aa r X i v : . [ m a t h . L O ] J u l Giant and illusionary giant Goodstein principles
Andreas WeiermannGhent UniversityDepartment of Mathematics: Analysis, Logic and Discrete MathematicsKrijgslaan 281 S89000 Ghent, Belgium email: [email protected]
Abstract
We analyze several natural Goodstein principles which themselves are definedwith respect to the Ackermann function and the extended Ackermann function.These Ackermann functions are well established canonical fast growing functionslabeled by ordinals not exceeding ε . Among the Goodsteinprinciples under con-sideration, the giant ones, will be proof-theoretically strong (being unprovable in PA in the Ackermannian case and being unprovable in ID in the extended Ack-ermannian case) whereas others, the illusionary giant ones, will turn out to becomparatively much much weaker although they look strong at first sight. Keywords—
Goodstein principles, independence results, first order arithmetic, Ack-ermann function, notation systems for natural numbers, ordinally informative prooftheory
This article is part of a general program on exhibiting natural independence results forfirst order Peano arithmetic PA , its subsystems and strong extensions, as for exam-ple, the theory ID of non iterated inductive definitions. (For readers more familiarwith reverse mathematics let us remark that he theory ID has the same proof-theoreticstrength as ACA + (Π − CA) − where (Π − CA) − refers to the scheme of light-face Π -comprehension.) We are mainly interested in natural combinatorial statementswhich have non trivial unprovability strength. The Goodstein principles serve as canon-ical examples since they provide prime examples of the intrinsic interplay betweenfinite and infinite numbers.Goodstein’s original result [9, 10] dealt with iterated exponential terms and theircanonical interpretation in the ordinals less than ε . The basic idea is as follows. Thereare two processes involved. One is making a given number under normal circumstancesgreater and the other making a given number smaller by subtracting a one. Beforeround a non negative integer is chosen. Then like in a two person game the followingmoves are carried out alternatingly. At round k ≥ the first player develops the1umber, which is already obtained in the game, completely with respect to base k + 2 as long as it is not yet zero. Here also the exponents are developed hereditarily. Thenhe replaces every occurrence of k + 2 by k + 3 . Then the second player makes anapparently innocent move. She just subtracts a one from the previous result. Then thegame moves over to round k + 1 with the new result. The second player wins thisspecific game if the number zero is reached after finitely many rounds. Surprisinglythe second player always wins but this fact is unprovable in PA . (See, for example,[11, 6] for a proof.) Goodstein’s theorem is an example for a giant Goodstein principlesince the underlying sequences become very (unprovably) long even for small startingvalues.In this paper we deal with Goodstein principles which are defined with respect tocanonical representations of natural numbers using the Ackermann function and theextended Ackermann function. The paper therefore splits naturally into two parts.In the fist part we deal with normal forms and Goodstein principles which are de-fined relative to the Ackermann function. Similar principles have already been studiedin [14] and [1] for the Ackermann function. These articles are based on a very compli-cated iterated sandwiching procedure and it is quite natural to ask what happens if thesandwiching is reduced to a one step approximation. This question has been investi-gated for the Ackermann function which starts at the bottom level with the exponentialfunction in [7]. It turned out that the strength of the resulting Goodstein principlesdropped considerably and we arrived in these cases at intermediate Goodstein princi-ples.In the first part we investigate normal forms which are based on a one step ap-proximation with respect to the Ackermann function which starts at the bottom levelwith the successor function. Surprisingly the resulting Goodstein principles is still notprovable from the axioms of first order arithmetic when base changes are carried out inthe two critical arguments of the Ackermann function which is involved in the numberrepresentations. This principle leads thus to a strong independence result and we callit a giant Goodstein principle. This reflects the fact that it will match in strengths withthe strongest possible Goodstein principle which is based on number representationsdefined in terms of the Ackermann function. In [8] it has been shown that the latterprinciple is equivalent to the one consistency of first order arithmetic PA .But we also consider restrictions where the base change is carried out only withrespect to one critical argument of the Ackermann function used to build up normalform representations under consideration. These lead to Goodstein principles whichlook strong at first sight but as a closer analysis reveals they are in fact of low proof-theoretic strength. One principle will be equivalent over primitive recursive arithmeticwith the assertion that ω ω + ω does not admit primitive recursive descending sequences.The other will be equivalent with the one consistency of primitive recursive arithmetic.We call the latter principle an illusionary giant Goodstein principle. These principles show similarities with a prominent figure from the literature, namely Tur Tur, knownfrom the story: Jim Button and Luke the Engine Driver written by Michael Ende. ”After a long and hazardousjourney, Jim Button and Luke the Engine Driver arrive in the Dragon City. Along the way, they make twonew friends, the giant Mr. Tur Tur (who is actually a ”Scheinriese”, an ”illusionary giant” he only appearsto be a giant from afar; when approached, it turns out he is actually of normal height), and Nepomuk, thehalf-dragon.” [Citation modelled after https://enacademic.com/dic.nsf/enwiki/2981127]
2n the second technically much more advanced part we consider Goodstein prin-ciples defined with respect to the extended Ackermann functions which are canonicalfast growing functions indexed by ordinals less than ε .It turns out that the resulting Goodstein principle becomes very strong when basechanges are performed in two critical arguments of the the extended Ackermann func-tion. In fact a termination proof necessarily involves a detour via uncountable ordinalssince it becomes unprovable in ID . For proving this we develop a novel theory ofmajorization properties of fundamental sequences. This theory allows us to prove that k -normal forms after base change are moved into k + 1 -normal forms.Moreover we employ the machinery of Buchholz’s collapsing function ψ . Wefound it very amazing that our ordinal mapping from natural numbers into ordinalshas the property that normal forms for numbers are moved by magic to ordinals inBuchholz ψ normal form and we believe that this underpins the naturality of our ap-proach.As far as we know this example will be the first example of a Goodstein principlefor natural numbers which has such a high unprovability strength.In analogy with the first part one would expect that the strength of the Goodsteinprinciple will drop seriously when base change is defined with respect to one critical ar-gument of the extended Ackermann function used to define the number representations.Quite surprisingly it turned out that a base change only in the first critical argument stillsuffices to generate a giant Goodstein principle.But a base change in the second critical argument only leads to an illusionary Good-stein principle which in strength is equivalent with the classical Goodstein principle. Let us define the Ackermann function A a ( k, b ) with respect to iteration parameter k <ω as follows: A ( k, b ) := b + 1 ,A a +1 ( k,
