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For question in Proof Theory, where "proofs" themselves are the object of mathematical investigation. It is not to be used to request a proof of some result.
6
votes
1
answer
271
views
Does $\text{ACA}_0$ + True Arithmetic prove the well-foundedness of every recursive ordinal?
As discussed in Noah Schweber's answer to What is the proof-theoretic ordinal of true arithmetic?, it is somewhat ambiguous what “the proof-theoretic ordinal of True Arithmetic” might mean. In one se …
7
votes
4
answers
1k
views
How can you formalize the metamathematics conventionally used to state Godel’s theorem?
Gödel’s incompleteness theorem states that for any sufficiently strong formal system $T$ there exists a statement $G$ such that if $T$ is consistent, then $G$ is true but not provable in $T$. But I’m …
3
votes
0
answers
845
views
What is the role of the (formalized) omega rule in Ramified Analysis?
In the 1960's, Feferman and Schutte did groundbreaking proof-theoretic work to find out the strength of predicative systems of second-order arithmetic. They used the ramified theory of types, a metho …
3
votes
2
answers
799
views
What is the proof-theoretic ordinal of Hyperarithmetical Comprehension?
As I discuss in my answer here, it seems to me that Solomon Feferman shows, on pages 10-11 of his seminal 1964 paper "Systems of Predicative Analysis", that if you consider predicative second-order ar …
4
votes
Did Gödel prove that the Ramified Theory of Types collapses at $\omega_1$?
I was reading Feferman's original 1964 paper "Systems of Predicative Analysis", and I think I've found an answer to my question about how much second-order arithmetic you can do if you let the ramifie …
12
votes
3
answers
644
views
Has the Ramified Theory of Types been applied to NBG?
Questions of predicativity are well-studied in the context of arithmetic. We have a base theory, first-order Peano arithmetic. Some people, like Edward Nelson (in chapter 1 of his book) and Charles …
5
votes
3
answers
881
views
Did Gödel prove that the Ramified Theory of Types collapses at $\omega_1$?
Second-Order Arithmetic is considered impredicative, because the comprehension scheme allows formulas with bound second-order variables that range over all sets of natural numbers, including the set b …
5
votes
1
answer
710
views
Can the Burgess-Hazen analysis of Predicative Arithmetic be extended to Transfinite Types?
Around page 300 of his book "Mathematical Thought and its Objects", Charles Parsons discusses the work of Edward Nelson, who believes that mathematical induction is impredicative, because it can be ap …