hairyfigment comments on Second-Order Logic: The Controversy - LessWrong

24 Post author: Eliezer_Yudkowsky 04 January 2013 07:51PM

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Comment author: Eliezer_Yudkowsky 05 January 2013 12:20:20AM 0 points [-]

You can substitute ZFC for ZF throughout the quoted paragraph and it will still be true. I don't know what you want me to get from the link given; it doesn't seem to contradict the quoted paragraph. The part where the link talks about Vk entailing ZFC for inaccessible k is exactly what the quoted paragraph is saying.

Comment author: hairyfigment 05 January 2013 02:06:45AM -1 points [-]

I'll make this more explicit, and note that at present I don't see it affecting anyone's argument:

ZF can't prove its own self-consistency and thus can't prove it has a model. You logician says that's as it should be, since not every model of ZF contains another model - which seems true provided we treat each model more as an interpretation than as an object in a larger universe - and then says that a model has to be "well-populated" - which AFAICT holds for standard models, but not for all other interpretations like those your logician used in the previous sentence.