in the Department SOME RESULTS ENUMERATION REDUCIBILITY LANCE GUTTERIDGE


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1 SOME RESULTS ENUMERATION REDUCIBILITY by LANCE GUTTERIDGE B.Sc., Univrsity of British Columbia, 1967 M.Sc., Univrsity of British Columbia, 1969 A DISSERTATION SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREZWNTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY in th Dpartmnt of Mathmatics LANCE GUTTERIDGE 1971 SIMON FR9SER 'VNIVERS ITY AUGUST 1971
2 APPROVAL Nam : Dgr : Titl of Dissrtation: Examining Committ: Lanc Guttridg Doctor of Philosophy Som rsults on numration rducibility,l  I ' r, Chairman: N. R. Rilly,' 7 1 A. H. Lachlan Snior Suprvisor A. R. Frdman G. E. Sacks Extrnal Examinr, Profssor, Massachustts Institut of Tchnology Cambridg, Mass. Dat ~pprovd: Sptmbr 24, 1971
3 ABSTRACT Enumration rducibility was dfind by Fridbrg and Rogrs in Mdvdv showd that thr ar partial dgrs which ar not total. Rogrs in his book Thory of Rcursiv Functions and Effctiv Computability givs all th basic rsults and dfinitions concrning numration rducibility and th partial dgrs. H mntions in this book that th xistnc of a minimal partial dgr is an opn problm. In this thsis it is shown that thr ar no minimal partial dgrs. This lads naturally to th conjctur that th partial dgrs ar dns. This thsis lavs this qustion unanswrd, but it is shown that thr ar no dgrs minimal abov a total dgr, and thr ar at most countably many dgrs minimal abov a nontotal dgr. J. W. Cas has provd svral rsults about th partial dgrs. H conjcturd that thr is no st in a total dgr whos complmnt is in a nontotal dgr. In this thsis that conjctur is disprovd. Finally, Cas's rsult that thr is a minimal pair of partial dgrs is strngthnd to show that thr is a minimal pair of partial dgrs which ar total and form a minimal pair of r.. dgrs.
4 Th contnt of this thsis has bn vastly influncd by Profssor A. H. Lachlan. Th author apprciats th mathmatical assistanc and prsonal ncouragmnt which h gav at all hours and through all criss. Th finishd form of this thsis ows its natnss and prsnta bility to th typing skill of Mrs. A. Grncsr. Without hr co opration th author could nvr hav mt his dadlins. Th author is also indbtd to th National Rsarch Council of Canada without whos financial aid in th form of a bursary th work on this thsis could not hav bn continud.
5 TABLE OF CONTENTS CHAPTER I INTRODUCTION AND TERMINOLOGY 51.1 INTrnDUCTION 51.2 TERMINOLOGY 1.3 ENUMERATION REDUCIBILITY 51.4 INFINITE GAMES PAGE 1 f CHAPTER 11 A SET IN A TOTAL DEGREE WHOSE COMPLEMENT IS IN A NONTOTAL DEGREE INTRODUCTION 52.2 DESCRIPTION OF THE GAME 52.3 AN EFFECTIVE WINNING STPATEGY 52.4 A PROOF THAT THE STRATEGY IS A WINNING STRATEGY 52.5 CONCLUSIONS CHAPTER 111 MINIMAL PAIRS OF PARTIAL DEGREES INTRODUCTION 53.2 DESCRIPTION OF THE GAME 53.3 BASIC IDEA EEHIND THE STRATEGY 53.4 A WINNING STRATEGY FOR THE GAME 53.5 PROOF THAT THE STRATEGY IS A WINNING ONE 53.6 CONCLUSIONS CHAPTER IV THERE ARE NO MINIMAL PARTIAL DEGREES 4. INTRODUCTION 54.3 STRATEGIES FOR T E GAMES Cv
6 PAGE CHAPTER V DEGREES MINIMAL ABOVE A NONTOTAL DEGREE INTrnDUCTION 55.2 EXTENDED ENUMERATION OPERATORS 55.3 AN ENUMERATION STRATEGY 55.4 CONCLUSIONS BIBLIOGRAPHY
7 CHAPTER I INTRODUCTION AND TERMINOLOGY INTRODUCTION In this thsis w considr th partial (numration) dgrs as dfind by Fridbrg and Rogrs [2]. In Chaptr I1 w disprov a conjctur by Cas [l] that thr is no st A in a total dgr whos complmnt is in a nontotal dgr. a minimal pair of partial dgrs. Cas [l] has shown that thr is In Chaptr I11 w strngthn this rsult to show that thr is a pair of cor.. sts whos partial dgrs form a minimal pair of partial dgrs; and hnc thir Turing dgrs form a minimal pair of r.. dgrs. In Chaptr IV w show that thr ar no minimal partial dgrs and dmonstrat that this rsult rlativizs to show that thr ar no dgrs minimal abov a total dgr. This still lavs opn th qustion as to whthr th partial dgrs ar dns. In Chaptr V w mov closr to a solution of th dnsity problm by showing that givn any partial dgr  a thr can b at most a countabl numbr of partial dgrs minimal abov.a.  In this sction w intnd to outlin all th basic notation and dfinitions usd in this thsis. For a dtaild xplanation and study of rcursion thory th radr is rfrrd to Rogrs [5].
8 W will dnot substs of N by uppr cas lttrs, with D, E, F bing rsrvd for finit substs. Mmbrs of N will b rprsntd by lowr cas lttrs xcpt for f, g, h which ar rsrvd for total functions. Partial functions (that is, functions whos domain is a subst of N) w ill b dnotd by $ and $. A will dnot th complmnt of A, C~ will b usd for th charactristic function of A and A A B will dnot th symmtric diffrnc of A and B. For notational convninc w will somtims us ~[n] to dnot CX(n). W will call a finit initial sgmnt of a charactristic function an initial function. A join B is th st W dfin th binary function T by, r(x1 y) = f(x2 + 2xy + y2 + 3x+ y), and T is a rcursiv, on to on mapping of N x N onto N (s k Rogrs [ 5, p For ach k w dfin th kary function T as follows : k M abbrviat T (xl,..., 5) by <XI,..f \>. For ach k w dfin th projction functions. 1 k, 1 5 i I kt by
9 k W will usually drop th suprscript on n i from th contxt. whn its valu is clar A st B is singlvalud if <n, j > B and <n, k> B+ j=k. A st B is total if for all n thr is a m such that <n, m> B. If D is finit. i.. D = {xl...., x < x <... < Al "k XI thn th canonical indx of D is , if D = p) w lt its canonical indx b 0. W dnot by Di th finit st whos canonical indx is i. W will oftn mak no distinction btwn a finit st and its canonical indx, as in th us of <E, x> for <it x>, whr Di = E. W will assum that th radr is familiar with th rcursiv partial functionals, (s [6. p.150]). w lt <o~>:=~ b an ffctiv numration of th rcursiv partial functionals. th For ach w dnot th partial rcursiv function (C n)] by O. Th function f is rcursiv if for som. B' 4 is total and f(n) = O(n) for all n. A st is rcursiv just if its charactristic function is rcursiv. A st is rcursivly numrabl if it is th rang of som parkial rcursiv function. W dnot th rang of O by W. W say f is Turingrducibl to g (f 5 g) if f = hn[o(g, n) ] for som. W xtnd this rducibility T to sts by dfining A 5 B if C < Two functions f and g T A T C ~.
