Predicatively unprovable termination of the Ackermannian Goodstein process

TL;DR

This paper investigates a variant of the Goodstein process based on the Ackermann function, demonstrating that while these sequences eventually terminate, their termination cannot be proven within predicative frameworks.

Contribution

It introduces an Ackermannian Goodstein process and proves its termination is unprovable in predicative systems, extending understanding of provability limits.

Findings

Ackermannian Goodstein sequences terminate

Termination is unprovable in predicative systems

Extends classical Goodstein results to Ackermannian case

Abstract

The classical Goodstein process gives rise to long but finite sequences of natural numbers whose termination is not provable in Peano arithmetic. In this manuscript we consider a variant based on the Ackermann function. We show that Ackermannian Goodstein sequences eventually terminate, but this fact is not provable using predicative means.

Equations

$$\max \{a,b\}<A_a(k,b)\le A_{a^{\prime }}(k^{\prime },b^{\prime }).$$

$$⟨\ell /k⟩m=A_{⟨\ell /k⟩a}⟨\ell /k⟩b+⟨\ell /k⟩c.$$

$$\big{[}A_{a_{1}}(b_{1}),A_{a_{1}}(b_{1}+1)\big{)}\supsetneq\ldots\supsetneq\big{[}A_{a_{n}}(b_{n}),A_{a_{n}}(b_{n}+1)\big{)}\ni m.$$

$$A_{a_j}{b}_j\le w<A_{a_j}(b_j+1).$$

$$A_{a}b+c=w<\min \{A_a(b+1),A_0{A}_{a}b\}.$$

$$A_{a-1}b\le c<A_{a}b,$$

$$b<c_{i-1}<c_i<m=A_{a-i}^k{c}_{i-1}<A_{a-i+1}{c}_{i-1}.$$

$$m-1{\approx }_k(k-1)\cdot c_a+(c_a-1).$$

$$⟨\lambda /k⟩m=A_{⟨\lambda /k⟩a}(\lambda ,⟨\lambda /k⟩b)\cdot p+⟨\lambda /k⟩q.$$

$$⟨\lambda ⟩m$$

$${>}^{\text{scih}}{B}_0⟨\lambda ⟩d\cdot (k-1)+⟨\lambda ⟩(k^d-1)=⟨\lambda ⟩(m-1).$$

$$⟨\lambda ⟩m$$

$${\ge }^{\text{scih}}⟨\lambda ⟩b\cdot k^{b-1}+⟨\lambda ⟩q\ge ⟨\lambda ⟩b\cdot p+⟨\lambda ⟩q=⟨\lambda ⟩(m-1),$$

$$⟨\lambda ⟩c_0$$

$${\le }^{\text{scih}}{B}_{⟨\lambda ⟩(a-1)}^{k(t-2)+k-1}⟨\lambda ⟩A_{a-1}b=B_{⟨\lambda ⟩(a-1)}^{k(t-2)+k-1}{B}_{⟨\lambda ⟩(a-1)}⟨\lambda ⟩b=B_{⟨\lambda ⟩(a-1)}^{k(t-1)}⟨\lambda ⟩b.$$

$$⟨\lambda ⟩b<⟨\lambda ⟩A_a^{k-1}{A}_{a_{-1}}{b}_{-1}=B_{⟨\lambda ⟩a}^{k-1}{B}_{⟨\lambda ⟩a_{-1}}⟨\lambda ⟩b_{-1},$$

$$B_{⟨\lambda ⟩a}⟨\lambda ⟩b$$

$$<B_{⟨\lambda ⟩a_{-1}-1}^{\lambda }{B}_{⟨\lambda ⟩a_{-1}}⟨\lambda ⟩b_{-1}=B_{⟨\lambda ⟩a_{-1}}(⟨\lambda ⟩b_{-1}+1),$$

$$B_{⟨\lambda ⟩a}$$

$$<B_{⟨\lambda ⟩a}^{tk+1}{B}_{⟨\lambda ⟩a}^{t-1}{B}_{⟨\lambda ⟩a+1}(⟨\lambda ⟩\check{m}-t)=B_{⟨\lambda ⟩a}^{t\lambda }{B}_{⟨\lambda ⟩a+1}(⟨\lambda ⟩\check{m}-t)=B_{⟨\lambda ⟩a+1}⟨\lambda ⟩\check{m}.$$

$$o_{i+1}m$$

$$<⟨\omega /i+3⟩⟨i+3⟩{\mathbb{G}}_{i}m=⟨\omega /i+2⟩{\mathbb{G}}_{i}m=o_{i}m,$$

$$[k]⟨\omega /k⟩m{\preccurlyeq }_1⟨\omega /k+1⟩(⟨k+1⟩m-1).$$

$$={\varphi }_{⟨\omega /\ell ⟩⟨\ell ⟩a}⟨\omega /\ell ⟩⟨\ell ⟩b+⟨\omega /\ell ⟩(⟨\ell ⟩c-1){\ge }^{\text{scih}}{\varphi }_{⟨\omega /k⟩a}⟨\omega /k⟩b+[k]⟨\omega /k⟩c$$

$$\displaystyle=[k]\big{(}{\phi}_{\langle{\omega/k}\rangle a}\langle{\omega/k}\rangle b+\langle{\omega/k}\rangle c\big{)}=[k]\langle{\omega/k}\rangle m.$$

$$⟨\omega /\ell ⟩(⟨\ell ⟩m-1)$$

$$\displaystyle=\big{(}\phi_{0}\langle{\omega/\ell}\rangle(\langle\ell\rangle b-1)\big{)}\cdot k\geq^{\text{\sc ih}}(\phi_{0}[k]\langle{\omega/k}\rangle b)\cdot k>[k]\langle{\omega/k}\rangle m.$$

$$⟨\omega /\ell ⟩(⟨\ell ⟩m-1)=⟨\omega /\ell ⟩(B_{⟨\ell ⟩a}0-1)=⟨\omega /\ell ⟩(B_{⟨\ell ⟩a-1}^{\ell }1-1)$$

$$>⟨\omega /\ell ⟩B_{⟨\ell ⟩a-1}^{k}1={\varphi }_{⟨\omega /\ell ⟩(⟨\ell ⟩a-1)}^{k}1{\ge }^{\text{scih}}{\varphi }_{[k]⟨\omega /k⟩a}^{k}1\ge [k]{\varphi }_{⟨\omega /k⟩a}0=[k]⟨\omega /k⟩m.$$

$$[k]{\varphi }_{⟨\omega /k⟩a}⟨\omega /k⟩b={\varphi }_{⟨\omega /k⟩a}[k]⟨\omega /k⟩b{\le }^{\text{scih}}{\varphi }_{⟨\omega /\ell ⟩⟨\ell ⟩a}⟨\omega /\ell ⟩(⟨\ell ⟩b-1)$$

$$=⟨\omega /\ell ⟩B_{⟨\ell ⟩a}(⟨\ell ⟩b-1)<⟨\omega /\ell ⟩(B_{⟨\ell ⟩a}⟨\ell ⟩b-1)=⟨\omega /\ell ⟩(⟨\ell ⟩m-1).$$

$$⟨\omega /\ell ⟩(⟨\ell ⟩m-1)$$

$$\ge ⟨\omega /\ell ⟩B_{⟨\ell ⟩a-1}^k⟨\ell ⟩b={\varphi }_{⟨\omega /\ell ⟩(⟨\ell ⟩a-1)}^k⟨\omega /\ell ⟩⟨\ell ⟩b$$

$${\ge }^{\text{scih}}{\varphi }_{[k]⟨\omega /k⟩a}^k⟨\omega /k⟩b\ge [k]{\varphi }_{⟨\omega /k⟩a}⟨\omega /k⟩b.$$

$${\phi}_{\alpha}{\beta}:=\begin{cases}\varphi_{\alpha}({\beta}+1)&\mbox{ if there exists a $\beta_{0}$ such that ${\beta}={\beta}_{0}+n$ and $\varphi_{\alpha}{\beta}_{0}={\beta}_{0}$},\\ \varphi_{\alpha}{\beta}&\mbox{ otherwise.}\end{cases}$$

Full text

1. Introduction

Among the greatest accomplishments of mathematical logic in the first half of the twentieth century was the identification of true arithmetical statements unprovable in Peano arithmetic ($\sf PA$): the consistency of $\sf PA$, due to Gödel [9], and transfinite induction up to the ordinal $\varepsilon_{0}$, due to Gentzen [8]. However, such statements do not clarify whether incompleteness phenomena should be pervasive in other disciplines such as combinatorics or number theory.

In contrast, Goodstein’s principle [11] is a purely number-theoretic statement simple enough to be understood by a high school student yet unprovable in $\sf PA$. While the statement makes no reference to ordinals, Goodstein’s proof uses the well-foundedness of $\varepsilon_{0}$ [10]. Kirby and Paris’ later independence proof shows that this use of $\varepsilon_{0}$ is, in some sense, essential [13]. These developments paved the road for the discovery of other combinatorial statements independent from $\sf PA$, including the Paris-Harrington theorem [14] and other Ramsey-style principles [7, 21].

A modern presentation of Goodstein’s result and proof consists of three ingredients. First, we need a notion of normal form for representing natural numbers. Say that a natural number $m>0$ has base-$k$ exponential normal form $k^{a}+b$ if $m=k^{a}+b$ and $k^{a}\leq m<k^{a+1}$. By iteratively applying this definition to $b$ we obtain a standard base-$k$ representation of $m$. Note that the exponents themselves can be recursively written in base $k$: for example, $21=2^{2^{2}}+2^{2}+2^{0}$ in base $2$.

