Section EXAMPLES.
Variables v2 v4 v5 : V.
Eval compute in insert 5 v5 (insert 2 v2 (insert 4 v5 empty_tree)).
(*      = T (T E 2 v2 E) 4 v5 (T E 5 v5 E) *)
Eval compute in lookup 5 (T (T E 2 v2 E) 4 v5 (T E 5 v5 E)).
(*      = v5 *)
Eval compute in lookup 3 (T (T E 2 v2 E) 4 v5 (T E 5 v5 E)).
(*      = default *)
Eval compute in elements (T (T E 2 v2 E) 4 v5 (T E 5 v5 E)).
(*      = [(2, v2); (4, v5); (5, v5)] *)
End EXAMPLES.
(* ################################################################# *)
(** * What Should We Prove About Search trees? *)
(** Search trees are meant to be an implementation of maps. That is, they have an [insert] function that corresponds to the [update] function of a map, and a [lookup] function that corresponds to applying the map to an argument.  To prove the correctness of a search-tree algorithm, we can prove: - Any search tree corresponds to some map, using a function or relation that we demonstrate. - The lookup function gives the same result as applying the map - The insert function returns a corresponding map. - Maps have the properties we actually wanted.  It would do no good to prove that searchtrees correspond to some abstract type X, if X didn't have useful properties!  What properties do we want searchtrees to have?  If I insert the binding [(k,v)] into a searchtree [t], then look up [k], I should get [v]. If I look up [k'] in [insert (k,v) t], where [k'<>k], then I should get the same result as [lookup k t].  There are several more properties. Fortunately, all these properties are already proved about [total_map] in the [Maps] module: *)
Check t_update_eq. (*  : forall (A : Type) (m : total_map A) (x : id) (v : A),
t_update m x v x = v   *)
Check t_update_neq.  (* : forall (X : Type) (v : X) (x1 x2 : id) (m : total_map X),
x1 <> x2 -> t_update m x1 v x2 = m x2    *)
Check t_update_shadow.  (* : forall (A : Type) (m : total_map A) (v1 v2 : A) (x : id),
t_update (t_update m x v1) x v2 = t_update m x v2    *)
Check t_update_same.  (* : forall (X : Type) (x : id) (m : total_map X),
t_update m x (m x) = m    *)
Check t_update_permute. (* forall (X : Type) (v1 v2 : X) (x1 x2 : id) (m : total_map X),
x2 <> x1 ->
t_update (t_update m x2 v2) x1 v1 =
t_update (t_update m x1 v1) x2 v2    *)
Check t_apply_empty. (* : forall (A : Type) (x : id) (v : A),
t_empty v x = v *)
(** So, if we like those properties that [total_map] is proved to have, and we can prove that searchtrees behave like maps, then we don't have to reprove each individual property about searchtrees.  More generally:  a job worth doing is worth doing well. It does no good to prove the "correctness" of a program, if you prove that it satisfies a wrong or useless specification. *)
(* ################################################################# *)
(** * Efficiency of Search Trees *)
(** We use binary search trees because they are efficient.  That is, if there are [N] elements in a (reasonably well balanced) BST, each insertion or lookup takes about logN time.  What could go wrong?  1.  The search tree might not be balanced.  In that case, each insertion or lookup will take as much as linear time.  - SOLUTION: use an algorithm, such as "red-black trees", that ensures the trees stay balanced.  We'll do that in Chapter [RedBlack].  2.  Our keys are natural numbers, and Coq's [nat] type takes linear time _per comparison_.  That is, computing (j <? k) takes time proportional to the _value_ of [k-j].  - SOLUTION: represent keys by a data type that has a more efficient comparison operator.  We just use [nat] in this chapter because it's something you're already familiar with.  3.  There's no notion of "run time" in Coq.  That is, we can't say what it means that a Coq function "takes N steps to evaluate."  Therefore, we can't prove that binary search trees are efficient.  - SOLUTION 1: Don't prove (in Coq) that they're efficient; just prove that they are correct.  Prove things about their efficiency the old-fashioned way, on pencil and paper.  - SOLUTION 2: Prove in Coq some facts about the height of the trees, which have direct bearing on their efficiency. We'll explore that in later chapters.  4.  Our functions in Coq aren't real implementations; they are just pretend models of real implementations.  What if there are bugs in the correspondence between the Coq function and the real implementation?  - SOLUTION: Use Coq's [extraction] feature to derive the real implementation (in Ocaml or Haskell) automatically from the Coq function.  Or, use Coq's [vm_compute] or [native_compute] feature to compile and run the programs efficiently inside Coq.  We'll explore [extraction] in a later chapter. *)
(* ################################################################# *)
(** * Proof of Correctness *)
(** We claim that a [tree] "corresponds" to a [total_map]. So we must exhibit an "abstraction relation" [Abs: tree -> total_map V -> Prop].  The idea is that [Abs t m] says that tree [t] is a representation of map [m]; or that map [m] is an abstraction of tree [t].  How should we define this abstraction relation?  The empty tree is easy:   [Abs E (fun x => default)].  Now, what about this tree?: *)
Definition example_tree (v2 v4 v5 : V) :=
T (T E 2 v2 E) 4 v4 (T E 5 v5 E).
(** **** Exercise: 2 stars (example_map)  *)
(* Fill in the definition of example_map with a total_map that you think example_tree should correspond to.  Use [t_update] and [(t_empty default)]. *)
Definition example_map (v2 v4 v5: V) : total_map V
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted. (** [] *)
(** To build the [Abs] relation, we'll use these two auxiliary functions that construct maps: *)
Definition combine {A} (pivot: key) (m1 m2: total_map A) : total_map A :=
fun x => if x <? pivot  then m1 x else m2 x.
(** [combine pivot a b] uses the map [a] on any input less than [pivot], and uses map [b] on any input [ >= pivot]. *)
Inductive Abs:  tree -> total_map V -> Prop :=
| Abs_E: Abs E (t_empty default)
| Abs_T: forall a b l k v r,
Abs l a ->
Abs r b ->
Abs (T l k v r)  (t_update (combine k a b) k v).
(** **** Exercise: 3 stars (check_example_map)  *)
(** Prove that your [example_map] is the right one. If it isn't, go back and fix your definition of [example_map]. You will probably need the [bdestruct] tactic, and [omega]. *)

Lemma check_example_map:
forall v2 v4 v5, Abs (example_tree v2 v4 v5) (example_map v2 v4 v5).