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Sep 26, 2013 at 15:06 comment added Keshav Srinivasan @AndreasBlass Thanks, I didn't notice that part. So now the question becomes, does $ATR_0$ exhaust the set of statements that are predicatively acceptable? In other words, is $ATR_0$ "maximally" locally predicatively justifiable, or is there a larger system that is maximal?
Sep 26, 2013 at 15:00 comment added Andreas Blass At the bottom of page 18, Feferman says that $ATR_0$ is locally predicatively justifiable, which, according to the definition on page 16, would imply that each individual theorem is predicatively justifiable, but not necessarily that a predicativist would see the whole system as justified.
Sep 26, 2013 at 14:57 comment added Keshav Srinivasan @AndreasBlass Yes, you may be right. So in that case do you think all the theorems of $ATR_0$ are predicatively acceptable?
Sep 26, 2013 at 14:49 comment added Andreas Blass @KeshavSrinivasan The passage you quoted isn't the whole sentence; Feferman precedes it with "It turns out in practice that". I suspect he is making a distinction here between mathematical facts (the subject of this sentence) and metamathematical facts like Con(PA) (or the well-foundedness of $\varepsilon_0$ or the existence of a truth predicate for first-order arithmetic). The next sentence reformulates the claim in terms of "everyday mathematics", and the sentence after that says that there is no theorem that can establish this.
Sep 26, 2013 at 14:39 comment added Keshav Srinivasan @AndreasBlass Then why does Feferman say "if a known mathematical result can be established predicatively, it can already be done in a system conservative over Peano Arithmetic (PA)" on page 19 of this paper: math.stanford.edu/~feferman/papers/predicativity.pdf
Sep 26, 2013 at 14:20 comment added Andreas Blass @KeshavSrinivasan It seems to me that a predicativist, who (as you assumed in the question) accepts all the ordinals below $\Gamma_0$ and thus in particular accepts $\varepsilon_0$, would also accept Gentzen's consistency proof for PA. In fact, I would expect predicativists to also accept the truth predicate for first-order sentences of arithmetic and thus to accept the "obvious" proof that PA is consistent because all its axioms are true.
Sep 26, 2013 at 12:55 comment added Keshav Srinivasan How does that quote answer my question? I want to know how to characterize the second-order statements that are predicatively provable. By the way, how can he say that $ATR_0$ proves the same first-order truths as a predicative system? Doesn't it prove $Con(ACA_0)$ and thus $Con(PA)$? And isn't it not possible to predicatively prove any statement of first-order arithmetic not already provable in $PA$?
Sep 26, 2013 at 6:31 history answered alexod CC BY-SA 3.0