The second ingredient is the base-change operation, which replaces every occurrence of $k$ by some $\ell>k$: formally, $\langle\ell/k\rangle 0=0$ and $\langle\ell/k\rangle(k^{a}+b)=\ell^{\langle\ell/k\rangle a}+\langle\ell/k\rangle b$. For brevity we write $\langle k+1\rangle m$ instead of $\langle k+1/k\rangle m$. With this, given a natural number $m$, we define a sequence $({\rm G}_{i}m)_{m<\alpha}$, where $\alpha\leq\omega$, recursively by ${\rm G}_{0}m=m$, ${\rm G}_{i+1}m=\langle i+3\rangle{\rm G}_{i}m-1$ if ${\rm G}_{i}m>0$, and setting $\alpha=i+1$ if ${\rm G}_{i}m=0$; if no such $i$ exists, $\alpha=\omega$. Goodstein’s principle is then the following.

Theorem 1.1** (Goodstein).**

For every $m\in\mathbb{N}$ there is $i\in\mathbb{N}$ such that ${\rm G}_{i}m=0$.

For the proof we need an additional ingredient: an interpretation of the Goodstein process in terms of ordinals below $\varepsilon_{0}$. Every non-zero $\xi<\varepsilon_{0}$ can be written uniquely in the form $\omega^{\alpha}+\beta$ with $\alpha,\beta<\xi$; this is the Cantor normal form of $\xi$. Given natural numbers $m,k$ with $m=k^{a}+b$ in base-$k$ exponential normal form and $k>1$, we may recursively define an ordinal $\langle{\omega/k}\rangle m=\omega^{\langle{\omega/k}\rangle a}+\langle{\omega/k}\rangle b$, setting $\langle{\omega/k}\rangle 0=0$. Induction on $m$ shows that $\langle{\omega/k}\rangle m<\varepsilon_{0}$ is well-defined. Now, let $({\rm G}_{i}m)_{i<\alpha}$ be the Goodstein process for $m$, and for $i<\alpha$ define $o_{i}m=\langle{\omega/i+2}\rangle{\rm G}_{i}m$. It is not too hard to check that $o_{0}m>o_{1}m>o_{2}m>\ldots.$ By the well-foundedness of $\varepsilon_{0}$ there can be no such infinite sequence, hence $\alpha<\omega$, as needed.

By coding finite Goodstein sequences as natural numbers in a standard way, Goodstein’s principle can be formalized in the language of arithmetic. However, this formalized statement is unprovable in $\sf PA$. This can be proven by showing that the Goodstein process takes at least as long as stepping down the fundamental sequences below $\varepsilon_{0}$; these are canonical sequences111We write $[n]\xi$ instead of the more standard $\xi[n]$ since it will ease notation later. $([n]\xi)_{n<\omega}$ such that $[n]\xi<\xi$ and for limit $\xi$, $[n]\xi\to\xi$ as $n\to\infty$. For standard fundamental sequences below $\varepsilon_{0}$, $\sf PA$ does not prove that the sequence $\xi>[2]\xi>[3][2]\xi>[4][3][2]\xi\ldots$ is finite.

Exponential notation falls short when attempting to represent large numbers that arise in combinatorics such as Graham’s number [12]. These numbers may instead be written in terms of fast-growing functions such as the Ackermann function $A_{a}(k,b)$; see Definition 2.1. Here $k$ is regarded as the ‘base’ and $a,b$ as parameters. There will typically be several ways to write a number in the form $A_{a}(k,b)+c$, so a suitable normal form is chosen in Definition 2.3. Our normal forms are based on iteratively approximating $m$ via a ‘sandwiching’ procedure. With this we can define the Ackermannian Goodstein process. This process always terminates, although the proof now uses the well-foundedness of the Feferman-Schütte ordinal $\Gamma_{0}$, regarded by Feferman [6] as representing the limit of predicative mathematics [22], which denies the existence of the real line as a completed totality. The ordinal $\Gamma_{0}$ is the proof-theoretic ordinal of the theory ${\sf ATR}_{0}$ of arithmetical transfinite recursion.

Let us give a brief description of ${\sf ATR}_{0}$; a more formal treatment can be found in [17].

Second-order arithmetic adds set-variables $X,Y,Z,\ldots$ and predicates $t\in X$ to the language of $\sf PA$. Quantifiers may range over first- or second-order variables. The theory ${\sf ACA}_{0}$ of arithmetical comprehension includes axioms for arithmetical operations along with the induction axiom and the comprehension scheme stating that $\{x\in\mathbb{N}:\Phi(x)\}$ is a set, where $\Phi$ is arithmetical, i.e. it contains no second-order quantifiers. Given a linear order $\prec$ and a set $X$ of ordered pairs $\langle\lambda,n\rangle$, let $X_{\lambda}=\{n\in\mathbb{N}:\langle\lambda,n\rangle\in X\}$, and $X_{{\prec}\lambda}=\{\langle\xi,n\rangle\in X:\xi\prec\lambda\}$. The theory ${\sf ATR}_{0}$ is ${\sf ACA}_{0}$ plus the scheme stating that if $\prec$ is a well-order, there exists a set $X$ such that \forall n,\lambda\big{(}n\in X_{\lambda}\leftrightarrow\Phi(n,X_{{\prec}\lambda})\big{)}, with $\Phi$ arithmetical. ${\sf ATR}_{0}$ is conservative over systems of predicative mathematics [5, 16].

2. Ackermannian Goodstein Sequences

In this section we establish the basic definitions needed to state our main results.

Definition 2.1**.**

Given $a,b,k\in\mathbb{N}$ with $k\geq 2$, we define $A_{a}(k,b)\in\mathbb{N}$ as follows. Fix $k$ and let us write $A_{a}b$ instead of $A_{a}(k,b)$. Define as an auxiliary value $A_{a}(-1)=1$. Then, $A_{a}b$ is given recursively by $A_{0}b=k^{b}$ and $A_{a+1}b=A_{a}^{k}A_{a+1}(b-1)$.

Note that $A_{0}(k,0)=1$ regardless of $k$. Aside from some trivial cases, the Ackermann function is strictly increasing on all parameters, as can be verified by an easy induction; we leave the details to the reader.

Lemma 2.2**.**

Let $a\leq a^{\prime}$, $b\leq b^{\prime}$, and $2\leq k\leq k^{\prime}$ be natural numbers. Then,

[TABLE]

If moreover $a+b+k<a^{\prime}+b^{\prime}+k^{\prime}$ and $a^{\prime}+b^{\prime}>0$, then $A_{a}(k,b)<A_{a^{\prime}}(k^{\prime},b^{\prime})$.

In order to canonically represent $m$ in the form $A_{a}b+c$, we first choose $a_{1}$ maximal such that $A_{a_{1}}0\leq m$, then $b_{1}$ maximal so that $A_{a_{1}}b_{1}\leq m$. However, $A_{a_{1}}b_{1}$ might be much smaller than $m$, even small enough that there is $a_{2}$ with $A_{a_{2}}A_{a_{1}}b_{1}\leq m$. In this case, $A_{a_{2}}b_{2}$ is a ‘better’ approximation to $m$, where $b_{2}\geq A_{a_{1}}b_{1}$ is maximal. Continuing this process, we reach the ‘best’ approximation $A_{a_{n}}b_{n}\leq m$.

Definition 2.3**.**

Fix $k\geq 2$ and let $A_{x}y=A_{x}(k,y)$. Given $m,a,b,c\in\mathbb{N}$ with $m>0$, we define $A_{a}b+c$ to be the $k$-normal form of $m$, in symbols $m\equiv_{k}A_{a}b+c$, if $m=A_{a}b+c$ and there exist sequences $a_{1},\ldots,a_{n}$ of sandwiching indices,, $b_{1},\ldots,b_{n}$ of sandwiching arguments and $m_{0},\ldots m_{n}$ of sandwiching values such that for $i<n$,

1. (1)

$m_{0}=0$; 2. (2)

$A_{a_{i+1}}m_{i}\leq m<A_{a_{i+1}+1}m_{i}$; 3. (3)

$A_{a_{i+1}}b_{i+1}\leq m<A_{a_{i+1}}(b_{i+1}+1)$; 4. (4)

$m_{i+1}=A_{a_{i+1}}{b_{i+1}}$; 5. (5)

$A_{0}m_{n}>m$, and 6. (6)

$a=a_{n}$ and $b=b_{n}$.

We denote the sequence of pairs $(a_{i},b_{i})$ by $(A_{a_{i}}b_{i})_{i=1}^{n}$ and call it the $k$-sandwiching sequence of $m$. If $m=0$, $m$ has empty sandwiching sequence and $k$-normal form [math].

We write simply sandwiching sequence when $k$ is clear from context. For our purposes $m\equiv_{k}A_{a}b+c$ is viewed as a relation between the numbers $a,b,c,m,k$; a different approach where $A_{a}b+c$ is regarded as a formal term is briefly discussed in Appendix B. Every positive integer has a unique $k$-sandwiching sequence and hence a unique normal form. This is because $a_{i+1}$ and $b_{i+1}$ are unique when defined since $A_{x}y$ is strictly increasing on both variables, and $(m_{i})_{i\leq n}$ is strictly increasing (see Lemma 3.1). Thus we must have $n\leq m$ and $m$ indeed has a finite sandwiching sequence. It is not hard to see that $1\equiv_{k}A_{0}0$ with sandwiching sequence $(A_{0}0)$.

Example 2.4**.**

Let us compute the $2$-normal form and the sandwiching sequence $(A_{a_{i}}b_{i})_{i=1}^{n}$ for $m=2^{2^{16}+1}$. Since $2^{16}=A_{1}1$, we may re-write $m$ as $A_{0}(A_{1}1+1)$. Note that $A_{1}1<m<A^{2}_{0}A_{1}1=A_{1}2<A_{1}A_{1}1=A_{2}0$. From this we obtain $A_{1}0<m<A_{2}0$ and hence $a_{1}=1$, as well as $A_{1}1<m<A_{1}2$ which yields $b_{1}=1$, so that $m_{1}=A_{1}1$. From $A_{0}A_{1}1<m<A_{1}2=A_{0}^{2}A_{1}1<A_{0}^{2}A_{1}(A_{1}1-1)=A_{1}A_{1}1$ we obtain $a_{2}=0$. Moreover, $A_{0}(2^{16}+1)=m<A_{0}(2^{16}+2)$ yields $b_{2}=2^{16}+1$. Thus we have that $m\equiv_{2}A_{0}(2^{16}+1)$. We may re-write this as $m\equiv_{2}A_{0}(A_{A_{0}0}A_{0}0+A_{0}0)$; the reader may verify that all subexpressions are also in normal form.

