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Jan 25, 2010 at 6:20 vote accept Ewan Delanoy
Jan 25, 2010 at 2:05 comment added François G. Dorais Well, proof theoretic ordinals and ordinal analysis only makes sense for recursively enumerable theories. I suppose you could go higher by relativizing, but proof theorists don't seem to like that sort of thing.
Jan 25, 2010 at 1:44 comment added Joel David Hamkins Are you saying that in PA or ZFC both we use the notations up to omega_1^CK using Kleene's O (or whatever), but that the weaker theories don't always prove the notations are well-ordered? Do any proof-theoretic ordinals go beyond omega_1^CK?
Jan 25, 2010 at 1:07 comment added François G. Dorais The notations are always the same. (In principle Kleene's O, but its often easier to work with different ones depending on context.) What differs is how well theories understand notations. Each notation t corresponds to an induction statement on notations: "induction on notations holds up to t" or simply "t is wellfounded". The proof theoretic ordinal of T is the "first" ordinal notation t such that T does not prove that t is wellfounded.
Jan 25, 2010 at 0:39 comment added Joel David Hamkins Are the "representable ordinals of a theory the same thing as the proof-theoretic ordinal? Is this what the proof-theorists mean?
Jan 25, 2010 at 0:25 comment added François G. Dorais PA is definitely easier to work with, but which ordinals are representable doesn't really change. (Albeit, some theories do much better than others at understanding certain types of notations, which does have consequences.) The point I was trying to make is to study notions in their appropriate settings. Con(ZFC) has no good semantics, so there is little point in studying it in the context of set theory. On the other hand, ZFC + "there is a wellfounded model of ZFC" is interesting to look at in set theory, but difficult to understand arithmetically.
Jan 24, 2010 at 23:32 comment added Joel David Hamkins But so what? Any large cardianl consistency strength statement is also arithmetic (since it is a consistency statement), but we still analyze them in ZFC rather than PA. My sense of the question is that since we believe ZFC, we naturally also want to include Con(ZFC) etc., as among the strongest theories that we really believe in. Perhaps the situations is that it is the negative results (e.g. on which ordinals are not representable) are stronger/better for PA. Does that sound right?
Jan 24, 2010 at 23:28 history edited François G. Dorais CC BY-SA 2.5
grammar
Jan 24, 2010 at 23:08 history edited François G. Dorais CC BY-SA 2.5
clarification
Jan 24, 2010 at 22:58 comment added François G. Dorais Yes, but you're still analyzing arithmetical statements. A more appropriate choice for T is PA + (all arithmetical consequences of ZFC), for example.
Jan 24, 2010 at 22:52 comment added Joel David Hamkins But isn't it true that if the process makes sense with alpha using PA or even second-order arithmetic, then it will also make sense with ZFC for this same ordinal?
Jan 24, 2010 at 22:29 history edited François G. Dorais CC BY-SA 2.5
correction
Jan 24, 2010 at 21:16 history answered François G. Dorais CC BY-SA 2.5