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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:

page 73 from Predicative Arithmetic page 74 from Predicative Arithmetic

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 $\forall n C[n] \rightarrow \forall n C[n]$?

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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
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There's a discussion about his project at golem.ph.utexas.edu/category/2011/09/…. –  Hans Lundmark Sep 30 '11 at 6:01

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

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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
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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
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@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

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