10 ar Turing quivalnt (f , 9) if f5 gandg5 f, It canbshown T T that E is an quivalnc rlation (s (5, p.1371). Th quivalnc T classs undr = of total functions (or sts by considring thir T charactristic function) ar calld th Turingdgrs. A Turingdgr that contains an r.. st is call6 an r.. dgr. W will us small undrlind lttrs for th Turingdgrs, for xampl 2, b, 2. Th jump of a st A, dnotd A', is th st I : O(CAf ) is dfind}. Th jump oprator is wll dfind for th dgrs and if a  is th dgr of A thn w dnot by 2' th dgr of A'. W us 2 to dnot th smallst dgr, that is th dgr of all rcursiv functions. W rsrv th symbol K for th st { : f w} and th symbol Kg for th st {<n, > : n w}. Both K and K ar of dgr 2' ENUMERATION REDUCIBILITY Enumration rducibility is dfind in Fridbrg and Rogrs [2] and discussd in Rogrs [ 5, p Intuitivly, a st A is numra tion rducibl to B is thr is an algorithm that will work on any numration of B (th input numration) and produc an numration of A (th output numration). Th formal dfinitions ar as follows: W(A1 = {x : for som D, D c A and <D, x> w} A is numration rducibl to B (A S B) if thr is an r.. st W such that A = W(B).
11 1.3.3 A = B if A 5 B and B 5 A A< B if A 5 B and B $ A. ~ U to S vry r.. st W thr is associatd a function ' N from 2* to 2. W call this function an numration oprator. In this thsis w shall not distinguish btwn an r.. st V and its associatd oprator m, for xampl is th numration oprator associatd with th r.. st W w will writ W (A) (A). Notic that th dfinition of W (A) in did not spcify that W had to b r.. In latr chaptrs for purposs of rlativization w shall somtims us nonr.. sts to oprat on othr sts. It is asy to show that E is an quivalnc rlation on th substs of N (s [ 5, p Th quivalnc classs ar calld numration dgrs. On of th rasons for studying numration rducibility, othr than th fact that it is a vry natural rlationship btwn numrations, is that w can gt a rduction btwn partial functions by dfining?+ L.. Th quivalnc classs of partial functions (or singlvalud sts) ar calld partial dgrs. A partial dgr is calld total if it has as a mmbr som total function f. W will us 5 to dnot th partial ordring of partial dgrs and numration dgrs inducd by 5. W will us undrlind lowr cas lttrs to dnot partial and numration dgrs as wll as
12 Turing dgrs. M writ  a < b  if  a 5  b and  b 2  a. Th structur obtaind by rstricting th partial dgrs to th total partial dgrs is ordr isomorphic to th Turing dgrs (s [5, p. 1531). Thr is rally no diffrnc btwn th structur of th numra tion dgrs undr 5 and th structur of th partial dgrs undr 5. Th map which taks to inducs an isomorphism btwn th partial dgrs and th numration dgrs. Th invrs of this iso morphism is inducd by a map which taks a st A into a constant function with domain A. Hnc in this thsis th trms partial dgr and numration dgr will b usd synonymously. W will us 2 to dnot th smallst numration dgr that is th dgr of all r.. sts (from a partial function viwpoint it is th dgr of all partial rcursiv functions). \ k Bcaus of th abov isomorphism btwn th numration and partial dgrs, and th isomorphic mbdding of th Turing dgrs into th partial dgrs, w ar jcstifid in using common notation, such as 5 and  0, for all thr structurs. On intrsting fact that thn Hnc th numration dgrs of cor.. sts hav a structur isomorphic to th r.. dgrs. Thus w will dnot by  0' th partial dgr For composition of oprators w will oftn omit th parnthss, for xampl w will us W(B) for W (V(I3) ). Also w will oftn us lowr cas lttrs in parnthss whr it would b mor usual to us subscripts as in V (k).
13 1 INFINITE GAMES Throughout this thsis w shall b using th mthod of infinit gams as xplaind in Lachlan [4]. In this sction w outlin som of th basic idas and trminology. W considr gams with two popl, on calld th playr and th othr th opponnt. Each prson numrats a squnc of sts. That is ach mmbr of th playr's squnc is a st that is numr atd during th gam, and similarly for th opponnt's squnc. To gain an intuitiv fl for this mthod it is hlpful to viw ths sts as rcptacls into which numbrs can b placd. Th playr and th opponnt tak altrnat turns, with th opponnt taking th first turn. For purposs of formal dfinition it is usual to insist that ach prson numrat at most on numbr in on st during his turn. Howvr in dscribing actual gams w shall only kp th opponnt to this rstriction and allow th plzyr any finit numbr of such actions in on turn (this dos not mak any chang to th ability of th playr to win a gam). Th gam nds aftr w turns. W shall call th conbind turns of th playr and th opponnt a stag. W numbr th stags starting with 0. Th nth stag of th gam is th portion of th gam that bgins at th start of th opponst st nt's n+l turn and nds whn th playr's n+l turn is compltd. If X is on of th sts bing numratd during th gam w will us xs to dnot X as it appars at th nd of stag s. W will normally us X to dnot th st as it appars at th nd of th gam.
14 Occasionally, hwvr, w will rfr to X as a rcptacl into which numbrs ar placd, as in th phras "th playr puts n into XI1. To spcify a gam w indicat th sts to b numratd by th playr and thos to b numratd by th opponnt. W also giv a rcursivly numrabl squnc of rquirmnts. Each rquirmnt stats a rlationship btwn som of th sts bing numratd. For xampl a rquirmnt could b A $ B whr th opponnt is numrating A and th playr is numrating B. A rquirmnt is said to b satisfid if it holds at th nd of th gam, in our xampl abov th rquirmnt would b satisfid it at th gam's conclusion th rsulting st A is diffrnt from th rsulting st B. Th playr is said to win th gam if at th nd of th gam all th rquirmnts ar satisfid. Aftr n stags of th gam, n < w, at most a finit numbr of th sts bing numratd will b nonmpty. A list of ths sts togthr with all th numbrs in thm and th stag ach numbr was put in is a gam situation. A stratgy is a map from gam situations into possibl movs. A stratgy S for th playr is said to b complt if no mattr what stratgy th opponnt follows th playr is always abl to follow S. A stratgy S For th playr is a winning stratgy if S is complt and vry play of th gam in which th playr follows S rsults in a win for th playr. By ncoding movs and gam situations into numbrs w can considr a stratgy as a function from N to N. Thus ffctiv stratgy, ffctiv winning
15 stratgy and th partial dgr of a stratgy ar dfind. Most of th gams considrd in this thsis will us rcursion thory notation in thir rquirmnts. It is possibl to rphras ths gams in purly gam thortic trms and liminat all notions spcial to rcursion thory. Although this would mphasiz Lachlan's rsult that rcursion thory can b don by strictly gam thortic mans [4], w hav sacrificd this intrsting point to mak th rsults mor concis. In th discussions and proofs about th gams w somtims idntify with th playr and occasionally rfr to th playr's actions as our own. All th stratgis givn for th gams in this thsis ar ffctiv. W shall not prov this for any stratgy as it is clar from thir dscription.
