partial function

such that

B ^ d ^°^(m)

and whose values are given by

is computable from B g, D g, where

follows from (Hi) and (H2) that

Bg

and

Dg

Dg

has received a B s(m) =

is the domain of

(m). Bg.

are each recursive in

It

§3.

UNCOUNTABLE SUBORDERINGS OP DEGREES

ako> ®ki * •••* aks' sive in

fact> D s

ako, akl, ..., a ^ .

is recursive.)

31

But then

g

is recur­

This last follows by an argument not essenti­

ally different from that given in the analysis of Case 2.3 of the second half of Theorem 3 of Section 2.

We note only that (H1)-(H3) provide us

with a picture of the stage of consists of the functions

at stage

akQ, a ^ , ..., a ^ ,

s,

a picture which

and finitely many remarks

about commitments. Thus we dispose of

d € D

as described in (^a), (^b), and (^c).

In Case (ha) and (^c), we make trivial (null) extensions of Bk ^ 0^ ^(dO)).

Case

ig Wi2ere tlie non-trivial extensions take place.

and After

all the members

of D

be the greatest

integer which has received a value during theabove series

of finite extensions;

have been exhausted in this manner, we define

that is,

z

those arguments of the functions

z

to

is the greatest argument among all

*{b ^|v = w (i ), w (2), ..., w(2q);

i > oj*

which received values during the above series of finite extensions. Remember that we include induced values. we take

z

to be

If all the extensions were trivial

u + 1; u was defined at the beginning of Case 4.

We de­

fine Rkq = {h^u < h ^ and rvalue received above at stage ^ k q g ^ = “jo otherwise for all of

D

h

and

g

such that

we assumed that

and (^c),

h € If

and q = 0,

and proceed directly to stage CASE 5.

such that

R^

s

is otherwise.

1 < g < 2m ^

,

i < s,

R^

that

B^(p^)

In our definition

we simply ignore (ha.), (^b), s + 1.

For each

if

i,

let

*n(i) > 0

m(i) be the least s,

and

u.±

be

otherwise.

If

and let

-1

• ! » . . « • ! ’ . ... . 4 ( i ) ■ S"'1’

n

•

we define "«1,

v(i)

1 < g < 2^.

s

then we write 6-

where

or prior to

has not been defined prior to stage

the greatest member of

For each

q>0.

s

is the least

■ {»'»!< » < P f l>*V o}

is recursive in

if

< nk

B^

by Case 2 ;

by Case 3.

we define

Rk = *jv|v € S & (Et) (t € V & at and

and

has cardinality equal to

of functions which satisfy (R1), (R2), and (R3) : B^ if

b£

recursive

in

&mt
z,

and k > n. where

has not been defined prior to stage

By the definition of

z,

s.

there must be a

V

§3. d € D

UNCOUNTABLE SUBORDERINGS OF DEGREES

33

such that {d(0), d(i)} = (u(q), v(q)}

.

The measures we took in Case it guarantee that either (6)

(S£(y), e2, a , y))

tj(njTj(B^(y), e,, m, y)) =

will be false for some

m

or at least one side of (6) will be undefined

for some

m

aks*

both sides of (6) are equal to

sive in

or the left side of (6) will be recursive in

a ^ , akl,

a^.

that each finite subset of member of m^. J; nk if

Mk

and

at;

a ^ , akl, ...,

consequently,

at

is recur­

It follows from the hypothesis of Theorem 1 Mk

has an upper bound in Mk.Let

with the property that for all m^ < nk, we must have

i < s,

be a

m ^ < m^.

Since

m^. J: mg. But this is absurd, because

a

is the element of A corresponding to m under the given order© © isomorphism between A and M, thena ^ is recursive in a^ for all i < s,

at

is recursive in

For each Since Let

V

k,

ag, and m^ < mg. Sk = jv|v € S & (Et)(t € V 8s B^ recursive in

let

has cardinality less than that of the continuum, so must

H = U{Rk U Sk lk > 0}.

continuum, it follows a sequence

(B^li > o)

Since

S -

H

H

is not empty.

For each v € S - H,we have

of functions which satisfy (Rl)-(R5).

to define a unique member

w

of

S - H

The main difficulty in our

argument arose from the fact that we were given If

M

M

M

as having cardinality

were countable, we could have pro­

ceeded quite simply in the vein of Section 2. bility of

It is easy

by induction.

That finishes the proof of Theorem 1.

less than that of the continuum.

S^.

has cardinality less than that of the

Since we allowed the possi­

being uncountable, we had to make use of a counting argument

based on Case ^ in order to satisfy (&b) and (R5). If

M,

and hence

V,

were countable, we could have satisfied (R^) and (R5) by a direct construc­ tion similar to those occurring in Section 2. We say a partially ordered set exists an ordinal that

cc and a collection

P = UCB^lr < a)

P

is completely normal if there

(B^lr < a)

of subsets of

and such that for each ordinal

y < a:

P

such

DEGREES OP UNSOLVABILITY

3b

(1)

U {B& |s < 7 ) has cardinality less than that of the continuum;

(2)

B

(3) (b)

is at most countable and is disjoint from U (Be l8 < 7 ) > 7 o no member of B^ is less than any member of U {Bc |8 < 7 ); 7 o for each n € B^, the set = {a|a < n & a e U CB& 16 < 7 }}

is at most countable and any two members of We say a partially ordered set

Q

completely normal, partially ordered set Note that if

P

P

l£

in which

is normal, then each member of

number of predecessors, and

P

have an upper bound in

L^

is normal if there exists a

P

Q

can be imbedded.

has at most a countable

has cardinality at most that of the con­

tinuum. LEMMA 2. Let P be a partially ordered set of cardi­ nality at most aleph-one. Then P is normal if and only if each member of P has at most a countable number of predecessors. PROOF. aleph-one

Let

P

be a partially ordered set of cardinality at most

such that each member of

predecessors.

For each

p € P,

P

has at most a countable number of

let

p* = (u|u c P & u < p} let

P* = (p'lp € P)

bers of

P*.

P*

and let

;

P ! be the set of all finite unions of mem­

is an upper semi-lattice partially ordered by set-lnclu-

sion and with set-theoretic union as its join operation; furthermore, is imbeddable in

P f,

of predecessors, and

each member of

all ordinals less than a one-one 7 < a,

P 1 has at most a countable number

P ! has cardinality at most aleph-one.

is completely normal.

Let a

a

P

We claim

be the least ordinal such that the set of

has the same cardinality as

map of the ordinals less than

a onto

P*.

P 1,

and let

f

For each ordinal

let Cr = (f(8 ) 16 < 7 }

,

= set of all finite unions of members of

Ar = CalaeP' & (Eb)(b € b 7 = a 7 - U (A 6 l 8 < 7 ) It is readily seen that

p*

{B^ 17 < or)

& a < b))

Cy

,

,

. is a collection of subsets of

that possesses the properties needed to show

P*

Pf

is completely normal.

be

§5.

