Ackermannian Goodstein sequences of intermediate growth
aa r X i v : . [ m a t h . L O ] A p r Ackermannian Goodstein sequences ofintermediate growth
David Fern´andez-Duque ∗ and Andreas Weiermann † Department of Mathematics: Analysis, Logic and DiscreteMathematics, Ghent UniversityApril 21, 2020
Abstract
The original Goodstein process proceeds by writing natural numbersin nested exponential k -normal form, then successively raising the baseto k + 1 and subtracting one from the end result. Such sequences al-ways reach zero, but this fact is unprovable in Peano arithmetic. In thispaper we instead consider notations for natural numbers based on theAckermann function. We define two new Goodstein processes, obtainingnew independence results for ACA ′ and ACA +0 , theories of second orderarithmetic related to the existence of Turing jumps. Goodstein’s principle [6] is arguably the oldest example of a purely number-theoretic statement known to be independent of PA , as it does not require thecoding of metamathematical notions such as G¨odel’s provability predicate [4].The proof proceeds by transfinite induction up to the ordinal ε [5]. PA doesnot prove such transfinite induction, and indeed Kirby and Paris later showedthat Goodstein’s principle is unprovable in PA [8].Goodstein’s original principle involves the termination of certain sequences ofnumbers. Say that m is in nested (exponential) base- k normal form if it is writtenin standard exponential base k , with each exponent written in turn in base k .Thus for example, 20 would become 2 + 2 in nested base-2 normal form.Then, define a sequence ( g k (0)) m ∈ N by setting g ( m ) = m and defining g k +1 ( m )recursively by writing g k ( m ) in nested base-( k + 2) normal form, replacing everyoccurrence of k + 2 by k + 3, then subtracting one (unless g k ( m ) = 0, in whichcase g k +1 ( m ) = 0). ∗ [email protected] † [email protected]
1n the case that m = 20, we obtain g (20) = 20 = 2 + 2 g (20) = 3 + 3 − + 3 · · g (20) = 4 + 4 · · − + 4 · · , and so forth. At first glance, these numbers seem to grow superexponentially.It should thus be a surprise that, as Goodstein showed, for every m there is k ∗ for which g k ∗ ( m ) = 0.By coding finite Goodstein sequences as natural numbers in a standard way,Goodstein’s principle can be formalized in the language of arithmetic, but thisformalized statement is unprovable in PA . Independence can be shown byproving that the Goodstein process takes at least as long as stepping downthe fundamental sequences below ε ; these are canonical sequences ( ξ [ n ]) n<ω such that ξ [ n ] < ξ for all ξ and for limit ξ , ξ [ n ] → ξ as n → ∞ . For stan-dard fundamental sequences below ε , PA does not prove that the sequence ξ > ξ [1] > ξ [1][2] > ξ [1][2][3] . . . is finite.Exponential notation is not suitable for writing very big numbers (e.g. Gra-ham’s number [7]), in which case it may be convenient to use systems of notationwhich employ faster-growing functions. In [2], T. Arai, S. Wainer and the au-thors have shown that the Ackermann function may be used to write naturalnumbers, giving rise to a new Goodstein process which is independent of thetheory ATR of arithmetical transfinite recursion; this is a theory in the lan-guage of second order arithmetic which is much more powerful than PA . Themain axiom of ATR states that for any set X and ordinal α , the α -Turing jumpof X exists; we refer the reader to [13] for details.The idea is, for each k ≥
