Accessible Independence Results

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ACCESSIBLE INDEPENDENCE RESULTS FORPEANO ARITHMETICLAURIE KIRBY ANDJE FF PARIS

Recently some interesting first-order statements independent of Peano Arithmetic(P) have been found. Here we present perhaps the first which is, in an informal sense,purely number-theoretic in character (as opposed to metamathematical orcombinatorial). The methods used to prove it, however, are combinatorial. We alsogive another independence result (unashamedly combinatorial in character) provedby the same methods.

The first result is an improvement of a theorem of Goodstein [2]. Let mand nbenatural numbers, n > 1. We define the base n representation of m as follows:

First write m as the sum of powers of n. (For example, if m = 266, n = 2, write266 = 2 8+23+ 21.) Now write each exponent as the sum of powers of n. (Forexample, 266 = 223+ 22 + 1 + 2 1 .) Repeat with exponents of exponents and so on untilthe representation stabilizes. For example, 266 stabilizes at the representation2* +l + 22 + l+2 l.We now define the number Gn(m ) as follows. If m = 0 set Gn(m) = 0. Otherwiseset Gn(m ) to be the number produced by replacing every n in the base nrepresentation of m by n+1 and then subtracting 1. (For example,G 2(266) = 333+1+33+ 1 + 2 ) .No w define the Goo dste in sequence for m startin g at 2 by

m 0 = m, mx = G2{m 0), m2 = G^mJ, m3 = G^m2), ... .So, for example,

266O = 266 = 222+1+22 + 1+ 2X= 333+1+33 + 1+2 ~ 1O38

2662 = 444+1+44 + 1 + l ~ 106162663 = 5s5+1+55 + 1 ~ 1010-000.

Similarly we can define the G ood stein sequence for m startin g at n for any n > 1.THEOREM 1. (i)(G oo dst ein [2 ]) Vm 3/cmk = 0. M ore generally for any m, n > 1the Goodsteinsequencefor m starting at n eventually hits zero.

(ii) Vm3k mk = 0(formalized in the language of first order arithmetic) is not provablein P.Received 1 Fe bru ary , 1982.The first-named author was supported by an SRC Research Fellowship.

Bull. London Math. Soc, 14 (1982), 285 -29 3.


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286 LAURIE KIRBY AND JEFF PARISSo, belying its early form, the sequence mk eventually hits zero. However despitethis fact being expressible in first order arithmetic we cannot give a proof of it in

Peano Arithmetic P. As we shall see later the reason for this is the immense time ittakes for the sequence mk to reach zero. (For example, the sequence 4k first reacheszero when k = 3 x24 0 2 - 6 5 3- 2 - 3 , which is of the order of io121-210-700.)Before pro ving Th eorem 1 we state our second result.

A hydra is a finite tree, which may be considered as a finite collection of straightline segments, each joining two nodes, such that every node is connected by a uniquepath of segments to a fixed node called the root. For example:

top node

rootA top node of a hydra is one which is a node of only one segment, and is not the

root. A head of the hydra is a top node together with its attached segment.A battle between Hercules and a given hydra proceeds as follows: at stage n(n ^ 1), Hercu les chops off one head from the hydra. The hydra then grow s n new

heads in the following manner:Fr om the node tha t used to be attach ed to the head which was jus t choppe d off,traverse one segment towards the root until the next node is reached. From this nodesprout n replicas of tha t part of the hydra (after decap itation) which is abo ve thesegment just traversed, i.e., those nodes and segments from which, in order to reachthe root, this segment would have to be traversed. If the head just chopped off hadthe root as one of its nodes, no new head is grown.Thus the battle might for instance commence like this, assuming that at eachstage Hercules decides to chop off the head marked with an arrow:

after stage 2 after stage 3


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ACCESSIBLE INDEPEN DENC E RESULTS FOR PEANO ARITHMETIC 2 8 7Hercules wins if after some finite number of stages, nothing is left of the hydra but

its root. Astrategy is a function which determines for Hercules which head to chopoff at each stage of any b attle. It is no t h ard to find a reaso nab ly fast winning strategy i.e. a strategy which ensures that Hercu les wins against any hydra ). M oresurprisingly, Hercules cannot help winning:

THEOREM 2. (i) Every strategy is a winning strategy.We can code hydras as numbers and thus talk about battles in the language offirst order arithmetic. We cannot formalise Theorem 2(i) as a statement of thislangua ge, as strategies are infinitary objects. How ever, we can if we restrict ourselvesto recursive strategies. In that case:THEOREM 2. (ii) The statement every recursive strategy is a winning strategy is

no tprovablefrom P .In proving the theorems we rely on work of Ketonen and Solovay [3] onordinals below e 0 , which in turn develops earlier work by the present authors,Harrington, Wainer, and others. Gentzen [1] showed that using transfinite inductionon ordinals below e0 one can prove the consistency of P, and the Ketonen-Solovaymachinery we use here can be viewed as illuminating in more detail the relationshipbetween e0 and P.(Note: Goodstein proved the following: if h :N -> N is a non-decreasing function,define an fi-Goodstein sequence bQ,b x,... by letting bi + l be the result of replacingevery h(i) in the base h(i) representation ofb{by h{ i +1), and subtracting 1. Then thestatem ent for every non-d ecreasin g h, every Ji-Goodstein sequence eventuallyreaches 0 is equivalent to transfinite inductio n below e0.)To prove T heo rem 1 we first define the base n representation more formally, atthe same time defining the ordinal on(m), in Cantor Normal Form, which results

from replacing every n in the base n representation of mby a> .Suppose m,ne N (the set of natural numbers), n > 1 andm = nkak +nklak_ l+ l 0

For x e iV or x = co, setkfm' (x) = a,-x/ ln x).

i = 0

(This definition is by induction on m, starting with / 0 > (x ) = 0.) The n form > 0, Gn(m ) = /m ' ( n + l ) - l and on(m ) = fm'n{(o). Set GB(0) = on(0) = 0. Finallyfor ne N w e define an o pe rat ion


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2 8 8 LAURIE KIRBY AND JEFF PARIS

LEMMA 3.(i) For m ^ 0, n > 1, if a = on +l{m) then on +1(m 1) = 1, (n) = on+ 1{G n{m)).

Proof, (i) Con sider the base n+1 representation of m: let

with 0 < a,- < w, and (since we may su ppo se m ^ 0) let j be minim al such tha tcij=/= 0. The result is clear if j = 0 so we may assume that; > 0 and that the resultholds for all 0 0. Then

and

B y ( i )and

o.+i((n + l)/ A l I ( + l ) - 1 -l) =


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ACCESSIBLE INDEPENDENCE RESULTS FOR PEANO ARITHMETIC 2 8 9

of ordinals. (E.g. in the example given before we would have

If we write (n l5n2,..., nk) for C-X(nfc) w e c a n write thissequence ason(b 0) = a , (n), ( n ,n + l ) , (n , n + 1 ,n+2), ... .

It is not hard to see that for any a < e0 and ne N,(n) < a if a > 0 .

Now we already have Theorem 1 (i) (following Go ods tein ). Fo r s uppo se m, nwere such that the G ood stein sequence for m startin g at n was always positive. Thenthe corresponding sequence of ordinals would be an infinite, strictly decreasingsequence which is an impossibility (by transfinite induction below e 0).In order to prove (ii) we introduce the Ketonen-Solovay machinery (see [3], [4]for detai ls). Fir st we define a no th er op erat io n {c-large is independent of P and is equivalentin P to C o n (P + Tt) where Tx is the set of the true I sentences.(iii) The functions gn(x) = least y x such that [x , y~\ iscon-largeare provably (in P)total recursive functions, and for any provably total recursive function f there existsne N such that f(x) < gn{x)for all sufficiently large x e N.