35

UNCOUNTABLE SUBORDERINGS OF DEGREES

THEOREM 3. Let A and B be sets of degrees such that A Is countable and B has cardinality less than that of the continuum; let D be the set of all degrees greater than every member of A and in­ comparable with every member of B. Let T be a non-empty, normal, partially ordered set. Then T is imbeddable in D if and only if no member of B is less than or equal to any finite union of members of A. PROOF.

If

T

is non-empty and imbeddable in

sity of the condition on normal.

Let

T

A

partially ordered set.

Let

collection of subsets of be normal.

B

a

T*

is clear.

T',

Let

A

T

T 1 is a completely normal,

be an ordinal, and let

be a

that has the properties necessary for

is a member of

A

and that

T T to

(A U B) n T*

T" = A U B U T 1. We define a partial ordering for

that each member of

ordinals less than

and of

B.

by

A

and incompar­

By means of a transfinite induction on the

cr, we construct a one-one function

the degrees which imbed

T”

is

T 1 and by specifying

T ! is greater than every member of

able with every member of

g

We show sufficiency for

where

retaining the partial orderings of A U B

and

then the neces­

Without loss of generality we assume that the least upper bound

of any two members of empty.

and

be imbeddable in

D,

T 1 in

D.

Suppose

y

g

from

T”

into

is an ordinal less than

a

is already defined on A U B U (U {Bo 16 < r))

in such A U B

is

a way that g

imbeds

U CB5 1 s < y)

the identity function.

in

D

and

g

restricted to

Inorder to extendg to

A U B U (U {Bo 15 — < y)) in such

U

< y)

a way that g

Imbeds

in

D,

weneed only verify:

(1) A U B U

(U {B^l5 < 7 )) has cardinalityless than that of

continuum; (2)

is at most countable and is disjoint from A U B U (U CB& la < y ) );

(3)

no member of

is less than any member of

A U B U (U {B^16 < r));

the

36

DEGREES OF UNSOLVABILITY (^)

for each

n e B^,

the set

-ja|a < n & a e A U B U and any two members of

=

(u (Bgle < L^

is countable

have an upper bound in

L^.

A direct application of Theorem 1 of the present section completes the proof. COROLLARY 1 . Let T be a partially ordered set of cardinality at most aleph-one. Then T is imbeddable in the upper semi-lattice of degrees if and only if each member of T has at most a countable number of predecessors. COROLLARY 2 . Let T be a partially ordered set of cardinality at most that of the continuum with the property that each member of T has at most alephone successors. Then T is imbeddable in the upper semi-lattice of degrees if and only if each member of T has at most a countable number of predecessors. PROOF.

The first corollary follows from the second.

Let

T

be

a partially ordered set of cardinality at most that of the continuum with the property that each member of at most

countably many predecessors.

tinuumhypothesis holds, then tinuum hypothesis is false. every member of T1

and

T

T2

T1

T If

has at most aleph-one successors and We show T

is normal.

is normal by Lemma T1

and

T2

2.

Suppose the con­

are subsets of

is incomparable with every member of

are incomparable subsets of

If the con­

T.

T

T2,

We claim that

T

such that then we say is the

union of a collection of disjoint, mutually incomparable subsets of

T

that each member of the collection has cardinality at most aleph-one. prove this last by transfinite induction. (T^lr < a) T

a

We

be an ordinal, and let

be a collection of disjoint, mutually incomparable subsets of

such that each

T^

has cardinality at most aleph-one; furthermore,

suppose every member of% T is a member of that let

Let

such

T^;

t € T - U (T^17 < a).

which is comparable with some member of some

finally, suppose Let

be the set of all members of

T

H(o) = (t);

T - U {T^lr < a) for each

n > o,

is non-empty, let

H(n+1 )

which are comparable with some member of

H(n). Let Ta = U (H(n) in > 0 )

.

T^

§3.

UNCOUNTABLE SUBORDERINGS OP DEGREES

Since each member of

T

has at most aleph-one successors and at most

countably many predecessors, it follows one.

Since each member of

Tr(r < °0

T

that each member of member of

Ta

has cardinality at most aleph-

which is comparable with some member of some

is a member of that

and incomparable with

37

T ,

it follows that

Ta

is disjoint from

U {T^lr < or} . It is clear from the definition of

T

which is

comparable

with some member of

Ta

H

is a

Ta .

Thus there exists

an ordinal

6

disjoint, mutually incomparable subsets of property that each

T^

CT^ Ir

and a collection T

whose union is

has cardinality at most aleph-one.

T

< 6)of

with the

By repeating

the argument of Lemma 2 of the present section, we can obtain for each

7 < 8 , a completely normal partially ordered set T^ imbeddable in

T^

and such that

T^

We can assume that the members of U {T^lr < 5).

T^

are disjoint. T1

Let

5,

every member of

T is imbeddable in

Then

is

every member of

T^ T !,

T* =

by retaining the par­

and by specifying that for each pair

distinct ordinals less than T^.

T^

has cardinality at most aleph-one.

(T^lr < or}

We define a partial ordering for

tial ordering of each

such that

(7 , a)

is incomparable with and T f is completely

normal. COROLLARY 3. Let A and B be sets of degrees such that A iscountable, B has cardinality less than that of thecontinuum, and no member of B is less than or equal to any finite union of members of A. Then there exists a degree d such that d is greater than every member of A and incomparable with every member of B. The notion of independent degrees was defined in Section 1 . We call the functions in a set independent if the degrees of the members of the set are independent. degrees are.

of

We say two functions are incomparable if their

Since Theorem k is a consequence solely of the methods used

in the proof of Theorem 1 , we are content to sketch its proof. THEOREM 4. Let A and B be sets of functions such that A iscountable and B has cardinality less than that of thecontinuum. Then (1 ) and (2 ) are equivalent: (1 ) there is a set C of independent functions such

DEGREES OP UNSOLVABILITY

38

that C has cardinality of the continuum, each member of C is incomparable with every member of B, and each member of A is recursive in every member of C; (2) no member of B is recursive in the members of any finite subset of A. PROOF. and let

S

We show (2) implies (1).

be defined as in Theorem 1.

Let

A = {aQ, a1, a2,

A set of integer-valued functions

{Q^ln > o & 1 < k < 2n} will be defined by stages. a non-empty, finite set least member ofRn+1 ber of

will

D

will have

arguments. For each

D

of functions such that

if and only if there is a member

o, the restriction of

f to

Rn

is

v

of

S

f

0 ^ ^ ) • After straightforward

commitments and finite extensions apply here. s

s.