2, to define a notion of Ackermannian normal formfor each m ∈ N . Having done this, we can define Ackermannian Goodsteinsequences analogously to Goodstein’s original version. The normal forms usedin [2] are defined using an elaborate ‘sandwiching’ procedure first introducedin [14], approximating a number m by successive branches of the Ackermannfunction. In this paper, we consider simpler, and arguably more intuitive, nor-mal forms, also based on the Ackermann function. We show that these giverise to two different Goodstein-like processes, independent of ACA ′ and ACA +0 ,respectively. As was the case for ATR , these are theories of second order arith-metic which state that certain Turing jumps exist. ACA ′ asserts that, for all n ∈ N and X ⊆ N , the n -Turing jump of X exists, while ACA +0 asserts that its ω -jump exists; see [13] for details. The proof-theoretic ordinal of ACA ′ is ε ω [1],and that of ACA +0 is ϕ (0) [9]; we will briefly review these ordinals later in thetext, but refer the reader to standard texts such as [10, 12] for a more detailedtreatment of proof-theoretic ordinals. Let us fix k ≥ efinition 2.1. For a, b ∈ N we define A a ( k, b ) by the following recursion.1. A ( k, b ) := k b ,2. A a +1 ( k,
0) := A ka ( k, · )(0) ,3. A a +1 ( k, b + 1) := A ka ( k, · )( A a +1 ( k, b )) . Here, the notation A ka ( k, · ) refers to the k -fold composition of the function x A a ( k, x ). It is well known that for every fixed a , the function b A a ( k, b )is primitive recursive and the function a A a ( k,
0) is not primitive recursive.We use the Ackermann function to define k normal forms for natural numbers.These normal forms emerged from discussions with Toshiyasu Arai and StanWainer, which finally led to the definition of a more powerful normal formdefined in [14] and used to prove termination in [2]. Lemma 2.2.
Let k ≥ . For all c > , there exist unique a, b, m, n ∈ N suchthat1. c = A a ( k, b ) · m + n ,2. A a ( k, ≤ c < A a +1 ( k, , 3. A a ( k, b ) ≤ c < A a ( k, b + 1) , and4. n < A a ( k, b ) . We write c = nf A a ( k, b ) · m + n in this case. This means that we havein mind an underlying context fixed by k and that for the number c we haveuniquely associated the numbers a, b, m, n . Note that it could be possible that A a +1 ( k,
