Lists.v

(** * Lists: Products, Lists and Options *)

(* $Date: 2011-02-09 15:47:56 -0500 (Wed, 09 Feb 2011) $ *)

(** The next line imports all of our definitions from the
    previous chapter. *)

Require Import Basics.
Open Scope nat_scope.
Open Scope bool_scope.
(** For it to work, you need to compile Basics.v into
    Basics.vo.  (This is like making a .class file from a .java file,
    or a .o file from a .c file.)
  
    Here are two ways to compile your code:
  
     - CoqIDE:
   
         Open Basics.v.
         In the "Compile" menu, click on "Compile Buffer".
   
     - Command line:
   
         Run [coqc Basics.v]

    In this file, we again use the [Module] feature to wrap all of the
    definitions for pairs and lists of numbers in a module so that,
    later, we can reuse the same names for improved (generic) versions
    of the same operations. *)

Module NatList.

(* ###################################################### *)
(** * Pairs of Numbers *)

(** In an [Inductive] type definition, each constructor can take
    any number of parameters -- none (as with [true] and [O]), one (as
    with [S]), or more than one, as in this definition: *)

Inductive natprod : Type :=
  pair : nat -> nat -> natprod.

(** This declaration can be read: "There is just one way to
    construct a pair of numbers: by applying the constructor [pair] to
    two arguments of type [nat]."

    Here are some simple function definitions illustrating pattern
    matching on two-argument constructors: *)
 
Definition fst (p : natprod) : nat := 
  match p with
  | pair x y => x
  end.
Definition snd (p : natprod) : nat := 
  match p with
  | pair x y => y
  end.

(** Since pairs are used quite a bit, it is nice to be able to
    write them with the standard mathematical notation [(x,y)] instead
    of [pair x y].  We can tell Coq to allow this with a [Notation]
    declaration. *)
 
Notation "( x , y )" := (pair x y) : list_scope.
Open Scope list_scope.

(** The new notation can be used both in expressions and in
    pattern matches (indeed, we've seen it already in the previous
    chapter -- this notation is provided as part of the standard
    library): *)

Eval simpl in (fst (3,4)).

Definition fst' (p : natprod) : nat := 
  match p with
  | (x,y) => x
  end.
Definition snd' (p : natprod) : nat := 
  match p with
  | (x,y) => y
  end.

Definition swap_pair (p : natprod) : natprod := 
  match p with
  | (x,y) => (y,x)
  end.

(** Let's try and prove a few simple facts about pairs.  If we
    state the lemmas in a particular (and slightly peculiar) way, we
    can prove them with just reflexivity (and its built-in
    simplification): *)

Theorem surjective_pairing' : forall (n m : nat),
  (n,m) = (fst (n,m), snd (n,m)).
Proof.
  reflexivity.  Qed.

(** But reflexivity is not enough if we state the lemma in a more
    natural way: *)

Theorem surjective_pairing_stuck : forall (p : natprod),
  p = (fst p, snd p).
Proof.
  simpl. (* Doesn't reduce anything! *)
Admitted.

(** We have to expose the structure of [p] so that [simpl] can
    perform the pattern match in [fst] and [snd].  We can do this with
    [destruct].

    Notice that, unlike for [nat]s, [destruct] doesn't generate an
    extra subgoal here.  That's because [natprod]s can only be
    constructed in one way.  *)

Theorem surjective_pairing : forall (p : natprod),
  p = (fst p, snd p).
Proof.
  intros p.  destruct p as (n,m).  simpl.  reflexivity.  Qed.

(** Notice that Coq allows us to use the notation we introduced
    for pairs in the "[as]..." pattern telling it what variables to
    bind. *)

(** **** Exercise: 1 star (snd_fst_is_swap) *)
Theorem snd_fst_is_swap : forall (p : natprod),
  (snd p, fst p) = swap_pair p.
Proof.
intro p. destruct p as (n,m). reflexivity. 
Qed.

(** **** Exercise: 1 star, optional (fst_swap_is_snd) *)
Theorem fst_swap_is_snd : forall (p : natprod),
  fst (swap_pair p) = snd p.
Proof.
intro p. destruct p. reflexivity.
Qed.

(* ###################################################### *)
(** * Lists of Numbers *)

(** Generalizing the definition of pairs a little, we can
    describe the type of _lists_ of numbers like this: "A list is
    either the empty list or else a pair of a number and another
    list." *)

Inductive natlist : Type :=
  | nil : natlist
  | cons : nat -> natlist -> natlist.

(** For example, here is a three-element list: *)

Definition l_123 := cons 1 (cons 2 (cons 3 nil)).

(** As with pairs, it is more convenient to write lists in
    familiar programming notation.  The following two declarations
    allow us to use [::] as an infix [cons] operator and square
    brackets as an "outfix" notation for constructing lists. *)

Notation "x :: l" := (cons x l) (at level 60, right associativity).
Notation "[ ]" := nil.
Notation "[ x , .. , y ]" := (cons x .. (cons y nil) ..).