0) := A a ( k, · ) k (0) ,A a +1 ( k, b + 1) := A a ( k, · ) k ( A a +1 ( k, b )) . Here the upper index denotes the number of function iterations. It is a routinematter to show that for any fixed k ≥ the function a, b A a ( k, b ) is not primitiverecursive whereas for fixed a the function b A a ( k, b ) is primitive recursive. It isalso easy to show that the function k, a, b A a ( k, b ) is strictly monotone in a, b and k ≥ . Convention.
From now on k denotes, if not stated otherwise, a positive integer notsmaller than . Lemma 1
For all m > there exist unique a, b, l < ω such that1. m = A a ( k, b ) + l, . a is maximal with A a (0) ≤ m (so that A a ( k, ≤ m < A a +1 ( k, ),3. b is maximal with A a ( b ) ≤ m (so that A a ( k, b ) ≤ m < A a ( k, b + 1) ). We write m = k -nf A a ( k, b ) + l in this case. If m = k -nf A a ( k, b ) + l and a = 0 then necessarily m = A ( k, · ) m (0) where m < k . If m = k then m = k -nf A ( k, .Indeed we see by an easy induction on b that A ( k, b ) = k · (1 + b ) . This yields inparticular that A a ( k, b ) > · b for a > . The latter estimate is used tacitly at severaloccasions.In the sequel we often write A a ( b ) for A a ( k, b ) and B a ( b ) for A a ( k + 1 , b ) when k is fixed in a given context. Lemma 2
1. If m = k -nf A a ( b ) + l and l > then A a ( b ) + l − is in k normalform, too.2. If m = k -nf A a ( b ) and b > then A a ( b − is in k normal form, too.3. If m = k -nf A a (0) then for < l < k A la (0) is in k normal form, too. Proof. All assertions are easy to see. Let us prove the last assertion. If < l < k then A a (0)) ≤ A la (0) < A a +1 (0) . This yields that A la (0) is in k normal form. (cid:3) Definition 1
We define the base change operations recursively as follows.1. If m = 0 then m [ k ← k + 1] := 0 . If m = k -nf A a ( k, b ) + l > then m [ k ← k + 1] := A a [ k ← k +1] ( k + 1 , b [ k ← k + 1]) + l .2. If m = 0 then m [[ k ← k + 1]] := 0 . If m = k -nf A a ( k, b ) + l > then m [[ k ← k + 1]] := A a ( k + 1 , b [[ k ← k + 1]]) + l .3. If m = 0 then m { k ← k + 1 } := 0 . If m = k -nf A a ( k, b ) + l > then m { k ← k + 1 } := A a { k ← k +1 } ( k + 1 , b ) + l . Lemma 3
Let m ′ := m [ k ← k + 1] , m ′′ := m [[ k ← k + 1]] , and m ′′′ := m { k ← k + 1 } . m ≤ m ′ , m ≤ m ′′ and m ≤ m ′′′ . Moreover A a ( k, b ) ≤ A a [ k ← k +1] ( k +1 , b [ k ← k + 1]) , A a ( k, b ) ≤ A a [[ k ← k +1]] ( k + 1 , b [[ k ← k + 1]]) , A a ( k, b ) ≤ A a { k ← k +1 } ( k + 1 , b { k ← k + 1 } ) even if A a ( k, b ) is not in k -normal form.2. If m ≥ k then m < m ′ , m < m ′′ and m < m ′′′ .3. If m > then ( m − ′ < m ′ , ( m − ′′ < m ′′ , and ( m − ′′′ < m ′′′ .4. If m = k -nf A a ( k, b ) + l then m ′ = k +1 − nf A a ′ ( k + 1 , b ′ ) + l , m ′′ = k +1 − nf A a ( k + 1 , b ′′ ) + l , and m ′′′ = k +1 − nf A a ′′′ ( k + 1 , b ) + l . m . Assume that m = k -nf A a ( b ) + l .Case 1. a = 0 .Then < m < k . Then m = m ′ .Case 2. a > . Then the induction hypothesis yields m = A a ( b )+ l ≤ B a ′ ( b ′ )+ l = m ′ . The other claims in the first assertion are proved similarly.The first claim of the second assertion can be seen by inspection of the proof of thefirst assertion since A a ( b ) < B a ( b ) for a > .The first claim of the third assertion is proved by induction on m .Assume that m = k -nf A a ( b ) + l .Case 1. a = 0 . Then < m < k . Then ( m − ′ = m − < m = m ′ .Case 2. a > .Case 2.1. l > .Then m − k -nf A a ( b )+ l − and ( m − ′ = A a ( b ) ′ + l − < A a ( b ) ′ + l = m ′ .Case 2.2. l = 0 .Case 2.2.1. b > .Then for some p we find m − k -nf A a ( b −
1) + p < m = k -nf A a ( b ) . If a = 1 then p < k hence p < k + 1 and then the induction hypothesis yields ( m − ′ = A a ( b − ′ + p ≤ B a ′ (( b − ′ ) + p ≤ B a ′ ( b ′ −
1) + p < B a ′ ( b ′ ) = m ′ .If a ≥ then ( m − ′ = A a ( k, b − ′ + p where p < A a ( b ) = A ka − ( A a ( b − ≤ B ka ′ − ( B a ′ ( b ′ − and we arrive at ( m − ′ = A a ( k, b − ′ + p < B ka ′ − ( B a ′ ( b ′ − · < B k +1 a ′ − ( B a ′ ( b ′ − B a ′ ( b ′ ) = m ′ .Case 2.2.2. b = 0 .Then m − A a (0) − A ka − (0) − A a − ( A k − a − (0) −
1) + p for some p < ω .If a = 1 then ( m − ′ = ( A (0) − ′ = k − < k + 1 = A (0) ′ . If a > then the induction hypothesis yields ( m − ′ = ( A ka − (0) − ′ =( A a − ( A k − a − (0) − ′ + p . We have p < A a (0) = A ka − (0) ≤ B ka ′ − (0) and the induc-tion hypothesis yields A a − ( A k − a − (0) − ′ ≤ B ( a − ′ (( A k − a − (0)) ′ ) ≤ B k ( a − ′ (0) ≤ B ka ′ − (0) . Hence ( m − ′ = ( A ka − (0) − ′ = A a − ( A k − a − (0) − ′ + p
5e define for technical reasons − m + α in the natural way. Thus − m + α is equalto α if α is infinite and we set − m + α be equal to the maximum of {− m + α, } if α < ω . Definition 2 ψ k ,2. If m = k -nf A a ( k, b ) + l then ψ k m := b + 1 if a = 0 ω · (1 + ψ k b ) + l if a = 1 ω ω +( − ψ k a ) + ω · ψ k b + l if a ≥ . Definition 3 χ k
2. If m = k -nf A a ( k, b ) + l then χ k m := b + 1 if a = 0 ω · (1 + χ k b ) + l if a = 1 ω ω +( − a ) + ω · χ k b + l if a ≥ . Definition 4 ξ k
2. If m = k -nf A a ( k, b ) + l then ξ k m := b + 1 if a = 0 ω · (1 + χ k b ) + l if a = 1 ω · ( − ξ k a ) + ω · b + l if a ≥ . Recall that we write m ′ := m [ k ← k + 1] , m ′′ := m [[ k ← k + 1]] , and m ′′′ := m { k ← k + 1 } when k is fixed in the context. Lemma 4 ψ k +1 m ′ = ψ k m , χ k +1 m ′′ = χ k m , and ξ k +1 m ′′ = ξ k m .2. If m > then ψ k ( m − < ψ k m , χ k ( m − < χ k m and ξ k ( m − < ξ k m . The first claim in the first assertion is proved by an easy induction on m . The claim isclear for m = 0 . Assume that m = k -nf A a ( b )+ l . If a = 0 then m < k and ψ k +1 m ′ = m = ψ k m . If a = 1 then ψ k +1 m ′ = ω · (1 + ψ k +1 b ′ )+ l = ω · (1 + ψ k b )+ l = ψ k m . If a > then ψ k +1 m ′ = ω ω +( − ψ k +1 a ′ ) + ω · ψ k +1 b ′ + l = ω ω +( − ψ k a ) + ω · ψ k b + l = ψ k m . The second and third claim in the first assertion follow similarly.The first claim in the second assertion is proved by induction on m .Assume that m = k -nf A a ( b ) + l .Case 1. a = 0 . Then m < k . Then ψ k ( m −
1) = m − < m = ψ k m .Case 2. a > .Case 2.1. l > . m − k -nf A a ( b ) + l − . Then we find ψ k ( m −
1) = ψ k ( A a ( b )) + l − < ψ k ( A a ( b )) + l = ψ k m .Case 2.2. l = 0 . 6ase 2.2.1. b > . Then for some p we find m − k -nf A a ( b −