0) = A a ( k, b ), so that we have to choose the right representation for thecontext; in this case, item 2 guarantees that a is chosen to take the maximalpossible value.By rewriting iteratively b and n in such a normal form, we arrive at theAckermann k -normal form of c . If we also rewrite a iteratively, we arrive atthe nested Ackermann k -normal form of c . The following properties of normalforms are not hard to prove from the definitions. Lemma 2.3. A ℓa ( k, is in k -normal form for every ℓ such that < ℓ ACA ′ ACA ′ . Wedo so by working with unnested Ackermannian normal forms. Such normalforms give rise to the following notion of base change. Definition 3.1. Given k ≥ and c ∈ N , define c [ k ← k + 1] by:1. k ← k + 1] := 0 .2. c [ k ← k +1] := A a ( k +1 , b [ k ← k +1]) · m + n [ k ← k +1] if c = nf A a ( k, b ) · m + n . With this, we may define a new Goodstein process, based on unnested Ack-ermannian normal forms. Definition 3.2. Let ℓ < ω . Put b ( ℓ ) := ℓ. Assume recursively that b k ( ℓ ) isdefined and b k ( ℓ ) > . Then b k +1 ( ℓ ) = b k ( ℓ )[ k + 2 ← k + 3] − . If b k ( ℓ ) = 0 ,then b k +1 ( ℓ ) := 0 . We will show that for every ℓ there is i with b i ( ℓ ) = 0. In order to provethis, we first establish some natural properties of the base-change operation. Lemma 3.3. Fix k ≥ and let c, d ∈ N . Then:1. c ≤ c [ k ← k + 1] .2. If c < d , then c [ k ← k + 1] < d [ k ← k + 1] .Proof. The first assertion is proved by induction on c . It clearly holds for c = 0.If c = nf A a ( k, b ) · m + n then the induction hypothesis yields c = A a ( k, b ) · m + n ≤ A a ( k, b [ k ← k + 1]) · m + n [ k ← k + 1] = c [ k ← k + 1] . The second assertion is harder to prove. The proof is by induction on d with a subsidiary induction on c . The assertion is clear if c = 0. Let c = nf A a ( k, b ) · m + n and d = nf A a ′ ( k, b ′ ) · m ′ + n ′ . We distinguish cases accordingto the position of a relative to a ′ , the position of b relative to b ′ , etc. Case 1 ( a < a ′ ). We sub-divide into two cases. Case 1 .1 ( A a +1 ( k, < d ). Then, the induction hypothesis applied to c
0) yields c [ k ← k + 1] < A a +1 ( k + 1 , < A a ′ ( k + 1 , b ′ [ k ← k + 1]) · m ′ + n ′ [ k ← k + 1] = d [ k ← k + 1]. Case 1 .2 ( A a +1 ( k, 0) = d ). In this case, a + 1 = a ′ , b ′ = 0, m ′ = 1, and n ′ = 0. We have A a ( k, b ) ≤ c < A a +1 ( k, 0) = A a ( k, A k − a ( k, · )(0)). For ℓ < k we have that A ℓa ( k, 0) is in k -normal form by Lemma 2.3. Thus the inductionhypothesis yields b [ k ← k + 1] < A k − a ( k + 1 , · )(0). The number A a ( k, b ) is in k -normal form and so the induction hypothesis applied to n < A a ( k, b ) yields4 [ k ← k + 1] < A a ( k + 1 , b [ k ← k + 1]). Moreover we have that m < A a +1 ( k, c [ k ← k + 1] = A a ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1] ≤ A a ( k + 1 , A k − a ( k + 1 , · )(0)) · A a +1 ( k, 0) + A a ( k + 1 , A k − a ( k + 1 , · )(0)) ≤ ( A ka ( k + 1 , · )(0)) + A ka ( k + 1 , · )(0) ≤ A a ( k + 1 , A ka ( k + 1 , · )(0)) = A a +1 ( k + 1 , , where the second inequality follows from A a +1 ( k, 0) = A ka ( k, · )(0) ≤ A ka ( k + 1 , · )(0)and the last from A a ( k + 1 , x ) ≥ A ( k + 1 , x ) ≥ x ≥ x + x. (1) Case 2 ( a ′ < a ). This case does not occur since then d < A a ′ +1 ( k, ≤ A a ( k, ≤ c . Case 3 ( a = a ′ and b < b ′ ). The induction