(** It is not necessary to fully understand these declarations,
    but in case you are interested, here is roughly what's going on.

    The [right associativity] annotation tells Coq how to parenthesize
    expressions involving several uses of [::] so that, for example,
    the next three declarations mean exactly the same thing: *)

Definition l_123'   := 1 :: (2 :: (3 :: nil)).
Definition l_123''  := 1 :: 2 :: 3 :: nil.
Definition l_123''' := [1,2,3].

(** The [at level 60] part tells Coq how to parenthesize
    expressions that involve both [::] and some other infix operator.
    For example, since we defined [+] as infix notation for the [plus]
    function at level 50,
[[
Notation "x + y" := (plus x y)  
                    (at level 50, left associativity).
]]
   The [+] operator will bind tighter than [::], so [1 + 2 :: [3]]
   will be parsed, as we'd expect, as [(1 + 2) :: [3]] rather than [1
   + (2 :: [3])].

   (By the way, it's worth noting in passing that expressions like "[1
   + 2 :: [3]]" can be a little confusing when you read them in a .v
   file.  The inner brackets, around 3, indicate a list, but the outer
   brackets are there to instruct the "coqdoc" tool that the bracketed
   part should be displayed as Coq code rather than running text.
   These brackets don't appear in the generated HTML.)

   The second and third [Notation] declarations above introduce the
   standard square-bracket notation for lists; the right-hand side of
   the third one illustrates Coq's syntax for declaring n-ary
   notations and translating them to nested sequences of binary
   constructors. *)

(** A number of functions are useful for manipulating lists.
    For example, the [repeat] function takes a number [n] and a
    [count] and returns a list of length [count] where every element
    is [n]. *)

Fixpoint repeat (n count : nat) : natlist := 
  match count with
  | O => nil
  | S count' => n :: (repeat n count')
  end.

(** The [length] function calculates the length of a list. *)

Fixpoint length (l:natlist) : nat := 
  match l with
  | nil => O
  | h :: t => S (length t)
  end.

(** The [app] ("append") function concatenates two lists. *)

Fixpoint app (l1 l2 : natlist) : natlist := 
  match l1 with
  | nil    => l2
  | h :: t => h :: (app t l2)
  end.

(** Actually, [app] will be used a lot in some parts of what
    follows, so it is convenient to have an infix operator for it. *)

Notation "x ++ y" := (app x y) 
                     (right associativity, at level 60).

Example test_app1:             [1,2,3] ++ [4,5] = [1,2,3,4,5].
Proof. reflexivity.  Qed.
Example test_app2:             nil ++ [4,5] = [4,5].
Proof. reflexivity.  Qed.
Example test_app3:             [1,2,3] ++ nil = [1,2,3].
Proof. reflexivity.  Qed.

(** Here are two more small examples of programming with lists.
    The [hd] function returns the first element (the "head") of the
    list, while [tail] returns everything but the first
    element.  Of course, the empty list has no first element, so we
    must pass a default value to be returned in that case.  *)

Definition hd (default:nat) (l:natlist) : nat :=
  match l with
  | nil => default
  | h :: t => h
  end.

Definition tail (l:natlist) : natlist :=
  match l with
  | nil => nil  
  | h :: t => t
  end.

Example test_hd1:             hd 0 [1,2,3] = 1.
Proof. reflexivity.  Qed.
Example test_hd2:             hd 0 [] = 0.
Proof. reflexivity.  Qed.
Example test_tail:            tail [1,2,3] = [2,3].
Proof. reflexivity.  Qed.

(** **** Exercise: 2 stars, recommended (list_funs) *)
(** Complete the definitions of [nonzeros], [oddmembers] and
    [countoddmembers] below.  *)

Fixpoint filter (f:nat -> bool) (l:natlist) : natlist :=
match l with
| [] => []
| h::t => if f h then h :: filter f t else filter f t
end.

Definition nonzeros (l:natlist) : natlist :=
filter (fun x:nat => (0 \= x)) l.

Example test_nonzeros:            nonzeros [0,1,0,2,3,0,0] = [1,2,3].
reflexivity. Qed.

Fixpoint oddmembers (l:natlist) : natlist :=
filter oddb l.

Example test_oddmembers:            oddmembers [0,1,0,2,3,0,0] = [1,3].
reflexivity. Qed.

Fixpoint countoddmembers (l:natlist) : nat :=
length (oddmembers l).