If

stage

Rn

and

then for each n < k,

prior to stage (H2)

s

for each

g

s

such that

are of the form

to a finite range of values.

p^+t^ ^

(m > o)

furthermore, for each then

has been defined prior to have also been defined

1 < g < 2n .

where j,

m > o

and

j

is restricted

if some natural number of the

has been committed to a closed classification prior to stage

natural number of the form

s,

p^,

For each

then there is a positive integer the form

R^

{Rn ln =* o, 1, 2, ...}

All natural numbers committed to a closed classification

prior to stage

p^

The induction hypothesis at

Only a finite number of the sets

have been defined prior to stage

form

concerning

of the construction is: (Hi)

s,

will be a

such that for all

modifications, the remarksmade in the proof of Theorem 1 above

stage

n, the

be the Immediate successor of the greatest mem­

Rn . We will define a set

member of n >

Rn as its domain of

Each

Qng(Pj+t(J) )

m,

t(j)

s,

such that every natural number of

has been similarly committed and such that no p^

(m < t(j))

if

p^+t^ ^

has been similarly committed;

has been put in

has been set equal to

a^(m)

Rn

prior to stage

for all

g

such that

1 < g < 2n . (H3)

For all

classification only if CASE 1. each member of

A

m

and

j,

p^

has been committed to a closed

2 • 3^ < s.

s = 2 • 3J .

The object of this case is to insure that

is recursive in every member of

D.

Similar to Case 2

§3.

UNCOUNTABLE SUBORDERINGS OP DEGREES

39

of Theorem 1. CASE 2.

(s)Q = 3,

(s)1 = e1 and

(s)2 = e2. The purpose of

this case is to guarantee that for each pair (u, v) of S,

if a function

f

is recursive in Du

in Dv

with Godel number

members of D

(s)Q =

independent. Let

not been defined prior to stage 1 < g < 2q = p,

define

finitely extend Dw ^

with Godel number

e1 and

f is recursive in the members of

Similar to Case k of Theorem 1. k and (s)1 = e. This case is needed to make the

some finite subset of A. CASE 3.

e2, then

of distinct members

w( g)

, Dw ^

q

be the least

s.

n

such that Rn

Let p = 2q . For each

as in Case

g

has

such that

k of Theorem 1. For each g we

, ..., Dw ^

in such a way that

Dw (8>(b ) = u(nyT^1'"-»1(Dw(l)(y), 0)

S » ^ ,)(y),

will be raise ror some some m

S^y),

or tne rignt siae or (i) will be undefined. ior

m

when the construction is complete.

finitely extended

In the process each Dw ^

2q times and the union of these consecutive

sions Is the needed finite extension at stage after Cases 3 and CASE

e, m, y))

s.

2q

Is

exten­

This case is modeled

k of Theorem 1.

k.

s is otherwise.

be the greatest member of r

Define

q

if q > o and

as in Case 3. -1

otherwise.

Let

u

We define:

Rq = {u + 1} 0

for all

g

(u+1)

such that

By Case members of D continuum.

f value assigned prior to stage s t o otherwise, 1 < g < 2q.

k, the members of D are well-defined. By Case 3, the

are independent and

Let F

D

has cardinality equal to that of the

be the set of all members of D

with some member of B. continuum and A

s

Since

B

which are comparable

has cardinality less than that of the

is countable, itfollows from Case 2 that

nality less thanthat of the continuum.

D

- F

F

has cardi­

is the desired set

COROLLARY 1. There exists a set of independent de­ grees whose cardinality is that of the continuum.

C.

DEGREES OF UNSOLVABILITY

IfO

COROLLARY 2. Let T be a partially ordered set of cardinality less than or equal to that of the con­ tinuum. If each member of T has only finitely many predecessors, then T is imbeddable in the upper semi-lattice of degrees. In Theorem k take

PROOF. G,

A

and

B

to be empty and thus obtain

a set of independent degrees whose cardinality is that of the continuum.

Let

T

be a partially ordered set as described in the hypothesis of Corol­

lary 2.

Let f

be a one-one map of

T

into

G. For each

t € T,

let

t* = {w|w € T & w < t } Since for each For each

t,

t e T,

{t”|t € T).

let

Then

means of the map

T

t 1isfinite, we write

t* = (uQ ,

,..., •

t" = f(uQ) U f(Uj) U ... U

Let

T" =

is order-iscmorphic to the set of degrees T"

by

t -► t".

Note that these last two corollaries are proved without any use of the axiom of choice. needed

In the proof of Theorem3, the axiom

of choice is

only to show F has cardinality less than that of the continuum.

In Corollaries 1 and

2, F is empty because

B

is empty.

Theorem 3 tells us that normality isa sufficient partially ordered set to be imbeddable in thedegrees.

In

condition for a Corollary 2 to

Theorem 3, we saw that it is possible to show certain partially ordered sets of cardinality of the continuum are normal without any use of the continuum hypothesis.

Unfortunately, the continuum hypothesis is closely tied to the

question of which partially ordered sets are normal. LEMMA 5. Let T be a partially ordered set of cardinality of the continuum such that any two members of Thave an upper bound in T and such that any member of T has at most countably many predecessors. Then T is normal if and only if the continuum hypothesis holds. PROOF. T

is normal.

mal, partially

If the continuum hypothesis holds, then by Lemma 2 above, Suppose T

is normal.

Then there exists

ordered set P and a one-one map

f

of

a completely nor­ T into

Pwhich

§3. imbeds

T

in P.

UNCOUNTABLE SCJBORDERINGS OF DEGREES

Let

a

be an ordinal and

of subsets

of P

normality.

We assume that each

CB^Ir < 7 *-

definition of complete normality tells us that in B^

for some

With the help

because

such that

B^

be such an upper t < n,

where

t e B^*,

This last is ruled out by clauses (2 ) and (3) of the

definition of complete normality.

bound

7

be the least Let

P,

and because any two elements

7 < s.

t

Then clause (k) of the and

d

have an upper

This last contradicts the definition of

8.

of Lemma 5, it is immediate that normality is a

necessary condition for a par-daily ordered set to be imbeddable in the upper semi-lattice of degrees only if the continuum hypothesis holds. After all, the upper semi-lattice of degrees itself is normal only if the continuum hypothesis holds.

Nonetheless, the notion of normality is useful,

since it made possible the proof of Corollary 2 to Theorem 3 without the

1*2 continuum hypothesis.

DEGREES OP TJNSOLVABILITY Corollary 2 of Theorem h provides us with very simple

partially ordered sets which are imbeddable in the degrees but which are not provably normal without the continuum hypothesis.

One such partially order­

ed set is the set of all finite subsets of real numbers ordered by setinclusion.

§1*.

THE PRIORITY METHOD OP FRIEDEERG AND MUCHNIK

In [1 7 ] Post raised but did not answer the following problem: do all non-recursive, recursively enumerable sets have the same degree of re­ cursive unsolvability?

The solution to Post!s problem was found almost

simultaneously by Friedberg [1 ] and Muchnik [1 3 ]; these authors independ­ ently discovered a new technique which we will call the priority method. We will use the term ’’priority method” somewhat ambiguously to designate any method of proof which owes a large portion of its inspiration to [1 ] and [1 3 ].