16 A SET IN A TOTAL DEGREE WHOSE COMPLEMENT IS IN A NONTOTAL DEGREE Cas [l, p.4261 has conjcturd that thr is no st A whos partial dgr is total and whos complmnt has a nontotal partial dgr. In this chaptr w will disprov this conjctur by shcwing that thr is a total function f, B it f 5 Kt such that for all T singlvalud B BC ~(f) and B total. +. BE DESCRIPTION OF THE GAME Considr a gam whr th playr dfins a st A and numrats th sts V 0, V 1,..., and whr th opponnt numrats th sts Th rquirmnts ar : 1): A is wll dfind, singl valud, and total R(O) : W(O) CA) singl valud and total + W(O) (A) A V(0) finit
17 R(2) : W(1) ('Ti) singl valud and total + W(1) (A) A V(1) finit R(3)  : A # W(1) 11 Th oppo~nt during his turn can put on numbr into on of his sts 0, 1.. Th playr during his turn can add numbrs to on of his sts 0, 1,.. Th playr is also allowd during his turn to rmov a finit nurnbr of mmbrs of A and rplac ach on with anothr numbr. W will allow th playr to chang th valu of CA(nl only finitly many tims for ach n. This is quivslnt to th playr S numrating a sris of initial functions, say <CA>s=O, 03 that convrg to A. If at th bginning of th gam A is rcursiv, th playr's stratgy is ffctiv, and th opponnt simultanously numrats all th r.. sts in an ffctiv mannr thn th Turing dgr of A will b lss than or qual to 0'. W will assum that at th start of th gam A = {<n, n> : n N). If all our changs to A that rmov a mmbr of th form <n, ml> also add a numbr of th form <n, m > 2 and occur blow som stag S(n) thn th rquirmnt R(1) will b satisfid. Throughout th gam th playr will b putting labls on rquirmnts. Ths ar a "bookkping" dvic which mak th stratgy mor concis. Th labls prsist until thy ar xplicitly rmovd. Th playr also has an infinit numbr of labls, dnotd 1labl, 2labl,..., with which to labl numbrs. If a numbr bcoms nlablld at som
18 stag s thn it is nlablld for all stags t, t 2 s AN EFFECTIVE WINNING STRATEGY At stag 0 all sts ar mpty xcpt for A, which is {<n, n> : n N}, and th playr dos nothing. Stratgy at stag s + 1: 1. If n (s + 1) = 2p thn for ach x ws (p) (A') th playr puts x into ~(p), and 1.2. if R(2p) is not lablld at th nd of stag s and for Som <Dl <n, m>> and <D*, <n, m*>> in ws (p) w hav m + m* and <x, y> (D U D*) fl A" + <x, y> is not <r, l>lablld for any r S 2p thn th playr taks th smallst such pair, say <El <j, k>> and <E*, <j, k*>>, and <2p, O>labls all mmbrs of E U E*, unlabls all lablld rquirmnts R(q), q 7 2pl labls R(2p), and for any <x, y> (E U E*) fi A' th playr rmovs <x, y> from A and puts th smallst <x, y'> such that <x, yl> j! E U E* and <x, yl> is not <r, O>lablld or <r, l>lablld for any r 5 2pr into A and <2p, l>labls <x, y> If l'r (s + 1) = 2p + 1, R(2p + 1) is not lablld at th nd 1 of stag SI and thr xists <n, m> f ws (p) such that
19 2.1. no numbr <n, m'> is <r, l>lablld for som r 5 2p + 1, and 2.2. <n, rn> is not <r, O>lablld for som r 5 2p + 1, thn th playr taks th smallst such numbr, say <j, k>, and 2.3. puts <j, k> into A, 2.4. <2p + 1, l>labls <jr k>r 2.5. rmovs and <2p + 11 O>labls all mmbrs of A of th form <jr k'> with kt # k, 2.6. unlabls all R(q), q > 2p + 1, and 2.7. labls R(2p + 1) A PROOF THAT THE STRATEGY IS A WINNING STRATEGY A rquirmnt R(p) lablld at th nd of stag s can b unlablld at stag s + 1 only if for som q < p, R(q) is not lablld at th nd of stag st but R(q) is lablld at th nd of stag s + 1. Thus for ach p thr xists an r(p) such that ithr R(p) is lablld at all stags 1 r(p) or R(p) is not lablld at any stag L r (p). Choos r 0, r 1,... such that r(0) 5 r (1) If <n, m(l)> is put into A at stag s (1) thn for som p (1), <n, m(l)> is <p(l), I>lablld at stag s (1). If <n, m(2)>, 2 # 1, is put into A at stag s(2) > s(l), thn <n, m(21> must b <p(2), 1> lablld at stag s(2) for som p(2) c p (1). Thrfor thr must b a stag s (k) and a numbr m(k) such that <n. m(k) > C At for all t E s (k). and consquntly A must b wlldfind. A is singlvalud and total sinc AS is singlvalud and total for vry s. It now rmains to b shown that R(qi is satisfid for vry
20 q L 0. Considr a rquirmnt of th typ R(2p), p 10, i.. W(p) 6) singlvalud and total + Vip) A W (p) (A) finit. Now suppos W (p) CX) is singlvalud and total. Lt L = Vr(2p) (p). Clarly L is finit. To show that R(2p) is satisfid it will suffic to show that V(p)  L C W (p) 6) and W (p) (x) c V(p). Considr any <n, m> that is put into s V(p) at a stag s + 1, s 1 r(2p). Thr must b a mmbr of W (p) of th form <Df <n, m>> whr D c AS. Now suppos, for proof by contradiction, that <n, m> p W (p) (A). As W (p) (A) is total thr must b a stag t + 1, with t > s and rl(t + 1) = 2p, such that t t for som D* and m* # n, <D*, <n, m*>> W (p) and D* C A. W claim that and hold for th pair <D, <n, m>> and <D*, <n, m*>> at stag t. Clarly m* f m, so holds. Now suppos <x, y> (D U D*) n At and <x. y> is <r, 1,lablld  t at stag t with r 9 2p. Now as D* c A' thn <x, y> D ll A.  Howvr D C AS. Hnc <x, y> must b put into A at a stag t',  t > t' > s > r(2p). This can occur at only two points in th playr's stratgy, or 2.3. Now if th playr puts <x, y> into A at stag t' on bhalf of , thn as t' > r(2p) w hav sl(tl) > 2p and <x, y> cannot b <r, l>lablld for any r 5 rl(tl), which contradicts our assumption about <x, y>. If th playr puts <x, y> into A at stag t' on bhalf of 2.3., thn by 2.1., nl(tl) c r 5 2p and ~(r~(t)) must bcom lablld at a stag t' > r (2p) P r IT^ (t) ) which is impossibl. Thrfor <D, <n, m>> and <D*, <n, m*>> satisfy and at stag t + 1. It follows that R(2p) is alrady lablld at stag t bcaus othrwis