hypothesis yields b [ k ← k + 1] c [ k ← k + 1] = A a ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1] ≤ A a ( k + 1 , b [ k ← k + 1])) · A a ( k, b + 1) + A a ( k + 1 , b [ k ← k + 1]) ≤ ( A ka − ( k + 1 , · )( A a ( k + 1 , b [ k ← k + 1]))) + A ka − ( k + 1 , · )( A a ( k + 1 , b [ k ← k + 1])) < A a ( k + 1 , b ′ [ k ← k + 1]) by (1),where the second inequality uses A a ( k, b + 1) = A ka − ( k, · )( A a ( k, b )) ≤ A ka − ( k + 1 , · )( A a ( k + 1 , b [ k ← k + 1])) . ase 4 ( a = a ′ and b ′ < b ). This case does not appear since otherwise d ≤ A a ( k, b ′ + 1) ≤ c . Case 5 ( a = a ′ and b ′ = b and m < m ′ ). Then the induction hypothesis yields c [ k ← k + 1] = A a ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1] < A a ( k + 1 , b [ k ← k + 1])) · m + A a ( k + 1 , b [ k ← k + 1]) ≤ A a ( k + 1 , b [ k ← k + 1])) · m ′ ≤ d [ k ← k + 1] . Case 6 ( a = a ′ and b ′ = b and m ′ < m ). This case is not possible given theassumptions. Case 7 ( a = a ′ and b ′ = b and m ′ = m ). Then n < n ′ and the inductionhypothesis yields c [ k ← k + 1] = A a ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1] < A a ( k + 1 , b [ k ← k + 1]) · m + n ′ [ k ← k + 1] = d [ k ← k + 1] . Thus, the base-change operation is monotone. Next we see that it alsopreserves normal forms. Lemma 3.4. If c = A a ( k, b ) · m + n is in k -normal form, then c [ k ← k + 1] = A a ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1] is in k + 1 normal form.Proof. Assume that c = nf A a ( k, b ) · m + n . Then, c < A a +1 ( k, c < A a ( k, b +1), and n < A a ( k, b ). Clearly, A a ( k + 1 , ≤ c [ k ← k + 1]. By Lemma 2.3, A a +1 ( k, 0) is in k -normal form, so that by Lemma 3.3, c < A a +1 ( k, 0) yields c [ k ← k + 1] < A a +1 ( k + 1 , A a ( k, b ) is in k -normal form, Lemma 3.3yields n [ k ← k + 1] < A a ( k + 1 , b [ k ← k + 1]). It remains to check that we alsohave c [ k ← k + 1] < A a ( k + 1 , b [ k ← k + 1] + 1).If a = 0, then c = nf A a ( k, b ) · m + n means that c = k b · m + n with m < k and n < k b . Then, m < k + 1 and n [ k ← k + 1] < ( k + 1) b [ k ← k +1] . Thus c [ k ← k + 1] = ( k + 1) b [ k ← k +1] · m + n [ k ← k + 1] < ( k + 1) b [ k ← k +1]+1 and thus c [ k ← k + 1] = nf ( k + 1) b [ k ← k +1] · m + n [ k ← k + 1] . In the remaining case, we havefor a > c [ k ← k + 1] = A a ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1] < A a ( k + 1 , b [ k ← k + 1]) · A a ( k, b + 1) + A a ( k + 1 , b [ k ← k + 1]) ≤ A a ( k + 1 , b [ k ← k + 1]) · A a ( k, b [ k ← k + 1] + 1) + A a ( k + 1 , b [ k ← k + 1]) ≤ ( A ka − ( k, · ) A a ( k + 1 , b [ k ← k + 1])) + A ka − ( k, · ) A a ( k + 1 , b [ k ← k + 1]) < A k +1 a − ( k + 1 , · ) A a ( k + 1 , b [ k ← k + 1]) by (1)= A a ( k + 1 , b [ k ← k + 1] + 1) . So A a ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1] is in k + 1-normal form.6hese Ackermannian normal forms give rise to a new Goodstein process.In order to prove that this process is terminating, we must assign ordinals tonatural numbers, in such a way that the