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2 9 LAURIE KIRBY AND JEFF PARIS

Writep -* aifand only iffor some j x , . . . , jk (;) Applying Lem ma 5 iv), (;) -*{}(;)Ifa = cos(P+1),5 limit, then byinductive hypothesis

7 WO )-


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ACCESSIBLE INDEPENDENCE RESULTS FOR PEANO ARITHMETIC 2 9 1

By Lemma 5(i), (oU) ->(Dm\ Thus

< a > ; ) = c o +coU)j + a)U)y{j) ->co5p +coU) byLemma 5(iv)

PROPOSITION 8. Let bo,b l,b 2,... be the Goodstein sequence for m starting at nand let k be minimal such that bk = 0. Then [n1,n+k~\ is on(m)-large.Proof. Consider the corresponding sequence ofordinals

on{m ) = on(b 0) = a

on+ k{b k) = on+ k{0) = 0 = (n,By Lemma6 and Proposition 7,

^ ^ < > , , ) = 0 .Hence [n1,n +/c] isa-large.Now toprove(ii) ofTheorem 1suppose we had

Ph Vm 3k mk= 0. (*)By Theorem 4 and themethodsof indicator theory (see [5]) we canfind M |= P andnonstandard ce M such that

M\= -i 3y ([1, y~\iscoc-large).(Briefly, this is done by taking a countable nonstandard model J of P andnonstandard c,aeJ such that Y(c,a)isnonstandard butless than c1, where y isasinTheorem 4(i). Now theindicator Y having nonstandard value on (c,a)meansprecisely that thereis aninitial segmentofJwhichis amodelof P an dlies betweencanda, that is,contains but nota.We can letM besuch an initial segment.)In M ,take d = 22 2with citerated exponentiations, so o2(d) = a>c.By (*) takee Msuch that de = 0.Sincetheproof ofP roposition 8 can becarriedout in P (seethe discussion preceding Lemma 4 in [4]), wehave in M

[1, 2+e] iswc-large,a contradiction.


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292 LAURIE KIRBY AND JEFF PARISWe sketch the proof of Theorem 2, which is similar to that of Theorem 1. Firstassign to each node in a given hydra an ordinal belowe0as follows:To each top node assign 0.To each other node assign ojai+ ... +(oa , where


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ACCESSIBLE INDEPEN DENCE RESULTS FOR PEANO ARITHMETIC 2 9 3(Note: A proof that Tis a winn ing strategy is equiva lent to a proof of e0-inductionwith respect to the predecessor function p(cc, n) = (a}(n) which am oun ts in turn to aproof that the function Xnx . gn(x) of Th eor em 4 (iii) is total recursiv e, and this ofcourse is impossible in P.)

REMARK. Let /If c denote Peano's axioms with induction restricted to Sk 'formulae. Then using results in [4] we can refine The orem 1 to give, for keN, k ^ 1,THEOREM 1'. (i) For each fixed peN, / Ik \- Vm,n > 1 {if m < n nP [where noccurs k times) then the Go odstein sequence for m starting at n eventually hits zero).

(ii) I'Lk \-f Vm,n > 1 (if m < n (where n occurs k + l times) then the Goodsteinsequencefor m starting at n eventually hits zero).Similarly if we restrict ourselves in Theorem 2 to hydras of height k + l (i.e. nonode is more than k+ l segments away from the root) then we cann ot prove that every recursive strategy is a winning strateg y using just /Xfc. In particular thefunction giving the lengths of battles for hydras of height 2 with strategy x is notprovably total in /Z x and hence is not primitive recursive.

References1. G. GENTZEN, 'Die Widerspruchsfreiheit der reinen Zahlentheorie', Math. Ann., 112 (1936), 493 -56 5.2. R. L.GOODSTEIN, 'On the restricted ordinal theorem', J. Symbolic Logic, 9 (1944), 3 3 - 4 1 .3. J. KETONEN and R. SOLOVAY, 'Rapidly growing Ramsey functions',Ann. of Math., 113 (1981), 267-314.4. J. PARIS, 'A hierarchy of cuts in models of arithmetic', Model theory of algebra and arithmetic,Proceedings, Karp acz, Polan d 1979. Lecture No tes in Mathem atics 834 (Springer, Berlin)

pp. 312-337.5. J. PARIS, 'Some independence results for Peano arithmetic', J. Symbolic Logic, 43 (1978), 725-731 .

Department of Mathematics,The University of Manchester,Manchester M13 9PL.

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Accessible Independence Results