21 R(2p) would bcom lablld at stag t + 1 contradicting t + 1 > r(2p). Lt u + 1 b th largst stag < t at which R(2p) bcoms lablld. Thsn thr must xist som last pair <E, < j, k>> and <E*, <j, k*>> in wu(~) that satisfy and at stag u + 1. Th playr <2p, O>labls vry mmbr of E U E* at stag u + 1 and rmovs vry mmbr of (E U E*) n AU from A at stag u Clarly thn <j, k>, <j, k*> W"~(~)(A~+'), and k # k*. If at som stag v + 1 > u + 1 a mmbr of E U E*, say <x, y>, is put into A thn, R(T (v + 1)) is lablld at stag v + 1 and as <x, y> is 1 <2p, O>lablld at stag u + 1 w hav rl(v + 1) c 2p. Thrfor r(ni(v + 1)) 2 v + 1 which contradicts v + 1 > r(2p). As no such stag v + 1 can xist w hav < j, k>, < j, k*> W (p) (A) and k # k*. This in turn contradicts our assumption that W(p) (Ti) is singlvalud. Hnc Now suppos <n, m> W (p) (A), thn for som s, <n, m> W (P (A for all t I s. Thr must b a stag u l s such that rl(u + 1) = 2pI u+ 1 hnc <n, m> v (p) c ~ (p) Clarly W (p) c V(p) and I(p) 6) h V(p) c L. Thrfor all th rquirmnts of th typ R(2p), p L 0 must b satisfid. Considr a rquirmnt of th typ R(2p + l), p 2 0, i.. t t As A is total and singlvalud, for any n thr is an m such that <n, m> A. Thus w may suppos without loss of gnrality that for all n thr is an m such that <n, m> W(p). Each tim a numbr bcoms <r, O>liblld or <r, l>lablld
22 th rquirmnt RCr) bcoms lablld. If r 5 2p + 1, thn R(r) cannot bcom lablld at any stag gratr than r(2p + 1). As only a finit st of numbrs bcom lablld at any stag, thr must b a n(p) such that if k > n(p) thn for all y, <k, y> is nvr <r, 0% lablld or <r, l>lablld for any r 5 2p + 1. Choos a <k, y> such that k > n(p) and <k, y> W(p). Thr must b a stag s + 1 > r(2p + 1) such that 'IT 1 (s + 1) = 2p + 1 and <k, y> W (p). Sinc k > n(p), 2.1. and 2.2 ar satisfid at stag s. Thus R(2p + 1) must b alrady lablld at stag s, as othrwis R(2p + 1) would bcom lablld at stag s + 1 contradicting s r(2p + 1). Lt t b th gratst stag < s + 1 at which R(2p + 1) bcoms lablld. Som <j, x> is put into A at stag t and <j, x> bcoms <2p + 1, l>lablld at stag t. Now if <j, x> j? A thn <j, x> must bcom <r, O>lablld at som stag t' > t whr r < 2p + 1. W would thn hav that t < t' < s + 1 and th labl on R(2p + 1) is rmovd at stag t', which contradicts our choic of S t. Thrfor t <j, x> A and as <j, x> W (p) c W(p) w hav W(p) fi A # pl. Thrfor # A CONCLUSIONS Considr a play of th gam in which th opponnt follows an ffctiv stratgy whrby 0 1,. is an numration of all th r.. sts and in which th playr follows his ffctiv winning stratgy,  thn all th sts V 0, 1 ,... will b r.. Now suppos B 5 A and th partial dgr of 3 is total. Thn thr must b a singlvalud total st B*, B* B. Thus for som j, B* = w ( j ) (%) and as
23 th playr's s2ratgy is a winning on, B* A V(j) is finit. Hnc B*E V(j) and 8 1 B* E B. Also x#w(j) for all j, whnc  A > 8. It follows that A > B, bcaus A is singlvalud and total. This complts th rsult claimd for this chaptr bcaus th partial dgr of A is total and nonzro, but thr is no nonzro partial  dgr, lss than or qual to th partial dgr of A, that is total. W also not that a thorm of Mdvdv is a corollary to th abov rsult. Corollary (Mdvdv) Thr is a non rcursiv such that for all f, f f rcursiv.
24 CHAPTER I11 MINIMAL PAIRS OF PARTIAL DEGREES 93.1 INTRODUCTION If r is a st partially ordrd by 5' with a last lmnt a, thn 8, y r ar said to b a minimal pair if B, y # a and V 6 I'(6 5'6 and 6 r'y t 6 =a). Cas [l] has shown that thr xists a minimal pair of partial dgrs.. Lachlan 131 and Yats [7] hav shown indpndntly that thr is a minimal pair of r.. dgrs. In this chaptr w will combin ths two rsults to show that thr ar two nonr.. but cor.. sts whos partial dgrs form a minimal pair. Th mthod usd is similar to th on in th papr hy Lachlan [3]. Whthr vry minimal pair of r,. dgrs forms a minimal pair of partial dgrs is an opn qustion DESCRIPTION OF THE GAM3 W considr a gam whr th playr numrats th sts A, B, VCO1, 1,..., and th opponnt numrats th sts W(0), W (11,..., U(O), U(1),... * Th rquirmnts of th gam ar: RC1) : W(O) (TI = U(0) (g) + 3i (~(i) A WCO) 61 is finit} R(2)t A# W(0)
25 53.3. BASIC IDEA BEHIND THE STRATEGY Bfor giving th actual stratgy and proof w outlin th main idas bhind th stratgy. Lt us first considr th problm of constructing two nonr.. but cor.. sts which satisfy th first rquirmnt, i.. At ach stag s of th gam w dfin Lvl (0, ST as th largst n < s such that W also considr th squnc s c s <..., dfind by 0 1 s = ~ ( C t l j < il (s. < x and Lvl CO, s ) i I j < Lvl (0, xlll. This squnc is ithr finit or infinit, If it is finit thn R(O1 is clarly satisfid. W rsrv a st V(0) for this rquirmnt. For ach i, w s  S i i numrat ach numbr of W (0)(A ) that 'is < ~vl (0, si) into V (0). At stag s w may wish to put som mmbr x into A to nsur  A # for som p. Putting x into A may rmov som k <  Lvl (0, s. ), with s. < s, from ws (0) CAS1. Dspit this w may rmoval from B would rmov k from us (0) (BSl. If Lvl CO, numrat x in A as long as w thn kp out of B all numbrs whos tl < k
26 2 0 for all t > s thn RCQL will b satkfid and th rstriction on B prsists from stag s through all subsqunt stags. If for som stag t > s, Lvl (0, t) L k thn th rstriction on B may b liftd at stag t and at stag t w may put any numbr y in A or B providd that an appropriat rstriction is placd on B or A. rspctivly. Th ffct of this procdur, if th squnc s < s < is infinit, is to nsur for all k and all t > s w hav k ut (0)(B~) so w only allow k  t t Howvr w want lim W (0) (A I[k] to b dfind, w  t t to lav W (0) (A ) whn w plac numbrs into A to nsur that A # W(q) whr q < k. This rstriction also nsurs that'th st of numbrs rstrictd from ntry to A on bhalf of k is finit. A similar mthod nsurs th sam for B. Now if th squnc so < sl<... is finit thn lt u b th maximum stag in th squnc. For ach k < Lvl CO, ul w hav sn that th st of all numbrs rstrictd from ntring A or B on bhalf of k is finit, This allows th playr to start playing th othr rquirmnts starting at stag u + 1 with only a finit intrfrnc from R(0). If th squnc is infinit thn any So< "' rstrictions on ntry into A or B will b vntually liftd and othr rquirmnts may wait for this rlaxation bfor thy act. At any stag s w assum that th squnc will b infinit, and i play a stratgy on ths stags that conforms with this assumption. On all othr stags w assum that w hav passd th last mmbr of th squnc and play accordingly, two stratgis bing playd. Thus for th rquirmnt R(1) w hav For ach stratgy w carry a distinct st from th squnc V 0, V 1,.... Whn w ar assuming that th