Thus we say that Theorem k of [2 3 ] and Theorem 2 of C1 83 were

proved with the help of the priority method. Our purpose in formulating Theorem 1 of this section is to sepa­ rate (insofar as is possible) the combinatorial aspects of the priority method as manifested in [1 ] from the recursion-theoretic aspects.

We do

not claim that Theorem 1 stands as a fundamental principle from which all results so far obtained by the priority method readily follow, but we do believe that Theorem 1 and its proof will be useful to anyone who wishes to develop an intuitive understanding of the workings of the priority method in all of its manifestations.

We will put Theorem 1 to practical use in

this section by deriving from it as corollaries the solution to Post's problem and the fact that every countable, partially ordered set can be imbedded in the upper semi-lattice of degrees of recursively enumerable sets. A requirement

R = {(Fi, H ^ l i cl)

is a collection (possibly

empty) of ordered pairs of disjoint, finite sets of natural numbers. T

of natural numbers meets requirement

Fi C T

and If

R

if there is an

i € I

A set

such that

H^ n T = 0 . L = (hQ, ht, ..., h^}

is a finite set of natural numbers, we h3

U

DEGREES OP ^SOLVABILITY

h0 define j (L) =2 + 2 map of

h-

h +...+2

. The function j

is a one-to-one

the set of all finite sets of natural numbers onto the set of all J(0 ) = 0 . Both

natural numbers, if it is understood that verse

j-1

and its in­

are effective.

Let

t be a function defined on the natural numbers with natural

numbers as values. s > 0,

j

We say

j”1((t(s))0)

and

t

enumerates requirements if for each

j”1((t(s)) ^

requirements enumerated by

t

are disjoint, finite sets.

are denoted by

RQ, R1, R 2, ...;

The

for each

Ek = {(j-’ Otf-Ho). J“1((t(3))i))|(t(s))2 = k} • If

t

enumerates requirements we denote

(t(s))2

by

F s, H s

and g(s)

j’1((t(s))0) , j”1 ((t(s)) 1)

respectively.

Thus for each

Rk = {(Fs, H s)|g(s) = k} A if

A

set

A

.

T

t

F* C Tr

s = 0. TQ = 0

Stage

s > 0 . Tg = Tg-1if (a) or there is an r < s

(1 Ts_, = 0

and

such that

there Is an r < s

and

if n Ts_, - 0;

(c)

Hs n T , ft o. s-i

and

k,

Fs £

T

s

we say Tg. If

such that

= T

Rk Rk

, U Fs s-i

is recursively

(b) or (c) is true: g(r) < g(s), r > 0 , F1* £ Tr_«|>

g(r) = g ( s ) , r > 0 , F r £ T

otherwise.

is met at stage

is met at stage

since clause (c) must be false at stage all

T

if n F s ? 0;

(b)

For each F s £ Tg-1

called the priority

t.

Stage

(a)

t

and meets every member of a certain subclass of the class

of requirements enumerated by

Fr C T r , f

T

is defined by stages, and as we shall see,

enumerable in

f

f . With each function

which enumerates requirements, we associate a set t.

k,

is said to be recursively enumerable in a function

is the range of a function recursive in

set of

and

s,

s

s,

T =

U T0 . s=0 s

if s > o, then

and since

,

k = g(s),

H3 n Tg = 0,

Hs n Fs = o

for

s. For each

r < s

k,

we say

such that Rk was met

Rk

is injured at stage

at stage

r,

s if there is an

i f n Tg-1 = o and

H1* n Tg ^ o.

THE PRIORITY METHOD OF FRIEDBERG AND MCJCHNIK

$k.

If

Rk

is met at stage

r

^

and is not injured at any stage after stage

r,

then

Fr C T and if fi T = o for all s > r. But then Fr C T and — r s — — H17 n T = 0 . Thus if Rk is met at stage r and is not injured at any succeeding stage, then For each there is an

k,

s > o

T

meets

we say

such that

R^.

Rk

is

t-dense if for each finite set Hs n T g_ 1 = o

F s $ Tg__1,

g(s) = k,

L,

and

LflF3 = 0 .

setting pair

The definition of

T

TQ - o.

s > 0,

(Fs, Hs)

At stage

may be described as follows. the function

of disjoint, finite sets.

t

We begin by

presents us with a

This pair is of interest to us

because it may represent a chance to make

T

meet

^

Rg(s) •

v e set

T S = T S—I - U Fs,7 and if we can manage to have T„ u — > s,7 U n Hs = 0 for all then T will meet Rg(s)clause (c) is true at stage s, then Hs n T g - 1 ^ o, Rg(s)-

and stage

s

clearly does not represent a chance to meet

If clause (b) is true, then there was a stage

such that

Rg(s)

was met

jured at any stage after

stage

r

and such that

rand before

with the hope of not injuring

Rg (s)

s;

Rg(r)

s

Rg(r )

Againwe set

T g- 1

U F s,

then

Rg(s)

s,

was met, and

g(r) < g(s);

and before s,

Rg(r)will be

Hg(s) at stage

have assigned a higher priority to

* T g_ 1

Tg

then there was a stage

Rg(r)

s

r

in addition,

but if

Tg

injured at stage

Tg = Tg__1, because we do not wish to injure

forthe sake of meeting

s

vas not in­

consequently we set

was not injured at any stage after r

is set equal to

s

at which

prior to stage

at any future stage.

If clause (a) is true at stage prior to stage

r

when

than to

Rg(r)

g(r)

s. atstage

< g(s). Thus

Rg(s)

when

we

g(r) < g(s).

LEMMA 1 . If r < s and R^ is met at stage r and at stage s, then there is a u such that r < u < s and Rk is injured at stage u. PROOF. follows that stage

s

Tp -1 = 0 ,

and

We have

g(r) = g(s) = k,

H 17 n T g __1 7^ 0 , R g(s)

Fr

Tr - 1

and

Fr £ Tr .

since otherwise clause (b) would be true at

would not be met at stage

s.

We know that

since otherwise clause (c) would be true at stage

would not be met at stage

r.

It

Then

H37 n T

= 0,

since

r

H 17 n and

H37 n Fr = 0

Rg(r) for

DEGREES OF ^SOLVABILITY

k6

all

r.Thus there is

and

H1* n Tu ^ o.

But then

LEMMA 2. at stage PROOF. i < k, 2*.

a unique u

such

Rk is injured atstage

Hr n T u l = o

u.

