countType
is a subclass of choiceType
Ana de Almeida Borges said:
Karl Palmskog said:
from what I can recall, one can get a
countType
from achoiceType
nearly automatically at any point, and I don't think anyfinType
is involvedHow? Are
choiceType
andcountType
actually interchangeable?
Without adding any axiom to Coq, in a closed context, I do not think there is a concrete choiceType
that couldn't be turned into a countType
.
However when adding a general choice axiom (as in mathcomp analysis) every type can be equipped with a choiceType
, while assuming any type could be countType
would easily lead to contradiction. In particular we can equip the classical reals with a choiceType
and use the normal mathcomp structures without any trouble.
Ana de Almeida Borges has marked this topic as resolved.
Last updated: Feb 08 2023 at 04:04 UTC