4
votes
1answer
336 views

Why is adopting Russell's Axiom of Reducibility as strong as eliminating the Ramified Hierarchy?

In order to respond to concerns of impredicativity, Bertrand Russell developed a system of ramified second-order logic, which is like regular second-order logic except the comprehension schema is ...
4
votes
2answers
363 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 ...
2
votes
0answers
361 views

Is there a notion of “predicative given the real numbers”?

A definition is called impredicative if it involves quantification over a domain that contains the thing being defined. For instance, if you define hereditary property to be a property which applies ...
4
votes
2answers
269 views

Does the Feferman-Schutte analysis give a precise characterization of Predicative Second-Order Arithmetic?

A definition is called impredicative if it involves quantification over a domain that contains the thing being defined. For instance, if you define hereditary property to be a property which applies ...
5
votes
6answers
230 views

Strength of Bishop style constructive mathematics vs $\mathsf{RCA}_0$

This question came out of this other MO question of mine. My question is Is there a formal comparison between $\mathsf{RCA}_0$ and $\mathsf{BISH}$ (Bishop style constructive mathematics as used ...
2
votes
0answers
206 views

Non-Computational classical subterms

Assume we have a proof term of the form $(a^{A\rightarrow^c B\rightarrow^{nc} C}b^Ac^B)^C$, where $c$ is classical (that is, contains free instances of duplex negatio affirmat). The extracted term ...