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There is a theorem that neither CH nor ~CH is provable in ZFC. Since it is a theorem is must have been proved. Is the proof possible within ZFC itself or do you need formal systems more powerful than ZFC to show it for ZFC?

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    The theorem needs the assumption that ZF is consistent, because an inconsistent theory proves everything. But except for that, much less than ZFC is needed; Peano arithmetic is more than enough. – Andreas Blass Sep 15 '22 at 00:23
  • You need the assumption that ZFC is consistent (since otherwise it proves everything), which (by the incompleteness theorem) is not provable in ZFC itself (unless it is inconsistent). – Andrés E. Caicedo Sep 15 '22 at 00:23
  • So under the assumption that ZFC is consistent the theorem could be proved in ZFC itself, right? –  Sep 15 '22 at 00:27
  • See the related Why is CH true if it cannot be proved? – Mason Sep 15 '22 at 00:42
  • @Pippen No, if ZFC could prove ZFC can’t prove something, it would prove ZFC is consistent, violating Godel (unless ZFC is inconsistent). On the other hand, ZFC +Con(ZFC) will prove it. – spaceisdarkgreen Sep 15 '22 at 01:03
  • ZFC proves [If Con(ZFC), then CH is independent from ZFC]. Equivalently: ZFC+Con(ZFC) proved [CH is independent from ZFC]. (Here, Con(ZFC) is the assertion that ZFC is consistent) – Akiva Weinberger Sep 15 '22 at 01:10
  • I suppose another way to phrase that is, ZFC proves that either neither CH nor ~CH have proofs in ZFC, or both do. – Akiva Weinberger Sep 15 '22 at 01:11
  • @spaceisdarkgreen: Is not your „ZFC + (Con)ZFC will prove it“ and my „under the assumption of ZFC‘s consistency the theorem can be proved in ZFC“ the same? –  Sep 15 '22 at 01:19
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    @Pippen There's a big difference between "ZFC+Con(ZFC) will prove it" and "If Con(ZFC) is true, then ZFC will prove it"! If Con(ZFC) is true, then ZFC will not prove it. Gödel's second theorem says if Con(ZFC) is true, then ZFC will not prove Con(ZFC). – Akiva Weinberger Sep 15 '22 at 02:18
  • @Pippen "Any reasonable metatheory + Con(ZFC)" proves "CH is independent of ZFC". But "any reasonable metatheory + Con(ZFC)" disproves "ZFC proves CH is independent of ZFC". (With, e.g, ZFC+Con(ZFC), ZFC, or PA being reasonable metatheories.) So "the assumption of Con(ZFC)" does not imply "the theorem can be proved in ZFC itself" (unless our metatheory is inconsistent with Con(ZFC)). Whereas "ZFC + Con(ZFC)" does prove "the theorem" (where "the theorem" = CH is independent of ZFC ). – spaceisdarkgreen Sep 15 '22 at 04:14

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