27 squnc SO c S1 <.." is infinit, that is whn w ar at a stag s j ' thn w play as if th only stags that hav occurrd ar S1< "' < s This policy is xtndd to all rquirmnts. Thus for a rquirmnt j ' R(p) thr ar p rquirmnts abov it and hnc thr ar 2' pos sibl assumptions. For ach of ths assumptions w rquir a distinct st from th squnc V (0), V (1) A WINNING STRATEGY FOR THE GAME W first giv som dfinitions usd in th stratgy, ~ inition f , Lvl (p, s) is th largst n < s such that 2, A pstat is a subst of {i : i < p}.3. Th pstats ar linarly ordrd by T* dfind by E 5* F +t W j ( j F  E + zi(i E  F and i < j)),4, W dfin E(s) C {i : i 5 s} and finit squncs <NU, s) I ii s> and <u(i, j, s] I izs, j  <N(i, s)> by th following stipulations: 4 ~(0, s) = s, u(0, j, s) =. j for j I NCO, s).4.2. k ECs) ++Lvl (kt s) 7 max Lvl k, uck, i, s)) i<n Ck, s).4.3. k j? ECs) and j i NCk, s).+. N(k + 1, s) = N(k, sl and
28 u(k + 1, j, S) < s.+, u(k + 1, j + 1, S) = u(k, y, S) whr y = pz(lv1 (k, u(k, z, s)) > Lvl (k, u(k + 1, j, s))).4.,6.. k ECs) and u(k+ 1, j, sl = s.+. N(k+ 1, s) = j,.5. W dfin Etp, s) as E(s) fl (i : i < p)..6. R(3p + 1) is frozn at stag s if for som m, m c wstp) n A"..7. R(3p + 2) is frozn at stag s if for som m, S m c us(p) n B. In th stratgy w will mak us of th auxiliary function L(p, EI s). It is assumd that if L(p, E, s + 1) is not st xplicitly it is qual to L(p, EI s). At stag 0 all sts ar mpty and L(p, El 0) = 0. At stag 0 th playr dos nothing. Stratgy at stag s + 1: 1, For ach p E(s) if thr xist n and D such that 1.4 n 5 Lvl Cp, s) thn th playr chooss th smallst such n and 1.5 puts n into V( p, E(p, s) ) and 1.6 sts L(p, E(p, s), s + 11 to th maximum of L(p, ECp, sl, s) and n, Th playr chooss th last p < s, if any, such that EITHER
29 2, RC3p + 11 is not frozn at stag s and thr xists m such 23 that 2,2. if q 5 p and for som qstat E w hav E:* ECq, s)  and furthr thr is a D such that <D, k> ws Cq) and D c AS for som k 5 L(q, E, s + 11, thn if D* is th last such D (not that at this point in th stratgy LCq, E, s + 1) will hav bn st, ithr xplicitly by part 1, or implicitly by our convntionl if q 5 p and k < p and if thr is a D such that D C AS and <D, k> C ws (q) thn if D* is th last such D w hav that 3, R(3p + 21 is not frozn at stag s and thr xists m such 3.2. if q Lp and for som qstat E w hav E T* ECq, s) and  furthr thr is a D such that <D, k> uscq1 and D c BS for som k 5 L(q, E, s + 11, thn if D* is th last such D Cnot that at this point in th stratgy LCq, E, s + 1) will hav bn st, ithr xplicitly by part 1, or implicitly by our convntion]  S if q 5 p and k < p and if thr is a D such that D c B and <D, U uscq) thn if D* is th last such D w hav
30 24 If p xists satisfying 2. thn th playr chooss th last m that satisfis 2, for p and puts <m, p, 1> into A. If p xists but not satisfying 2. thn th playr chooss th last m that satisfis 3. for p and puts <m, p, 2> into B PROOF THAT THE STPATEGY IS A WINNING ONE For ach p l 0 w dfin th final pstat, dnotd by F P ' as th smallst pstat undr S*, such that th squnc of all stags s with s) = F P I say is infinit, Now it is clar from th dfinition of ECs] that F C P Fp+l W dfin th final gam stat, dnotd by Fa as 8 F. p=o P Considr a rquirmnt of th typ R(3p + i], i = 1, 2, W lt rc3p + i) b th last stag s such that RC3p + i) is frozn at stag s, if no such stag xists w lt rc3p + il b 0. Lmma If for som p, p L 0 w hav Wn 3s W t C t > s +Lvl (p, t] L n) '. Proof: Suppos p satisfis th hypothsis but p R Fa. Clarly thn Considr th infinit squnc For any. k, k > 0, thr must b a j > k such that
31 2 5 Lvl (p, + 1, j)) > k. Thrfor thr is an inflnit subsqunc suchthat E(p+ 1, t(i)) = F U (PI<* Fp+l, p+l i=0, 1, 2,.... This contradicts our dfinition of F and our rsult is shown, p+l Lmma If for som p 2 0, p = 1 ) thn Proof t Considr p > 0, and suppos 6) = (c]. NOW for a proof by contradiction assum that thr is an n such that for all s thr is a stag t' (s) > s such that Lvl (p, t' (s) ) < n. Clarly thr must b an infinit squnc <t(i) >m such that for som m i=o for all i.  i = 1, 2. Now ithr m C Wt(j'(pl rnt(j)) or m t tcj) Choos j sufficintly larg so that tcj1 > r(3q + il, q 5 m, Suppos th formr, thn for som st D, D c A tcj1 and <D, C PIt (j) (p). Thrfor thr must b som st D* such that D* is th  last st such that for som s 2 t(j) D* 'c' A' and <D*, m> ws (p). If D* % thn at som last stag r > s a numbr of th form <n, q', 1> is put into A, Hnc R(3qt + 11 bcoms frozn at stag r and as r> s 2 t(j1 > rc3q + 11, q Lm, i = 1, 2, w must hav qe >m, Hawvr n q, 1 must satisfy 2.. of th playr's stratgy at stag r and hnc m > qe by part 2,3, jf th playr rs stratgy,
32 "rh.is is a cont.:adiction so w must hav D* C r, Now as <D*, m> C ws (p) C w hav m 6 1. Howvr = Hnc m 6) n UCpl 6 1 and for som k w t (k) t (k) ), ut 'k' (B hav m E K (A t'k)l. similarly for m c u t(j) (Bt(j)). OD This contradicts our choic of th squnc <t(i)>io  and stablishs our lmma. Lmma For vry p I 0 thr is a stag z(pl such that Prooft As thr ar only finitly many pstats it suffics to shw that for ach pstat E, E <* F that thr is a stag tce) such P ' that This is immdiat from th dfinition of F P. For conomy of notation w not that thr is a stag zcp1 such that > and z(p) > rc3q + il, q 5 p, i = 1, 2. Lmma 3,5.4 For all q l 0 thr is a bcql suchthat LCq, E, sl 5 bcql for all stags s and all qstats E such that E c* F' = Fq U {q}. Q Proof! Considr a qstat E with E <* F' It suffics to show that q' for som n, L(q, E, s) < n for all s. Suppos for proof by contra diction that for all n thr is a stag s such that LCq, E, sl > n. Hnc thr muat b an infinit squnc of stags say
