Skip to main content

All Questions

Filter by
Sorted by
Tagged with
5 votes
1 answer
128 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{HA}...
Christopher King's user avatar