For each k, the set CslRv,is injured lr s) has cardinality less than 2 . By induction on

the set

that r < u < s,

{s|Ri

k. Let

k > o,

is injured at stage

and suppose that for each

s)has cardinality less than

Then the set R = (sI(Ei)(i < k & R1 ^

has cardinality less than

- k.

is injured at stage Let

S = (s|(Ei)(i < k & R.^ With the help of Lemma 1, dinality of stage

s).

is met at stage

s)}

we proceed to obtain an upper bound on the car­

S. Let m. be the cardinality of the set {s|R^ is met at ^ v k-i 1 Then the cardinality of S is at most / i_Q . By Lemma 1^ = (slR^^

1, the cardinality of the set least

s)}

- 1.

Since for each

i < k,

is injured at stage

s)

is at

Ii has cardinality less than

2^

is must be that

Thus the cardinality of

k-i £ (n^ - 1) < 2k - k i=0 If S is less than % .

Wenow show that if i < k set

such that ts|Rk

Ri

s

is injured at stage

and suppose

such that

R^

Thus wehave Fs / o.

Rk

S

0
0}

s1

be such that

s and

n such that

be the greatest member of

s" > s 1 and

j < yi#(s,

n, e*).

It follows

e*)|s > 0} is infinite.

Suppose that the set

m ! be its greatestmember.

all

’

is finite;

s' > s* s>

{m^fs, e*)|s > s').

s' Let

let

andc(s, n) and n < m*. s"

be such

m^# (s", e*) = m” < m f.

We now show by induction on

s > s”

that

=

m^Cs, e*) = m"

and

DEGREES OP UNSOLVABILITY

6k

Kj*(s>

e*)*=&j*(3"> e*) e*) = m"

and

for a11

3 > 3"* Suppose then that

Kj*(s, e*) » Kj*(sl!,e*). By Lemma 3, 7 j*(s, n, e*) = y±*(s+i, n, e*)
s",

n < m^fs, n, e*),

since

s

s > s" > s ! > s*.

This means that

P±*(s, n, e*) = P±*(s+i, n, e*) = c(s, n) for all

n < m^fs, e*). But

n < m" < m f,

since

s > s*.

c(s+i, n) = c(s, n) = c(n)

for all

Hence

pA.( s+i, n, e*) = c(s+i, n) for all

n < mn * m^fs, e*),

and consequently,

follows from the definition of

m”

that

m" < m^# (s+i, e*). It

m^# (s+i, e*) = m ” . But then by

remark (R2), y±*(s, n, e*) = y1#(B+i, n, e*) < Kj*(s, e*) for all

n < rn^Cs+i,e*) * m 11 = m^fs, e*). Prom this last and

nition of

Kj*,

it is clear that

Thus we have shown

Kj*(s+1, e*) = Kj*(s, e*).

Kj*(s, e*) * K^Cs", e*)

but this is absurd, since by definition of { K ^ s , e*) Is > 0}

e*

and

for all i*,

the set

is infinite.

LEMMA 5. If s > s* and c(n) * P±*(s, n, e*). PROOF. Let

s > a*

and

n < m^Cs, e*),

then

n < m^Cs, e*). By Lemma 3,

y±*(s, n, e*) = y^fs', n, e*) < s for all

By Lemma k ,

s* > s.

there is an

s" > s such that

(s", e*) > ^*(3, e*)

and

c(s", n) = lim3 c(s, n) = c(n) But then

c(s", n) - Pi#(s", n, e*) * u(yi#(sM, n, e*)),

n < m^# (sfl,e*)

and

y^*(sfl,

n, e*) < s < s".

c(n) = u(yi#(s", n, since

t(n)

since

It follows that

e*)) = P±*(s, n, e*)

y1#(s", n, e*) = y±*(s, n, e*) < s. Let

the defi­

denote the partial function nsfs > s* & n < m^Cs, e*) ]

;

,

s > sMj

AN EXISTENCE THEOREM FOR R.E. DEGREES

§5. by Lemma k 9

t(n)

is defined for all

n.

65

By Lemma 5,

c(n) = P±*(t(n), n, e*) for all

n.

Since each of the functions

recursive in

B,

it must be that

our argument that for each is finite. D±

This

for any

odd.

i < 2

i #di(n)

Let

Cy(n) In < m}.

Let

n < m.

n< m

and all

j

B U Di

is

only

since

B U

Di

for any

b • d^. e)|s > o}.

c(n)

for some

Then e, n, j ) '

3

for all

n < m.

s be such that

< y.

B,

,

is either undefined or unequal to

c(s, n) * c(n) for all

B UD^

m be thelargest member of (K^Cs,

y(n) “ nyrj( II J< 7

member of

{K^(s, e)|s > 0}

C is not recursive in

C is notrecursive in

TJ(y(n)) = c(n)

is

That completes

& n € B U D^)

Suppose this last is not the case.

is defined and

the set

are each recursive in

The representing function of

We now show

e,

Pi.Cs, n, e)

B.

contains only even numbers and C

1 ► (n is odd

Thus it is sufficient to show

i

and

«— ► (n is even & n € B U D i)

n €^

n < m.

and each

and

n € B

Fix

m^fs, e)

is recursive in

is all we need to know to show

This means that

i < 2.

C

Let

y

be the largest

y < s,

and d± (s, j) * d± (j) Then y ^ s , n, e) * y(n) < s for all

It follows that P± (s, n, e) = u(yi (s, n, e)) = u(y(n)) = c(n) * c(s, n)

for all

n < m.This means

y(m) < K^Cs, e) < m.

m < m^fs,e) . But then by remark (R2),

Thislast is absurd, since

It remains only to see that It is clear that

Di ^ 0,

For each

D1

would equal

i < 2,

if

^((n^,

(n),) = 0

gj/n) - { nt(t € Di)

otherwise

D

and

B. C

we define a function

and which enumerates (n),

GND.

is recursively enumerable in

since otherwise

would be recursive in B, which is recursive in B

Di

m < y(m) by

.

g^^

66

DEGREES OF UNSOLVABILITY We need the following elementary lemma to obtain some corollaries

to Theorem 1. LEMMA 6. If AQ for each i < 2, Aq U A 1, then the least upper bound PROOF. A0 U A 1

Let

and

A1;

in both,

and

Clearly

n € AQ .

no, then n t AQ . Aq

aQ, a1

respectively.

wish toknow if

and A 1 are disjoint sets and if A1 is recursively enumerable in degree of AQ U A 1 equals the of the degrees of AQ and A 1

If the

eventually

£g

be the degrees of

Aq, A 1

U a^. We show

< a.

a
£0 U £1

b < £, < £.

then

= £•

We saw in Section h that there exists a recursively enumerable degree

d

such that

Corollary 3, than

d.

d

£ o)

such that

i,

.

Our construction is quite similar to that of Theorem 1.