Example test_countoddmembers1:    countoddmembers [1,0,3,1,4,5] = 4.
reflexivity. Qed.
Example test_countoddmembers2:    countoddmembers [0,2,4] = 0.
reflexivity. Qed.
Example test_countoddmembers3:    countoddmembers nil = 0.
reflexivity. Qed.
(** [] *)

(** **** Exercise: 2 stars (alternate) *)
(** Complete the definition of [alternate], which "zips up" two lists
    into one, alternating between elements taken from the first list
    and elements from the second.  See the tests below for more
    specific examples.

    Note: one natural way of writing [alternate] will fail to satisfy
    Coq's requirement that all [Fixpoint] definitions be "obviously
    terminating."  If you find yourself in this rut, look for a
    slightly more verbose solution that considers elements of both
    lists at the same time. *)

Fixpoint alternate (l1 l2 : natlist) : natlist :=
match l1, l2 with
| nil, _ => l2
| _, nil => l1
| h1::t1, h2::t2 => h1 :: h2 :: alternate t1 t2
end.

Example test_alternate1:        alternate [1,2,3] [4,5,6] = [1,4,2,5,3,6].
reflexivity. Qed.
Example test_alternate2:        alternate [1] [4,5,6] = [1,4,5,6].
reflexivity. Qed.
Example test_alternate3:        alternate [1,2,3] [4] = [1,4,2,3].
reflexivity. Qed.
Example test_alternate4:        alternate [] [20,30] = [20,30].
reflexivity. Qed.
(** [] *)

(* ###################################################### *)
(** ** Bags via Lists *)

(** A [bag] (or [multiset]) is like a set, but each element can appear
    multiple times instead of just once.  One reasonable
    implementation of bags is to represent a bag of numbers as a
    list. *)

Definition bag := natlist.  

(** **** Exercise: 3 stars (bag_functions) *)
(** Complete the following definitions for the functions
    [count], [sum], [add], and [member] for bags. *)

Definition count (v:nat) (s:bag) : nat := 
length (filter (fun x => x==v) s).

(** All these proofs can be done just by [reflexivity]. *)

Example test_count1:              count 1 [1,2,3,1,4,1] = 3.
reflexivity. Qed.
Example test_count2:              count 6 [1,2,3,1,4,1] = 0.
reflexivity. Qed.

(** Multiset [sum] is similar to set [union]: [sum a b] contains
    all the elements of [a] and of [b].  (Mathematicians usually
    define [union] on multisets a little bit differently, which
    is why we don't use that name for this operation.)
    For [sum] we're giving you a header that does not give explicit
    names to the arguments.  Moreover, it uses the keyword
    [Definition] instead of [Fixpoint], so even if you had names for
    the arguments, you wouldn't be able to process them recursively.
    The point of stating the question this way is to encourage you to
    think about whether [sum] can be implemented in another way --
    perhaps by using functions that have already been defined.  *)

Definition sum : bag -> bag -> bag := app.

Example test_sum1:              count 1 (sum [1,2,3] [1,4,1]) = 3.
reflexivity. Qed.

Definition add := cons.

Example test_add1:                count 1 (add 1 [1,4,1]) = 3.
reflexivity. Qed.
Example test_add2:                count 5 (add 1 [1,4,1]) = 0.
reflexivity. Qed.

Definition member (v:nat) (s:bag) : bool := count v s \= 0.

(*
Fixpoint member (v:nat) (s:bag) : bool := 
match s with
| nil => false
| h::t => if h==v then true else member v t
end.
*)

Example test_member1:             member 1 [1,4,1] = true.
reflexivity. Qed.
Example test_member2:             member 2 [1,4,1] = false.
reflexivity. Qed.
(** [] *)

(** **** Exercise: 3 stars, optional (bag_more_functions) *)
(** Here are some more bag functions for you to practice with. *)

Fixpoint remove_one (v:nat) (s:bag) : bag :=
  (* When remove_one is applied to a bag without the number to remove,
     it should return the same bag unchanged. *)
match s with
| nil => nil
| h::t => if v==h then t else h :: remove_one v t
end.

Example test_remove_one1:         count 5 (remove_one 5 [2,1,5,4,1]) = 0.
reflexivity. Qed.
Example test_remove_one2:         count 5 (remove_one 5 [2,1,4,1]) = 0.
reflexivity. Qed.
Example test_remove_one3:         count 4 (remove_one 5 [2,1,4,5,1,4]) = 2.
reflexivity. Qed.
Example test_remove_one4:  count 5 (remove_one 5 [2,1,5,4,5,1,4]) = 1.
reflexivity. Qed.

Fixpoint remove_all (v:nat) (s:bag) : bag :=
filter (fun x => v \= x) s.

Example test_remove_all1:          count 5 (remove_all 5 [2,1,5,4,1]) = 0.
reflexivity. Qed.
Example test_remove_all2:          count 5 (remove_all 5 [2,1,4,1]) = 0.
reflexivity. Qed.
Example test_remove_all3:          count 4 (remove_all 5 [2,1,4,5,1,4]) = 2.
reflexivity. Qed.
Example test_remove_all4:          count 5 (remove_all 5 [2,1,5,4,5,1,4,5,1,4]) = 0.
reflexivity. Qed.