1) + p 1) = ω · (1 + ψ k ( b − p < ω · (1 + ψ k b ) = ψ k m .If a ≥ then the induction hypothesis yields ψ k ( m − 1) = ψ k ( A a ( b )) + ω · ψ k ( b − 1) + p < ψ k ( A a ( b )) + ω · ψ k ( b ) = ψ k m .Case 2.2.2. b = 0 . If a = 1 then ψ k ( m − 1) = ψ k ( A (0) − 1) = ψ k ( k − 1) = k − < ω = ψ k ( A (0)) = ψ k m. If a = 2 then for certain p , . . . , p k we find ψ k ( m − 1) = ψ k ( A k (0) − 1) = ω · ( ψ k ( A k − (0) − 1) + p ) = ω k · p k + . . . + ω · p < ω ω = ψ k m. If a > then for certain p , . . . , p k − the induction hypothesis yields ψ k ( m − 1) = ψ k ( A ka − (0) − ψ k ( A a − (( A k − a − (0) − p )= ω ω +( − ψ k ( a − + ω · ψ k ( A k − a − (0) − 1) + p ≤ ω ω +( − ψ k ( a − · k + ω k − · p k − + · · · + ω · p < ω ω +( − ψ k ( a )) = ψ k m. The second claim in the second assertion is proved by a similar induction on m .The third claim in the second assertion is proved by induction on m . Let us justconsider the case m = k -nf A a ( b ) + l where a ≥ , b = 0 , l = 0 . Then the inductionhypothesis yields ξ k ( m − 1) = ξ k ( A ka − (0) − ξ k ( A a − (( A k − a − (0) − p )= ω · ( − ξ k ( a − ω · ( A k − a − (0) − 1) + p < ω ( − ξ k ( a ))= ξ k m. (cid:3) Definition 5 Let m < ω .1. Put m := m. Assume recursively that m l is defined and m l > . Then m l +1 = m l [ l + 3 ← l + 4] − . If m l = 0 then m l +1 := 0 .2. Put ˜ m := m. Assume recursively that ˜ m l is defined and ˜ m l > . Then ˜ m l +1 =˜ m l [[ l + 3 ← l + 4]] − . If ˜ m l = 0 then ˜ m l +1 := 0 .3. Put m := m. Assume recursively that m l is defined and m l > . Then m l +1 = m l { l + 3 ← l + 4 } − . If m l = 0 then m l +1 := 0 . heorem 1 1. For all m < ω there exists an l < ω such that m l = 0 . This isprovable in PRA + TI( ε ) .2. For all m < ω there exists an l < ω such that ˜ m l = 0 . This is provable in PRA + TI( ω ω + ω ) .3. For all m < ω there exists an l < ω such that m l = 0 . This is provable in PRA + TI( ω ω ) . Proof. Define o ( m, l ) := ψ l +3 ( m l ) . If m l +1 > then by the previous lemmata o ( m, l + 1) = ψ l +4 ( m l +1 )= ψ l +4 ( m l [ l + 3 ← l + 4] − < ψ l +4 ( m l [ l + 3 ← l + 4])= ψ l +3 ( m l )= o ( m, l ) This proves the first assertion. The second and assertion are proved similarly by nowusing χ k ( ξ k resp.) instead of ψ k .Let us now prove the independence results. Recall that the standard system offundamental sequences for the ordinals less than ε is defined recursively as follows.If α = 0 then α [ x ] := 0 . If α = β + 1 then α [ x ] := β . If α = ω α + · · · + ω α n where α ≥ . . . ≥ α n and if α n is limit then α [ x ] = ω α + · · · + ω α n [ x ] . If α = ω α + · · · + ω α n +1 where α ≥ . . . ≥ α n + 1 then α [ x ] = ω α + · · · + ω α n · x . Thenfor limit ordinals λ < ε we have λ [ x ] < λ [ x + 1] < λ and λ = sup { α [ x ] : x < ω } .Moreover note that − x = ω [ x − and − λ [ x ] ≤ ( − λ )[ x ] holds for λ > ω .It is easy to show that these fundamental sequences fulfill the so called Bachmannproperty: If α [ x ] < β < α then α [ x ] ≤ β [1] . (See, e.g., [5], for a proof and furtherdiscussion.)Let α ≥ l β iff there exist α , . . . , α m such that α = α , β = α m and α i +1 = α i [ l ] for all i < m . If α [ x ] < β < α then the Bachmann property yields β ≥ α [ x ] .Moreover the Bachmann property yields that α ≥ l β implies α ≥ l +1 β . Lemma 5 Assume that m > .1. ψ k m > ψ k +1 ( m ′ − ≥ ( ψ k m )[ k − .2. χ k m > χ k +1 ( m ′′ − ≥ ( χ k m )[ k − .3. ξ k m > ξ k +1 ( m ′′′ − ≥ ( ξ k m )[ k − . Proof. Let us first prove the first assertion. Clearly ψ k m = ψ k +1 ( m ′ ) > ψ k +1 ( m ′ − . So let us prove the second inequality.If < m < k then ψ k +1 ( m ′ − 1) = m − ψ k m )[ k − . Assume that m = k -nf A a ( k, b ) + l ≥ k . Then a > .Case 1. l > . Then m − k -nf A a ( k, b ) + l − . Then ( ψ k m )[ k − 2] =( ψ k ( A a ( k, b ))+ l )[ k − 2] = ψ k ( A a ( k, b ))+ l − and ψ k +1 ( m ′ − 1) = ψ k +1 ( B a ′ ( b ′ )+ l − 1) = ψ k ( A a ( k, b )) + l − . 8ase 2. l = 0 .Case 2.1. b > .If a = 1 then the induction hypothesis yields ( ψ k m )[ k − 2] = ( ω (1+ ψ k b ))[ k − ≤ ω (1 + ( ψ k b )[ k − k ≤ ω (1 + ( ψ k +1 ( b ′ − k = ψ k +1 ( B a ′ ( b ′ − 1) + k ≤ ψ k +1 ( B a ′ ( b ′ ) − 1) = ψ k +1 ( m ′ − .If a > then the induction hypothesis yields ( ψ k m )[ k − 2] = ( ω ω +( − ψ k a ) + ω · ψ k b )[ k − ≤ ω ω +( − ψ k a ) + ω · (( ψ k b )[ k − k ≤ ω ω +( − ψ k +1 a ′ ) + ω · ( ψ k +1 ( b ′ − k = ψ k +1 ( B a ′ ( b ′ − 1) + k ≤ ψ k +1 ( B a ′ ( b ′ ) − 1) = ψ k +1 ( m ′ − .Case 2.2. b = 0 . If a = 1 then m = A (0) = k and ( ψ k m )[ k − 2] = ω [ k − ≤ k = ψ k +1 ( k + 1 − 1) = ( ψ k +1 ( m ′ − .If a = 2 then ( ψ k m )[ k − 2] = ( ω ω +( − ψ k (2) ) + ω · k − 2] = ω ω [ k − ≤ ω k = ω (1 + ψ k +1 ( B k − (0)) = ψ k +1 ( B k (0)) ≤ ψ k +1 ( B k +11 (0) − 1) = ( ψ k +1 ( m ′ − .Note that ψ k +1 B l (0) = ω l holds by induction on l for < l ≤ k .If a = k then ( ψ k m )[ k − 2] = ( ω ω +( − ψ k a ) )[ k − ≤ ( ω ω +( − ω ) )[ k − ω ω +( − k ) )= ( ω ω +( − ψ k +1 ( a ′ − ) · k = ψ k +1 ( B ka ′ − (0)) ≤ ψ k +1 ( B a ′ (0) − ψ k +1 ( m ′ − . If a > and a = k then ( − ψ k a )[ k − 2] = − ψ k a )[ k − and ( ψ k m )[ k − 2] = ( ω ω +( − ψ k a ) )[ k − ≤ ( ω ω +( − ψ k a )[ k − ) · k ≤ ( ω ω +( − ψ k +1 ( a ′ − ) · k = ψ k +1 ( B ka ′ − (0)) ≤ ψ k +1 ( B a ′ (0) − ψ k +1 ( m ′ − . Note that ψ k +1 ( B la ′ − (0)) = ( ω ω +( − ψ k +1 ( a ′ − ) · l holds by induction on l for < l ≤ k .The second assertion is proved by a similar induction on m . An even stronger resultis proved later, see assertion 2 of Lemma 15.Let us finally prove the third assertion. Clearly ξ k m = ξ k +1 ( m ′′′ ) > ξ k +1 ( m ′′′ − . So let us prove the second inequality.If < m < k then ξ k +1 ( m ′′′ − 1) = m − ξ k m )[ k − . Assume that m = k -nf A a ( k, b ) + l ≥ k . Then a > .Case 1. l > . Then m − k -nf A a ( k, b ) + l − . Then ( ξ k m )[ k − 1] =( ξ k ( A a ( k, b ))+ l )[ k − 1] = ξ k ( A a ( k, b ))+ l − and ξ k +1 ( m ′′′ − 1) = ξ k +1 ( B a ( b ) ′′′ )+ l − ξ k ( A a ( k, b )) + l − . 9ase 2. l = 0 .Case 2.1. b > .If a = 1 then the induction hypothesis yields ( ξ k m )[ k − 1] = ω (1 + b )[ k − ≤ ω (1+ b [ k − k ≤ ω (1+( b − k = ξ k +1 ( B a ′′′ ( b − k ≤ ξ k +1 ( B a ′′′ ( b ) − 1) = ξ k +1 ( m ′′′ − .If a > then the induction hypothesis yields ( ξ k m )[ k − 1] = ( ω ( − ξ k a ) + ω · b )[ k − ≤ ( ω ( − ξ k a )+ ω · b )[ k − k ≤ ω ( − ξ k +1 a ′′′ )+ ω · ( b − k = ξ k +1 ( B a ′′′ ( b − 1) + k ≤ ξ k +1 ( B a ′′′ ( b ) − 1) = ξ k +1 ( m ′′′ − .Case 2.2. b = 0 . If a = 1 then m = A (0) = k and ( ξ k m )[ k − 1] = ω [ k − ≤ k = ξ k +1 ( k + 1 − 1) = ( ξ k +1 ( m ′′′ − .If a = 2 then ( ξ k m )[ k − 1] = ( ω ( − ξ k (2)) + ω · k − 1] = ω [ k − ≤ ω · k ≤ ω (1 + ( B k − (0)) = ξ k +1 ( B k (0)) ≤ ξ k +1 ( B k +11 (0) − 1) = ( ξ k +1 ( m ′′′ − .Note that ξ k +1 ( B l (0)) ≥ ω · l holds by induction on l for < l ≤ k .If a = k then ( ξ k m )[ k ] = ( ω ( − ξ k a ))[ k − ≤ ( ω )[ k − ω · ( k − ω ( − ξ k +1 ( a ′′′ − ≤ ξ k +1 ( B a ′′′ − (0)) ≤ ξ k +1 ( B a ′′′ (0) − ξ k +1 ( m ′′′ − If a > and a = k then ( ξ k m )[ k − 1] = ω ( − ψ k a ))[ k − ≤ ω ( − ξ k a )[ k − ω · k ≤ ω ( − ψ k +1 ( a ′′′ − ω · B k − a ′′′ − (0)= ξ k +1 ( B ka ′′′ − (0)) ≤ ξ k +1 ( B a ′′′ (0) − ξ k +1 ( m ′′′ − (cid:3) For α < ε let PRA + TI( < α ) be the union of the theories PRA + TI( β ) where β < α . Theorem 2 PA ( ∀ m )( ∃ l )[ m l = 0] .2. PRA + TI( < ω ω + ω ) ( ∀ m )( ∃ l )[ ˜ m l = 0] .3. PRA ( ∀ m )( ∃ l )[ m l = 0] . m (1) := A (3 , and m ( r + 1) := A m ( r ) (3 , .Let ω := ω and ω r +1 := ω ω r . Then ψ ( m ( r )) = ω r +1 for r ≥ . We claim that o ( m ( r ) , l ) ≥ ω r +1 [1] . . . [ l ] . Proof of the claim. Write m for m ( r ) . For o ( m, l ) > Lemma 5 yields o ( m, l ) > o ( m, l +1) = ψ l +4 ( m l [ l +3 ← l +4] − ≥ ( ψ l +3 ( m l ))[ l +1] = ( o ( m, l )[ l +1] . The Bachmann property yields o ( m, l +1) ≥ ( o ( m, l )[ l +1] . Theinduction hypothesis yields o ( m, l ) ≥ ω r +1 [1] . . . [ l ] . This yields o ( m, l )[ l + 1] ≥ ( o ( m, l )[ l + 1] ≥ ω r +1 [1]] . . . [ l ][ l + 1] .Therefore the least l such that o ( m, l ) = 0 is at least as big as the least l such that ω r +1 [1] . . . [ l + 1] = 0 . The result follows from PA 6⊢ ∀ r ∃ l ( ω r +1 )[2] . . . [ l + 1] = 0 .The second assertion follows similarly.Let m (( r )) := A r +2 (3 , . Then χ ( m ( r )) = ω ω + r . The result follows from PRA 6⊢ ∀ r ∃ l ( ω ω + r )[1] . . . [ l ] = 0 .The third assertion follows similarly by employing ξ k and m ( r ) . (cid:3) It is somewhat surprising that one obtains a PA independence via ∀ m ∃ lm l = 0 without using the nested sandwiching procedure from [14]. We consider this Goodsteinprinciple for the Ackermann function as a giant Goodstein principle.Note that the assertion ∀ m ∃ l ˜ m l = 0 is considerably weaker than the correspondingassertion in [14] where the base change in the second argument of the Ackermannfunction led to independence from IΣ . We consider this Goodstein principle thereforeas an intermediate Goodstein principle.The assertion ∀ m ∃ lm l = 0 is an example for an illusionary giant Goodstein prin-ciple. In strength it does not exceed the axioms needed for proving the totality of thefunctions which are involved in the definition of the underlying normal forms. But thisprinciple is still non trivial since it is independent of PRA .An even weaker Goodstein principle can be obtained by performing a trivial basechange in the iteration parameter k . For m = k -nf A a ( k, b ) + l ≥ k let m ′′′′ := A a ( k + 1 , b ) + l . A Goodstein principle base on this definition becomes even provablein in a weak meta theory using the ordinal assignment ord ( m ) := ω · a + ωb + l . Notethat the graph of the Ackermann function is elementary and so it makes sense to speakabout writing m in k normal form even in a very weak base theory. But defining thebase change already requires some tricky machinery. Let us agree that small Greek letters denote ordinals less than ε .Let us recall the definition of the standard assignment of fundamentals sequencesfor the ordinals below ε . We agree on ω α +1 ( β + 1)[ x ] = ω α +1 β + ω α · x and ω λ ( β + 1)[ x ] = ω λ β + ω λ [ x ] . We also assume for convenience that ( α + 1)[ x ] := α and x ] := 0 .Using these fundamental sequences we define the extended Ackermann functions α, b A α ( k, b ) for α < ε and k, b < ω recursively as follows:11 ( k, b ) := b + 1 ,A α +1 ( k, 0) := A α ( k, · ) k (0) ,A α +1 ( k, b + 1) := A α ( k, · ) k ( A α +1 ( k, b )) ,A λ ( k, 0) := A λ k,k, ( k, · ) k (0) ,A λ ( k, b + 1) := A λ k,k,Aλ ( k,b ) ( k, · ) k ( A λ ( k, b )) . where for a limit λ the ordinal λ l,k,b is defined recursively by λ ,k,b := λ [ b ] and λ l +1 ,k,b := λ [ A λ l,k,b ( k, b )] .For the rest of this article by k will denote a positive integer not smaller than .In the sequel we often write A α ( b ) for A α ( k, b ) and B α ( b ) for A α ( k + 1 , b ) and λ l,b for λ l,k,b when k is fixed in a given context.The functions A α come along with natural monotonicity properties due to the Bach-mann property of the system of fundamental sequences. Recall that his property statesthat α [ l ] < β < α yields α [ l ] ≤ α [1] . Moreover recall that ≤ l is the transitive andreflexive closure of { ( α [ l ] , α ) : α < ε } .In general the function α A α ( b ) is not monotone in α but it shows decentmonotonicity with respect to the relation ≤ l .Furthermore, for α = ω α · m + · · · + ω α n · m n with α > . . . > α n and < m , . . . , m n < ω let mc ( α ) := max { mc ( α ) , . . . , mc ( α n ) , m , . . . , m n } where mc (0) := 0 . We call mc ( α ) the maximal coefficient of an ordinal α . This maximalcoefficient plays an important role in bounding values of A α ( b ) . Lemma 6 A α ( b ) < A α ( b + 1) ,2. α [ l ] < β < α yields A α [ l ] ( b ) < A β ( b ) ,3. α ≤ l β yields A α ( b ) ≤ A β ( b ) for all b ≥ l ≥ ,4. mc ( α ) < A α ( b ) ,5. α < β and mc ( α ) ≤ b yields A α ( b ) ≤ A β ( b ) . Proof. This is easy. One can e.g. consult [5] or [12] if needed. (cid:3) Lemma 7 For any m there will be at most finitely many α < ε such that A α ( k, ≤ m . Proof. First note that for a given m there will be finitely many α < ε such that N α ≤ m where N α is the number of occurrences of ω in the Cantor normal form of α (cf., e.g., [5]). An easy induction on α yields that A α ( k, b ) ≥ N α + b . Putting thingstogether the Lemma follows. (cid:3) Lemma 8 For all m > there exist unique α < ε and b, l < ω such that1. m = A α ( k, b ) + l, . α is maximal with A α ( k, ≤ m (so that A α ( k, ≤ m < A α +1 ( k, ),3. b is maximal with A α ( k, b ) ≤ m (so that A α ( k, b ) ≤ m < A α ( k, b + 1)) . We write m = k -nf A α ( k, b ) + l in this case and call A α ( k, b ) + l the k normal formof m . This normal form is uniquely determined. If m = k -nf A α ( k, b ) + l and a = 0 then necessarily m = A ( k, · ) m (0) where m < k . If m = k then m = k -nf A ( k, and A ( k, b ) = k · (1 + b ) . The yields in particular that A α ( k, b ) > · b for α > .The latter estimate is used tacitly at several occasions. Lemma 9 1. If m = k -nf A α ( b ) and b > then A α ( b − is in k normal form,too.2. A lα (0) is in k normal form for < l < k .3. If m = k -nf A λ (0) and λ is a limit then A λ l, (0) is in k normal form for < l Definition 6 We define the base change operations recursively as follows.1. If m = 0 then m [ k ← k + 1] := 0 . If m = k -nf A α ( k, b ) + l > then m [ k ← k + 1] := A α [ k ← k +1] ( k + 1 , b [ k ← k + 1]) + l . If α = ω β · m + γ is in Cantornormal form then α [ k ← k + 1] = ω β [ k ← k +1] · m [ k ← k + 1] + γ [ k ← k + 1] . 2. If m = 0 then m [[ k ← k + 1]] := 0 . If m = k -nf A α ( k, b ) + l then m [[ k ← k + 1]] := A α ( k + 1 , b [[ k ← k + 1]]) + l .3. If m = 0 then m { k ← k + 1 } := 0 . If m = k -nf A α ( k, b ) + l then m { k ← k + 1 } := A α { k ← k +1 } ( k + 1 , b ) + l . If α = ω β · m + γ is in Cantor normalform then α { k ← k + 1 } = ω β { k ← k +1 } · m { k ← k + 1 } + γ { k ← k + 1 } . It is highly non trivial to show that the base change operations preserve monotonic-ity and normal forms. To show these properties we develop some new machinery aboutfundamental sequences.For an ordinal context λ [[ · ]] with exactly one occurrence of the placeholder [[ · ]] we define an ordinal context λ ∗ [[ · ]] , the truncation of λ [[ · ]] as follows. If λ [[ · ]] = [[ · ]] then λ ∗ [[ · ]] = [[ · ]] . If λ [[ · ]] = ω α + · · · + ω α i + [[ · ]] + ω α i +1 + · · · + ω α n then λ ∗ [[ · ]] = ω α + · · · + ω α i + [[ · ]] . If λ [[ · ]] = ω α + · · · + ω α i [[ · ]] + ω α i +1 + · · · + ω α n then λ ∗ [[ · ]] = ω α + · · · + ω α ∗ i [[ · ]] . So we basically cut off hereditarily terms after theplaceholder. (We tacitly assume here as usual that α ≥ . . . ≥ α n .) Lemma 10 If α < β then β = α +1 or α < β [1] or there exists a context λ , an ordinal γ and a natural number r such that α = λ [[ ω γ · r ]] and β = λ ∗ [[ ω γ +1 ]] . Moreover wehave α < λ ∗ [[ ω γ · ( r + 1)]] . α = ω α + · · · + ω α m and that β = ω β + · · · + ω β n are