Remark 2.5*.*

Definition 2.3 is not the only reasonable definition for normal forms, but it has advantages over other obvious candidates; see Appendix B.

Next we define the base-change operation based on our normal forms.

Definition 2.6**.**

Given $k,\ell\geq 2$ and $m\in\mathbb{N}$ we define the base change operation $\langle\ell/k\rangle m$ recursively by setting $\langle\ell/k\rangle 0=0$ and, for $m\equiv_{k}A_{a}b+c$ setting

[TABLE]

Sometimes we abbreviate $\langle\ell/k\rangle m$ by $\langle\ell\rangle m$, in which case it is assumed that $k=\ell-1$ unless a different value for $k$ is specified.

Example 2.7**.**

Let us write $A_{x}y$ for $A_{x}(2,y)$ and $B_{x}y$ for $A_{x}(3,y)$. Recall from Example 2.4 that $2^{16}+1\equiv_{k}A_{1}1+1$. Then, $\langle 3/2\rangle(2^{16}+1)=B_{1}1+1$. We have that $B_{1}1=B_{0}^{6}1$, and hence $\langle 3/2\rangle(2^{16}+1)=3^{3^{3^{3^{27}}}}+1$.

With this we are ready to define the Ackermannian Goodstein process.

Definition 2.8**.**

Given a natural number $m$ we define a sequence $({\mathbb{G}}_{i}m)_{i<{\xi}}$ where ${\xi}\leq\omega$, by the following recursion.

1. (1)

${\mathbb{G}}_{0}m=m$; 2. (2)

if ${\mathbb{G}}_{i}m>0$ then ${\mathbb{G}}_{i+1}m=\langle i+3\rangle{\mathbb{G}}_{i}m-1$; 3. (3)

if ${\mathbb{G}}_{i}m=0$ then ${\xi}=i+1$; if no such $i$ exists, $\xi=\omega$.

The sequence $({\mathbb{G}}_{i}m)_{i<\xi}$ is the Ackermannian Goodstein sequence starting on $m$.

Our main results are then the following.

Theorem 2.9**.**

Given any $m\in\mathbb{N}$ there is $i\in\mathbb{N}$ such that ${\mathbb{G}}_{i}m=0$.

Theorem 2.10**.**

Theorem 2.9 is not provable in ${\sf ATR}_{0}$.

The rest of this article will be devoted to proving these results. The proofs use properties of the ordinal $\Gamma_{0}$ reviewed in Appendix A.

3. Properties of Normal Forms

In this section we elaborate on the properties of normal forms as given by Definition 2.3. Fix $k$ and write $A_{x}y$ for $A_{x}(k,y)$. The intuition is that we obtain the normal form of $m$ by ‘sandwiching’ it in smaller and smaller intervals, so that

[TABLE]

The following basic properties can readily be verified by the reader.

Lemma 3.1**.**

Let $(A_{a_{i}}b_{i})_{i=1}^{n}$ be the sandwiching sequence for $m$ and $(m_{i})_{i\leq n}$ be its sandwiching values.

1. (1)

If $0\leq i<n$ then $m_{i}\leq b_{i+1}<m_{i+1}$. 2. (2)

If $0<i<n$ then $a_{i}>a_{i+1}$.

Thus we have that $A_{a_{i}}b_{i}<A_{a_{i+1}}b_{i+1}$, so in order to see that (3.1) holds it remains to check that also $A_{a_{i}}(b_{i}+1)\geq A_{a_{i+1}}(b_{i+1}+1)$.

Lemma 3.2**.**

Let $(A_{a_{i}}b_{i})_{i=1}^{n}$ be the $k$-sandwiching sequence for some $m>0$. Then, for any $i\in[1,n)$,

1. (1)

$b_{i+1}<\min\{A_{a_{i+1}}^{k-1}A_{{a}_{i+1}+1}(m_{i}-1),A_{a_{i}-1}^{k-1}m_{i}\}$; 2. (2)

$A_{{a}_{i+1}}(b_{i+1}+1)\leq A_{{a}_{i+1}+1}m_{i}$, and 3. (3)

if $i>0$ then $A_{{a}_{i+1}}(b_{i+1}+1)\leq A_{{a}_{i}}(b_{i}+1)$.

Proof.

(1) That $b_{i+1}<A_{a_{i+1}}^{k-1}A_{{a}_{i+1}+1}(m_{i}-1)$ follows from $A_{a_{i+1}}^{k}A_{{a}_{i+1}+1}(m_{i}-1)=A_{a_{i+1}+1}m_{i}>m\geq A_{a_{i+1}}b_{i+1}$ and the monotonicity of $A_{a_{i+1}}$.

For the other inequality, if $a_{i+1}<a_{i}-1$ then the definition of $a_{i+1}$ yields $m<A_{a_{i}-1}m_{i}$ and there is nothing to prove. Otherwise, by Lemma 3.1.2 we would have that $a_{i+1}=a_{i}-1$. But then we observe that $A_{a_{i+1}}^{k}A_{a_{i}}b_{i}=A_{a_{i}}(b_{i}+1)>m\geq A_{a_{i+1}}b_{i+1}$, so that $b_{i+1}<A_{a_{i+1}}^{k-1}A_{a_{i}}b_{i}=A_{a_{i+1}}^{k-1}m_{i}$.

(2) By item (1), $A_{a_{i+1}+1}m_{i}=A_{{a}_{i+1}}A_{{a}_{i+1}}^{k-1}A_{a_{i+1}+1}(m_{i}-1)\geq A_{{a}_{i+1}}(b_{i+1}+1)$.

(3) Once again by item (1), $A_{a_{i}}(b_{i}+1)=A_{a_{i}-1}^{k}m_{i}\geq A_{a_{i+1}}(b_{i+1}+1)$. ∎

The following two propositions provide the basic techniques for computing the normal form of $w$ assuming $m\equiv_{k}A_{a}b+c$, where $w$ is “not too different” from $m$.

Proposition 3.3*.*

Suppose that $m$ has $k$-sandwiching sequence $(A_{a_{i}}b_{i})_{i=1}^{n}$ and $w$ and $1\leq j\leq n$ are such that

[TABLE]

Then, $(A_{a_{i}}b_{i})_{i=1}^{j}$ is an initial segment of the $k$-sandwiching sequence for $w$.

If moreover $A_{0}m_{j}>w$, then $(A_{a_{i}}b_{i})_{i=1}^{j}$ is the full $k$-sandwiching sequence for $w$ and hence $w\equiv_{k}A_{a_{j}}b_{j}+c$ for suitable $c$.

Proof.

Assume that $A_{a_{j}}b_{j}\leq w<A_{a_{j}}(b_{j}+1)$ and let $(A_{d_{i}}e_{i})_{i=1}^{r}$ be the $k$-sandwiching sequence for $w$. In view of Lemmas 3.1 and 3.2 we have for $i<j$ that $A_{a_{i+1}}m_{i}\leq A_{a_{i+1}}b_{i+1}\leq A_{a_{j}}b_{j}\leq w<A_{a_{j}}(b_{j}+1)\leq A_{a_{i+1}}(b_{i+1}+1)\leq A_{a_{i+1}+1}m_{i}$, so that $(A_{a_{i}}b_{i})_{i=1}^{j}$ satisfies the recursion of Definition 2.3 applied to $w$. It follows that $r\geq j$ and for $0<i\leq j$ that $a_{i}=d_{i}$ and $b_{i}=e_{i}$. If moreover $A_{0}m_{j}>w$ then the sandwiching halts and $r=j$, witnessing that $w\equiv_{k}A_{a_{j}}b_{j}+c$ for some $c$. ∎

As a special case, if $1\leq j$ we can set $w=m_{j}$ and see that $(A_{a_{i}}b_{i})_{i=1}^{j}$ is the sandwiching sequence for $m_{j}$, hence $m_{j}\equiv_{k}A_{a_{j}}b_{j}$. Note that $(A_{a_{i}}b_{i})_{i=1}^{0}$ is also the sandwiching sequence for $m_{0}$, since $0=m_{0}$ has empty sandwiching sequence.

Proposition 3.4*.*

Let $m\equiv_{k}A_{a}b$ with sandwiching sequence $(A_{a_{i}}b_{i})_{i=1}^{n}$, $j\leq n$ and $w=A_{d}e$. Suppose that

1. (1)

$A_{d}m_{j}\leq w<A_{d+1}m_{j}$, and 2. (2)

if $j>0$ then $w<A_{a_{j}}(b_{j}+1)$.

Then, $w\equiv_{k}A_{d}e$ and has sandwiching sequence $(A_{d_{i}}e_{i})_{i=1}^{j+1}$, where $d_{i}=a_{i}$ and $e_{i}=b_{i}$ for $i\in[1,j]$, $d_{j+1}=d$, and $e_{j+1}=e$.

Proof.

Let $(A_{d_{i}}e_{i})_{i=1}^{r}$ be the sandwiching sequence for $w$. We claim that $r\geq j$ and for $1\leq i\leq j$, $d_{i}=a_{i}$ and $e_{i}=b_{i}$. When $j=0$ this is vacuously true, otherwise this follows from the inequality $A_{a_{j}}b_{j}=m_{j}<A_{d}m_{j}\leq w<A_{a_{j}}(b_{j}+1)$ and Proposition 3.3. In particular, $w_{j}=m_{j}$. Now, the inequality $A_{d}m_{j}\leq w<A_{d+1}m_{j}$ yields $d_{j+1}=d$, and since $w=A_{d}e<A_{0}A_{d}e$ the sandwiching halts and $w\equiv_{k}A_{d}e$. ∎

Corollary 3.5*.*

Let $m\equiv_{k}A_{a}b$ with sandwiching sequence $(A_{a_{i}}b_{i})_{i=1}^{n}$, $1\leq x<k$, and $j\leq n$ be such that if $j\geq 1$ then $d<a_{j}$. Then, $A_{d}^{x}m_{j}$ is in normal form.

