(* The "?" of cons and eq should be inferred *)
Variable list:Set -> Set.
Variable cons:(T:Set) T -> (list T) -> (list T).
Check (n:(list nat)) (EX l| (EX x| (n = (cons ? x l)))).

(* Examples provided by Eduardo Gimenez *)

Definition c [A;Q:(nat*A->Prop)->Prop;P] :=
   (Q [p:nat*A]let (i,v) = p in (P i v)).

(* What does this test ? *)
Require PolyList.
Definition list_forall_bool  [A:Set][p:A->bool][l:(list A)] : bool := 
 (fold_right ([a][r]if (p a) then r else false) true l).

(* Checks that solvable ? in the lambda prefix of the definition are harmless*)
Parameter A1,A2,F,B,C : Set.
Parameter f : F -> A1 -> B.
Definition f1 [frm0,a1]: B := (f frm0 a1).

(* Checks that solvable ? in the type part of the definition are harmless *)
Definition f2 : (frm0:?;a1:?)B := [frm0,a1](f frm0 a1).