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Jan 31, 2017 at 19:36 history edited Nik Weaver CC BY-SA 3.0
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Jan 31, 2017 at 19:35 comment added Nik Weaver @AndreasBlass: you're right, I was sloppy. I'll fix it.
Jan 31, 2017 at 19:25 comment added Andreas Blass If one can prove Con(SEAR) in ZFC${}^+$, then one cannot prove Con(SEAR)$\implies$Con(ZFC) in Peano arithmetic, because, by combining the two along with Con(ZFC)$\implies$Con(ZFC${}^+$), we'd get that ZFC${}^+$ proves its own consistency.
Jan 31, 2017 at 18:23 comment added Nik Weaver @JohannesHahn: you can't prove in ZFC${}^+$ that ZFC has any model, transitive or not, assuming ZFC is consistent. You can only prove that any finite fragment of ZFC has a model (which can already be done in ZFC). (This is the idea of why ZFC and ZFC${}^+$ are equiconsistent.)
Jan 31, 2017 at 18:12 comment added Johannes Hahn Wait, what? Isn't the existence of a transitive model of ZFC strictly stronger than the existence of just any model?
Jan 31, 2017 at 15:58 history answered Nik Weaver CC BY-SA 3.0