Why does coq's typechecker reject my map definition? - coq

Question

I try to experiment with list's definition.
For example let's see this definition:
Inductive list1 : Type -> Type := nil1 : forall (A : Type), list1 A
| cons1 : forall (A : Type), A -> list1 A -> list1 A.
You might think that the definition above is equivalent to this:
Inductive list0 (A : Type) : Type := nil0 : list0 A
| cons0 : A -> list0 A -> list0 A.
Why this map:
Fixpoint map0 {A : Type} {B : Type} (f : A -> B) (xs : list0 A) : list0 B :=
match xs with
nil0 _ => nil0 B
| cons0 _ v ys => cons0 B (f v) (map0 f ys)
end.
accepted, but this one is not:
Fail Fixpoint map1 {A : Type} {B : Type} (f : A -> B) (xs : list1 A) :=
match xs with
nil1 _ => nil1 B
| cons1 _ v ys => cons1 B (f v) (map1 f ys)
end.
?

Answer 1

This is indeed a confusing aspect of datatype definitions. The problem is that list1 is not equivalent to list0, because of how indices and parameters are treated in these definitions. In Coq jargon, an "index" means an argument declared to the right of the colon, as in list1. A "parameter", by contrast, is an argument declared to the left of the colon, as A in list0.
When you use an index, the return type of match expressions must be generic with respect to the index. This can be seen in the type of list1_rec, a combinator for writing recursive definitions on lists:
Check list1_rec.
list1_rec
: forall P : forall T : Type, list1 T -> Set,
(forall A : Type, P A (nil1 A)) ->
(forall (A : Type) (a : A) (l : list1 A), P A l -> P A (cons1 A a l)) ->
forall (T : Type) (l : list1 T), P T l
This type says that given a generic type P indexed by lists and an element l : list1 A, you can produce a result of type P A l by telling Coq what to return on nil1 and cons1. However, the type of the cons1 branch (the third argument of list1) says that the branch must work not only for the A that appears in the type of l, but also for all other types A'. Compare this to the type of list0_rec:
Check list0_rec.
list0_rec
: forall (A : Type) (P : list0 A -> Set),
P (nil0 A) ->
(forall (a : A) (l : list0 A), P l -> P (cons0 A a l)) ->
forall l : list0 A, P l
The branch of cons0 does not have the forall A bit, which means that the branch only has to work for the type A in l : list0 A.
This makes a difference when writing a function such as map. In map0, we are allowed to apply f : A -> B because we know that the argument of cons0 has type A. In map1, the argument of cons1 has a different generic type A', leading to this error message:
Fail Fixpoint map1 {A : Type} {B : Type} (f : A -> B) (xs : list1 A) :=
match xs with
nil1 A' => nil1 B
| cons1 A' v ys => cons1 B (f v) (map1 f ys)
end.
(* The term "v" has type "A'" while it is expected to have type "A". *)

Answer 2

To be complete, you can define function map over list1 :
Fixpoint map1 {A : Type} {B : Type} (f : A -> B) (xs : list1 A) :=
match xs with
| nil1 A' => fun _ => nil1 B
| cons1 A' v ys => fun f => cons1 B (f v) (map1 f ys)
end f.
This is an example of the so-called convoy pattern. Usually, one needs to add a return clause to the match construct so that it typechecks, but here Coq is smart enough to infer it.
However, I certainly discourage using this definition of lists as it will be cumbersome to use in similar cases.