MrMind comments on How to offend a rationalist (who hasn't thought about it yet): a life lesson - Less Wrong

9 Post author: mszegedy 06 February 2013 07:22AM

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Comment author: MrMind 11 February 2013 10:22:44AM 1 point [-]

it is possible to axiomatize the set of all arithmetical truths by taking as your axioms the set of all arithmetical truths.

Yes, I've thought to add "recursively" to the original statement, but I felt that the word "axiomatize" in the OP carried the meaning of somehow reducing the number of statement, so I decided not to write it. But of course the trivial axiomatization is always possible, you're totally correct.

it is very likely that Peano arithmetic is consistent, but this isn't a proposition I would assign probability 1.

Heh, things get murky really quickly in this field. It's true that you can prove arithmetic consistent inside a stronger model, and it's true that there are non-standard submodel that think they are inconsistent while being consistent in the outer model. There are also models (paraconsistent in the meta-logic) that can prove their own consistency, avoiding Goedel theorem(s). This means that semantically, from a formal point of view, we cannot hope to really prove anything about some true consistency. I admittedly took a platonist view in my reply.

Comment author: Qiaochu_Yuan 11 February 2013 05:05:40PM 3 points [-]

we cannot hope to really prove anything about some true consistency.

Sure we can. If we found a contradiction in Peano arithmetic, we'd prove that Peano arithmetic is inconsistent.