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    $\begingroup$ "For example, it is conceivable that under these hypotheses, ZFC + universes could prove ¬Con(ZFC)." That's definitely false: ZFC + universes proves Con(ZFC), so if ZFC + universes is consistent it can't prove $\neg$Con(ZFC). $\endgroup$ Commented Jun 21, 2019 at 2:43
  • $\begingroup$ Whoops! I mean that ZFC + universes could prove ¬Con(ZFC+universes). $\endgroup$ Commented Jun 21, 2019 at 2:44
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    $\begingroup$ "No meta-theorem of this sort is known" Or, indeed, possible, since Con(ZFC) is a $\Pi^0_1$ statement already! $\endgroup$ Commented Jun 21, 2019 at 12:55
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    $\begingroup$ @PaceNielsen : I would suggest Torkel Franzén's book Inexhaustibility: A Non-Exhaustive Treatment, especially Chapter 7, which gives a detailed explanation of arithmetical truth. $\endgroup$ Commented Jun 21, 2019 at 17:17
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    $\begingroup$ By the way, in case you missed it, this MO question might interest you: mathoverflow.net/questions/331441/… $\endgroup$ Commented Jun 24, 2019 at 15:04