All Questions
2 questions
5
votes
1
answer
145
views
Mathematical strength of the statement "Heyting Arithmetic admits Markov's rule"
Consider the following theorem about Heyting arithmetic (HA):
For every arithmetical formula $\phi$ whose only free variable is $n$, if $\text{HA} \vdash \forall n. \phi \lor \lnot \phi$ and $\text{...
8
votes
3
answers
2k
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 ...