### Extraction erases Props

Extraction in Coq works by erasing Props. For example, consider the following definition of div:

div expects a proof that its second argument is non-zero. Indeed, in coq, it is impossible for div to divide by zero. However, when this code is extracted to OCaml, the n <>0 prop is erased (Coq renames our div to div0 to avoid name clash with NPeano’s div):

#### Sumbool

Coq’s sumbool type is another good example to demonstrate the proof erasing nature of extraction. sumbool is defined in Specif module as following:

sumbool is usually the return type of equality decision procedures of various types. For example, the string_dec, the string equality function has the type:

Consider a type id defined as:

A decision procedure ids can be constructed from string_dec as following:

 Theorem eq_id_dec : forall id1 id2 : id, {id1 = id2} + {id1 <> id2}.
Proof.
intros id1 id2.
destruct id1 as [n1]. destruct id2 as [n2].
destruct (string_dec n1 n2) as [Heq | Hneq].
Case "n1 = n2".
left. rewrite Heq. reflexivity.
Case "n1 <> n2".
right. intros contra. inversion contra. apply Hneq. apply H0.
Defined


Extracting the sumbool and eq_id_dec generates the following:

OCaml’s sumbool has no arguments because coq’s sumbool has only Prop arguments. The advantage of using sumbool instead of bool is that it can be used seamlessly in proofs and computations. When used in a computation, sumbool simply tells whether a property is true or false, but when used in a proof, sumbool also tells why a property is true or false.

#### Theorems can also be extracted

Observe that eq_id_dec has been written as a theorem. Theorem can be used to assert the existence of a witness to, not only Props, but also Sets and Types. For example:

  Theorem there_is_a_nat : nat.
Proof.  apply 0.  Defined.

Extraction there_is_a_nat.
(*
let there_is_a_nat = O
*)


Why do we say Defined instead of Qed whenever we are doing extraction? Qed defines proof/definition as opaque, whereas Defined defines it as transparent. If we used Qed instead of Defined extraction would produce:

(** val there_is_a_nat : nat **)

let there_is_a_nat =
failwith "AXIOM TO BE REALIZED"


Note that theorems can only be extracted if the statement is either Set or a Type, not Prop. The following two examples should demonstrate this point. Example 1:

  Theorem there_is_plus: forall (n1 n2 : nat), exists (n :nat), n = n1 + n2.
Proof.  intros.  exists (n1+n2).  reflexivity.
Defined.

Check (forall (n1 n2 : nat), exists (n :nat), n = n1 + n2). (* Prop *)

Extraction there_is_plus.
(*
(** val there_is_plus : __ **)

let there_is_plus =
__
*)


Contrast Example 1 with the following Example 2:

  Inductive plus_defined (n1 n2: nat) : Set :=
| PlusT : forall(n:nat), (n=n1+n2) -> plus_defined n1 n2.

Extraction plus_defined.
(*
type plus_defined =
nat (* singleton inductive, whose constructor was PlusT *)
*)

Theorem there_is_plus_t: forall (n1 n2 : nat), plus_defined n1 n2.
Proof.  intros.
apply PlusT with (n:=n1+n2); reflexivity.
Defined.

Extraction there_is_plus_t.
(*
let there_is_plus_t n1 n2 = plus n1 n2
*)


Why would anyone want to write a Set or a Type term as a proof of a theorem, rather than a Definition or a Fixpoint? If the Set or Type term expects Prop witnesses (like sumbool), then its better to write it as a proof. Sometimes, it may not even be possible to write the term otherwise. For example, here is a failed attempt at defining eq_id_dec as a Definition:

  Inductive id : Type :=
T : string -> t.

Theorem eq1 : forall (s1 s2: string), s1=s2 -> (T s1) = (T s2).
Proof.
intros.
subst. reflexivity.
Qed.

Theorem neq1 : forall (s1 s2 : string), s1<>s2 -> (T s1) <> (T s2).
Proof.
unfold not.
intros.
apply H; inversion H0. subst; reflexivity.
Qed.

Definition id_eq (id1 id2 : id) : ({id1 = id2}+{id1 <> id2}) :=
match (id1,id2) with
| (T s1,T s2) => match string_dec s1 s2 with
| left A => left (eq1 s1 s2 A)
| right B => right (neq1 s1 s2 B)
end
end.


The approach fails because the term left (eq1 s1 s2 A), which has the type {T s1 = T s2} + {?66}, fails to type check agains the required type {id1 = id2} + {id1 <> id2}. The problem is that while pattern matching under a definition, coq does not populate the context with equalities between scrutinees and patterns. Although we know that id1 = T s1 and id2 = T s2, we have no way of obtaining it from the context. Recall that we did not face this problem when proving eq_id_dec. This is why we sometimes prefer writing Set or Type terms as proofs.