process gives rise to a decreasing(hence finite) sequence. For each k , we define a function ψ k : N → Λ, where Λis a suitable ordinal, in such a way that ψ k m is computed from the k -normalform of m . Unnested Ackermannian normal forms correspond to ordinals belowΛ = ε ω , as the following map shows. Definition 3.5. For k ≥ , define ψ k : N → ε ω as follows:1. ψ k .2. ψ k c := ω ε a + ψ k b · m + ψ k n if c = nf A a ( k, b ) · m + n . Lemma 3.6. If c < d < ω then ψ k c < ψ k d .Proof. Proof by induction on d with subsidiary induction on c . The assertionis clear if c = 0. Let c = nf A a ( k, b ) · m + n and d = nf A a ′ ( k, b ′ ) · m ′ + n ′ . Wedistinguish cases according to the position of a relative to a ′ , the position of b relative to b ′ , etc. Case 1 ( a < a ′ ). We have n < c < A a +1 ( k, ≤ A a ′ ( k, 0) and, since A a ′ ( k, ≤ d , the induction hypothesis yields ψ k n < ω ε a ′ + ψ k = ε a ′ . We have b < c
0) and the induction hypothesis yields ψ k b < ω ε a ′ + ψ k = ε a ′ . It follows that ε a + ψ k b < ε a ′ , hence ψ k c = ω ε a + ψ k b · m + ψ k n < ε a ′ ≤ ψ k d. Case 2 ( a > a ′ ). This case is not possible since this would imply that d b ′ ). This case is not possible since this would imply d m ′ ). This case is not possible since this would imply d = A a ( k, b ) · m ′ + n ′ ≤ A a ( k, b ) · m ≤ c < d . Case m = m ′ ). The inequality c < d yields n < n ′ and the inductionhypothesis yields ψ k n < ψ k n ′ . Hence ψ k c = ω ε a + ψ k b · m + ψ k n < ω ε a + ψ k b · m + ψ k n ′ = ψ k d . 7ur ordinal assignment is invariant under base change, in the following sense. Lemma 3.7. ψ k +1 ( c [ k ← k + 1]) = ψ k c .Proof. Proof by induction on c . The assertion is clear for c = 0 . Let c = nf A a ( k, b ) · m + n . Then, c [ k ← k + 1] = nf A a ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1],and the induction hypothesis yields ψ k +1 ( c [ k ← k + 1]) = ψ k +1 ( A a ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1])= ω ε a + ψ k +1 ( b [ k ← k +1]) · m + ψ k +1 ( n [ k ← k + 1])= ω ε a + ψ k b · m + ψ k n = ψ k c. It is well-known that the so-called slow-growing hierarchy at level ϕ ω PA + TI( ϕ ω ϕ ω ε ω = ϕ ω . Theorem 3.8. For all ℓ < ω , there exists a k < ω such that b k ( ℓ ) = 0 . This isprovable in PA + TI( ε ω ) .Proof. Define o ( ℓ, k ) := ψ k +2 b k ( ℓ ) . If b k ( ℓ ) > 0, then, by the previous lemmata, o ( ℓ, k + 1) = ψ k +3 b k +1 ( ℓ ) = ψ k +3 ( b k ( ℓ )[ k ← k + 1] − < ψ k +3 ( b k ( ℓ )[ k ← k + 1]) = ψ k +2 ( b k ( ℓ )) = o ( ℓ, k ) . Since ( o ( ℓ, k )) k<ω cannot be an infinite decreasing sequence of ordinals, theremust be some k with o ( ℓ, k ) = 0, yielding b k ( ℓ ) = 0.Now we are going to show that for every α < ε ω , PA +TI( α ) 6⊢ ∀ ℓ ∃ k b k ( ℓ ) = 0 . This will require some work with fundamental sequences. Definition 3.9. Let Λ be an ordinal. A system of fundamental sequences onΛ is a function · [ · ] : Λ × N → Λ such that α [ n ] ≤ α with equiality holding if andonly if α = 0 , and α [ n ] ≤ α [ m ] whenever n ≤ m . The