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Jun 22, 2016 at 9:40 comment added Noah Schweber @ThomasKlimpel Yes, that is right.
Jun 22, 2016 at 8:53 vote accept Thomas Klimpel
Jun 22, 2016 at 8:53 comment added Thomas Klimpel Even if "Con(T)" is not in A, A still provides a model of T in case T is consistent, but it no longer provides a proof that T is consistent (or that the provided model is consistent). The provided model is consistent if and only if T is consistent, but A remains unable to prove that T is consistent.
Jun 20, 2016 at 17:47 history answered Noah Schweber CC BY-SA 3.0