Question

In COQ the type prod (with one constructor pair) corresponds to cartesian product and the type sig (with one constructor exist) to dependent sum but how is described the fact that the cartesian product is a particular case of dependent sum? I wonder there is a link between prod and sig, for instance some definitional equality but I don’t find it explicitely in the reference manual.

Answer 1

As a matter of fact, the type prod is more akin to sigT than sig:

Inductive prod (A B : Type) : Type :=
  pair : A -> B -> A * B

Inductive sig (A : Type) (P : A -> Prop) : Type :=
  exist : forall x : A, P x -> sig P

Inductive sigT (A : Type) (P : A -> Type) : Type :=
  existT : forall x : A, P x -> sigT P

From a meta-theoretic point of view, prod is just a special case of sigT where your snd component does not depend on your fst component. That is, you could define:

Definition prod' A B := @sigT A (fun _ => B).

Definition pair' {A B : Type} : A -> B -> prod' A B := @existT A (fun _ => B).

Definition onetwo := pair' 1 2.

---

They can not be definitionally equal though, since they are different types. You could show a bijection:

Definition from {A B : Type} (x : @sigT A (fun _ => B)) : prod A B.
Proof. destruct x as [a b]. auto. Defined.

Definition to {A B : Type} (x : prod A B) : @sigT A (fun _ => B).
Proof. destruct x as [a b]. econstructor; eauto. Defined.

Theorem fromto : forall {A B : Type} (x : prod A B), from (to x) = x.
Proof. intros. unfold from, to. now destruct x. Qed.

Theorem tofrom : forall {A B : Type} (x : @sigT A (fun _ => B)), to (from x) = x.
Proof. intros. unfold from, to. now destruct x. Qed.

But that’s not very interesting… 🙂