system of fundamentalsequences is convergent if λ = lim n →∞ λ [ n ] whenever λ is a limit, and has the Bachmann property if whenever α [ n ] < β < α , it follows that α [ n ] ≤ β [1] . It is clear that if Λ is an ordinal then for every α < Λ there is n such that α [1][2] . . . [ n ] = 0, but this fact is not always provable in weak theories. TheBachmann property that will be useful due to the following. Proposition 3.10. Let Λ be an ordinal with a system of fundamental sequencessatisfying the Bachmann property, and let ( ξ n ) n ∈ N be a sequence of elements of Λ such that, for all n , ξ n [ n +1] ≤ ξ n +1 ≤ ξ n . Then, for all n , ξ n ≥ ξ [1][2] . . . [ n ] . roof. Let (cid:22) k be the reflexive transitive closure of { ( α [ k ] , α ) : α < ϕ (0) } .We need a few properties of these orderings. Clearly, if α (cid:22) k β , then α ≤ β .It can be checked by a simple induction and the Bachmann property that, if α [ n ] ≤ β < α , then α [ n ] (cid:22) β . Moreover, (cid:22) k is monotone in the sense that if α (cid:22) k β , then α (cid:22) k +1 β , and if α (cid:22) k β , then α [ k ] (cid:22) k β [ k ] (see, e.g., [11] fordetails).We claim that for all n , ξ n (cid:23) n ξ [1] . . . [ n ], from which the desired inequalityimmediately follows. For the base case, we use the fact that (cid:23) is transi-tive by definition. For the successor, note that the induction hypthesis yields ξ [1] . . . [ n ] (cid:22) n ξ n , hence ξ [1] . . . [ n + 1] (cid:22) n +1 ξ n [ n + 1]. Then, consider threecases. Case 1 ( ξ n +1 = ξ n ). By transitivity and monotonicity, ξ [1] . . . [ n + 1] (cid:22) n +1 ξ [1] . . . [ n ] (cid:22) n ξ n = ξ n +1 yields ξ [1] . . . [ n + 1] (cid:22) n +1 ξ n +1 . Case 2 ( ξ n +1 = ξ n [ n + 1]). Then, ξ [1] . . . [ n + 1] (cid:22) n +1 ξ n [ n + 1] = ξ n +1 . Case 3 ( ξ n [ n + 1] < ξ n +1 < ξ n ). The Bachmann property yields ξ n [ n + 1] (cid:22) ξ n +1 , and since ξ [1] . . . [ n + 1] (cid:22) n +1 ξ n [ n + 1], monotinicity and transitivityyield ξ [1] . . . [ n + 1] (cid:22) n +1 ξ n +1 .Let ω ( α ) := α and ω k +1 ( α ) = ω ω k ( α ) . Let us define the standard funda-mental sequences for ordinals less than ϕ α = ω β + γ with 0 < γ < α , then α [ k ] := ω β + γ [ k ].2. If α = ω β > β , then we set α [ k ] := 0 if β = 0, α [ k ] := ω γ · k if β = γ + 1,and α [ k ] := ω β [ k ] if β ∈ Lim.3. If α = ε β > β , then α [ k ] := ω k (1) if β = 0, α [ k ] := ω k ( ε γ + 1) if β = γ + 1,and α [ k ] := ε β [ k ] if β ∈ Lim.This system of fundamental sequences enjoys the Bachmann property [11].In view of Proposition 3.10, the following technical lemma will be crucial forproving our main independence result for ACA ′ . Lemma 3.11. Given k, c < ω with k ≥ , ψ k +1 ( c [ k ← k + 1] − ≥ ( ψ k c )[ k ] .Proof. We prove the claim by induction on c . Let c = nf A a ( k, b ) · m + n. Case 1 ( n > ψ k +1 ( c [ k ← k + 1] − 1) = ω ε a + ψ k +1 ( b [ k ← k +1]) · m + ψ k +1 ( n [ k ← k + 1] − ≥ ω ε a + ψ k ( b ) · m + ( ψ k ( n ))[ k ] = ( ω ε a + ψ k ( b ) · m + ψ k ( n ))[ k ]= ( ψ k ( A a ( k, b ) · m + n ))[ k ] = ( ψ k c )[ k ] . ase 2 ( n = 0 and m > ψ k +1 ( c [ k ← k + 1] − ψ k +1 ( A a ( k + 1 , b [ k ← k + 1]) · ( m − 1) + ψ k +1 ( A a ( k + 1 , b [ k ← k + 1]) − ≥ ψ k ( A a ( k, b ) · ( m − ψ k ( A a ( k, b )))[ k ] = ( ψ k ( A a ( k, b ) · m ))[ k ] = ( ψ k c )[ k ] . Case 3 ( n = 0 and m = 1). We consider several sub-cases. Case 3 .1 ( a > b > ψ k +1 ( c [ k ← k + 1] − 1) = ψ k +1 ( A a ( k + 1 , b [ k ← k + 1]) − ≥ ψ k +1 ( A a ( k + 1 , ( b [ k ← k + 1]) − · k ) = ω ε a + ψ k +1 ( b [ k ← k +1] − · k ≥ ω ε a +( ψ k ( b ))[ k ] · k ≥ ( ω ε a + ψ k ( b ) )[ k ] = ( ψ k c )[ k ] , since A a ( k + 1 , ( b [ k ← k + 1]) − · k is in k + 1 normal form by Lemma 2.3 andLemma 3.4. Case 3 .2 ( a > b = 0). Then, the induction hypothesis yields ψ k +1 ( c [ k ← k + 1] − 1) = ψ k +1 ( A a ( k + 1 , − 1) = ψ k +1 ( A k +1 a − ( k, · )(0) − ψ k +1 ( A a − ( k + 1 , A ka − ( k + 1 , · )(0) − ≥ ψ k +1 ( A ka − ( k + 1 , · )(0)) = ω ε a − + ψ k +1 (( A k − a − ( k +1 , · )(0))) ≥ ω ψ k +1 (( A k − a − ( k +1 , · )(0))) ≥ ω ω k − ( ε a − +1) = ( ε a )[ k ] = ( ψ k ( A a ( k, k ] = ( ψ k c )[ k ] , since A ℓa − ( k + 1 , · )(0) is in k + 1 normal form for ℓ ≤ k by Lemma 2.3 andLemma 3.4. Case 3 .3 ( a = 0 and b > Case 3 .1: ψ k +1 ( c [ k ← k + 1] − 1) = ψ k +1 ( A ( k + 1 , b ) − ψ k +1 (( k + 1) ( b [ k ← k +1] − · k + · · · + ( k + 1) · k ) ≥ ψ k +1 (( k + 1) ( b [ k ← k +1] − · k ) ≥ ω ψ k +1 ( b [ k ← k +1] − · k ≥ ω ( ψ k b )[ k ] · k ≥ ( ψ k c )[ k ] , since ( k + 1) ( b [ k ← k +1] − · k is in k + 1 normal form. Case 3 .4 ( a = 0 and b = 0). The assertion follows trivially since then c = 1.10 heorem 3.12. Let α < ε ω . Then PA + TI( α ) 6⊢ ∀ ℓ ∃ k b k ( ℓ ) = 0 . Hence ACA ′ 6⊢ ∀ ℓ ∃ kb k ( ℓ ) = 0 . Proof. Assume for a contradiction that PA + TI( α ) ⊢ ∀ ℓ ∃ k b k ( ℓ ) = 0 . Then PA +TI( α ) ⊢ ∀ ℓ ∃ k b k ( A ℓ (2 , . Recall that o ( A ℓ (2 , , k ) = ψ k +2 ( b k ( A ℓ (2 , o ( A ℓ (2 , , 0) = ε n . Lemma 3.11 and Lemma 3.6 yield o ( A ℓ (2 , , k )[ k +1] ≤ o ( A ℓ (2 , , k +1) < o ( A ℓ (2 , , k ), hence Proposition 3.10 yields o ( A ℓ (2 , , k ) ≥ o ( A ℓ (2 , . . . [ k ] . So the least k such that b k ( A ℓ (2 , k such that ε ℓ [1] . . . [ k ] = 0. But by standard results in proof theory[3], PA + TI( α ) does not prove that this k is always defined as a function of ℓ .This contradicts PA + TI( α ) ⊢ ∀ ℓ ∃ k b k ( A ℓ (2 , . ACA +0In this section, we indicate how to extend our approach to a situation where thebase change operation can also be applied to the first argument of the Acker-mann function. The resulting Goodstein principle will then be independent of ACA +0 . The key difference is that the base-change operation is now performedrecursively on the first argument, as well as the second. Definition 4.1. For k ≥ and c ∈ N , define c [ k ← k + 1] by:1. k ← k + 1] := 0 c [ k ← k + 1] := A a [ k ← k +1] ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1] if c = nf A a ( k, b ) · m + n . Note that in this section, c [ k ← k + 1] will always indicate