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Backgrounds

The part that goes beyond ZFC is complete in Cantor’s Attic. The portion below Second order arithmetic is complete in Table of ordinal analyses. However the parts in between, especially the various subsystems of ZFC/$\sf Z^-$/MAC/CIC/$\sf CC^\omega$/... , no one seems to have systematically organized the ordering of consistency strengths.

Question

I would like a summary of the strength of agreement between Minimal transitive model of ZFC and Second order arithmetic (especially compared to the beth number). Any diagram or a compendium will work.

asked Apr 8 at 2:28
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  • $\begingroup$ I don't think the precise consistency strength of CIC is known in terms of classical theories. CIC with the excluded middle is equiconsistent with $\mathsf{ZFC} + \{\text{there are $n$ inaccessibles}\}_{n<\omega},ドル and I guess CIC itself is also equiconsistent with the latter. $\endgroup$ Commented Apr 8 at 13:00
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    $\begingroup$ Even for the classical theories, I have not seen any tables lining up the theories from $\mathsf{Z}_2$ and $\mathsf{ZFC}$ in increasing consistency strength. You may make your own, and you should check the results of Mathias and Rathjen (and also, some papers about determinacy in second-order arithmetic, like Montalban-Shore and relatives) for the new table. $\endgroup$ Commented Apr 8 at 13:04
  • $\begingroup$ @HanulJeon According to "Sets in Coq, Coq in Sets", pCIC is not weaker than IZF + Grothendieck universes. However, all such proofs require the introduction of Universe polymorphism. There is only one universe where CIC should be weaker than IZF. $\endgroup$ Commented Apr 10 at 13:42
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    $\begingroup$ @HanulJeon Just because of section 5 "Conclusion" of Sets in "Types, Types in Sets". In short, the lower limit on the strength of consistency of the proof in question should be viewed as an engineering expedient, not as a truth with absolutes. For example, if Lean4/Rocq(Coq) decides to use Agda's $ω*2$ univers Levels, or even the $ε_0$ universes, it would be natural to be able to buy more inaccessibles. $\endgroup$ Commented Apr 10 at 22:47
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    $\begingroup$ I agree with that this discussion is off-topic and I don't think one can persuade the other. I will stop here. $\endgroup$ Commented Apr 12 at 8:02

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