We omit the proof, which consists of checking that the required inequalities for Proposition 3.4 hold. Now let us introduce some additional notation for sandwiching sequences. If $(x_{i})_{i=n_{0}}^{n}$ is any sequence with $0\leq n_{0}<n$, we denote its second-to-last element by $x_{-1}$, i.e. $x_{-1}:=x_{n-1}$. If $m>0$ and $(m_{i})_{i=0}^{n}$ are its sandwiching values, we define $\check{m}=m_{-1}$; note in particular that if $(A_{a_{i}}b_{i})_{i=1}^{n}$ is the sandwiching sequence for $m$ and $n>1$ then $\check{m}=A_{a_{-1}}b_{-1}$.

Lemma 3.6**.**

Let $a,b,e$ be natural numbers and let $m\equiv_{k}A_{a}b$ and $w=A_{a-1}e$. Then, $w\equiv_{k}A_{a-1}e$ if either

1. (1)

$A_{a-1}\check{m}\leq w<A_{a}\check{m}$, or 2. (2)

$A_{a-1}m\leq w<A_{a}(b+1)$.

Proof.

Let $(A_{a_{i}}b_{i})_{i=1}^{n}$ be the sandwiching sequence for $m$. In the first case, if $n=1$ this is an instance of Proposition 3.4. Otherwise, note that $\check{m}=A_{a_{-1}}b_{-1}$, and since $a<a_{-1}$ we obtain $A_{a-1}\check{m}\leq w<A_{a}\check{m}<A_{a_{-1}-1}^{k}A_{a_{-1}}b_{-1}=A_{a_{-1}}(b_{-1}+1)$. In the second we instead have $A_{a-1}m\leq w<A_{a}(b+1)<A_{a}A_{a}b=A_{a}m$. Thus in both cases we can apply Proposition 3.4. ∎

Normal forms can be divided into the cases where $b=\check{m}$ and $b>\check{m}$, as each behaves in a different manner. The following lemma makes this precise.

Lemma 3.7**.**

Let $(A_{{a}_{i}}b_{i})_{i=1}^{n}$ be the $k$-sandwiching sequence for $m\equiv_{k}A_{a}b$. Then, exactly one of the following cases occurs:

1. (a)

$n=1$* and $\check{m}=b=0$,* 2. (b)

$n>1$* and $0<\check{m}=b\equiv_{k}A_{a_{-1}}b_{-1}$, or* 3. (c)

$n\geq 1$* and $\check{m}<b\equiv_{k}A_{d}e+s$ with $d\leq a_{-1}$ and either $d\leq a$ or $s>0$.*

Proof.

Assume that (a) fails, so that $n>1$ or $b>0$. Let $(A_{d_{i}}e_{i})_{i=1}^{r}$ be the $k$-sandwiching sequence for $b$. Note that $(A_{{a}_{i}}b_{i})_{i=1}^{n-1}$ is a (possibly empty) initial segment of this sequence: if $n=1$ this holds trivially, while for $n>1$ we have that $A_{a_{-1}}b_{-1}\leq b\leq A_{a_{-1}}(b_{-1}+1)$, hence we can apply Proposition 3.3. Thus $r\geq n-1$ and for $1\leq i<n$ we have that $d_{i}=a_{i}$ and $e_{i}=b_{i}$. Now, consider two cases.

1. Case 1

($r=n-1$). This would witness that $b\equiv_{k}A_{{a}_{-1}}b_{-1}+s$ for some $s$. Hence ${d}={a}_{-1}$ and $b_{-1}=e$, so that either $b=\check{m}$ and (b) holds, or $b>\check{m}$ and $s>0$, so that (c) holds. 2. Case 2

($r\geq n$). In this case we claim that ${d}_{n}\leq{a}$. Indeed, $A_{a+1}A_{d_{-1}}e_{-1}=A_{a+1}\check{m}>A_{a}b>b$. Since $A_{{d}_{n}}A_{a_{-1}}e_{-1}\leq b$, we conclude that ${a}\geq d_{n}\geq d_{r}=d$. This shows that $b\equiv_{k}A_{d_{r}}e_{r}+s$ for some $s$, and (c) holds.∎

Lemma 3.8**.**

Let $a,b,c,m,w$ be natural numbers such that

[TABLE]

Then, $w\equiv_{k}A_{a}b+c$ if either

1. (1)

$m\equiv_{k}A_{a}(b+1)>A_{a}\check{m}$, or 2. (2)

$m\equiv_{k}A_{a}b+c^{\prime}$* for some $c^{\prime}$.*

Proof.

In the first case, it is not hard to check using Proposition 3.3 that the sandwiching sequence for $w$ is the same as that for $m$, except that the last item is replaced by $A_{a}b$. In the second, $w$ has the same sandwiching sequence as $m$. ∎

Note that if $m\equiv_{k}A_{a}b+c$ we may still have that $c\geq A_{a}b$. For such cases we define the extended $k$-normal form of $m$ to be $A_{a}b\cdot p+q$, in symbols $m\cong_{k}A_{a}b\cdot p+q$, if $m\equiv_{k}A_{a}b+c$ for some $c$, $m=A_{a}b\cdot p+q$, and $0\leq q<A_{a}b$; note that $p$ and $q$ are uniquely defined.222The operations $A_{x}y$ and $\langle z/y\rangle x$ are always assumed to be performed before multiplication. If $m\cong_{k}A_{a}b\cdot p+q$ and $d=A_{a}b$ we write $m\approx_{k}d\cdot p+q$ and call it the simplified $k$-normal form of $m$.

We state two useful corollaries of Lemma 3.8. Both are proven by checking (2); in the first we set $c=0$ and $c^{\prime}=A_{a}b\cdot(p-1)$, and in the second $c=d\cdot p+q$ and $c^{\prime}=d\cdot(p-1)+q$. In both cases, the required inequality $w<\min\{A_{a}(b+1),A_{0}A_{a}b\}$ is not hard to verify.

Corollary 3.9*.*

If $A_{a}b$ is in normal form and $0<p<k$ then $A_{a}b\cdot p$ is in extended normal form.

Corollary 3.10*.*

If $m\approx_{k}d\cdot(p+1)+q$ with $p>0$ then $d\cdot p+q$ is in simplified normal form.

As we will see in the next section, if $m\equiv_{k}A_{a}b$ it is usually not easy to compute the normal form of $m-1$. However, the case where $a=0$ is somewhat simpler.

Lemma 3.11**.**

Fix $k\geq 0$ and write $A_{x}y=A_{x}(k,y)$. Suppose that $m\equiv_{k}A_{0}b$.

1. (1)

If $b>\check{m}$ then $m-1\approx_{k}A_{0}(b-1)\cdot(k-1)+(k^{b-1}-1)$. 2. (2)

If $b=\check{m}>0$ then $m-1\approx_{k}b\cdot p+q$ for some $p<k^{b-1}$ and some $q<b$.

Proof.

We prove (2); item (1) is similar. Suppose that $m\equiv_{k}A_{0}b$ and let $(A_{a_{i}}b_{i})_{i=1}^{n}$ be the sandwiching sequence for $m$. By Lemma 3.7, $n>1$ and $b=A_{a_{-1}}b_{-1}$. It is then easy to check using Proposition 3.3 that $m-1\equiv_{k}A_{a_{-1}}b_{-1}+c$ for some $c$, hence $m-1\approx_{k}b\cdot p+q$ for some $p$, $q$. Since $a_{-1}>a_{n}$ we must have that $a_{-1}\geq 1$. It follows that $b\geq A_{1}0>k$, so that if $p\geq k^{b-1}$ we would have that $bp>k^{b-1}\cdot k=A_{0}b$, which is impossible as $bp\leq m-1$. That $q<b$ follows from the definition of simplified normal forms. ∎

4. Left and Right Expansions

In this section we develop operations for computing normal forms using the information that $m\equiv_{k}A_{a}b+c$, but where the new normal form may vary drastically from that of $m$. We begin with right expansions of $A_{a}b$.

Lemma 4.1**.**

Let $m=A_{a}b$ with ${a},b>0$, and let $s\in[1,b+1]$. Then:

1. (1)

$A_{a}b=A^{sk}_{a-1}A_{a}(b-s).$** 2. (2)

Let $\ell\in[1,k]$ and $c=A^{\ell}_{a-1}A_{a}(b-s)$. If $m\equiv_{k}A_{a}b$, $b=\check{m}$, and

[TABLE]

it follows that $c$ is in normal form as written.

Proof.

The first claim follows by a simple induction on $s$ and the definition of $A_{a}$ and the second is an instance of Lemma 3.6.1. ∎

The following is a variant of the ordinal predecessor function of Cichon [2], which works by expanding $A_{a}b$ on the left, useful for computing the normal form of $m-1$.

Definition 4.2**.**

Let $A_{a}b$ be in normal form with $a>0$ and define a sequence $c_{0},\ldots,c_{a}$ by recursion as follows:

1. (1)

$c_{0}=A_{a}(b-1)$; 2. (2)

$c_{i}=A_{a-i}(A^{k-1}_{a-i}c_{i-1}-1)$ if $i>0$.

We call the sequence $(c_{i})_{i\leq a}$ the left expansion sequence for $A_{a}b$.

The following chain of inequalities summarizes some basic properties of left expansion sequences and can be easily checked using induction on $i$.

Lemma 4.3**.**

If $m\equiv_{k}A_{a}b$ with $a>0$, $(c_{i})_{i\leq a}$ is the left expansion sequence for $m$, and $i\in[1,a]$, then

[TABLE]

In fact, the inequality $c_{i}<m$ can immediately be sharpened.

Corollary 4.4*.*

Suppose that $m\equiv_{k}A_{a}b$ with $a>0$ and let $(c_{i})_{i\leq a}$ be the left expansion sequence for $m$. Then, if $i\leq a$, it follows that $m\geq kc_{i}$, with equality holding if and only if $i=a$.

Proof.

For $i=a$, note that $m=A_{0}^{k}c_{a-1}=kA_{0}(A_{0}^{k-1}c_{a-1}-1)=kc_{a}$. Otherwise, $kc_{i}<kc_{a}=m$. ∎

Lemma 4.5**.**

Let $m\equiv_{k}A_{a}b$ with $a>0$.

1. (1)

If $0<\ell<k$ and $0<i\leq a$ then $A_{a-i}^{\ell}c_{i-1}$ is in normal form. 2. (2)

If $i\leq a$ and either $i>0$ or $b>\check{m}$, then $c_{i}$ has normal form as written in Definition 4.2.

Proof.

Proceed by induction on $i$, considering several cases.

1. Case 1

($i=0$ and $b>\check{m}$). The first claim does not apply with $i=0$ and the second is an instance of Lemma 3.8.1, using the condition $b>\check{m}$. 2. Case 2

($i>0$). We will prove both claims uniformly. In order to do this, let $d$ be either $A_{a-i}^{\ell-1}c_{i-1}$ for the first claim or $A_{a-i}^{k-1}c_{i-1}-1$ for the second. Note that it follows from Lemma 4.3 that $A_{a-i}d<m$. We will consider two sub-cases.

1. Case 2.1

($b=\check{m}$ and $i=1$). This is an instance of Lemma 4.1.2. 2. Case 2.2

($b>\check{m}$ or $i>1$). By the induction hypothesis we have that $c_{i-1}\equiv_{k}A_{a-i+1}e$ for some $e$, where $e=b-1$ if $i=1$ and $e=A^{k-1}_{a-i+1}c_{i-2}-1$ if $i>1$. In either case $A_{a-i+1}(e+1)=m$, in the first case since $m=A_{a}b$ and in the second by Lemma 4.3. Moreover, $c_{i-1}\leq d$, so $A_{a-i}c_{i-1}\leq A_{a-i}d<m=A_{a-i+1}(e+1)$. Thus we may use Lemma 3.6.2 to conclude that $A_{a-i}d$ is in normal form.∎

With this we can describe the normal form of $m-1$ when $m\equiv_{k}A_{a}b$.

Lemma 4.6**.**

If $m\equiv_{k}A_{a}b$ with $a>0$ and left expansion sequence $(c_{i})_{i\leq a}$ then

[TABLE]

Proof.

Let $d=A_{0}^{k-1}c_{a-1}-1$. We know that $c_{a}\equiv_{k}A_{0}d$ and $A_{0}d=c_{a}<m-1<m=A_{0}(d+1)\leq A_{0}A_{0}d$. Thus Lemma 3.8.2 implies that $m-1\equiv_{k}A_{0}d+s$ for some $s$, hence $m-1\approx_{k}c_{a}\cdot p+q$ for some $p,q$ with $q<c_{a}$. But by Corollary 4.4, $m-1=kc_{a}-1=c_{a}\cdot(k-1)+(c_{a}-1)$, so that $p=k-1$ and $q=c_{a}-1$. ∎

5. The Base Change Operation

Next we elaborate on the base change operation as given by Definition 2.6. We begin by extending it to base $\omega$ as follows:

1. (1)

If $m=0$ then $\langle{\omega/k}\rangle m=0$. 2. (2)

If $m\equiv_{k}A_{a}b+c$ then $\langle{\omega/k}\rangle m={\phi}_{\langle{\omega/k}\rangle a}\langle{\omega/k}\rangle b+\langle{\omega/k}\rangle c$.

Here, ${\phi}_{\alpha}$ denotes the fixed point-free Veblen functions, reviewed in Appendix A. We regard ${\phi}_{\alpha}$ as an analogue of the Ackermann function with base $\omega$. To stress this, we set $A_{\alpha}(\omega,\beta):={\phi}_{\alpha}\beta$, so that Definition 2.6 uniformly extends to $\ell\leq\omega$. The following is shown by induction on $p$ using Corollary 3.10.

Lemma 5.1**.**

If $m\cong_{k}A_{a}(k,b)\cdot p+q$ and $1<k\leq\lambda\leq\omega$ then

[TABLE]

Next we prove that the base change operation is strictly increasing, which will be a crucial ingredient in the proofs of our main results.

Proposition 5.2*.*

If $n<m$ and $1<k\leq\lambda\leq\omega$ then $\langle\lambda/k\rangle n<\langle\lambda/k\rangle m$.

Proof.

By induction on $m$. In this proof we write $\langle\lambda\rangle x$ instead of $\langle\lambda/k\rangle x$. We remark that if $x<m$ our induction hypothesis yields $x\leq\langle\lambda\rangle x$, since the map $x\mapsto\langle\lambda\rangle x$ is strictly increasing below $m$. Without loss of generality we may assume that $n=m-1$. Let $A_{x}y=A_{x}(k,y)$ and $B_{\xi}\zeta=A_{\xi}(\lambda,\zeta)$. Write $m\equiv_{k}A_{a}b+c$, so that $\langle\lambda\rangle m=B_{\langle\lambda\rangle a}\langle\lambda\rangle b+\langle\lambda\rangle c$.

We then consider several cases.

1. Case 1

($a=b=c=0$). Then $m=1$ so $\langle\lambda\rangle n=n=0<\langle\lambda\rangle m$. 2. Case 2

($c>0$). In this case we have by Lemma 3.8.2 that $m-1\equiv_{k}A_{a}b+(c-1)$, so that $\langle\lambda\rangle(m-1)=B_{\langle\lambda\rangle a}\langle\lambda\rangle b+\langle\lambda\rangle(c-1)<^{\text{\sc ih}}B_{\langle\lambda\rangle a}\langle\lambda\rangle b+\langle\lambda\rangle c=\langle\lambda\rangle m$. 3. Case 3

($a=c=0$ and $\check{m}<b$). Write $b=d+1$. Lemma 3.11.1 then yields $m-1\cong_{k}A_{0}d\cdot(k-1)+(k^{d}-1)$, hence

[TABLE] 4. Case 4

($a=c=0$ and $\check{m}=b$). Use Lemma 3.11.2 to see that $m-1\approx_{k}b\cdot p+q$ for some $p<k^{b-1}$ and some $q<b$. Then,

[TABLE]

where for $\lambda<\omega$ the first inequality follows from the fact that $x^{y-1}y+y\leq(x+1)^{y}$ whenever $x,y$ are natural numbers with $x>0$. 5. Case 5

($a>0$ and $c=0$). Let $c_{0},\ldots,c_{a}$ be the left expansion sequence for $m$. If $\lambda<\omega$ let $\delta_{0},\ldots,\delta_{\langle\lambda\rangle a}$ be the left expansion sequence for $\langle\lambda\rangle m$, and if $\lambda=\omega$ define $\delta_{i}=\langle\lambda\rangle m$ for all $i$. If $\lambda<\omega$ note by the induction hypothesis that $a-i\leq\langle\lambda\rangle(a-i)\leq\langle\lambda\rangle a-i$ whenever $i\leq a$. We claim that for each $i\leq a$, $\langle\lambda\rangle c_{i}<\max\{2,\delta_{i}\}$. The theorem will then follow, since by Lemma 4.6 we see that $m-1\approx_{k}c_{a}\cdot(k-1)+(c_{a}-1),$ hence $\langle\lambda\rangle(m-1)=\langle\lambda\rangle c_{a}\cdot(k-1)+\langle\lambda\rangle(c_{a}-1)<^{\text{\sc ih}}\langle\lambda\rangle c_{a}\cdot k<\langle\ell\rangle m$, where the last inequality follows from Corollary 4.4 if $\lambda<\omega$.

We proceed to prove that $\langle\lambda\rangle c_{i}<\max\{2,\delta_{i}\}$ by a secondary induction on $i$. Let us first do the inductive step: $\langle\lambda\rangle c_{i+1}=\langle\lambda\rangle A_{a-i-1}(A^{k-1}_{(a-i-1)}c_{i}-1)=B_{\langle\lambda\rangle(a-i-1)}\langle\lambda\rangle(A^{k-1}_{(a-i-1)}c_{i}-1)<B_{\langle\lambda\rangle(a-i-1)}\langle\lambda\rangle A^{k-1}_{(a-i-1)}c_{i}=B^{k}_{\langle\lambda\rangle(a-i-1)}\langle\lambda\rangle c_{i}<\delta_{i+1}$, where the first inequality follows from $A^{k-1}_{(a-i-1)}c_{i}<m$ and the main induction hypothesis and the last from B^{k}_{\langle\lambda\rangle(a-i-1)}\langle\lambda\rangle c_{i}<B_{\langle\lambda\rangle a-i-1}\big{(}B^{k}_{\langle\lambda\rangle a-i-1}\delta_{i}-1\big{)}=\delta_{i+1} when $\lambda<\omega$ and from $\langle\ell\rangle m$ being closed under $B^{k}_{\langle\lambda\rangle(a-i-1)}$ when $\lambda=\omega$.

Thus it remains to prove that $\langle\lambda\rangle c_{0}\leq\delta_{0}$.

1. Case 5.1

($b=0$). We have that $\langle\lambda\rangle c_{0}=c_{0}=1<2$. 2. Case 5.2

($b>\check{m}$). We have that $\langle\lambda\rangle c_{0}=B_{\langle\lambda\rangle a}\langle\lambda\rangle(b-1)<\delta_{0}$, where if $\lambda<\omega$ we use the induction hypothesis to see that $\langle\lambda\rangle(b-1)\leq\langle\lambda\rangle b-1$ and if $\lambda=\omega$ we use $\langle\lambda\rangle(b-1)<\langle\lambda\rangle b$ and the monotonicity of $B_{\langle\lambda\rangle a}$. 3. Case 5.3

($b=\check{m}>0$). We use Lemma 4.1 to see that for $s\in[1,b]$, $A_{a}(b-1)=A^{k(s-1)}_{{a}-1}A_{a}(b-s)$. Since $A_{a}(-1)=1\leq b$ we have that there is a least $t\in[2,b+1]$ such that $A_{a}(b-t)\leq b$. Similarly, there is a greatest $r<k$ such that $u:=A^{r}_{{a}-1}A_{a}(b-t)\leq b$. Note that $A_{a-1}u>b$, and hence $A^{2}_{a-1}u>A_{a-1}b=A_{a-1}\check{m}$. Hence by Lemma 4.1.2 we have that $A_{{a}-1}^{v}u$ is in normal form whenever $1<v\leq k(t-2)+k-r.$ Similarly, by Corollary 3.5, $A_{{a}-1}b$ is in normal form, since by assumption $b=\check{m}$. Then,

[TABLE]

If $\lambda=\omega$ then $\langle\lambda\rangle(a-1)<\langle\lambda\rangle a$ yields $B^{k(t-1)}_{\langle\lambda\rangle(a-1)}\langle\lambda\rangle b<B_{\langle\lambda\rangle a}\langle\lambda\rangle b=\langle\lambda\rangle m$. If $\lambda<\omega$, then $B^{k(t-1)}_{\langle\lambda\rangle(a-1)}\langle\lambda\rangle b<B^{k(t-1)}_{\langle\lambda\rangle(a-1)}B^{t-1}_{\langle\lambda\rangle(a-1)}B_{\langle\lambda\rangle a}(\langle\lambda\rangle b-t)\leq B^{\lambda(t-1)}_{\langle\lambda\rangle(a-1)}B_{\langle\lambda\rangle a}(\langle\lambda\rangle b-t)\leq B^{\lambda(t-1)}_{\langle\lambda\rangle a-1}B_{\langle\lambda\rangle a}(\langle\lambda\rangle b-t)=B_{\langle\lambda\rangle a}(\langle\lambda\rangle b-1)=\delta_{0}$, as claimed.∎

6. Normal Form Preservation

Our goal now is to show that the base-change operation preserves normal forms: if $A_{a}(k,b)+c$ is in $k$-normal form and $\lambda\geq k$, then $A_{\langle\lambda/k\rangle a}(\lambda,\langle\lambda/k\rangle b)+\langle\lambda/k\rangle c$ is in $\lambda$-normal form. Throughout this section we fix $2\leq k<\lambda\leq\omega$, and write $A_{x}y=A_{x}(k,y)$, $B_{\xi}\zeta=A_{\xi}(\lambda,\zeta)$.

Proposition 6.1*.*

If $m\equiv_{k}A_{a}b+c$ then $\langle\lambda\rangle m\equiv_{\lambda}B_{\langle\lambda\rangle a}\langle\lambda\rangle b+\langle\lambda\rangle c$.

Proof.

Assume that $m\equiv_{k}A_{a}b+c$. The proposition is immediate from Proposition 5.2 when $\lambda=\omega$, since the only condition for $\langle\omega\rangle m={\phi}_{\langle\omega/k\rangle a}\langle\omega/k\rangle b+\langle\omega/k\rangle c$ to be in normal form is for $\langle\omega/k\rangle a,\langle\omega/k\rangle b,\langle\omega/k\rangle c<\langle\omega/k\rangle m$, which holds since $a,b,c<m$. Thus we assume that $\lambda<\omega$ and without loss of generality that $\lambda=k+1$. Let $(A_{a_{i}}b_{i})_{i=1}^{n}$ be the $k$-sandwiching sequence for $m$. We will prove by induction on $m$ that $(B_{\langle\lambda\rangle a_{i}}\langle\lambda\rangle b_{i})_{i=1}^{n}$ is the $\lambda$-sandwiching sequence for $\langle\lambda\rangle m$.

1. Case 1

($c>0$). In this case $m-1\equiv_{k}A_{0}b+(c-1)$, so that by the induction hypothesis $(B_{\langle\lambda\rangle a_{i}}\langle\lambda\rangle b_{i})_{i=1}^{n}$ is the $\lambda$-sandwiching sequence for $\langle\lambda\rangle(m-1)$. Consider two sub-cases.

1. Case 1.1

($a=0$). Since $B_{0}B_{0}\langle\lambda\rangle b\geq B_{0}(\langle\lambda\rangle b+1)$, to apply Proposition 3.3 we need only check that $B_{0}(\langle\lambda\rangle b+1)>\langle\lambda\rangle m$. Write $m\cong_{k}A_{0}b\cdot p+q=k^{b}p+q$ with $q<k^{b}$. From $m<A_{0}(b+1)$ we obtain $p<k$. However, $B_{0}(\langle\lambda\rangle b+1)>B_{0}\langle\lambda\rangle b\cdot k\geq\langle\lambda\rangle A_{0}b\cdot p+\langle\lambda\rangle A_{0}b>\langle\lambda\rangle A_{0}b\cdot p+\langle\lambda\rangle q=\langle\lambda\rangle m$. 2. Case 1.2

($a>0$). Since $B_{\langle\lambda\rangle a}(\langle\lambda\rangle b+1)=B_{\langle\lambda\rangle a-1}^{\lambda}B_{\langle\lambda\rangle a}\langle\lambda\rangle b>B_{0}B_{\langle\lambda\rangle a}\langle\lambda\rangle b$, now it suffices to show $B_{0}B_{\langle\lambda\rangle a}\langle\lambda\rangle b>\langle\lambda\rangle m$ to conclude that $(B_{\langle\lambda\rangle a_{i}}\langle\lambda\rangle b_{i})_{i=1}^{n}$ is also the sandwiching sequence for $\langle\lambda\rangle m$. We know that $A_{0}A_{a}b>m.$ Since $a>0$ we have that $A_{0}A_{a}b$ is in normal form by Corollary 3.5, hence by Proposition 5.2, $B_{0}B_{\langle\lambda\rangle a}\langle\lambda\rangle b=\langle\lambda\rangle A_{0}A_{a}b>\langle\lambda\rangle m$, as claimed. 2. Case 2

($c=0$). This is the critical case in the proof. By Proposition 3.3, $\check{m}$ has $k$-sandwiching sequence $(A_{a_{i}}b_{i})_{i=1}^{n-1}$. Thus by the induction hypothesis, $\langle\lambda\rangle\check{m}$ has $\lambda$-sandwiching sequence $(B_{\langle\lambda\rangle a_{i}}\langle\lambda\rangle b_{i})_{i=1}^{n-1}$. Since we already have that $\langle\lambda\rangle\check{m}\leq\langle\lambda\rangle b$, in view of Proposition 3.3, it suffices to show that

1. (a)

$B_{\langle\lambda\rangle a}\langle\lambda\rangle b<B_{\langle\lambda\rangle a+1}\langle\lambda\rangle\check{m}$, and 2. (b)

$B_{\langle\lambda\rangle a}\langle\lambda\rangle b<B_{\langle\lambda\rangle a_{-1}}(\langle\lambda\rangle b_{-1}+1)$ if $n>1$.

From this and Proposition 3.4 we may conclude that $(B_{\langle\lambda\rangle a_{i}}\langle\lambda\rangle b_{i})_{i=1}^{n}$ is the $\lambda$-sandwiching sequence for $\langle\lambda\rangle m$. Consider two sub-cases.

1. Case 2.1

($n>1$ and $a+1=a_{-1}$). In this case we have the inequality

[TABLE]

where the inequality is by Lemma 3.2.1 and Proposition 5.2 and the equality holds because $A_{a}^{x}A_{a_{-1}}b_{-1}$ is in normal form for all $x<k$, given Proposition 3.4. Hence B_{\langle\lambda\rangle a+1}\langle\lambda\rangle\check{m}=B_{\langle\lambda\rangle a}^{\lambda}B_{\langle\lambda\rangle a+1}\big{(}B_{\langle\lambda\rangle a_{-1}}\langle\lambda\rangle b_{-1}-1\big{)}>B_{\langle\lambda\rangle a}^{\lambda}B_{\langle\lambda\rangle a_{-1}}\langle\lambda\rangle b_{-1}>B_{\langle\lambda\rangle a}\langle\lambda\rangle b, where the last inequality uses (6.1), thus establishing Case 2(a). For Case 2(b),

[TABLE]

where the first inequality uses (6.1) and the second $k<\lambda$ and $\langle\lambda\rangle a\leq\langle\lambda\rangle a_{-1}-1$. 2. Case 2.2

($n=1$ or $a+1<a_{-1}$). Let us establish Case 2(a) first. The argument is similar to that for Case 5Case 5.3 of Proposition 5.2. Let $t\in[1,\check{m}+1]$ be least with the property that $A_{a+1}(\check{m}-t)<A_{a}\check{m},$ and similarly $r<k$ be greatest such that $u:=A^{r}_{a}A_{a+1}(\check{m}-t)<A_{a}\check{m}$. Since $a+1<a_{-1}$, $A_{a}\check{m}$ and $A_{a+1}\check{m}$ are in normal form by Corollary 3.5. By Lemma 4.1 we have that $A_{a+1}\check{m}=A_{a}^{tk}A_{a+1}(b-t)$, hence if $1\leq s<tk-r$, $A_{a}\check{m}\leq A_{a}^{s}u<A_{a+1}\check{m}$, so that by Lemma 3.6.1, $A_{a}^{s}u$ is in normal form. Since $A_{a}b<A_{a+1}\check{m}=A^{tk-r}_{a}u$, we have that $b<A_{a}^{tk-r-1}u.$ Then,

[TABLE]

In the case that $n>0$, Case 2(b) immediately follows, since $B_{\langle\lambda\rangle a_{-1}}(\langle\lambda\rangle b_{-1}+1)=B_{\langle\lambda\rangle a_{-1}-1}^{\lambda}B_{\langle\lambda\rangle a_{-1}}\langle\lambda\rangle b_{-1}>B_{\langle\lambda\rangle a+1}\langle\lambda\rangle\check{m}>B_{\langle\lambda\rangle a}\langle\lambda\rangle b$.∎

7. Proofs of the Main Theorems

In this section we put our results together to prove Theorems 2.9 and 2.10. The next lemma is the final ingredient we need for the former.

Lemma 7.1**.**

If $k<\ell<\omega$ and $m<\omega$ then $\langle{\omega/k}\rangle m=\langle{\omega/\ell}\rangle\langle\ell/k\rangle m$.

Proof.

By induction on $m$. The claim is trivial if $m=0$, so we assume otherwise. Let $m\equiv_{k}A_{a}b+c$ and write $\langle\ell\rangle x$ for $\langle\ell/k\rangle x$. Then $\langle\ell\rangle m$ has the $\ell$-normal form $B_{\langle\ell\rangle a}\langle\ell\rangle b+\langle\ell\rangle c$, and the induction hypothesis yields $\langle{\omega/\ell}\rangle\langle\ell\rangle m={\phi}_{\langle{\omega/\ell}\rangle\langle\ell\rangle a}\langle{\omega/\ell}\rangle\langle\ell\rangle b+\langle{\omega/\ell}\rangle\langle\ell\rangle c=^{\text{\sc ih}}{\phi}_{\langle{\omega/k}\rangle a}\langle{\omega/k}\rangle b+\langle{\omega/k}\rangle c=\langle{\omega/k}\rangle m$. ∎

With this we are ready to prove termination for the Ackermannian Goodstein process, based on the fact that the sequence $o_{i}m$, as defined below, is a decreasing sequence of ordinals.

Definition 7.2**.**

Given $m,i\in\mathbb{N}$, we define $o_{i}m=\langle{\omega/i+2}\rangle{\mathbb{G}}_{i}m$.

Proof of Theorem 2.9.

Let $({\mathbb{G}}_{i}m)_{i<{\alpha}}$ be the Ackermannian Goodstein sequence starting on $m$ and $i$ be such that $i+1<{\alpha}$. Then,

[TABLE]

where the inequality follows from Proposition 5.2 and the second-to-last equality from Lemma 7.1. Hence $(o_{i}m)_{i<{\alpha}}$ is decreasing below $\varepsilon_{0}$, so ${\alpha}$ must be finite. ∎

In Appendix A we review the relations $\preccurlyeq_{k}$, fundamental sequences for $\Gamma_{0}$, and Theorem A.5, which is unprovable in ${\sf ATR}_{0}$. We show that Theorem 2.9 is also unprovable by deriving Theorem A.5 from it, using the following key lemma.

Lemma 7.3**.**

If $0<m<\omega$ and $1<k<\omega$ then

[TABLE]

Proof.

Let $\ell=k+1$ and proceed by induction on $m$. We write $A_{a}b$ for $A_{a}(k,b)$ and $B_{a}b$ for $A_{a}(\ell,b)$. It suffices to show $[k]\langle{\omega/k}\rangle m\leq\langle{\omega/\ell}\rangle(\langle\ell\rangle m-1)$. The refined inequality follows from the Bachmann property (Proposition A.4), since Proposition 5.2 yields $\langle{\omega/k}\rangle m=\langle{\omega/\ell}\rangle\langle\ell\rangle m>\langle{\omega/\ell}\rangle(\langle\ell\rangle m-1)$. Write $m\equiv_{k}A_{a}b+c$. Note that by Proposition 6.1, $\langle\ell\rangle m\equiv_{\ell}B_{\langle\ell\rangle a}\langle\ell\rangle b+\langle\ell\rangle c$. We consider several cases.

1. Case 1

($a=b=c=0$).   $\langle{\omega/\ell}\rangle(\langle\ell\rangle m-1)=0=[k]\phi_{0}0=[k]\langle{\omega/k}\rangle m$. 2. Case 2

($c>0$).   $\langle{\omega/\ell}\rangle(\langle\ell\rangle m-1)=\langle{\omega/\ell}\rangle(B_{\langle\ell\rangle a}\langle\ell\rangle b+\langle\ell\rangle c-1)$

[TABLE] 3. Case 3

($a=c=0$ and $b>\check{m}$). Lemma 3.7 and Proposition 5.2 imply that $\langle{\omega/k}\rangle b\not\in{\rm Fix}(0)$, since $b=A_{d}e+s$ with either $d\leq 0$ or $s>0$, which means that $\langle{\omega/k}\rangle b={\phi}_{\langle{\omega/k}\rangle d}\langle{\omega/k}\rangle e+\langle{\omega/k}\rangle s$ with $\langle{\omega/k}\rangle d=0$ or $\langle{\omega/k}\rangle b>\langle{\omega/k}\rangle s>0$. Therefore, $[k]{\phi}_{0}\langle{\omega/k}\rangle b={\phi}_{0}[k]\langle{\omega/k}\rangle b$. By Lemma 3.8.1, $B_{0}(\langle\ell\rangle b-1)$ is in normal form, hence by Corollary 3.9, $B_{0}(\langle\ell\rangle b-1)\cdot k$ is in extended $\ell$-normal form, so that

[TABLE] 4. Case 4

($a=c=0$ and $b=\check{m}>0$). Lemma 3.7 yields $b\equiv_{k}A_{d}e$ for some $d>a$, so that $\langle{\omega/k}\rangle b\in{\rm Fix}(0)$ and $[k]\langle{\omega/k}\rangle m=\langle{\omega/k}\rangle b\cdot k$. By Corollary 3.9, $\langle\ell\rangle b\cdot k$ is in simplified $\ell$-normal form and hence $\langle{\omega/\ell}\rangle(\langle\ell\rangle m-1)=\langle{\omega/\ell}\rangle(\ell^{\langle\ell\rangle b}-1)>\langle{\omega/\ell}\rangle({\langle\ell\rangle b}\cdot k)=\langle{\omega/k}\rangle b\cdot k=[k]\langle{\omega/k}\rangle m$, where the inequality follows from $(x+1)^{y}>xy$. 5. Case 5

($a>0$ and $b=c=0$). By Lemma 4.5, $B_{\langle\ell\rangle a-1}^{j}1$ is in $\ell$-normal form for $1\leq j\leq k$. Then Proposition 5.2 and the induction hypothesis yield

[TABLE] 6. Case 6

($a>0$, $b>\check{m}$ and $c=0$). Once again by Proposition 5.2 and Lemma 3.7 one can check that $\langle{\omega/k}\rangle b\not\in{\rm Fix}(\langle{\omega/k}\rangle a)$ and thus

[TABLE] 7. Case 7

($a>0$, $b=\check{m}$ and $c=0$). Note that in this case $\langle{\omega/k}\rangle b\in{\rm Fix}(\langle{\omega/k}\rangle a)$ so that $[k]{\phi}_{\langle{\omega/k}\rangle a}\langle{\omega/k}\rangle b={\phi}_{[k]\langle{\omega/k}\rangle a}^{\nicefrac{{k}}{{\langle{\omega/k}\rangle}}a}\langle{\omega/k}\rangle b.$ Moreover, note that $B_{\langle\ell\rangle a-1}^{j}\langle\ell\rangle b$ is in $\ell$-normal form for $j\leq k$ by Corollary 3.5. Then Proposition 5.2 and the induction hypothesis yield

[TABLE]

With this we may show that $o_{k}m$ decreases at least as slowly as $\llbracket k+1\rrbracket o_{0}m$.

Lemma 7.4**.**

If $m\geq 0$ and $k\geq 2$ then $o_{k}m\succcurlyeq_{k+1}\llbracket k+1\rrbracket o_{0}m$.

Proof.

The assertion holds for $k=0$. Assume now that $o_{k}m\succcurlyeq_{k+1}\llbracket k+1\rrbracket o_{0}m.$ Then $o_{k}m\succcurlyeq_{k+2}\llbracket k+1\rrbracket o_{0}m$ and hence $[k+2]o_{k}m\succcurlyeq_{k+2}[k+2]\llbracket k+1\rrbracket o_{0}m$. Moreover Lemma 7.3 yields $o_{k+1}m=\langle{\omega/k+3}\rangle{\mathbb{G}}_{k+1}m=\langle{\omega/k+3}\rangle(\langle k+3\rangle{\mathbb{G}}_{k}m-1)\succcurlyeq_{1}[k+2]\langle{\omega/k+2}\rangle{\mathbb{G}}_{k}m=[k+2]o_{k}m$, hence $o_{k+1}m\succcurlyeq_{k+2}[k+2]o_{k}m$. Putting things together we arrive at $o_{k+1}m\succcurlyeq_{k+2}\llbracket k+2\rrbracket o_{0}m.$ ∎

Proof of Theorem 2.10.

Let $\gamma_{n}$ be as in Theorem A.5. Moreover let $a_{0}:=0$ and $a_{n+1}:=A_{a_{n}}(2,0).$ Recall that $o_{\ell}a_{m}=\langle{\omega/\ell+2}\rangle{\mathbb{G}}_{\ell}a_{m}$. Then $o_{0}a_{m}=\langle{\omega/2}\rangle a_{m}=\gamma_{m}$ for $m\geq 3$. Lemma 10 yields $\llbracket k+1\rrbracket o_{0}a_{m}\preccurlyeq_{k+1}o_{k}a_{m}$.

Assume that ${\sf ATR}_{0}\vdash\forall m\exists\ell\ ({\mathbb{G}}_{\ell}m=0).$ The function $m\mapsto a_{m}$ is provably computable in ${\sf ATR}_{0}$ and therefore ${\sf ATR}_{0}\vdash\forall m\exists\ell\ ({\mathbb{G}}_{\ell}a_{m}=0).$ From this we obtain ${\sf ATR}_{0}\vdash\forall m\exists\ell\ (o_{\ell}a_{m}=0).$ But then by formalizing the proof of Lemma 7.4, we see that ${\sf ATR}_{0}\vdash\forall m\exists\ell\ (\llbracket\ell\rrbracket\gamma_{m}=0),$ contradicting Theorem A.5. ∎

Appendix A The Feferman-Schütte Ordinal

In this appendix we briefly review the ordinal $\Gamma_{0}$. For a more detailed treatment see e.g. [15]. Given an ordinal $\alpha$ recall that $\varphi_{\alpha}$ is defined recursively so that $\varphi_{0}\beta:=\omega^{\beta}$ and for $\alpha>0$, $\varphi_{\alpha}{\beta}$ is the $\beta$-th member of $\{\eta:(\forall\xi<{\beta})[\varphi_{\xi}\eta=\eta]\}$.

We will use the fixed point-free variant given by

[TABLE]

Then, $\Gamma_{0}$ is the first non-zero ordinal closed under $(\alpha,\beta)\mapsto{\phi}_{\alpha}\beta$. The technical benefit of the modified hierarchy is witnessed by the following.

Proposition A.1*.*

If $0<\xi<\Gamma_{0}$, there exist unique $\alpha,\beta,\gamma<\xi$ such that $\xi={\phi}_{\alpha}\beta+\gamma$.

We will call this the Veblen normal form of $\xi$ and write $\xi\equiv_{\omega}{\phi}_{\alpha}\beta+\gamma$. Proposition A.1 does not hold for $\varphi$, as for example $\varphi_{0}\varphi_{1}0=\varphi_{1}0<\Gamma_{0}$. The order relation between elements of $\Gamma_{0}$ can be computed recursively on their Veblen normal form. Below we consider only $\xi,\zeta>0$, as clearly $0<{\phi}_{\alpha}\beta+\gamma$ regardless of $\alpha,\beta,\gamma$.

Lemma A.2**.**

Given $\xi,\xi^{\prime}<\Gamma_{0}$ with $\xi={\phi}_{\alpha}{\beta}+{\gamma}$ and $\xi^{\prime}={\phi}_{\alpha^{\prime}}{\beta^{\prime}}+{\gamma^{\prime}}$ both in Veblen normal form, $\xi<\xi^{\prime}$ if and only if

1. (1)

$\alpha=\alpha^{\prime}$, $\beta=\beta^{\prime}$ and $\gamma<\gamma^{\prime}$; 2. (2)

${\alpha}<{\alpha}^{\prime}$* and ${\beta}<{\phi}_{\alpha^{\prime}}\beta^{\prime}$;* 3. (3)

${\alpha}={\alpha}^{\prime}$* and ${\beta}<{\beta}^{\prime}$, or* 4. (4)

${\alpha}^{\prime}<{\alpha}$* and ${\phi}_{\alpha}{\beta}\leq{\beta}^{\prime}$.*

The above properties of modified Veblen functions will suffice to prove that the Ackermannian Goodstein process always terminates on finite time. In order to prove independence from ${\sf ATR}_{0}$, we also need to review fundamental sequences. Let ${\rm Fix}({\alpha}):=\{{\phi}_{\beta}\gamma:\gamma>{\alpha}\}$; these are the fixed points of $\varphi_{\alpha}$. Fundamental sequences will be defined separately at such points.

Definition A.3**.**

For $\xi<\Gamma_{0}$ and $x<\omega$, define $[x]{\alpha}$ recursively as follows. First we set $\nicefrac{{x}}{{\alpha}}=x$ if $\alpha$ is a successor, $\nicefrac{{x}}{{\alpha}}=1$ otherwise. Then, define:

1. (1)

$[x]0=0$. 2. (2)

$[x]({\phi}_{\alpha}{\beta}+\gamma)={\phi}_{\alpha}{\beta}+[x]\gamma$ if $\gamma>0$. 3. (3)

$[x]\phi_{0}0=0$ (note that ${\phi_{0}0}=1$). 4. (4)

$[x]\phi_{0}\lambda=\lambda\cdot x$ if $\lambda\in{\rm Fix}(0)$. 5. (5)

$[x]{\phi}_{\alpha}0:={\phi}^{\nicefrac{{x}}{{\alpha}}}_{[x]{\alpha}}1$ if ${\alpha}>0$. 6. (6)

$[x]{\phi}_{\alpha}({\beta}+1):={\phi}^{\nicefrac{{x}}{{\alpha}}}_{[x]{\alpha}}{\phi}_{\alpha}{\beta}$ if ${\alpha}>0$. 7. (7)

$[x]{\phi}_{\alpha}\lambda:={\phi}_{{\alpha}}[x]\lambda$ if $\lambda$ is a limit and $\lambda\not\in{\rm Fix}({\alpha})$. 8. (8)

$[x]{\phi}_{\alpha}\lambda:={\phi}^{\nicefrac{{x}}{{\alpha}}}_{[x]{\alpha}}\lambda$ if $\lambda\in{\rm Fix}({\alpha})$ and ${\alpha}>0$.

For an ordinal $\xi<\Gamma_{0}$ and $n\geq 2$ we define inductively $\llbracket 2\rrbracket\xi=[2]\xi$ and $\llbracket n+1\rrbracket\xi=[n+1]\llbracket n\rrbracket\xi$. Define $\prec_{k}$ to be the transitive closure of $\{({\alpha},{\beta}):{\alpha}=[k]{\beta}\}$, and $\preccurlyeq_{k}$ to be its reflexive closure. The ordering $\preccurlyeq_{1}$ satisfies the Bachmann property:

Proposition A.4*.*

If $\alpha,\beta<\Gamma_{0}$ and $k<\omega$ are such that $[k]{\alpha}<\beta<{\alpha}$, then $[k]{\alpha}\preccurlyeq_{1}{\beta}$.

See, for example, [3], [18] and [19] for more details. It is further well-known that ${\alpha}\preccurlyeq_{k}{\beta}$ yields ${\alpha}\preccurlyeq_{k+1}{\beta}$. Our unprovability result for ${\sf ATR}_{0}$ follows from Theorem A.5 below; see any of [1, 4, 20] for a proof.

Theorem A.5**.**

By recursion on $n$ define $\gamma_{n}<\Gamma_{0}$ given by $\gamma_{0}:=0$ and $\gamma_{n+1}:={\phi}_{\gamma_{n}}0.$ Then, $\forall m\exists\ell\ (\llbracket\ell\rrbracket\gamma_{m}=0)$ is not provable in ${\sf ATR}_{0}$.

Appendix B Alternative Normal Forms

There are other notions of normal form that we may consider aside from those of Definition 2.3. For example, we may skip the sandwiching procedure and choose $a_{0},b_{0}$ so that $A_{a_{0}}{b_{0}}$ is maximal with the property that $A_{a_{0}}b_{0}\leq m$, then choose $a$ maximal such that there exists $b\geq 0$ with $A_{a}b=A_{a_{0}}{b_{0}}$; call this the alternative $k$-normal form of $m$.

These alternative normal forms can be rather inefficient. Fix a base $k$. We claim that for any $j>0$ and any $a,b$, if $A_{a}b=A^{j}_{1}2$ then $a\leq 1$. It suffices to show that $A^{j}_{1}2$ is not in the range of $A_{2}$. Proceed by induction on $j$, and towards a contradiction, assume that $A^{j}_{1}2=A_{2}d$. Then $A^{j}_{1}2=A^{k}_{1}A_{2}(d-1)$. Now consider two cases. If $j\leq k$ then $2=A^{k-j}_{1}A_{2}(d-1)$. If $d=k-j=0$ then $A^{k-j}_{1}A_{2}(d-1)=A_{2}(-1)=1<2$, otherwise $A^{k-j}_{1}A_{2}(d-1)\geq A_{1}1>2$. If $j>k$ then $A^{j-k}_{1}2=A_{2}(d-1)$, contradicting the induction hypothesis if $d>0$ and false if $d=0$ since $A_{2}(-1)=1$. Thus under such normal forms $m=A^{j}_{1}2$ would have to be written with $j$ instances of $A_{1}$, even if $m$ is larger than (say) $A_{100}0$. Our sandwiching procedure is tailored to avoid such a situation.

A third approach would be to choose the normal form of $m$ to be the shortest possible with respect of the number of symbols used. However, we currently do not know if given an expression $\tau$, there is a primitive recursive procedure to compute the shortest $\tau_{\ast}$ with the same value as $\tau$.

If $\tau$ is a formal expression built from [math], $A_{x}y$, and $+$, define $\|\tau\|$ by $\|0\|=1$ and $\|A_{\pi}\sigma+\rho\|=\|\pi\|+\|\sigma\|+\|\rho\|$. Neither the original nor the alternative $k$-normal forms produce minimal norms. Let $m=A_{0}A_{1}0-1$. Then $m\cong_{2}A_{1}0\cdot p+q$ with a large $p$ of about $k^{A_{1}0-\log_{k}A_{1}0}$. However, $m=\sum_{i=0}^{A_{1}0-1}A_{0}i$ which has norm about $(A_{1}0)^{2}$. Now let $n=A_{1}(A_{1}^{k-1}A_{2}(k-1)+1)$. Then $n$ is in alternative normal form with norm about $4k$, since $k-1$ can only be written as $A_{0}0\cdot(k-1)$. But $n=A_{0}^{k}A_{2}A_{0}1$ which has norm about $k$.

Nevertheless, Goodstein sequences for such normal forms could be of interest and we leave their study as an open line of research. We conjecture that Ackermannian Goodstein sequences will terminate for any ‘reasonable’ notion of normal form.