How does one build the list of only true elements in Coq using dependent types?

Question

I was going through the Coq book from the maths perspective . I was trying to define a dependently typed function that returned a length list with n trues depending on the number trues we want. Coq complains that things don't have the right type but when I see it if it were to unfold my definitions when doing the type comparison it should have worked but it doesn't. Why?

Code:

Module playing_with_types2.
  Inductive Vector {A: Type} : nat -> Type :=
  | vnil: Vector 0
  | vcons: forall n : nat, A -> Vector n -> Vector (S n).

  Definition t {A: Type} (n : nat) : Type :=
    match n with
    | 0 => @Vector A 0
    | S n' => @Vector A (S n')
    end.
  Check t. (* nat -> Type *)
  Check @t. (* Type -> nat -> Type *)

  (* meant to mimic Definition g : forall n: nat, t n. *)
  Fixpoint g (n : nat) : t n :=
    match n with
    | 0 => vnil
    | S n' => vcons n' true (g n')
    end.
End playing_with_types2.

Coq's error:

In environment
g : forall n : nat, t n
n : nat
The term "vnil" has type "Vector 0" while it is expected to have type
 "t ?n@{n1:=0}".
Not in proof mode.

ie t?n@{n1:=0} is Vector 0 ...no?

Answer 1

In this case, it looks like Coq does not manage to infer the return type of the match expression, so the best thing to do is to give it explicitly:

Fixpoint g (n : nat) : t n :=
  match n return t n with
  | 0 => vnil
  | S n' => vcons n' true (g n')
  end.

Note the added return clause.

Then the real error message appears:

In environment
g : forall n : nat, t n
n : nat
n' : nat
The term "g n'" has type "t n'" while it is expected to have type "Vector n'".

And this time it is true that in general t n' is not the same as Vector n' because t n' is stuck (it does not know yet whether n' is 0 or some S n'' ).