33 such that L(q, E, tti) +11 >L(q, E, tciil, i 0 1, 2,.. Clarly thn, as th auxiliary function L can b incrasd only on bhalf of part 1, of th playr's stratgy, ECq, t(i)l = E, i = 0, 1, 2,..., Thrfor by our dfinition of F F' I* E, This contradicts our 9' 9 assumption that E <* F' q and stablishs our Lmma. Lmma If q < p and q Fa thn for all k > 0. a 03 Proofr As kl>k=o is a subsqunc of <vcq, kl>ko  it suffics to show that for all k >_ 0 1. LCq, Fq, vcq, k1 + ll 5 Lvl (4, vcq, kl). Now L Cq, Fq, v(q, 0) + 1) ithr is qual to LCq, Pq, vcq, 01) which has valu 0 or is st to th maximum of L Cq, F vcq, 0) l and 4' som n 9 Lvl (q, v(q, 01). Clarly thn LCq, Fq, vcq, ) 9 Lvl (q, v (q, 0) l. Now suppos 1. is tru for k = j. Now L(q, F, vcq, j ) ithr quals L(q, F, v(q, j st to thmaximumof L(q, F v(q, j and som n 5 9' Lvl Cq, vcq, j + 121, To show that 1, holds for k = j + 1 it clarly suffics to show that or is
34 LCq, Fq, vcq, jl + 1) whnc by our induction hsothsis This complts our proof by induction of th rquird rsult. W now show that rquirmnts of th typ RC3p) ar satisfid. Now suppos n W(pl (A?. W can choos s sufficintly larg so tfmt if kc n and k f W(pl(ji) thn t 2 s + k W (p)u 1 n vt and for  S som D C A, <Dl n> W (p). Now by Lmma and Lmma 3.5.1w hav p Fm. AS s was chosn arbitrarily high w can assum that s = v (p + 1, il for som if hnc p E (s). Also by Lmma w can choos s so that n 5 Lvl (p, s). Thrfor 1.1, 1.3. and 1,4., t t of th playr's stratgy hold for n and D at stag s + 1. Thrfor S ithr 1.2. dos not hold for n and n V, or 1.2. holds for n s+l and n V. Clarly n V, which complts our proof by induction Suppos n V  L, Clarly thr is a stag s(1) such that n ws (l) (A' and D c A ). Lt D (1) b th smallst szt D such that <D, n> W s (1) 00. In th following construciion of th squncs <D (i)>il, 
35 <s (i)>il , <w (i)> w will assum that if w(i) = 1 and s(i+l) I s > i= s (i) thn D (i) is th last st D with <Dl n> ws (AS) and D c AS, similarly for w(i)  2 by rplacing A with B. Clarly this assumption is quivalnt to dropping an argumnt s from ach D(i). w5at w want to show, W call this a trminating cas, Cas 2, DUI 9 A. Lt us suppos th lattr cas holds. Now a mmbr of DCll can b put into A only on bhalf of part 2. of th playr's stratgy. Lt s (21 b th smallst stag > s (1) such that a mmbr of D (1), say <m (2), q (2), w (2) > whr w(2) = 1, is placd into A at tkis stag. Lt t (2) b th stag immdiatly prcding s (2). Now s (2) > s (1) > z(p) as n W s (1) )  L, hnc as R(3q(2) + 1) bcoms frozn at stag s(2) w hav q(2) > p. Now by Lmma 3.5.3, F 5* E(p, tc21). P Also Hnc by part 2.2. of th playr's stratgy, That is ithr Now if 1. is tru w can rpat th abov argumnt. Eithr this rptition will continu without nd or it will trminat finitly, i., ithr w can gt an infinit squnc
36 s(1) < s(2) < sc3' <... or a finit squnc s(l) < sc21 <... < scj) such that n C U (j ' (8' 'j) ). Suppos th lattr (this cas includs 2., as th spcial cas j = 21, thn thr must b som smallst st D(j), D(j) C ElS('), such that <D(j), n> C Us(') (p). Now as bfor w hav two cass. Cas 1. D (j) C B, in which cas n = ~(pl 6) which is wkat'w want to show. This is also a trminating cas. Cas 2. DCj) % Lt us suppos cas 2. holds, thn thr must b a smallst stag s(j + l), s(j + 1) > s(j), such that ~(j) $ B (j+ll. som nunhr <m(j + l), qcj +I), w(j +I)>, w(j + 1) = 2, must bput into B at stag s(j + 1). Now s(j + 1) > s(1) > z(p), so q(j + 1) > p., Lt t (j + 1) b th stag immdiatly prcding s (j + 1). By Lmma 3,5.,3, F 5* t (.j + 1) 1. Thrfor by part 3.2. of th P playr's stratgy which w can apply as n 5 Fp, ~(11) 5 L4, Fpf t(j + 1)) w hav That is ithr
37 Now if 1. holds w can rpat this argumnt and continu th squnc. As bfor w gt ithr an infinit squnc or a finit squnc such that n ws 'k) (A' 'k) 1. In th lattr cas (2. is a spcial subcas with k = j + 1) w can apply th argumnt usd on s(l1 to continu th squnc. Thus ithr on of th trminating cass will hold and n W(p1 (A1 or w gt an infinit squnc with th associatd squncs Considr i 2 1. Now if w(i + 1) = 1, thn DfiI is th smallst  st such that DCi) C A S(i' and <DCi1, n> W ~ ~ I I. By part 2.3. of  th playr ts stratgy q(i + 1) 5 n, Now if wci + 1) = 2, thn DCiI is th smallst st such that D(i) C B s(i' and <D(il1n>CU s (il (pl. By part 3.3. of th playr's stratgy q(i + 1) 5 n. Now r (3q(il + w(i) ) quals scil for all i L 2, hnc as at most on rquirmnt bcoms
38 froznat any stag, 3qCil + wci1 # 3qCjl +wcj1 for all if j, 3 2 i, 9 L 2, This is a contradiction, hnc on of our trminating cass must hold. W hav shown that W(p) (x) c V and V  L c W(p) (3, thrfor (A) A V c L, and all th rquirmnts RC3pl, p 2 0, ar satisfid. Now lt us considr a rquirmnt of th typ RC3p + 11, i,. If R(3p + 1) is frozn at som stag t thn thr is an x such t that x A t r) W (p), and A f W (p). Now suppos for proof by contradiction that R(3p + 1) is nvr frozn at any stag, Clarly thn for all m, <m, p, 1>,f A. If at th gamts conclusion thr is an m such that <m, p, 1>,f W(p) thn <m, p, 1>  and  A f W(p). W ar lft with th cas whn for all m, <m, p, 1> H will first show that thr is a numbr and a function dl such that if m > dl and s > hl(m) thn <n, p, 1> will satisfy hl part 2.1, of th playr's stratgy at stag s, Lt dl = 0 and dfin hl(m) to b th smallst stag s such S <m, p, 1> W (pl whnc 2.1. is satisfid. W will now show that thr ar numbrs d3' 3 such that if m>d and s > thn <m, p, 1> satisfis part 2.3. of th 3 3 playr's stratgy at stag s.