We define six

§5. functions, and

t(s),

K(i, s, e),

f (s)

AN EXISTENCE THEOREM FOR R.E. DEGREES

d(i, s, n),

y(i, s, n, e), P(i, s, n, e), m(i, s, e)

simultaneously by induction on

into just one of the D1 ,s

D.

will be the representing function of D^.

member of

B

B.

s

d(o, o, f(o)) = o. We set

Stage

D^s

will

limg d(i, s, n)

Each of our six functions will

1 for

P(i, o, n, e)

= 2, d(i+i,

and m(i, o, e) = K(i, o,

s > o. We define

z(s) =

e) « o

^ °J

We set

all

for

n ^ f(o), o, n) « all i,

and we t(o) *

e and

n.

t(s) and z(s):

< t(s) &

z(s) < f(s).

is odd.

d(o, o, n) =

& f(s) < K(i, s-1, e))J

t(s) = iixx < f (s^(El)(Ee)(x = p^+e

Note that

i,

s we put

We assume without any loss of generalitythat every

= o. We set

y(i, o, n, e) = i

At stage

For each

is even and thatevery member of D

Stage set

s.

thereby guaranteeing that the

be disjoint and that their union will be

be recursive in

71

;

•

d(z(s), s, f(s)) » o

and

d(i, s, n) = d(l, s-1, n) for all f(s)

i

into

and

n such that

1 ^ z(s)

or

n ^ f(s). Thus we have

put

D Z(S). Let d1 ^ , J) =

,11

d(k, s, j)

k^i & k < s Wedefiney(i, s, n, e)

and

P(l, s, n, e)

y(i, s, n, e) * uyy < s[T](

forall

i,

n

Pj(J) * d

u(y(i, s, n, e))

if

and

e:

e, n,y)] y(i, s, n, e) < s

P(i, s, n, e) = | s+2 Before we define

m(i, s, e),

otherwise.

we observe that

(i)(e)(Et)Ct < s & d(s, t) ? P(i, s, t, e)] This last is clear, since = s+ 2

for all

i

and

d(s, s) = 1 ,

and since by

P(i, s, s, e)

e:

m(i, s, e) = nt[d(s, t) ^ P(i, s, t, e)] " K(i, s-i, e) K(i, s, e) =
0}

is not recursive in

e

such that the set

B, D 1

is finite, and then use this for any

i*

and

Suppose there is

{K(i, s, e) Is > 0} is infinite.

t* = ut(Ei) (Ee)£t = p^+e &{K(i,s, e)|s > Let

i.

0}

is infiniteJ

Let .

e*be such that t* = p±*

All that is needed now is to repeat the arguments of Lemmas 1-5 in order to show

D

is recursive in

B,

which is impossible, since

b < d.

There

would be little profit in actually repeating these arguments, since the only real difference between the constructions of Theorems 1 and 2 resides in the assignment of priorities; all other differences are merely notational. We content ourselves with proving the counterpart of Lemma 1 and stating the counterparts of Lemmas b and 5. LEMMA 7. z(s) = i* PROOF. and

of

Let

and

repetitions.) S and a

S

Since t**

S

exist

for all i** and

e**

ssuch that

(Recall that

f

z(s)

t(s) « t**

for all

f(s) > t* > t(s) = t**

enumerates

D

without

for all

s € R. t(s)

such that

and f(s)
s 1,

be an infinite set such that for all

f(s) > t* > t(s).

Thus we have > t(s)

such that for all

Suppose there are infinitely many

t* > t(s)*.

z(s) ^ i*

There is an s' or t* < t(s).

3 -1 , e**)

Since

f(s)

that there

§5. for all since

s c R.

R

AN EXISTENCE THEOREM FOR R.E. DEGREES

But then the set

{K(i**, s, e**)ls > 0}

is infinite, and since

This means that t(s) = t*

t* < t**,

for all

s € R.

Let

consequently

But

then

= i*

s1

n < m(i*,

s, e*),then

Theorem 1, that m(i*, s, e*) that

D

D

and

is recursive in

B,

P(i*, s, n, e*)

becomes: if

e,

the set

D

is not recursive in

B, D1

It remains only to see that the

and D

recursive in B

For each

and enumerates the if*

| (nt(d(i, D± *

of Lemma 6.

then

taneously enumerate the D^s,

Di ,s

since otherwise

D1

The function

would

g(i, n)

is

simultaneously:

d(l, (n)Q,(n),) - o

n i D .. J

otherwise

;

. D,

n € By

we repeat the argument

Ve first ask if

n € D.

If the answer is yes, then we simul-

and eventually

n

will turn up in exactly

thereby answering our initial question.

have just shown that

the D ^ s are uniformly recursive in

there is a recursive

function h

Godel number

is

are simultaneously recur­

B, B^.

is recursive in

D^'s,

Since we are given

easily repeated in order

(t)Q , (t)n). o))1

Supposewe wish to know if

If the answer is no,

one of the

D^'s

Cg(i, n) In > 0 }

To see that each

B.

for any i.

i, Di ^ o,

would be recursive in

| (n))

as in

{K(i, s, e)|s > o}

to show

D,

s > s* and

that completes our argument by reductio

The concluding argument of Theorem l is

equal

plays the

since each of the functions

finite.

B.

t* < f(s)

s*

There is now no difficulty in

is recursive in

for each i and

sively enumerable in

s > s !,

P(i*, s, n, e*). It then follows,

is not recursive in B,

ad absurdum that

s € R,

Lemma k becomes: the set

Lemma 5

d(n) =

for all

The natural number

mimicing the arguments of Lemmas 2-5. s, e*)|s > 0)is infinite.

It follows that

s e S.

such that for all

ort* < t(s).

is infinite,

without repetitions.

t* = t**.

for all

same role it did in the proof of Theorem l.

(m(i*,

D

z(s) = i** = i*

z(s) ^ i*

s*be the least z(s)

enumerates

and

which is impossible, since

and either

f

73

such that D 1

Actually, we D;

that is,

is recursive in

D

h(i). By a similar argument, it follows that for each

with i,

DEGREES OF ^SOLVABILITY

Ik

the members of the sequence D0> D i> •••' Di-i* Di+1’ ••• are uniformly recursive in For each

i,let

cause every member of dition the F^>s

F^BUD^

all

m, vn|F(m, n) s > o,

B.

Let

D is

let

F1

xmn|F(m, n)

In ad­

be a function such that for

n € U (D1 1i < n))

recursive in xmnlF(m, n),

F^'s

is odd.

and are simultaneously re­

Fm - Recall that for

This means

(n) (n € D «— ► n

i,

D

f(s) e D z(s)> 5111(1 since

s > o,

We show that the

is recursive in F b e ­

is the representing function of

(n) (n € D

that

B

B is even and every member of

z(s) < f(s).

since for all

Then

are uniformly recursive in

cursively enumerable in each

D1 .

, € D o*

follow,s

since

is odd &(Ei) (i < n & F(i, n) = 0))

are recursively independent (in sequence).

For each

denote the function xmn|F(m + sg((m+i) — i), n) .

