On Gentzen’s First Consistency Proof for Arithmetic

Abstract

If nowadays “Gentzen’s consistency proof for arithmetic” is mentioned, one usually refers to [3] while Gentzen’s first (published) consistency proof, i.e. [2], is widely unknown or ignored. The present paper is intended to change this unsatisfactory situation by presenting [2, IV. Abschnitt] in a slightly modified and modernized form.

Appendix

In this appendix we will show how Gentzen’s original ordinal assignment [2, Sect. 15] can be transformed into the assignment which we have used in Sect. 5. This transformation consists in essentially four steps.

Step 1::

We do not use exactly the same set of decimal fractions as Gentzen did. Gentzen defined his set of Ordnungszahlen (let’s call it $\mathcal{O}_{G}$) by: $\mathcal{O}_{G}:=\{ n.u: n \in \mathbb{N}\;\&\;u \in \mathcal{M}_{n}\}$ where $\mathcal{M}_{0}:=\{ 1,\, 11,\, 111,\ldots, 2\}$, $\mathcal{M}_{n+1}:=\{ u_{0}0^{n+1}u_{1}0^{n+1}\ldots 0^{n+1}u_{l}: l \geq 0\;\&\;u_{0},\ldots,u_{l} \in \mathcal{M}_{n}\;\&\;0.u_{l} <_{\mathbb{R}}\cdots <_{\mathbb{R}}0.u_{0}\}$. This corresponds to representing ordinals in base 2 Cantor normal form, while here we shall use base ω. Instead of $\mathcal{O}_{G}$ we define the set $\mathcal{O}:=\{ n.u: n \in \mathbb{N}\;\&\;u \in M_{n}\}$, where M ₀: = { 1}, $M_{n+1}:=\{ u_{0}0^{n+1}u_{1}0^{n+1}\ldots 0^{n+1}u_{l}: l \geq 0\;\&\;u_{0},\ldots,u_{l} \in M_{n}\;\&\;0.u_{l} \leq _{\mathbb{R}}\cdots \leq _{\mathbb{R}}0.u_{0}\}$.

Step 2::

We define an embedding of $(\mathcal{O},<_{\mathbb{R}})$ into the set theoretic ordinals, namely for each ‘Ordnungszahl’ $n.u \in \mathcal{O}$ we define an ordinal | n. u | ∈ On such that $\forall n.u,m.\mathit{v} \in \mathcal{O}(n.u <_{\mathbb{R}}m.\mathit{v} \Rightarrow \vert n.u\vert <\vert m.\mathit{v}\vert )$ (Lemma 3).

Step 3::

We modify Gentzen’s assignment of ‘Ordnungszahlen’ to derivations [2, Sect. 15.2] according to the alterations made in step 1. For each derivation d we define its numerus $\rho (d) \in \mathbb{N}$, mantissa $\mu (d) \in \bigcup _{n\in \mathbb{N}}M_{n}$, and ‘Ordnungszahl’ $\mathrm{Ord}(d):=\rho (d).\mu (d) \in \mathcal{O}$. Actually we only consider the crucial case where d ends with a chain-rule inference.

Step 4::

We show how the ordinal | Ord(d) | can be defined directly by recursion on the build-up of d, without referring to the decimal fraction Ord(d). Then we compare the involved recursion equations with the corresponding equations in the definition of $\tilde{\mathsf{o}}(d)$, o(d) in Sect. 5.

Step 1.

Let {0, 1}⁺ denote the set of all finite nonempty words u over the alphabet {0, 1}, and let $\{0, 1\}^{(+)}:=\{ u \in \{ 0, 1\}^{+}: \mbox{ the first and the last letter of $u$ is 1}\}$.

Further, let 0ⁿ denote the word consisting of n zeros. Each expression n. u (with $n \in \mathbb{N}$ and u ∈ { 0, 1}⁽⁺⁾) will be identified with the real number denoted by it in the usual way.

Definition of $M_{n} \subseteq \{ 0, 1\}^{(+)}$

• 1. M ₀: = { 1};

• 2. $M_{n+1}:=\{ u_{0}0^{n+1}u_{1}0^{n+1}\ldots 0^{n+1}u_{l}: l \geq 0\;\&\;u_{0},\ldots,u_{l} \in M_{n}\;\&\;0.u_{l} \leq _{\mathbb{R}}\cdots \leq _{\mathbb{R}}0.u_{0}\}$.

Further we set $M:=\bigcup _{n\in \mathbb{N}}M_{n}$. The elements of M are called mantissas.

Definition

$\mathrm{h}: M \rightarrow \mathbb{N}$, $\mathrm{h}(u):=\min \{ n: u \in M_{n}\}$.

Remark

$M_{n} \subseteq M_{n+1}$, and h(u) is the maximal number of consecutive zeros in u.

Lemma 1

If $u = u_{0}0^{n+1}\ldots 0^{n+1}u_{l} \in M_{n+1}$ and $\text{v} = \text{v}_{0}0^{n+1}\ldots 0^{n+1}\text{v}_{k} \in M_{n+1}$ with $u_{0},\ldots,u_{l},\text{v}_{0},\ldots,\text{v}_{k} \in M_{n}$ , then $0.u <_{\mathbb{R}}0.\text{v}$ if, and only if, $l <k\;\&\;\forall i \leq l(u_{i} = \text{v}_{i})$  or  $\exists j \leq \min \{ l,k\}\big(\forall i <j(u_{i} = \text{v}_{i})\;\&\;0.u_{j} <_{\mathbb{R}}0.\text{v}_{j}\big)$ .

Proof

Straightforward. □ 

Definition

$\mathcal{O}:=\{ n.u: n <\omega \; \&\;u \in M_{n}\}$ (Ordnungszahlen)

Step 2. :

Definition of |u| _n ∈ On for u ∈ M _n

• 1.  | 1 | ₀: = 0.

• 2. If $u = u_{0}0^{n+1}\ldots 0^{n+1}u_{l} \in M_{n+1}$, then $\vert u\vert _{n+1}:=\omega ^{\vert u_{0}\vert _{n}} + \cdots +\omega ^{\vert u_{l}\vert _{n}}$.

As usual we set $\omega _{0}(\alpha ):=\alpha$, $\omega _{n+1}(\alpha ):=\omega ^{\omega _{n}(\alpha )}$.

Lemma 2

For u ∈ M _n the following holds:

• (a) $\vert u\vert _{n+k} =\omega _{k}(\vert u\vert _{n})$ ,

• (b) $\omega _{n}(0) \leq \vert u\vert _{n} <\omega _{n+1}(0)$ .

Definition

For $n.u \in \mathcal{O}$ let | n. u | : = | u |  _n  ∈ On.

Lemma 3

$n.u \in \mathcal{O}\;\&\;m.\text{v} \in \mathcal{O}\;\&\;n.u <_{\mathbb{R}}m.\text{v}\; \Rightarrow \;\vert n.u\vert <\vert m.\text{v}\vert$ .

Proof by induction on the length of u: Case n < m: Then $\vert n.u\vert = \vert u\vert _{n} <\omega _{n+1}(0) \leq \omega _{m}(0) \leq \vert \text{v}\vert _{m} = \vert m.\text{v}\vert$. Case n = m: Then $0.u <_{\mathbb{R}}0.\text{v}$ and u, v ∈ M _n with n > 0. Hence $u = u_{0}0^{n}\ldots 0^{n}u_{l} \in M_{n}$ and $\text{v} = \text{v}_{0}0^{n}\ldots 0^{n}\text{v}_{k} \in M_{n}$ with $u_{0},\ldots,u_{l},\text{v}_{0},\ldots,\text{v}_{k} \in M_{n-1}$. By Lemma 1 it follows that one of the following two cases applies:

• (i) $l <k\;\&\;\forall i \leq l(u_{i} = \text{v}_{i})$: Then trivially $\vert u\vert _{n} <\vert \text{v}\vert _{n}$.

• (ii) $\forall i <j(u_{i} = \text{v}_{i})\;\&\;0.u_{j} <_{\mathbb{R}}0.\text{v}_{j}$ for some $j \leq \min \{ l,k\}$:Then $\forall i \in \{ j,\ldots,l\}(0.u_{i} <_{\mathbb{R}}0.\text{v}_{j})$ and thus, by IH, $\forall i \in \{ j,\ldots,l\}(\vert u_{i}\vert _{n-1} <\vert \text{v}_{j}\vert _{n-1})$. Hence $\vert u\vert _{n} =\omega ^{\vert \text{v}_{0}\vert _{n-1}} + \cdots +\omega ^{\vert \text{v}_{j-1}\vert _{n-1}} +\omega ^{\vert u_{j}\vert _{n-1}} + \cdots +\omega ^{\vert u_{l}\vert _{n-1}} <\omega ^{\vert \text{v}_{0}\vert _{n-1}} + \cdots +\omega ^{\vert \text{v}_{j}\vert _{n-1}} \leq \vert \text{v}\vert _{n}$.

Step 3. :

The following are more or less Gentzen’s own words (in [2, 15.2])—of course with some alterations enforced by the modifications made in step 1.

To each given derivation d we assign an ‘Ordnungszahl’ $\mathrm{Ord}(d):=\rho (d).\mu (d) \in \mathcal{O}$ according to the following recursive rule: (…) If the endsequent of d is the conclusion of a ‘chain-rule’ inference (i.e., if $d = \mathsf{K}_{\Pi }^{r}d_{0}\ldots d_{l}$ ) we consider the mantissas u _i = μ(d _i ) of the ‘Ordnungszahlen’ of the derivations d _i ; suppose that ν is the maximum number of consecutive zeros in all of these mantissas (i.e., $\nu =\max _{i\leq l}\mathrm{h}(u_{i})$ ). The mantissas are written down from left to right according to their size (the largest one first) and any two successive mantissas are seperated by ν+1 zeros. (It may well be that several successive mantissas are equal.) The result is the mantissa μ(d) of the ordinal number for the whole derivation; i.e., $\mu (d):= u_{\sigma (0)}0^{\nu +1}u_{\sigma (1)}0^{\nu +1}\ldots 0^{\nu +1}u_{\sigma (l)}$ where $\sigma$ is an appropriate permutation of {0,…,l}, and u _i = μ(d _i ). As the numerus ρ(d) we take the least natural number ρ whose excess over the maximum number of consecutive zeros in the mantissa is ≥ 0 and, firstly, is not more than 1 less than the corresponding excess in any of the ordinal numbers for the derivations of the premises and, secondly, is not less than the rank of the succedent formula of any one of the premises preceding the major premise (14.25). W.l.o.g. we may assume here that l ≥ 1 and therefore $\mathrm{h}(\mu (d)) =\nu +1$. So ρ(d) is the least number ρ such that (i) $\rho -(\nu +1) \geq \rho (d_{i}) -\mathrm{h}(u_{i}) - 1$ for i = 0, …, l, and (ii) $\rho -(\nu +1) \geq r$, which amounts to: $\rho (d) -\mathrm{h}(\mu (d)) =\max (\{\rho (d_{i}) -\mathrm{h}(\mu (d_{i})) - 1: i \leq l\} \cup \{ r\})$.

Step 4. :

Let h(d): = h(μ(d)), $\mathrm{exc}(d):=\rho (d) -\mathrm{h}(d)$, and $\widehat{\mathsf{o}}(d):= \vert \mu (d)\vert _{\mathrm{h}(d)}$

Then

• (1) $\vert \mathrm{Ord}(d)\vert =\omega _{\mathrm{exc}(d)}(\widehat{\mathsf{o}}(d))$, 

  and for $d = \mathsf{K}_{\Pi }^{r}d_{0}\ldots d_{l}$ we have the recursion equations

• (2) $\mathrm{h}(d) =\max _{i\leq l}\mathrm{h}(d_{i}) + 1$, and

• (3) $\mathrm{exc}(d) =\max (\{\mathrm{exc}(d_{i}) - 1: i \leq l\} \cup \{ r\})$.

• (4) $\widehat{\mathsf{o}}(d) =\omega ^{\alpha _{0}}\,\#\,\cdots \,\#\,\omega ^{\alpha _{l}}$ with $\alpha _{i}:=\omega _{\nu -\mathrm{h}(d_{i})}(\widehat{\mathsf{o}}(d_{i}))$ and $\nu:=\max _{i\leq l}\mathrm{h}(d_{i})$.

Proof of (1) and (4):

• (1) $\vert \mathrm{Ord}(d)\vert = \vert \rho (d).\mu (d)\vert = \vert \mu (d)\vert _{\rho (d)} =\omega _{\rho (d)-\mathrm{h}(d)}(\widehat{\mathsf{o}}(d)) =\omega _{\mathrm{exc}(d)}(\widehat{\mathsf{o}}(d))$.

• (4) By definition, $\mu (d) = u_{\sigma (0)}0^{\nu +1}\ldots 0^{\nu +1}u_{\sigma (l)}$ with $u_{i} =\mu (d_{i})$ and $\nu =\max _{i\leq l}\mathrm{h}(d_{i})$. Hence $\nu +1 = \mathrm{h}(\mu (d)) = \mathrm{h}(d)$, $\widehat{\mathsf{o}}(d) = \vert \mu (d)\vert _{\nu +1} =\omega ^{\vert u_{0}\vert _{\nu }}\,\#\,\cdots \,\#\,\omega ^{\vert u_{l}\vert _{\nu }}$, and $\vert u_{i}\vert _{\nu } = \vert \mu (d_{i})\vert _{\nu }\stackrel{L.2a}{=}\omega _{\nu -\mathrm{h}(d_{i})}(\vert \mu (d_{i})\vert _{\mathrm{h}(d_{i})})$.

Observation: In case that $\mathrm{h}(\mu (d_{0})) = \cdots = \mathrm{h}(\mu (d_{l}))$ we have

(5) $\widehat{\mathsf{o}}(d) =\omega ^{\widehat{\mathsf{o}}(d_{0})}\,\#\,\cdots \,\#\,\omega ^{\widehat{\mathsf{o}}(d_{l})}$.

Now compare (1), (3), (5) with the corresponding clauses in the definitions of o(d), dg(d), $\tilde{\mathsf{o}}(d)$ in Sect. 5:

• (1)’ $o(d) =\omega _{\mathrm{dg}(d)}(\tilde{\mathsf{o}}(d))$

• (3)’ $\mathrm{dg}(d) =\max (\{\mathrm{dg}(d_{i})-1: i \leq l\} \cup \{ r\})$

• (5)’ $\tilde{\mathsf{o}}(d) =\omega ^{\tilde{\mathsf{o}}(d_{0})}\#\cdots \#\omega ^{\tilde{\mathsf{o}}(d_{l})}$