39 that Considr any q 5 p and k < p. Suppos thr is a st D such  t for som stag t, t 2 z(k), D c At and <D, k> W (q). Lt 3 3 D (k, q) b th smallst such st D, and lt t(k, q) b th smallst stag such that D (k, q) c A t(krq' and <D&, q), k> C w t(krq) (q). If D&, ql $x thn som <j, qt, l>,<j, qt, 1> D(k, q), must b put into A at som stag t > t (k, q). Now as <j, q', 1> satisfis part 2.3. of th playr's stratgy at stag t w hav qt < k. Now R(3q1 + 1) bcoms frozn at stag t, and w hav t > t (k, q) > z (kl > r(3qr + 1). Thus w must hav DCk, q) c A,   Now if thr is no such t st D such that for som t 2 z(k), D c At and <D, k> W Cq), thn'w lt D(k, q) = and t(k, ql = zck1. W lt d3 b th maximum of all m such that <m, p, 1> DCk, ql for som k < p, q 5 p, and w lt b th maximum of all th stags t Ck, q), 3 kc p and qip, Now suppos m > d3 and s > 3 ' Considr any q 5 p and k< p such that thr is a D, D c AS and <D, k> tis Cq). Now s > tck, q) hnc if D* is th smallst such D, D* = D C ~, ql. Clarly thn, by our choic of d <m, p, 1> j? D*, Thrfor 3' dsird proprty. d3 and 3 hav th W will now dmonstrat that thr ar numbrs d2 and 2 that if m > d2 and for som such i, v(pmii) > 2. P' = P f 11 thn <m. PI 1> satisfis part 2.3. of th playr's stratgy at stag V(P ' ~i) + 1. Considr any q 5 p and k I b(q1. w can construct D(k, q) and t(k, ql xactly as abov. Lt d2 b th maximum of all m such that <m, p, 1> D (k, q), k I b (q), q 5 p, and lt b th maximum of 2
40 Lt m > d and lt i b sufficintly larg so that v (pf, i) > 2 Assum for proof by contradiction that part 2.2. of th playr's 2  stratgy dos not hold for <m, p, 1> at stag v(pl,i) + 1. Hnc thr must b a q 5 p, a qstat E with E C* ~(p'ri)) = F P and a k I LCq, E, v(p', i) + 1), such that if D* is th smallst st D* and k p U v(p., i) (Bv(pfIi) 1. Now if E c* F U{q), thn by Lmma q k 5 b(q), and as ~(~l,i) > t(k, q), w must hav D* = DCk, q). Hnc <m, p, 1> p D* as m > d 2 ' This lavs th cas E = F and q E Fa. By 9 Lmma 3.5.5, L(q, E, v(p8,i) + 1) 5 Lvl (q, v(pf,i)), hnc k 5 Lvl (q, v(pf I ill. Now as w hav u vcp'ti) and 2 v(pfii) CB ) which contradicts our assumption about k.. Hnc 2 hav th dsird proprty. Now choos m such that m is gratr than any of dl, d2 and d3, and choos an i such that i) > z (pl, v 4', i] > h(m) and v(p' I il is gratr than ithr or Now <m, p, 1> must 2 3 ' satisfy 2.1,, 2.2., and 2.3. at stag vcpl,il + 1. By our assumption that R(3p + 1) is nvr frozn thr must b a q c p such that ithr R(3q + 1) or R(3q + 2) hcoms frozn at stag vcpf, il + 1 This contradicts our choic of v(pl,i) > zcp). Hnc w hav shown that all rquirmnts of th form R(3p + I), p ' 0 ~ ar satisfid. W can show that th rquirmnts of th form R(.3p + 21, p L 0 ar satisfid by a similar argumnt. 3,6 CONCLUSIONS Considr a play of th gam whr th opponnt follows an ffctiv
41 stratgy so te.at { i, U i 1 : i C N} fs th st of all pairs of r.. sts. Thrfor as th playr's stratgy is ffctiv A, B and all th sts V (0), V (1),... will b r., Furthrmor A > 9 and B > pl as A # ~(i) and B # U(i), i = 0 1, 2,.. Vow if C 5 A and  C 5 B thn for som pair of r.. sts W i, U i 1, w will hav C = WCi) 61 = UCi) (El. As RC3i) is satisfid, thr is a st vc~) such that C A Vck) is finit, and as V(k) is r.. C E PI. Thus th   partial dgrs of A and B form a minimal pair. Now A and B ar cor.. hnc thir partial dgrs ar total. Thrfor th Turing   dgrs of A and B form a minimal pair of r.. dgrs.
42 CHAPTER IV THERE ARE NO MINIMAL PARTIAL DEGREES dgrs. In this chaptr w will show that thr ar no minimal partial Dfinition l. A partial dgr  a is minimal if  0 <  a and  b < a+  b = A st B is minimal if its partial dgr is minimal. Mntion of this problm occurs in Rogrs [5, p.2821 and Cas [l]. Th proof will b in two parts. First w will show that a minimal dgr must b 50' and thn w shall show that thr is an r.. st V such  that for all BI jl c B 5 K, w hav pl c V (B) c B TWO GAMES stratgy. In this sction w dscrib two gams that hav an ffctiv winning W considr th following gam whr th playr numrats V and th opponnt numrats ki 0 I W (1) I.... W will assum that th partial dgr of th opponnt's stratgy is lss than or qual to th partial dgr of a givn st A. If th playr's strztgy is ffctiv thn any st W(i) that th opponnt numrats will b such that
43 3 7 W(i) I A. Also if th playr's stratgy is ffctiv w will hav V 5 A. Th rquirmnts of th gam ar: : for ach n thr xists an m L 1 such that for all j Rl <n, j> V(X) + j 5 m for all X, X > A O < j<m+<n, j> VCX) forall X, X A > n X * <n, m> VCX) for all X, X > A Now if w hav an ffctiv winning stratgy for this gam it will work whn th opponnt simultanously ffctivly numrats all th r.. sts. In th following discussion w will assum that this is th cas. W can tak A to b jd and hnc th rquirmnts will b satisfid by any non r.. st X. Also th st V th playr numrats will b r.. Now suppos B is minimal. W must hav B r jdr hnc  B is non r.. and B satisfis all th rquirmnts of th abov gam. By dfinition fl 5 V(B) S B, hnc ithr V(B) Z B or V(B)!if.
44 V (B) f B wo~ld imply that B 5 V (B), hnc for som r.. st W (p) w would hav B = W ( p)~ 3 8 (B). This is impossibl as B satisfis R(p). Thrfor, for vry minimal B w must hav V(B) r.., that is V(B) = W(q) for som q. Now if 2 is th partial dgr of th join of th charactristic function of V and th charactristic function of W (q), thn w hav by Rl that th partial dgr of C 5 b. Having B  an oracl for V would allow us to comput for ach n th m whos xistnc is assurd by. w can thn consult an oracl for W(q) Rl to s whthr <n, m> is in ~(q) or not. In th formr cas n C B, in th lattr n j? B. Now clarly b is a total dgr and  b 5  O'r hnc w hav th following: Thorm If X is minimal thn X has Tdgr 50'. As V (B) = W (q), an numration of V and W (q) will giv us a squnc of initial functions {cs lm which will convrg to. For B s=o c~ givn partial numrations of V and w (q), say vs and ws (q), w s tak CB(n) to b 1, if thr is <n, j> C ws (q) such that S <n, k> W (q) + k 5 j and k < j * <n, k>> vs, and 0 othrwis. Now rcall that V satisfis Rl and thus for ach n thr xists an S m L 1 such that <n, j> C WS(q) t j 5 m and j < m t <n, j> W (q) and <n, m> C V(B) * n B. For any n ithr <n, m> will appar t in ' W (q) for som s > n in which cas t 2 s t CB (n) = 1, or thr will b an s > n such that t l s +cr(n) = 0. As W(q) = V(B) and B <n, m> V(B) ++ n B ths final valus will agr with of ths initial functions S CB CB. Now ach is finit and hnc has a canonical indx, thus w can considr th squnc {cslrn as a function from N to N, B s=o
45 say g. As gam on is playd w ar ssntially gtting an nmration of countably many such functions, and if a minimal B xistd at last on of ths functions would hav to convrg to B in th mannr dscribd abov. To show that a non r.. st B is not minimal it is sufficint to find an ffctiv procdur that, applid to th abov function g numratd by gam on, will numrat a st V such that B<, V(B) c B. Lt us considr th following gam whr th playr numrats V and th opponnt numrats W 0, W (1),.... As bfor w assum that th opponnt's stratgy has partial dgr lss than or qual to th partial dgr of a givn st A. W will also assum that at ach stag of th gam w hav an initial function ps such that CB B = 1% CB, S B > A and th function that ncods this squnc of initial fanctions has partial dgr lss than or qual to th partial dgr of A. S Now suppos that w hav an ffctiv winning stratgy for th abov gam. Applying that stratgy whn th opponnt sim~ltanously ffctivly numrats all th r.. sts such that WC2p) = W(2p + 1) and A = B