First we observe that (m)(n)jF(m + sg((m+l) - i), r.) = 0— It follows that

F*

n e B v [n e Dm+gg( (m+,^ is recursive in B, D1, D0>

D

F^

D*.

is not recursive in

cursive in F1 ,

since

is not recursive in

D

since the sets

,##> Di-i* Di+l'

are uniformly recursive in since

n > m ♦ Sg((m+1) - l)] } .

But then D B, D*.

is not recursive in

This means

is recursive in

xmn|F(m, n)

xmn|F(m, n).

F1 ,

is not re­

It follows that

F*. recursive, universal, partial ordering relation

described in the proof of Corollary 3 of Section k . For each

i,

let

Bi = {pl+ni» « F±} Then the

B^'s

are uniformly recursive in

D,

simultaneously recursively

§5 .

enumerable in

B

AN EXISTENCE THEOREM FOR R . E . DEGREES

and recursively independent (in sequence); also

recursive in each

B^.

For each

u,

Cu

u, B

is recursive in

is recursively enumerable in

B

is

let

Cu - u CB± |i < R u) Then for each

75

Cu ,

Cu

. is recursive in D

and

B.

We conclude the proof of our theorem in exactly the same manner Section k.

that we concluded theproof of Corollary 3 of

It is necessary

to show (u)(v)(u < R v «— ► Cu If

u { R v,

then

independence of the

Cu

is recursive in

Is not recursive in

Bi ,s.

If

because of the recursiveness of

u < R v, 0 [h ^ 1 < 4h(u)& P? = It is clear

that the predicate

less than or equal to

o ’.

i ■ t] }

(Ey)K((y)Q, (y).,, e, d,

h, c)has degree

It follows the predicate

(Ey)K((y)0, (y),, e, d, h, c(e)) is recursive

in the function c,

since

c > 0/.

Our plan is to define three functions, simultaneously by induction on sive in

c.

e;

y(e), h(e),

and

d(e, i),

each of these functions will be recur­

In addition, the function

d(e, i)

will have the property

that d(i, i) = d(e, i) whenever

e> i.

The desired function

d(i)

will be definedby

d(i) = d(i, i) Stage all

e * o.

We set

.

y(o) = h(o) * i.

We define

for

i: c(0) f c(

d(0,

L l Stage

e > 0.

if

(Em)(pQ = i & m > o)

otherwise

We set:

nyK((y)0 , (y),,e, (2T(e, h(e - i)))e_,, y(e) *

if such a 1

otherwise ;

Note that d(e, i) for

- 1),c(e))

-

mind, we define

i:

c(e)

lf

h(e - 1) < i 0 (i = p“ )

d(e - 1, i)

otherwise

That concludes the construction. Since > e

h(e

exists,

£h((y(e))Q) < h(e) . With this lastin

all

d(e, 1) =

y

h(e) a h(e-i) + y(e).

' (y(e))o,l

h(e)

d(o, i)

for all e. Since

.

y(e) > o

h(e - 1) < h(e)

for all

whenever

e,

e > o,

wehave it

follows

§6 . THE JUMP OPERATOR d(e,

i)=

d(e - 1, i)

whenever

79

i < e -1 . But then

d(e, i) = d(i, i) whenever

e > i.

We define

d(i) = d(i, i). Let

d be the degree of

d(i). LEMMA 1 . PROOF.

c < d*.

It is clear from the construction that d(t, p“ ) = c(t)

whenever

h( t)
0.

We show

d(e, p*) = c(t) whenever and

m

h(t) < p!j?,

m > o

be such that

and

h(t) < p“ ,

t < e

by an induction on

m > o and

e > t,

e.

Let t, e

and suppose

d(e - 1, p“) = c(t) Let

p® = i.

since

Now

i ^ p^.

d(e, i)

Suppose

equals either

d(e, i)

d(e - 1, i)

or

h(e - 1) < i < *h((y(e))Q) & p “ = i & t < e & since the first caseof the definition of follows from the definition of

K

and

d(e, i)

y(e)

y(e) ^ 1

must hold.

,

But

then it

that

(r(®>)o,i ■ c(t) since

(y(e))0

^ d(e - l, i). Then we have

*

y(e) ^ 1. Thus

d(e, p^) = c(t)

whenever

h(t)
0

and

t < e.

It follows d(p^) = d(p“ , p“) = c(t) whenever

h(t) < p^

and

m > o,

since

P t > t for all

t.

Then we have

(t)(En)[(m)m > n (d(p“ ) = o) V (m)m ^ n (d(p“) = i)] sincec(t) < 1 n.

It

for all t;

isclear that

for each

t,

let

k(t)

be the least

k has degree less than or equal to c(t) = d(pt(t))

, such

d! , and that

80

DEGREES OF ^SOLVABILITY LEMMA 2.

d 1 < c.

PROOF.

Since

y

is recursive in

c,

it is sufficient to show

(Ey)T](cf(y), e, e, y) «-► y(e) ^ 1 for

all

e > o. Fix

e > o

and suppose

y(e) ^ 1.

all

i
eand

i < £h((y(e))0) < y(e) < h(e) < h(t - 1) since

t> e

nition of

t

If

is non-decreasing. that

that

d(t, i) = d(e, i)

Since

then

y(e) ^ 1 ,

i < h(e - 1) < h(e),

,

But then it follows from the defi­

d(t, i) = d(t - 1, i).

h(e - 1) < i,

i < h(e - 1). Since

h

d(t, i)

duction on d(e, i).

and

for

for all

Thus we have shown by in­ t > e.

It follows

(y(e))0 ^ . have (y(e))0 ±

d(e, i) =

we must

we must have

d(i) =

Suppose * ± . Since

y(e) ^ 1,

t ]((y(e))0,

it follows

e, e, (yfe)),) & (y(e)), < £h((y(e))Q)

We knew d(i) = (y(e))0 ±

for all

.

i < £h((y(e))0). It follows

(Ey)T’(d(y), e, e, y)

.

That completes the first half of the proof of Lemma 2. Now suppose y

has the property that

show

y(e) ^ 1.

Since

T^(d(y), e, e, y). 0

Again let

e > 0.

We wish to

is not the Godel number of a deduction, it will

be sufficient to show K ^ y ) , y, e, (cfte, h(e - i)))e_,, h(e - i),c(e))

.

But then it will be enough to prove: (1)

(l)[i < h(e - 1) - d(l) = d(e - i, 1)]

(2)

(D(t)t < e (m)m > 0 [h(e - 1) < 1 < y & p“ = 1

To prove (1),

suppose

andconsequently, d(e - 1, i).

i < h(e - 1).

d(t, i) = d(t - 1,i)

To prove (2), suppose

Then

;

i < h(t - 1)

for all

t

- d(i) =c(t)]

.

for all

t > e,

> e.It follows

d(i)

h(e - 1) < i < y,

t < e,

p* = i

and

§6 . THE JUMP OPERATOR m >

0 . Then

h( t) < p*£,

Lemma 1 that

d(p“ ) = c(t)

By Theorem 1 , £ ! > £ !.

since

t < e.

whenever

It was shown in the proof of

h(t) < p“

the equation

x’ = £

has a solution if and only if have a unique solution? c;

showed that the answer is no regardless of the value of

his result states: for each degree b < b ^ b < b1

such that

m > 0»

and

x* = c

It is now natural to ask: does

Spector [2 5 ]

81

b,

there are degrees

and

b«j

bg U b 1 = b 1 = b^ = b • We combine the

and

idea of his proof with the system of priorities of Theorem 1 of Section 5 in order to prove Theorem 2 of the present section.

