Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Edward Nelson has started writing out the details of a proof of the inconsistency of a version of arithmetic. I'm an undergraduate trying to read through this slowly and carefully.

I've run into a problem, however, with one of his arguments on page 11. It's a proof that "there is a specific number that is not a finite number", which he attributes to Simon Kochen. I did a little digging, and found he had written up a better version of the argument on page 74 of his 1986 book "Predicative Arithmetic" (available on his website). For your convenience, here are just the relevant pages:

$\textbf{Chapter 18}$

$\textbf{An impassable barrier}$

$\quad$ Let us pause to examine from an impredicative point of view what we are doing. Take a strong theory $\rm T$ containing $\rm 0$ and $\rm S$, say an extension by definitions of Peano Arithmetic $I$ or even of $\sf ZFC$ (Zermelo-Fraenkel set theory with the axiom of choice). Let $\rm\hat T$ be the theory obtained by adjoining a unary predicate symbol $\phi$ and the axiom $$\text{Fin. }\ \phi(0)\ \& \ (\phi(x)\to\phi({\rm S}x))\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$ This adjunction does not increase the power of the theory in any way; we can for example interpret $\phi(x)$ by $x=x$. If $\rm T$ is an axiomatization of arithmetic, say an extension by definitions of $I$, then by a specific number we mean a variable-free term $\rm b$ of $\rm T$; if $\rm T$ is an extension by definitions of $\sf ZFC$ containing the constant $\omega$ denoting the set of all natural numbers, then by a specific number we mean a variable-free term $\rm b$ of $\rm T$ such that $\vdash_{\rm T}{\rm b}\in\omega$. We say that a specific number $\rm b$ is a finite number in case $\vdash_{\rm\hat T}\phi(\rm b)$.

$\quad$ Now consider a specific number $\rm b$, such as $10000$ or $10\uparrow10\uparrow10$, and try to prove $\phi(\rm b)$. The initial reaction of some mathematicians to such a problem is a failure to see the difficulty: they suggest proving $\phi(\rm b)$ by induction. But even if the induction principle is an axiom scheme of $\rm T$, we do not have induction available for $\phi(x)$, because the axioms of $\rm T$ say nothing about formulas containing $\phi$. A second reaction is that the problem is trivial because there is an obvious proof in $\rm b$ steps. This observation manages to be both meaningless and incorrect; meaningless because of the ubiquitous pun confusing the formal and genetic concepts of number, and incorrect because there are specific numbers $\rm b$ for which one can show that there is no proof in $\rm\hat T$ of $\phi(b)$. That is, there is a specific number that is not a finite number.

$\quad$ To see this, we will let $\rm T$ be an extension by definitions of $I$. We assume that $I$ is consistent; otherwise there is indeed a proof in $\rm\hat T$ of $\phi(b)$. Consider Gödel's construction of a closed formula $\forall n\mathrm{C}[n]$ that is unprovable even though $\mathrm{C}[0],\mathrm{C}[1],\mathrm{C}[2],...$ are all provable. Let $\rm D$ be the unary formula $\mathrm{C}[n]\to\forall n C[n]$. Then $\exists n{\rm D}[n]$ is provable, since it is equivalent to the tautology $\forall n\mathrm{C}[n]\to\forall n\mathrm{C}[n]$, but ${\rm D}[0],{\rm D}[1],{\rm D}[2],...$ are all unprovable, since from a proof of one of them and the proof of the corresponding theorem among $\mathrm{C}[0],\mathrm{C}[1],\mathrm{C}[2],...$ we would immediately obtain a proof of $\forall n {\rm C}[n]$. Now consider the constant $N$ with the defining axiom $$1.\ \ \ \ N=n\ \leftrightarrow\ \mathrm{D}[n]\ \&\ \forall m(m<n\ \to \ \neg\mathrm{D}[m]).\qquad\qquad\quad\qquad\qquad\qquad$$ The existence condition holds by the least number principle and the uniqueness condition is obvious. We assume that $(1)$ is an axiom of $\rm T$. We claim that $\phi(N)$ is not a theorem of $\rm\hat T$. To see this, take a model of the consistent theory obtained by adjoining $\neg{\rm D}[0],\neg{\rm D}[1],\neg{\rm D}[2],...$ to $\rm T$ and represent $\phi$ by membership in the smallest subset of the universe of the model containing the individual representing $0$ and closed under the function representing $\rm S$. Then $\phi(N)$ is not valid in the model, so $\phi(N)$ is not a theorem of $\rm\hat T$. I am grateful to Simon Kochen for this example.

My problem is specifically these two sentences:

Let $D$ be the unary formula $C[n] \rightarrow \forall n C[n]$. Then $\exists n D[n]$ is provable, since it is equivalent to the tautology $\forall n C[n] \rightarrow \forall n C[n]$.

How in God's name is $\exists n D[n]$ equivalent to $\forall n C[n] \rightarrow \forall n C[n]$?

share|cite|improve this question
Proving Peano arithmetic inconsistent is an ambitious goal indeed. Possibly that enterprise would be more convincing if it resulted in a machine-readable formal proof of $1=0$ rather than in English prose ... but that's probably not your fault. – Henning Makholm Sep 30 '11 at 5:54
There's a discussion about his project at…. – Hans Lundmark Sep 30 '11 at 6:01
up vote 12 down vote accepted

$$\exists n.(C[n]\to\forall n.C[n])$$ $$\exists n.(C[n]\to\forall m.C[m])$$ $$\exists n.(\neg C[n] \lor \forall m.C[m])$$ $$(\exists n.\neg C[n]) \lor (\exists n.\forall m.C[m])$$ $$(\neg\forall n.C[n]) \lor (\forall m.C[m])$$ $$(\forall n.C[n]) \to (\forall m.C[m])$$ $$(\forall n.C[n]) \to (\forall n.C[n])$$ because $(A\to B)\leftrightarrow (\neg A\lor B)$ and $A\leftrightarrow \exists x.A$ when $A$ does not contain $x$.

share|cite|improve this answer
Ah, perfect! I was using the heuristic translation "There exists a man such that if he is bald, all men are bald." I see now I was missing the obvious! If there is a non-bald man, Joe, then it's trivially true that "If Joe is bald, all men are bald".If all men are bald, then it's also trivially true. – James Moody Sep 30 '11 at 6:02
Well, it is at least somewhat non-intuitive -- enough that the intuitionistic movement in the early 1900s denied that this reasoning is valid. – Henning Makholm Sep 30 '11 at 6:55
@James Raymond Smullyan presents this confusing bit of reasoning in the following form: "There exists a person $P$ such that, if $P$ drinks, then everybody drinks." For if $P$ is a nondrinker, the claim is vacuously true, and if there is no such $P$, then indeed everybody drinks. – MJD May 24 '12 at 4:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.