46 w s that thr xists an r.. st V such that all th rquirmnts ar satisfid. By dfinition  5 V(B). If % V(B) thn for som r.. st W(2r) w hav V(B) = W (2r), which would imply that B dos not satisfy R(2r). If V(B) B thn for som r.. st W(2r + 1) w hav B = W (2r + 1) V (B) and B dos not satisfy R(2r + 1). Both of ths assumptions contradict th xistnc of an ffctiv winning 40 stratgy. Hnc if w can find an ffctiv winning stratgy for both gams w hav: Thorm For vry B > 9 thr is an r.. st V such that pl c V(B) < B. W notic that th V dpnds upon th choic of B and hnc th rsult is not uniform. Now suppos B > C and th partial dgr of C is total (w may thn rgard C as bing th graph of a charactristic function). W now rxamin th gams and this tim in th first gam lt th opponnt play so that his stratgy has partial dgr lss than or qual to th partial dgr of C, and N(i) = U (i) (C) i = 0, 1, 2,... whr U 0, U 1,... ar all th r.. sts. W also tak A t o b C. Hnc as our winning stratgy is ffctiv and C is total V 5 C, that is for som r.. st UO, V = UO(C). This is not ncssarily th cas if th partial dgr of C, say  c, is not total. As 5 is total w can assum that th ordr of numration oi {<n, j> : n C W(j) (C)) is fixd. Bcaus givn any numration of C thr is an ffctiv procdur that will output a fixd numration of C. Hnc as our gam taks a fixd numration of {<n, j> : n C W(j) (C)) and outputs an numration of a st V, w
47 can combin ths two ffctiv procdurs and gt an ffctiv procdur such that for any numration of C it will oiltput an numration of V. Howvr, if  c is nontotal w cannot ffctivly xtract a fixd numration of C from an arbitrary numration of C, s Cas [l], and so w cannot conclud that V S C Now V(B) = UO (C) (B) but B > C C = U1 (B). Thrfor V (B) = UO (U1 (B) ) unlss 2 is total. hnc for som r.. st ul' (B) thus thr is an r.. st U2 such that V(B) = U (B) 5.B. Fina ~lly thr must b som r.. st 2 U3 such that U (B) = V(B) join C. Now suppos thr is no st strictly 3 btwn C and B. If U3(B) E B, thn for som r.. st W U (B) = B and so W1(V(B) join C) = B. Thrfor for som q, 1 3 W(q)V(B) = B. W1 ' This would contradict th fact that an ffctiv winning stratgy xists for gam on. Hnc U3 (B) E C, that is U (B) = W(q) 3 for som q. Mow U (B) = V(B) join C, hnc w can xtract a squnc 3 of initial functions as bfor, th only diffrnc bing that this tim th function that ncods thm will hav a partial dgr that is lss than or qual to th partial dgr of C. Hnc w can apply th scond gam to this squnc whr th opponnt's stratgy is 5 C and whr W(2i + 1) = W(2i) = U(i)(C), i = 0 1, 2,.. Again w tak A to b C. Now considr v(b). Clarly  V 5 C, hnc as in th analysis of gam on, thr is an r.. st U* such that u*(b) = V(B) jcin C, and C 5 U*(B) 5 B. If u*(b) B thn for som r.. st W*, W*U* (B) = 5. Howvr, W*U* (B) = W* (V(B) join C) thus as B > C thr is a q such that W*U* (B) = W (2q + 1) V (B). hr for B = W*U* (B) = W(2q + l)v(e) which contradicts R(2q + 1).
48  42 Th only othr altrnativ is U*CB) C, that is for som r.. st Wf, W(C) = U*(B) = V(B) join C. Thus thr must b an r.. st w** such that w** (C) = V(B) but W** (C) = W(2q) for som q and hnc W (2q) = V (B). This is impossibl by R (2q) hnc our assumption that thr is no st strictly btwn B and C must b fals. W hav Thorm If  a and  b ar partial dgrs with 5 total and  a c b thn thr is a partial dgr  c such that ac  c c  b STRATEGIES FOR THE GAMES W now prsnt an ffctiv stratgy for th first gam and prov that it is a winning on. Th plziyr will construct a function f(n, s) at ach stag s. It is intndd that th valu of f(n, s) should rmain fixd for all sufficintly larg stags. This final valu of f(n, S) will b th m rquird for n by Rl. If, during th construction th playr dos not xplicitly st th valu of f(n, s + 1) it is assumd to b f(n, s). At stag 0, all sts ar mpty and f (n, 0) = 1 for all n. A suprscript s on a st bing numratd will indicat that st as it appars aftr s stags of th gam. Th basic ida bhind th stratgy is that ach numbr n is issud a flag <n, 1> by putting <{n), <n, 1>> into V. (At stag s th flag of n will b <n, f(n, s)>). W call <n, 1> a flag sinc th prsnc of <n, 1> in V(X) would indicat th prsnc of n in X. To satisfy RO w nsur that th only solution of X = w (O)V(X) is 0 ( W(O)V)~(~), whr (w(o)v)~ is W(O)V and (W(O)V)"+~ is n=o W(O)V( (W(0)V)n). For th first instruction of th form <El O> that is placd in W(0) w put <g, <m, j>> into V for any <m, j> C E.
49 indicat th prsnc of m in XI SO w issu a nw flag <m, j + 1> by putting <{m}, <m, j + 1>> into V. For th first instruction of th form <E, 1> in W(0) w procd as w would for 0 but w do not disturb any flags of 0. ~hus 1 C x * 1 C W(O)V(~I) il (w(o)v)'(%). ~t n is clar that w can nsur that n C jljo (W (O)V(O) ) j (j3) * n X. Notic also that this stratgy rquirs that n has at most n + 1 flags, as w only issu a nw flag for n to handl a numbr lss than or qual to n. This will satisfy R(0). For R(k), k > 0, th mthod is th sam xcpt that w nvr rpal a flag of any n c k. This mans that any solution of X = W (k)v(x) hs.s th form 0! w(~)v)~(f) whr n=o is a subst of 0 1,.. k  1 and hnc will b 5 A. F Playr's stratgy at stag s + 1: 1. For vry mmbr <Dl r> of ws (IT (s) ) and for v 1 <FoI...! F > with F c {x : x c IT~(s)} such that r i <rl (s) 1.1. D c?(n) 1.2. <m, k> C D and r (s) 5 m < r t <m, k> vs(j3) or 1 Fm>lablld 1.3. r is not <IT1 (s), Fr>lablld 1.4. Fr ={n< IT~(S) : <n, j> C D  vs(j3) for som j} U /I m : IT 1 (s) 5 m < r and <m, k> C D  vs (PI) for som k) th playr <IT1(~)l Fr>labls r and for ach <nl j> D such that n E IT~(S) and n 2 r th playr 1.5. puts <n, j>> and <In), <n, j + l>> into V
50 Ul procds through th mmbrs <D. r> of L? (s) ) and th substs Fof..., Fr of In : n < li (s) 1 in numrical ordr with rspct to th 1 indx <Fo, F2,..., F~ <D, r>>.) 2. Th playr puts { n s, n s, 1 into V and sts f(nl(s). s + 1) = 1 if f(nl(s), s) = 0. Now considr any XI X >. A. W show that R is satisfid. 1 Considr any n N. Clarly by our construction if f(n, s + 1) > f (n, S) L 1 thn n (s) 5 n and thr is an r 5 n that bcoms 1 <nl (s), E>lablld at stag s + 1 for som E c {x : x < ITl (s) 1. Howvr a numbr may bcom <nl(s), E>lablld at most onc. Hnc for all t, f (n, t) 5 2" (n + 1) + 1. Thus thr must b a smallst such that s 1 t t f(n, s) = f(n, tn). W dnot tn n f (n). By induction on s : f (n, tn) by 1. 3t(t < s and?r (t) = s) t f(n, s) L <n, j> vs(x) t j 5 f (n, S) Thrfor w can conclud: Thus Rl is satisfid for all X. Now suppos for som X > A and som W(q) w hav W (q)v(x) = X W dfin U as follows: