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Sep 6, 2012 at 21:42 comment added Andreas Blass Joel, you're right. In fact, it seems that the second formulation of B is actually equivalent (in PA or appropriate weaker theories) to $\neg$Con(ZF).
Sep 6, 2012 at 20:27 comment added Joel David Hamkins I think your second formulation actually is consistent with ZF, since it is consistent with ZF that $\neg\text{Con}(\text{ZF})$, and in such a model, proofs from ZF are easy to come by. That is, if there are any models of ZF at all, then there are models of ZF that have a definable well-ordering of the reals such that, in that model, there is a proof from ZF that this definition does define a well-ordering of the reals.
Sep 6, 2012 at 19:54 history answered Andreas Blass CC BY-SA 3.0