Assuming I have some T : eqType
together with a proof of Finite.axiom Ts
for some Ts : seq T
. Is there a direct way of turning T
into a finite type, that does not involve explicitly mentioning any pickle
/unpickle
function to get the required countType
instance?
I think that's a missing piece indeed...
Okay, in my case, Ts
is actually a concrete nonempty list enumerating an inductive without parameters , so I can do:
Lemma foo: cancel (index^~ Ts) (nth t0 Ts). Proof. by case. Qed.
followed by the usual litany of instances for countType
, choiceType
, and finType
. Oh well...
Last updated: Jan 29 2023 at 19:02 UTC