We could, if we wished,

obtain Theorem 2 as a ccmplicated corollary of Theorem 1 of Section k ; how­ ever, the price of such elegance would be a total lack of clarity. THEOREM 2 . Let A and B be sets such that B is recursively enumerable in A. Then there exist dis­ joint sets Bq and B 1 such that BQ U B 1 = B and such that for each i < 2 , (B.^ U A) 1 is recursive in A f, and Bi is recursively enumerable in A. PROOF. is

B -A.

b^s,

Let f

be a one-one function recursive in

We will define six functions simultaneously by induction on

n), y^s, n) and

be recursive in

A.

(i = 0 , 1 ).

t^s)

sand

n.

sively enumerable in we put

f(s)(the

We assume

all

n.We set

We set all

A,

namely,

sth member of

Ci . At stage B

s = 0 . We set bi (0 , n) =

bQ (o, f(0 )) = 1

i < 2

and all

Stage

+ 1 ,n) > 0

is the representing function of a set recur­

B - A is infinite, since Stage

Each of these functions will

It follows from this last (cf. Section 5) that for

i < 2 , lims bi (s, n)

each

s:

i < 2 , we will have

For each

1 > b± (s, n) > b ^ s

for all

Awhose range

and

1

- A)

s

in CQ

or

of the construction, C1

but not in both.

otherwise there is nothing to prove.

y^Co, n) = t^o) = 0 for all

i < 2

for all

and all n

i < 2

t A U (f(o)}.

b ^ o , f(o)) = 0 . We set bi (o,n) = 0

n c A.

s > 0 . For each

i < 2 , we define

t± (s) = nn^ < g[f (s) < yi (s-i, n)]

.

and

for

8a

DEGREES OP nNSOLVABILITY

Let

z(s)

. 1 if tQ(s)

b^(s, n)

< t^s), and let

forall i < 2

z(s) = o

otherwise.

We define

and all n:

b,(s, n) =

r o if i = z(s) & n = f(s) -j *• b^(s - 1, n) otherwise.

We conclude the construction by defining _1 /

yi (s, n) for all

TT

Pi

o

n:

. n » n> y)

v J o

and

A 1.

i

and

n

be k

does not exist and 2m + k
n + 1 & yi(s, n) ^ y ^ s - 1, n) > o since

i < 2.

lims yi (s, n)

does not exist.

i. There must be an infinite set

forall

have BQ n B 1 = o and i (B^ U A) 1 = ^ is recursive in

and each

limg yi(s, n)

b1 -

we

We use the method of infinite descent.

such that

2n +

be the set whose representing function is

TT

yi (s - 1, n) > o.

bj (S-1 ,j) V Pj , n, n, j )

There must be a

,

j < yi(s - i, n)

§ 6 . THE JUMP OPERATOR such that

hi(s - 1, J) ^ ^ ( s , j),

But then

j * f(s),

because

f(s) < y ^ s - 1, n)

83

y^fs, n) ^ yi (s - 1, n).

and

b± (s, f(s)) t b ^ s - i, f(s)) This last can happen only if Since

f(s) < y ^ s - 1, n) Since

i = z(s).

and

o < t1-i(s) < n

infinite subset

T

of

S

.

It follows

n < s,

^ ^ ( s ) < t^Cs).

we must have

for all

t^s) < n.

s € S, there is an

m

and an

such that Vi(s) = m < n

for all then

s

2m +

t^s) = n

€ T.

Letk = l - i.

k < 2m + 1< 2n < 2n + i. for all

We saw above that and

We show

2m + k < 2n +

Suppose m = n.

s€ T,and consequently,

z(s) = i

for all

i. If

Then

z(s) » i

m < n,

t1_l (s) =

for all

s € T.

i = 1,

k = 0,

s e S. It follows

2m + k < 2n + i. It remains onlyto see that

suffices to show the set

limg y^Cs, m)

(yk (s, m)|s > o}

does not exist.

is infinite.

For each

It s € T,

we have tk (s) = m < n < s But

then f(s) < yk (s, m)

for all

infinite, it follows the set Now we fix

i < 2

s e T.

.

Since

(yk (s, m)|s > o) and show

f

isone-one and

T

is

is infinite.

is recursive in

A f.

Lemma 3

tells us that (n)(Et)(s)a> t(y1 (s, n) = y± (s + 1, n))

.

We define: t(n) - nt(s)a> t(y± (s, n) - y ^ s + i, n)) y^n) - y^tCn), n) The function

y^(n)

is recursive in

A.

.

is recursive in A' Clearly,

j

because the function

yi (n) = limg yi (s, n)

for all

y^(s, n) n.

defined in Section 1; it is the representing set of the predicate (Ey)T](bi (y), n, n, y)

.

C£ was

DEGREES OP UNSOLVABILITY

8b

We show

is recursive in

A 1 by showing

(EyiT^b^y), n, n, y) «— y^n) > 0 for all

n. Fix

n.

First suppose

y^n) > o.

Then

y±(s + i, n) « y(s, n) > o for all

s > t(n).

Since

o

is not the Godel number of a deduction, it

follows v i~

y^n) = y ^ s , n) » yi(t(n), n) = for all

s > t(n). Let

s f be so large that

’ ’ n ’ n ' y)

s* > t(n)

and

b± (s', J) = b± Q) for all

J < y^(n). Then we have Ti(t>l(7i (n))> n, n, y1(n))

Now suppose

.

(Ey)T] (b^Cy), n, n, y). Let y ’ = ►‘y T i ^ C y ) , n, n, y) > 0

Let

t

be so large that

t > y*

and

b± (t, i ) for all

.

= b± (3)

i o

for all

s > t.

This last means

y^(n)

= y 1 > o.

COROLLARY 1. Let a and b be degrees such that a < b and b is recursively enumerable in a. Then there exist degrees bg and b1 such that ^ U b 1 = b , a ’ = b^ = bj, and such that for each i < 2, a < b^ and b^ is recursively enumerable in a. PROOF. has degree

b,

bers and B B1

B

A

and

BQ U B 1 = B A 1,

B

be sets such that A

is recursively enumerable in

has only odd.

such that

recursive in

Let

and

Bi

A, A

has degree

a, B

has only even mem­

By Theorem 2, there are disjoint sets and such that for each i < 2, is recursively enumerable in

BQ and

(Bi U A) 1 is A.

For each

i