bothwritten in Cantor normal form.Case 1. Assume m < n and α i < β i for all i ≤ m . Then of course β = α + 1 or α < β [1] .Case 2. There exists an i ≤ min { m, n } such that α i < β i and ∀ j < i ( α j = β j ) . If n > i then of course α < β [1] . So assume i = n . Let us write α = ω α + · · · + ω α i · s + ξ with ξ < ω α i .Case 2.1. Assume that β i = δ + 1 .If δ > α i then of course α < β [1] . If δ = α i + 1 then put λ := ω α + · · · + ω α i − +[[ · ]] + ω α i +1 + · · · + ω α m , γ = α i and r := s . Then α = λ [[ ω γ · r ]] , β = λ ∗ [[ ω γ +1 ]] and α < λ ∗ [[ ω γ · ( r + 1)]] .Case 2.2. Assume β i ∈ Lim . We have α i < β i . The case β i = α + 1 is impossibleand so by induction hypothesis there are two cases.Case 2.2.1. α i < β i [1] . Then α < β [1] .Case 2.2.2. β i = µ ∗ [[ ω γ +1 ]] and α i = µ [[ ω γ · r ]] where α i < µ ∗ [[ ω γ · ( r + 1)]] . Let λ := ω α + · · · + ω α i − + ω µ [[ · ]] + ω α i · ( s − 1) + ω α i +1 + · · · + ω α m . Then α = λ [[ ω γ · r ]] , β = λ ∗ [[ ω γ +1 ]] and α < λ ∗ [[ ω γ · ( r + 1)]] . (cid:3) For notational reasons we agree on A α ( k, − 1) := 0 . Lemma 11 Assume α < β . Moreover, assume for all λ , γ and r : if α = λ [[ ω γ · r ]] and δ := λ ∗ [[ ω γ +1 ]] then r < A δ k − ,Aδ ( b − ( A δ ( b − . Then we obtain either α + 1 = β or α + 1 ≤ β [ A β k − ,Aβ ( b − ( A β ( b − . Moreover A α +1 ( b ) ≤ A β ( b ) . Proof. The assertion is clear for β = α + 1 . If α < β [1] then of course α +1 ≤ β [ A β k − ,Aβ ( b − ( A β ( b − . Finally assume by Lemma 10 that α = λ [[ ω γ · r ]] and β = λ ∗ [[ ω γ +1 ]] . Then β [ l ] = λ ∗ [[ ω γ · l ]] for every l and β k,A β ( b − = β [ A ( λ ∗ [[ ω γ +1 ]]) k − ,Aβ ( b − ( A β ( b − . By assumption we obtain r < A ( λ ∗ [[ ω γ +1 ]]) k − ,Aβ ( b − ( A β ( b − . Lemma 10 yields α = λ [[ ω γ · r ]] < λ ∗ [[ ω γ · ( r + 1)]] ≤ λ ∗ [[ ω γ · A ( λ ∗ [[ ω γ +1 ]]) k − ,Aβ ( b − ( A β ( b − β [ A ( λ ∗ [[ ω γ +1 ]]) k − ,Aβ ( b − ( A β ( b − β [ A β k − ,Aβ ( b − ( A β ( b − . The second claim follows from the first claim by induction on β . Indeed, the claimis obvious when β is α + 1 . Otherwise α < β [ A β k − ,Aβ ( b − ( A β ( b − and theinduction hypothesis yields A α +1 ( k ) ≤ A β [ A βk − ,Aβ ( b − ( A β ( b − ( A β ( b − ≤ A β ( b ) . (cid:3) To prove the preservation of normal forms after performing a base change operationthe following Lemma will be of key importance (together with Lemma 11).14 emma 12 Assume that there is no δ > α with A δ ( b ) ≤ A α ( b ) . Then for all λ , γ and r : if α = λ [[ ω γ · r ]] and β = λ ∗ [[ ω γ +1 ]] then r ≤ A β k − ,Aβ ( b − ( A β ( b − . Proof. Assume that α = λ [[ ω γ · r ]] . Assume for a contradiction that r > A ( λ ∗ [[ ω γ +1 ]]) k − ,Aβ ( b − ( A β ( b − . Let β := λ ∗ [[ ω γ +1 ]] . Then β > α . But A β ( b )= A kβ k,Aβ ( b − ( A β ( b − A kλ ∗ [[ ω γ · A ( λ ∗ [[ ωγ +1]]) k − ,Aβ ( b − ]] ( A β ( b − < A λ ∗ [[ ω γ · r ]] ( b ) ≤ A λ [[ ω γ · r ]] ( b )= A α ( b ) . where we used that λ ∗ [[ ω γ · r ]] (cid:22) λ [[ ω γ · r ]] . This contradicts the maximality propertyof α . (cid:3) Lemma 13 Let m ′ := m [ k ← k + 1] , m ′′ := m [[ k ← k + 1]] , and m ′′′ := m { k ← k + 1 } . m ≤ m ′ , α ≤ α ′ , m ≤ m ′′ , m ≤ m ′′′ and α ≤ α ′′′ .2. A α ( b ) ′ ≤ B α ′ ( b ′ ) , A α ( b ) ′′ ≤ B α ′′ ( b ′′ ) and A α ( b ) ′′′ ≤ B α ′′′ ( b ′′′ ) even if A α ( b ) is not in normal form.3. If m ≥ k then m < m ′ , m < m ′′ , m < m ′′′ .4. If m > then ( m − ′ < m ′ , ( m − ′′ < m ′′ , and ( m − ′′′ < m ′′′ 5. If m = k -nf A α ( k, b ) + l then m ′ = k +1 − nf A α ′ ( k + 1 , b ′ ) + l , m ′′ = k +1 − nf A α ( k + 1 , b ′′ ) + l , and m ′′′ = k +1 − nf A α ′′′ ( k + 1 , b ) + l . The first three assertions are easy to prove using the last two assertions of Lemma 6.The first claim of the fourth assertion is proved by induction on m .Assume that m = k -nf A α ( b ) + l .Case 1. α = 0 . Then < m < k . Then ( m − ′ = m − < m = m ′ .Case 2. α > .Case 2.1. l > . m − k -nf A α ( b ) + l − . Then we find ( m − ′ = A α ( b ) ′ + l − < A α ( b ) ′ + l = m ′ .Case 2.2. l = 0 .Case 2.2.1. b > . Then we find m − k -nf A α ( b − 1) + p < m = k -nf A α ( b ) . Case 2.2.1.1. α = 1 . Then p < k hence p < k + 1 and then the induction hypothesisyields ψ k ( m − 1) = A α ( b − ′ + p ≤ B α ′ (( b − ′ ) + p ≤ B α ′ ( b ′ − 1) + p yields mc (( α l,A α ( b − ) ′ ) ≤ B α ′ l − ,k +1 ,Bα ′ ( b ′− ( B α ′ ( b ′ − . Moreover we have mc (( α ,A α ( b − ) ′ ) ≤ B α ′ ( b ′ − . Then Lemma 6 yields p < A α ( b ) = A kα k,Aα ( b − ( A α ( b − ≤ B kα ′ k,Aα ( b − ( B α ′ ( b − ′ ) ≤ B kα ′ k +1 ,k +1 ,B ′ α ( b ′− ( B α ′ ( b ′ − and we arrive at ( m − ′ = A α ( k, b − ′ + l < B k +1 α ′ k +1 ,k +1 ,Bα ′ ( b ′− ( B α ′ ( b ′ − B α ′ ( b ′ ) .Case 2.2.2. b = 0 .If a = 1 then ( m − ′ = ( A (0) − ′ = k − < k + 1 = A (0) ′ . If a > then the induction hypothesis yields for some p that ( m − ′ = ( A kα (0) − ′ = A α k, ( A k − α k, (0) − ′ + p . We have p < A α (0) = A kα k, (0) ≤ B k ( α k, ) ′ (0) ≤ B kα ′ k,k +1 , (0) and A α k, ( A k − α k, (0) − ′ ≤ B kα ′ k +1 ,k +1 , (0) . Hence ( m − ′ = ( A kα k, (0) − ′ = ( A α k, ( A k − α k, (0) − ′ + p < B kα ′ k +1 ,k +1 , (0) · < B k +1 α ′ k +1 ,k +1 , (0) = B α ′ (0) .Let us now prove the first claim in the fifth assertion.Assume that m = k -nf A α ( b )+ l . Then A α (0) is in k normal form. Then there is no δ > α with A δ (0) ≤ A α (0) . We obtain by Lemma 12 that for all contexts λ , ordinals γ and natural numbers r : If α = λ [[ ω γ · r ]] and β = λ ∗ [[ ω γ +1 ]] then r ≤ A β k − , (0) .Assume now that α ′ = ˜ λ [[ ω ˜ γ · ˜ r ]] . Then there exist λ, γ, r such that ˜ λ = λ ′ , ˜ γ = γ ′ and ˜ r = r ′ and α = λ [[ ω γ · r ]] . Let β := λ ∗ [[ ω γ +1 ]] and ˜ β = ( λ ∗ ) ′ [[ ω γ ′ +1 ]] . To applyLemma 11 we have to show r ′ < B ˜ β k,k +1 , (0) .We obtain ˜ β [ k ′ ] = ( λ ∗ ) ′ [[ ω γ ′ · k ′ ]] = ( β [ k ]) ′ . This yields by induction on l < k that ˜ β l,k +1 , = ( β l, ) ′ . Indeed we find ˜ β ,k +1 , = ( λ ∗ ) ′ [[ ω γ ′ · (0) ′ ]] = ( β , ) ′ . Moreover ˜ β l +1 ,k +1 , = ( λ ∗ ) ′ [[ ω γ ′ · B ˜ β l,k +1 , ]] = ( λ ∗ [[ ω γ · A β l, ]]) ′ = ( β l +1 , ) ′ . As desired r ≤ A β k − , (0) yields r ′ ≤ ( A β k − , (0)) ′ ≤ B ( β k − , ) ′ (0) < B ˜ β k,k +1 , (0) . We claim that there is no δ > α ′ with B δ (0 ′ ) ≤ B α ′ ( b ′ ) + l . Assume α ′ < δ .We claim that B δ (0) > m ′ . Indeed we find m ′ = B α ′ ( b ′ ) + l ≤ B α ′ ( B k − α ′ (0)) + B kα ′ (0) ≤ B kα ′ (0) · ≤ B k +1 α ′ (0) = B α ′ +1 (0) ≤ B δ (0) by Lemma 11. So α ′ fulfillsthe maximality condition and m ′ is in k + 1 -normal form. (cid:3) For denoting ordinals below the Howard Bachmann ordinal we use Buchholz’s ψ function from [3].Let Ω be the first uncountable ordinal. Let C ( α ) be the least set C such that1. { , Ω } ⊂ C ,2. If β = β + . . . + β n and if β , . . . , β n ∈ C are additive principal and if β ≥ . . . ≥ β n then β ∈ C ,3. If β = Ω γ · δ + η and γ, δ, η ∈ C then β ∈ C .4. β ∈ C ∩ α ⇒ ψβ ∈ C . 