the operation ofDefinition 4.1. We can then define a Goodstein process based on this new basechange operator. Definition 4.2. Let ℓ < ω . Put c ( ℓ ) := ℓ. Assume recursively that c k ( ℓ ) isdefined and c k ( ℓ ) > . Then, c k +1 ( ℓ ) = c k ( ℓ )[ k + 2 ← k + 3] − . If c k ( ℓ ) = 0 ,then c k +1 ( ℓ ) := 0 . Termination and independence results can then be obtained following thesame general strategy as before. We begin with the following lemmas, whoseproofs are similar to those for their analogues in Section 3. Lemma 4.3. If c < d and k ≥ , then c [ k ← k + 1] < d [ k ← k + 1] . Lemma 4.4. If c = A a ( k, b ) · m + n is in k -normal form, then c [ k ← k + 1] = A a [ k ← k +1] ( k + 1 , b [ k ← k + 1]) · m + n [ k ← k + 1] is in k + 1 normal form. 11t is well-known that the so-called slow-growing hierarchy at level Γ matchesup with the functions which are elementary in the Ackermann function, so onemight expect that the corresponding Goodstein process can be proved termi-nating in PA + TI(Γ ). This is true but, somewhat surprisingly, much less isneeded here. Indeed, nested Ackermannian normal forms are related to themuch smaller ordinal ϕ (0) by the following mapping. Definition 4.5. Given k ≥ , define a function χ k : N → ϕ (0) given by:1. χ k .2. χ k c := ω ε χka + χ k b · m + ψ k n if c = nf A a ( k, b ) · m + n . As was the case for the mappings ψ k , the maps χ k are strictly increasingand invariant under base change, as can be checked using analogous proofs tothose in Section 3. Lemma 4.6. Let c, d, k < ω with k ≥ .1. If c < d , then χ k c < χ k d .2. χ k +1 ( c [ k ← k + 1]) = χ k c . Theorem 4.7. For all ℓ < ω , there exists a k < ω such that c k ( ℓ ) = 0 . This isprovable in PA + TI( ϕ . Next, we show that for every α < ϕ PA + TI( α ) 6⊢ ∀ ℓ ∃ k c k ( ℓ ) = 0 . Forthis, we need the following analogue of Lemma 3.11. Lemma 4.8. χ k +1 ( c [ k ← k + 1] − ≥ ( χ k c )[ k ] .Proof. We proceed by induction on c . Let c = nf A a ( k, b ) · m + n . Let usconcentrate on the critical case m = 1 and n = 0, where a > b = 0.The induction hypothesis yields χ k +1 ( c [ k ← k + 1] − 1) = χ k +1 ( A a ( k + 1 , − χ k +1 ( A k +1 a [ k ← k +1] − ( k + 1 , · )(0) − ≥ χ k +1 ( A ka [ k ← k +1] − ( k + 1 , · )(0))= ω ε χk +1( a [ k ← k +1] − + ω χk +1( Ak − a [ k ← k +1] − k +1 , · )(0)) ≥ ω k ( ε χ k +1 ( a [ k ← k +1] − + 1) ≥ ω k ( ε ( χ k a )[ k ] + 1) ≥ ( ε χ k a )[ k ] = ( χ k ( A a ( k, k ] , since A ka [ k ← k +1] − ( k + 1 , · )(0) is in k + 1 normal form.The remaining details of the proof of the theorem can be carried out similarlyas before. Theorem 4.9. For every α < ϕ , PA + TI( α ) 6⊢ ∀ ℓ ∃ k c k ( ℓ ) = 0 . Hence ACA +0 6⊢ ∀ ℓ ∃ kc k ( ℓ ) = 0 . eferences [1] B. Afshari and M. Rathjen. Ordinal analysis and the infinite Ramsey theo-rem. In How the World Computes - Turing Centenary Conference and 8thConference on Computability in Europe, CiE 2012, Cambridge, UK, June18-23, 2012. Proceedings , pages 1–10, 2012.[2] T. Arai, D. 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