16et ψα := min { ξ : ξ C ( α ) } . Then ψα < Ω . Moreover ψα ∈ Lim for α > , ψ and ψ ( α + 1) = ψα · ω . Then ψω = ω ω . We write β = NF ψα if β = ψα and α ∈ C ( α ) . Then β = NF ψα and δ = NF ψγ and α < γ yield β < δ .Let OT be defined as follows.1. { , Ω } ⊂ OT ,2. If β = β + . . . + β n and if β , . . . , β n ∈ OT are additive principal and if β ≥ . . . ≥ β n then β ∈ OT ,3. If β = Ω γ · δ + η and γ, δ, η ∈ OT then β ∈ OT .4. If β ∈ OT and β ∈ C ( β ) then ψβ ∈ OT .It is well known that OT ∩ Ω = ψε Ω+1 . Let η := ψε Ω+1 be an abbreviation forthe Howard Bachmann ordinal.For the termination proof it will be essential to work with ψ terms in normal form.For α ∈ OT define Gα ⊆ OT as follows.Let G ∅ , G (Ω α · β + γ ) := Gα ∪ Gβ ∪ Gγ and Gψα := Gα ∪ { α } . Then ψα is in ψ normal form iff α ∈ C ( α ) iff Gα < α . Therefore G can be used to single outnormal forms and to prove that OT is a primitive recursive set. By restricting to termsin ψ normal form there will be no chains of terms in OT like ( ψ Ω , ψψ Ω , ψψψ Ω , . . . ) .To deal with the third Goodstein principle for the extended Ackermann function wework also with a slight variant OT ′ of OT .To define it let us first consider a modification of C ( α ) . Let C ′ ( α ) be the least set C such that1. { , Ω } ⊂ C ,2. If α = β + . . . + β n + ω · b + l and if β , . . . , β n ≥ ω ∈ OT are additiveprincipal and if β ≥ . . . ≥ β n and b, l < ω then α ∈ C 3. If α = Ω β · γ + δ and β, γ, δ ∈ C and δ < Ω β and γ < Ω then α ∈ C .4. < β ∈ C ∩ α ⇒ Ψ β ∈ C .For α > let Ψ α := min { ξ : ξ C ( α ) } . Then Ψ α < Ω . Moreover Ψ α ∈ Lim , ψ ω and Ψ( α + 1) = Ψ α + ω . We write β = NF Ψ α if β = Ψ α and α ∈ C ′ ( α ) .Then β = NF Ψ α and δ = NF Ψ γ and α < γ yield β < δ .1. { , Ω } ⊂ OT ′ ,2. If α ≥ ω and b, l < ω then α + ω · b + l ∈ OT ′ 3. If α = Ω β · γ + δ and β, γ, δ ∈ OT ′ and δ < Ω β and γ < Ω then α ∈ OT ′ .4. If < β ∈ OT ′ and β ∈ C ′ ( β ) then Ψ β ∈ OT ′ .Again it is well known OT ′ ∩ Ω = Ψ ε Ω+1 Moreover η = Ψ ε Ω+1 . (See, forexample, Buchholz’s contributions in [2] for a proof.)17 efinition 7 ψ k ,2. If m = k -nf A α ( k, b ) + l then ψ k m := b + 1 if α = 0 ω (1 + ψ k b ) if α = 1 ψ ( ω + ( − ψ k α )) + ωψ k b + l if α ≥ . Moreover for α = ω β · m + γ in normal form put ψ k α := Ω ψ k β · ψ k m + ψ k γ . Definition 8 χ k 2. If m = k -nf A α ( k, b ) + l then χ k m := b + 1 if a = 0 ω (1 + χ k b ) if α = 1 ω ω +( − α ) + ω · χ k b + l if α ≥ . Definition 9 ξ k 2. If m = k -nf A α ( k, b ) + l then ξ k m := b + 1 if a = 0 ω (1 + b ) if α = 1Ψ( − ξ k α ) + ω · b + l if α ≥ . Moreover for α = ω β · m + γ in normal form put ξ k α := Ω ξ k β · ξ k m + ξ k γ . In the sequel we write m ′ := m [ k ← k + 1] , m ′′ := m [[ k ← k + 1]] and m ′′′ := m { k ← k + 1 } when k is fixed in the context. Lemma 14 ψ k +1 m ′ = ψ k m , χ k +1 m ′′ = χ k m , and ξ k +1 m ′′ = ξ k m ,2. If m > then ψ k ( m − < ψ k m , χ k ( m − < χ k m , and ξ k ( m − < ξ k m .3. If m = A α ( b ) + l and α > then ψ k m = ψ ( ω + ( − ψ k α ) + ω · ψ k b + l isin ψ normal form and in Cantor normal form. Similarly ξ k m = Ψ( − ξ k α ) + ω · ψ k b + l is in Ψ normal form and in Cantor normal form for α > . The first assertion is proved by induction on m .The first claim in the second assertion is proved simultaneously with the first claimin the third assertion by induction on m . (We use the third assertion implicitly.)Assume that m = k -nf A α ( b ) + l .Case 1. α = 0 . Then m < k . Then ψ k ( m − 1) = m − < m = ψ k m .Case 2. α > .Case 2.1. l > . m − k -nf A α ( b )+ l − . Then for some β we find ψ k ( m − 1) = β + l − < β + l = ψ k m .Case 2.2. l = 0 . 18ase 2.2.1. b > . Then for some p we find m − k -nf A α ( b − 1) + p 1) = ψ ( ω + ( − ψ k α )) + ω · ( ψ k ( b − p < ψ ( ω + ( − ψ k α )) + ω · ψ k ( b ) .Case 2.2.2. b = 0 . If α = 1 then for some p we find ψ k ( m − 1) = ω · ψ k ( b − 1) + p < ω · ψ k b = ψ k m . Now consider the case α ≥ . Then for some p wefind m − A kα k, (0) − A α k, (( A k − α k, (0) − p where the latter is in k normal form. The induction hypothesis yields ψ k ( A α k, ( A k − α k, (0) − 1) + p ) = ψ ( ω + ( − ψ k α k, ) + ω · ψ k ( A k − α k, (0) − 1) + p < ψ ( ω + ( − ψ k α )) = ψ k m since ψ k α k, < ψ k α and ψ k α is an additive principal number which is in ψ normalform. The third claim in the second assertion is proved simultaneously with the secondclaim in the the third assertion by a similar induction on m .Assume that m = k -nf A α ( b ) + l .Case 1. α = 0 . Then m < k . Then ξ k ( m − 1) = m − < m = ξ k m .Case 2. α > .Case 2.1. l > . m − k -nf A α ( b )+ l − . Then for some β we find ξ k ( m − 1) = β + l − < β + l = ξ k m .Case 2.2. l = 0 .Case 2.2.1. b > . Then for some p we find m − k -nf A α ( b − 1) + p 1) = Ψ( − ξ k α ) + ω · ( b − 1) + p < Ψ( − ξ k α ) + ω · b .Case 2.2.2. b = 0 . If α = 1 then for some p we find ξ k ( m − 1) = ω · ( b − 1) + p <ω · b = ξ k m . Assume now that α ≥ .Then for some p we find m − A kα k, (0) − A α k, (( A k − α k, (0) − p wherethe latter is in k normal form. The induction hypothesis yields ξ k ( A α k, ( A k − α k, (0) − 1) + p ) = Ψ( − ψ k α k, ) + ω · ( A α k, ) k − (0) − 1) + p < Ψ( − ψ k α ) = ψ k m since ψ k α k, + ω ≤ ψ k α because ψ k α is in Ψ normal form.The second claim in the second assertion is proved by a much simpler induction on m . Assume that m = k -nf A α ( b ) + l .Case 1. α = 0 . Then m < k . Then χ k ( m − 1) = m − < m = χ k m .Case 2. α > .Case 2.1. l > . m − k -nf A α ( b )+ l − . Then for some β we find χ k ( m − 1) = β + l − < β + l = χ k m .Case 2.2. l = 0 .Case 2.2.1. b > . Then for some p we find m − k -nf A α ( b − 1) + p 1) = ω · (1 + χ k ( b − p < ω · (1 + χ k ( b )) = χ k m . If α > then for some β we find χ k ( m − 1) = β + ω · χ k ( b − 1) + p < β + ω · χ k ( b )) = χ k m .Case 2.2.2. b = 0 . Then for some p we find m − A kα k, (0) − A α k, (( A k − α k, (0) − p where the latter is in k normal form. If α = 0 or α = 1 we can argue as inthe proof of the first claim of the second assertion. If α = 2 then χ k ( m − 1) = χ k ( A ( A k − (0) − p ) = ω · (1+ χ k ( A k − (0) − p ) < ω ω = χ k ( A (0)) = χ k m .If α > then χ k ( m − 1) = ψ k ( A α k, ( A k − α k, (0) − 1) + p ) = ω ω +( − α k, ) + ω · ω ω +( − α k, ) + . . . ) < ω ω +( − α ) ) = χ k m since α k, < α .19roof of the first claim in the third assertion by induction on m . Here we use thefirst claim of the first assertion implicitly.If l > then the claim follows by applying the induction hypothesis to m − .Case 1. b > . Then, by induction hypothesis, n := A α ( b − is in k normal formand so ψ k n = ψ ( ω + ( − ψ k α )) + ω · ψ k ( b − 1) + p is in ψ normal form. Hence G ( ω +( − ψ k α )) < ω +( − ψ k α ) . Hence ψ k m = ψ ( ω +( − ψ k α ))+ ω · ψ k b + p is in ψ normal form. Let us show that ψ k m is also in Cantor normal form. Assumethat b has normal form A β ( c ) + q . Then β ≤ α since m is in k -normal form. Thismeans that ψ k ( ω + ( − ψ k β )) ≤ ψ k ( ω + ( − ψ k α ) . By induction hypothesis ψ k b = ψ k ( ω + ( − ψ k β )) + ωψ k c + q is in Cantor normal form. This yields that ψ k m is in Cantor normal form.Case 2. b = 0 and α = β + 1 is a successor.Write β + 1 = λ + r with λ = 0 or λ a limit. Then ψ k ( β + 1) = ψ k λ + ψ k r and ψ k β = ψ k λ + ψ k ( r − .We have to show that G ( ω + ( − ψ k ( β + 1)) < ω + ( − ψ k ( α ) .We find G ( ω + ( − ψ k ( β + 1)) = { } Gψ k λ ∪ Gψ k r . Since A β (0) is in k normalform the induction hypothesis yields that ψ ( ω + ( − ψ k β )) is in normal form. Hence Gψ k λ ⊆ Gψ k β ⊆ G ( ω + ( − ψ k β )) < ω + ( − ψ k β ) < ω + ( − ψ k α ) .Now write r = A γ ( c ) + d in k normal form. Then ψ k r = ψ ( ω + ( − ψ k γ )) + ω · ψ k c + d is in ψ normal form and in Cantor normal form by induction hypothesis.Therefore Gψ k c ≤ Gψ ( ω + ( − ψ k γ ) ≤ ω + ( − ψ k γ . Since m is in k normalform we find γ ≤ α . Since γ is a strict subterm of α we have γ < α We find Gψ k r ≤ ω + ( − ψ k γ ) < ω + ( − ψ k α ) . Hence ψ k m is in ψ normal form.Let us show that ψ k m is also in Cantor normal form. Assume that b has normalform A δ ( e ) + q . Then δ ≤ α since m is in k -normal form. This means that ψ k ( ω +( − ψ k δ )) ≤ ψ k ( ω + ( − ψ k α ) . By induction hypothesis ψ k b = ψ k ( ω + ( − ψ k δ )) + ωψ k c + q is in Cantor normal form. This yields that ψ k m is in Cantor normalform.Case 3. b = 0 and α = λ [[ ω β +1 · r ]] is a limit where α [ x ] = λ [[ ω β +1 · ( r − 1) + ω β · x ]] . Then n := A α k − , (0) is in k normal form and so ψ k n is in ψ nor-mal form by induction hypothesis. We have n = A λ [[ ω β +1 · ( r − ω β · A αk − , (0)]] (0) ≥ max { mc ( α [0]) , mc ( β + 1) , r } .Let p := mc ( α ) Then p = max { mc ( α [0]) , mc ( β + 1) , mc ( r ) } and p ≤ n .We find ψ k p ≤ ψ k n = ψ ( ψ k α k − , ) where the letter is in ψ normal form byinduction hypothesis so that G ( ψ ( ψ k α k − , )) ≤ ψ k α k − , .This yields G ( ω + ( − ψ k α )) ≤ G ( ψ k p ) ≤ G ( ψ k n ) = G ( ψ ( ψ k α k − , )) ≤ ψ k α k − , < ψ k α . Hence ψ k m is in ψ normal form.Proof of the second claim in the third assertion by induction on m . The details aresimilar to the details in the proof of the first claim of the third assertion. (cid:3) We define the corresponding Goodstein sequences similarly as before. Definition 10 Let m < ω .1. Put m := m. Assume recursively that m l is defined and m l > . Then m l +1 = m l [ l + 3 ← l + 4] − . If m l = 0 then m l +1 := 0 . . Put ˜ m := m. Assume recursively that ˜ m l is defined and ˜ m l > . Then ˜ m l +1 =˜ m l [[ l + 3 ← l + 4]] − . If ˜ m l = 0 then ˜ m l +1 := 0 .3. Put m := m. Assume recursively that m l is defined and m l > . Then m l +1 = m l { l + 3 ← l + 4 } − . If ˜ m l = 0 then m l +1 := 0 . We are going to prove that the first principle is a giant Goodstein principle, that thesecond principle is an illusionary giant Goodstein principle, and that somewhat sur-prisingly t the third principle is a giant Goodstein principle. Theorem 3 1. For all m < ω there exists an l < ω such that m l = 0 . This isprovable in PRA + TI( η ) .2. For all m < ω there exists an l < ω such that ˜ m l = 0 . This is provable in PRA + TI( ε ) .3. For all m < ω there exists an l < ω such that m l = 0 . This is provable in PRA + TI( η ) . Proof. Define o ( m, l ) := ψ l +3 ( m l ) . If m l +1 > then by the previous lemmata o ( m, l + 1) = ψ l +4 ( m l +1 )= ψ l +4 ( m l [ l + 3 ← l + 4] − < ψ l +4 ( m l [ l + 3 ← l + 4])= ψ l +3 ( m l )= o ( m, l ) This proves the first assertion. The second and third assertion are proved similarly bynow using χ k ( ξ k resp.) instead of ψ k .Let us now prove the independence results. For this we use canonical fundamentalsequences for the elements in OT which go back to Buchholz [3]. We put x ] := 0 and ( β + 1)[ x ] := β . We put ( ψ x ] := 0 and ( ψ ( β + 1))[ x ] := ψβ · x . If λ isof cofinality ω the we recursively put ( ψλ )[ x ] := ψ ( λ [ x ]) . Then ( ψω )[ x ] = ω x . If λ is of cofinality Ω then we put ( ψλ )[ x ] := ψλ x, where λ , := λ [0] and λ l +1 , := λ [ ψλ l, ] . We hereby assume that for α ≥ Ω we agree by recursion on the following.If α = Ω β · δ + γ and γ is a limit then the cofinality of α is the cofinality of γ and α [ ξ ] := Ω β · ξ + γ [ ξ ] . If α = Ω β · δ and δ is a limit then the cofinality of α is thecofinality of δ and α [ ξ ] := Ω β · δ [ ξ ] . If α = Ω β ( δ + 1) and β is a limit then thecofinality of α is the cofinality of β and α [ ξ ] := Ω β · δ + Ω β [ ξ ] . If α = Ω β +1 ( δ + 1) then α has cofinality Ω and α [ ξ ] = Ω β +1 δ + Ω β · ξ. These fundamental sequences have the Bachmann property [13]. Moreover, by [13]we have that if ψα is in ψ normal form then ( ψα )[ x ] is in ψ normal form, too. Thesetwo results go back to Buchholz [3].The fundamental sequences for OT ′ are defined in complete analogy. The onlydifference is the clause Ψ( α + 1)[ x ] = Ψ α + ω · x . As before one can prove that if Ψ α is in Ψ normal form then (Ψ α )[ x ] is in Ψ normal form, too.We call a natural number m of k -successor type if m = k -nf A α ( b ) + l where α = 0 , or α > and l > . For those m the ordinal ψ k m is a successor ordinal. We21all a natural number m of k -limit type if m = k -nf A α ( b ) with α ≥ . For those m the ordinal ψ k m is a limit ordinal of countable cofinality.For technical reasons we need a specific description of ordinals below ε in termsof certain place holders. Definition 11 We define by recursion on α a context λ k ( α ) for < α < ε which arenot of the form β + q with β ∈ Lim ∪ { } and q is of k successor type. Assume that α = ω α · m + · · · + ω α n · m n where α > . . . > α n and < m , . . . , m n .Case 1. m n is of k limit type. Then λ k ( α ) := ω α · m + · · · + ω α n − · m n − + [[ · ]] .Then λ k ( α )[[ ω α n · m n ]] = α .Case 2. m n is of k successor type. We have excluded the case α n = 0 by assump-tion.Case 2.1. α n = β + q with β ∈ Lim ∪ { } and q is of k successor type.Then λ k ( α ) := ω α · m + · · · + ω α n − · m n − + [[ · ]] . Then λ k ( α )[[ ω α n · m n ]] = α .Case 2.2. α n ∈ Lim or α n = β + q with β ∈ Lim ∪ { } and q is of k limit type.By recursion we can assume that λ k ( α n ) is defined and that λ k ( α )[[ ω γ · p ]] = α where p is of k limit type, or p is of k successor type and γ = β + r with β ∈ Lim ∪ { } and r is of k successor type. λ k ( α ) := ω α · m + · · · + ω α n · ( m n − 1) + ω λ k ( α n ) . Then λ k ( α )[[ ω γ · p ]] = α . Then λ k ( α )[[ ω γ · p ]] = α where p is of k limit type, or p is of k successor type and γ = β + q with β ∈ Lim ∪ { } and q is of k successor type. It is easy to show that ( λ k ( α )[[ ω γ · p ]]) ′ = ( λ k ( α )) ′ [[ ω γ ′ · p ′ ]] and ψ k ( λ k ( α )[[ ω γ · p ]]) = ( ψ k ( λ k ))[[Ω ψ k γ · ψ k p ]] .This is because if m is of k successor type then ( m − ′ +1 = m ′ and ψ k ( m − ψ k m .If λ k ( α )[[ ω γ · p ]] = α and p is of k limit type then ψ k α is a limit of countablecofinality and ( ψ k α )[ x ] = ( ψ k ( λ k ))[[Ω ψ k γ · ( ψ k p )[ x ]]] . Lemma 15 . Assume that m > .1. ψ k m > ψ k +1 ( m ′ − ≥ ( ψ k m )[ k − ,2. χ k m > χ k +1 ( m ′ − ≥ ( χ k m )[ k − ,3. ξ k m > ξ k +1 ( m ′ − ≥ ( ξ k m )[ k − . Proof. Clearly ψ k m = ψ k +1 ( m ′ ) > ψ k +1 ( m ′ − . This argument works also for χ k and ξ k so that we only need to show the second inequality in every assertion.So let us first prove the second inequality of the first assertion. If < m < k then ψ k +1 ( m ′ − 1) = m − ψ k m )[ k − . Assume that m = k -nf A α ( b ) + l ≥ k . Then α > .Case 1. l > . Then m − k -nf A α ( b ) + l − . Then ( ψ k m )[ k − 2] =( ψ k ( A α ( b )) + l )[ k − 2] = ψ k ( A α ( b )) + l − and ψ k +1 ( m ′ − 1) = ψ k +1 ( B α ′ ( b ′ ) + l − 1) = ψ k ( A α ( b )) + l − .Case 2. l = 0 .Let us first consider the case α = 1 . If b = 0 then The ψ k A (0)[ k − 2] = ω [ k − ≤ k = ψ k +1 ( k + 1 − 1) = ψ k +1 ( m ′ − . 22f b > then ψ k A ( b )[ k − 2] = ( ω (1+ ψ k b ))[ k − ≤ ω (1+ ψ k b [ k ])+ k . Moreoverfor some p ≥ k the induction hypothesis yields ψ k +1 ( B ( b ′ ) − ≥ ψ k +1 ( B ( b ′ − 1) + p ) ≥ ω (1 + ψ k +1 ( b ′ − p ≥ ω (1 + ψ k b [ k − k .Let us second consider the case α = 2 . If b = 0 then The ψ k A (0)[ k ] = ψ ( ω + ( − ψ k k − 2] = ( ψω )[ k − 2] = ψ ( ω [ k − ≤ ω k ≤ ψ k +1 ( B k (0)) ≤ ψ k +1 ( B k +11 (0) − 1) = ψ k +1 ( m ′ − .If b > then ψ k A ( b )[ k − 2] = ( ψω + ω ( ψ k b ))[ k − ≤ ψω + ω (1 + ψ k b [ k − k . Moreover for some p ≥ k the induction hypothesis yields ψ k +1 B ( b ′ ) − ≥ ψ k +1 ( B ( b ′ − 1) + p ) ≥ ψω + ω ( ψ k +1 ( b ′ − p ≥ ψω + ω (1 + ψ k b [ k − k .From now on we assume that α ≥ .Case 2.1. b > . Then for some l ≥ k m ′ − B α ′ ( b ′ ) − B α ′ ( b ′ − 1) + l where the latter is in k + 1 normal form.Here the induction hypothesis yields ( ψ k m )[ k − 2] = ( ψ ( ω + ( − ψ k α )) + ω · ( ψ k b ))[ k − ≤ ψ ( − ψ k α + ω · ( ψ k b [ k − k ≤ ψ ( ω + ( − ψ k +1 ψ k +1 α ′ )) + ω · ( ψ k +1 ( b ′ − p = ψ k +1 ( B α ′ ( b ′ − p ) = ψ k +1 ( B α ′ ( b ′ ) − 1) = ψ k +1 ( m ′ − .Case 2.2. b = 0 .Then B α ′ (0) − B α ′ k +1 , (0) k +1 − > B kα ′ k, (0) where the latter is in k + 1 normal form.Case 2.2.1. α is a limit.We now analyse the specific forms for α according to Definition 11.Case 2.2.1.1. α = λ k ( α )[[ ω β + p · q ]] where p and q are of k successor type and β isa limit or zero. Then ψ k α = ( ψ k λ k ( α ))[[Ω ψ k β + ψ k p · ψ k q ]] has uncountable cofinalitysince ψ k p = ψ k ( p − 1) + 1 and ψ k q = ψ k ( q − 1) + 1 .Then ( ψ k m )[ k ] = ( ψ ( ω + ( − ψ k α )))[ k ] = ( ψ ( ψ k α ))[ k ] = ψ (( ψ k ( α )) k, ) and ψ k +1 ( m ′ − 1) = ψ k +1 ( B α ′ (0) − 1) = ψ k +1 ( B k +1( α ′ ) k +1 , (0) − ≥ ψ k +1 ( B ( α ′ ) k, (0)) = ψ ( ω + ( − ψ k +1 ( α ′ k, ))) = ψ ( ψ k +1 ( α ′ k, )) since B ( α ′ ) k, (0) is in k + 1 normalform because B α ′ (0) is in k + 1 normal form.We claim that ( ψ k α ) l, ) = ψ k +1 ( α ′ l,k +1 , ) for l ≤ k . This then yields the assertionin this case.We prove this by induction on l . Note first that α ′ = ( λ k ( α )) ′ [[ ω β ′ + p ′ · q ′ ]] where p ′ = ( p − ′ +1 and q ′ = ( q − ′ +1 . Further note that ψ k α = ( ψ k λ k ( α ))[[Ω ψ k β + ψ k p · ψ k q ]] .Assume that l = 0 . We find α ′ , = α ′ [0] = ( λ k ( α ) ′ )[[ ω β ′ + p ′ · ( q − ′ ]] and ( ψ k α )[0] = ( ψ k λ k ( α ))[[Ω ψ k β + p · ψ k ( q − . This yields ( ψ k α ) , = ( ψ k α )[0]= ( ψ k λ k ( α ))[[Ω ψ k β + ψ k p · ψ k ( q − ψ k +1 ( λ k ( α )) ′ )[[Ω ψ k +1 β ′ + ψ k +1 p ′ · ψ k +1 ( q − ′ ]]= ψ k +1 ( α ′ [0])= ψ k +1 ( α ′ ,k +1 , ) . l . Then ( ψ k α ) l +1 , = ψ k λ k ( α ))[[Ω ψ k β + ψ k p · ψ k ( q − 1) + Ω ψ k β + ψ k ( p − · ψ (( ψ k α ) l, )]]= ( ψ k +1 ( λ k ( α )) ′ )[[Ω ψ k +1 β ′ + ψ k +1 p ′ · ψ k +1 ( q − ′ + Ω ψ k +1 β ′ + ψ k +1 ( p − ′ · ψ ( ψ k +1 ( α ′ l,k +1 , ))]]= ψ k +1 ( λ k ( α ) ′ [[ ω β ′ + ψ k +1 p ′ · ψ k +1 ( q − ′ + ω β ′ + ψ k +1 ( p − ′ · B α ′ l,k +1 , (0)]]= ψ k +1 ( α ′ l +1 ,k +1 , ) . Case 2.2.1.2. α = λ k ( α )[[ ω γ · q ]] with q of k limit type. Then q ′ of k + 1 limittype Then ψ k α = ( ψ k λ k ( α ))[[Ω ψ k γ · ψ k q ]] is a limit of countable cofinality and α ′ =( λ k ( α )) ′ [[ ω γ ′ · q ′ ]] where q ′ ≥ k + 1 .Since α is a limity we have ( − ψ k α )[ k − 2] = − ψ k α [ k − and we find ( ψ k m )[ k ]= ( ψ ( ω + ( − ψ k α ))[ k ]= ψ ( ω + ( − ψ k α ))[ k ])= ψ ( ω + ( − ψ k λ k ( α )[[Ω ψ k γ · ( ψ k q ))[ k − ≤ ψ ( ω + ( − ψ k +1 ( λ k ( α ) ′ [[Ω ψ k +1 γ ′ · ψ k +1 ( q ′ − ψ k +1 ( B λ k ( α ) ′ [[ ω γ ′ · ( q ′ − (0)) ≤ ψ k +1 ( B α ′ (0) − since λ k ( α ) ′ [[ ω γ ′ · ( q ′ − (cid:22) α ′ .Case 2.2.2. α is a successor say α = β + r with r > and β is zero or a limit. If β = 0 then r ≥ .If α = k then ( ψ k m )[ k − 2] = ψ k ( ω + ( − ψ k k ))[ k − 2] = ψ ( ω + ω )[ k − 2] = ψ ( ω + ( − k )) = ψ ( ω + ( − ψ k +1 ( α ′ − ψ k +1 B α ′ − (0) ≤ ψ k +1 ( B α ′ (0) − . Now assume α = k so that ( − ψ k α )[ k − 2] = − ψ k α [ k − .Here the induction hypothesis yields ( ψ k m )[ k − ψ ( ω + ( − ψ k ψ k α ))[ k − ≤ ψω + ( − ψ k ( β ) + ψ k ( r )[ k − · k ≤ ψ ( ω + ( − ψ k +1 ( β ′ ) + ψ k +1 ( r ′ − · k = ψ ( ω + ( − ψ k +1 ( β ′ + r ′ − · k ≤ ψ k +1 ( B kβ ′ + r ′ − (0)) ≤ ψ k +1 ( B k +1 α ′ k +1 , (0) − < ψ k +1 ( B α ′ (0) − 1) = ψ k +1 ( m ′ − . The second equality in the third assertion is proved by a similar induction on m .The differences will be very small and so we skip most of the proof. Let m = k -nf A α ( k, b ) + l . The case l > is as before. The case b > is also similar as before. But24or b > we use the fact that Ψ( γ + 1) = (Ψ γ ) + ω to model the ξ k interpretation.The fact that Ψ( γ + 1) = (Ψ γ ) + ω is also used to model the ξ k interpretation in thecase α = β + r with r > and β is zero or a limit. The fact that we use k − asargument of the fundamental sequence in the assertion has to do with the degeneratecase α = k, b = 0 , l = 0 where we need ( ω · ( − ω ))[ k − 1] = ω · ( k − 1) = ω · ( − ψ k +1 ( k ′ )) .Let us now prove second equality in the second assertion.If < m < k then χ k +1 ( m ′′ − 1) = m − χ k m )[ k − . Assume that m = k -nf A α ( k, b ) + l ≥ k . Then α > .Case 1. l > . Then m − k -nf A α ( k, b ) + l − . Then ( χ k m )[ k − 2] =( χ k ( A α ( k, b ))+ l )[ k − 2] = χ k ( A α ( k, b ))+ l − and χ k +1 ( m ′′ − 1) = χ k +1 ( B α ( b ′′ )+ l − 1) = χ k ( A α ( k, b )) + l − .Case 2. l = 0 .Case 2.1. b > .If α = 1 then the induction hypothesis yields ( χ k m )[ k − 2] = ω (1 + χ k b )[ k − ≤ ω (1 + ( χ k b )[ k − k ≤ ω (1 + ( χ k +1 ( b ′′ − k = χ k +1 ( B α ′ ( b ′′ − 1) + k ≤ χ k +1 ( B α ( b ′′ ) − 1) = χ k +1 ( m ′′ − .If α > then the induction hypothesis yields ( χ k m )[ k − 2] = ( ω ω +( − α ) + ω · χ k b )[ k − ≤ ω ω +( − α ) + ω · (( χ k b )[ k − k ≤ ω ω +( − α ) + ω · (( χ k +1 b ′′ − k = χ k +1 ( B α ( b ′′ − 1) + k ≤ χ k +1 ( B α ( b ′′ ) − 1) = χ k +1 ( m ′′ − .Case 2.2. b = 0 . If α = 1 then m = A (0) = k and ( χ k m )[ k − 2] = ω [ k − ≤ k = χ k +1 ( k + 1 − 1) = ( χ k +1 ( m ′′ − .If α = 2 then ( χ k m )[ k ] = ( ω ω +( − ) + ω · k − 2] = ω ω [ k − ≤ ω k = ω (1 + χ k +1 ( B k − (0)) = ψ k +1 ( B k (0)) ≤ χ k +1 ( B k +11 (0) − 1) = χ k +1 ( m ′′ − .Note that χ k +1 B l (0) = ω l holds by induction on l for < l ≤ k .If α > then ( χ k m )[ k ] = ( ω ω +( − α ) )[ k − ≤ ( ω ω + − α )[ k ]) ) · k ≤ ( ω ω +( − α k, ) ) · k = χ k +1 ( B kα k, (0)) ≤ χ k +1 ( B α (0) − ≤ χ k +1 ( B α (0) ′′ − χ k +1 ( m ′′ − Note that ψ k +1 ( B lα k, (0)) = ω ω +( − α k, ) · l holds by induction on l for < l ≤ k . Theorem 4 ID ( ∀ m )( ∃ l )[ m l = 0] .2. PA ( ∀ m )( ∃ l )[ ˜ m l = 0] .3. PA ( ∀ m )( ∃ l )[ m l = 0] . Proof. Let ω := ω and ω r +1 := ω ω r . Let m ( r ) := A ω r (3 , . Let Ω := Ω and Ω r +1 := Ω Ω r . 25hen ψ ( m ( r )) = ψ (Ω r ) for r ≥ . We claim that o ( m ( k ) , l ) ≥ ψ (Ω r )[1] . . . [ l ] .Proof of the claim. Write m for m ( r ) . For o ( m, l ) > we have o ( m, l ) > o ( m, l +1) = ψ l +4 ( m l [ l + 3 ← l + 4] − ≥ ( ψ l +3 ( m l ))[ l + 1] = ( o ( m, l )[ l + 1] . The Bach-mann property yields o ( m, l + 1) ≥ ( o ( m, l )[ l + 1] . The induction hypothesis yields o ( m, l ) ≥ ψ (Ω r )[1] . . . [ l ] hence o ( m, l )[ l + 1] ≥ ψ (Ω r )[2] . . . [ l ][ l + 1] . Therefore o ( m, l + 1) ≥ ( o ( m, l )[ l + 1] ≥ ψ (Ω r )[1] . . . [ l ][ l + 1] .Therefore the least l such that o ( m, l ) = 0 is at least as big as the least l such that ψ (Ω r )[1] . . . [ l ] = 0 . The result follows from ID