Fixpoint map (f:nat->nat) (l:natlist) : natlist :=
match l with
| nil => nil
| h::t => f h :: map f t
end.

Fixpoint subset (s1:bag) (s2:bag) : bool :=
match s1 with
| nil => true
| h::t => if member h s2 then subset t (remove_one h s2) else false
end.

Example test_subset1:              subset [1,2] [2,1,4,1] = true.
reflexivity. Qed.
Example test_subset2:              subset [1,2,2] [2,1,4,1] = false.
reflexivity. Qed.
(** [] *)

(** **** Exercise: 3 stars, recommended (bag_theorem) *)
(** Write down an interesting theorem about bags involving the
    functions [count] and [add], and prove it.  Note that, since this
    problem is somewhat open-ended, it's possible that you may come up
    with a theorem which is true, but whose proof requires techniques
    you haven't learned yet.  Feel free to ask for help if you get
    stuck!
[]
 *)

(*
add = cons
count = how many times some elem is in some list
*)
Theorem add_x_incs_count_x : forall x:nat, forall xs:bag,
count x (x :: xs) = count x xs + 1.
Proof.
intro x. induction xs.
 simpl. unfold count. simpl. rewrite<- beq_nat_refl. reflexivity.
 unfold count. simpl. rewrite<- beq_nat_refl.
 set (is_x := (fun x0 : nat => x0 == x)). set (c := n==x).
 destruct c; rewrite plus_comm; reflexivity.
Qed.

(* ###################################################### *)
(** * Reasoning About Lists *)

(** Just as with numbers, simple facts about list-processing
    functions can sometimes be proved entirely by simplification. For
    example, the simplification performed by [reflexivity] is enough
    for this theorem... *)

Theorem nil_app : forall l:natlist,
  [] ++ l = l.
Proof.
   reflexivity.  Qed.

(** ... because the [[]] is substituted into the match position
    in the definition of [app], allowing the match itself to be
    simplified. *)

(** Also, as with numbers, it is sometimes helpful to perform case
    analysis on the possible shapes (empty or non-empty) of an unknown
    list. *)

Theorem tl_length_pred : forall l:natlist,
  pred (length l) = length (tail l).
Proof.
  intros l. destruct l as [| n l'].
  Case "l = nil".
    reflexivity.
  Case "l = cons n l'". 
    reflexivity.  Qed.

(** Here, the [nil] case works because we've chosen to define
    [tl nil = nil]. Notice that the [as] annotation on the [destruct]
    tactic here introduces two names, [n] and [l'], corresponding to
    the fact that the [cons] constructor for lists takes two
    arguments (the head and tail of the list it is constructing). *)

(** Usually, though, interesting theorems about lists require
    induction for their proofs. *)

(* ###################################################### *)
(** ** Micro-Sermon *)

(** Simply reading example proofs will not get you very far!  It is
    very important to work through the details of each one, using Coq
    and thinking about what each step of the proof achieves.
    Otherwise it is more or less guaranteed that the exercises will
    make no sense. *)

(* ###################################################### *)
(** ** Induction on Lists *)

(** Proofs by induction over datatypes like [natlist] are
    perhaps a little less familiar than standard natural number
    induction, but the basic idea is equally simple.  Each [Inductive]
    declaration defines a set of data values that can be built up from
    the declared constructors: a boolean can be either [true] or
    [false]; a number can be either [O] or [S] applied to a number; a
    list can be either [nil] or [cons] applied to a number and a list.

    Moreover, applications of the declared constructors to one another
    are the _only_ possible shapes that elements of an inductively
    defined set can have, and this fact directly gives rise to a way
    of reasoning about inductively defined sets: a number is either
    [O] or else it is [S] applied to some _smaller_ number; a list is
    either [nil] or else it is [cons] applied to some number and some
    _smaller_ list; etc. So, if we have in mind some proposition [P]
    that mentions a list [l] and we want to argue that [P] holds for
    _all_ lists, we can reason as follows:

      - First, show that [P] is true of [l] when [l] is [nil].

      - Then show that [P] is true of [l] when [l] is [cons n l'] for
        some number [n] and some smaller list [l'], asssuming that [P]
        is true for [l'].

    Since larger lists can only be built up from smaller ones,
    eventually reaching [nil], these two things together establish the
    truth of [P] for all lists [l].  Here's a concrete example: *)

Theorem app_ass : forall l1 l2 l3 : natlist, 
  (l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).   
Proof.
  intros l1 l2 l3. induction l1 as [| n l1'].
  Case "l1 = nil".
    reflexivity.
  Case "l1 = cons n l1'".
    simpl. rewrite IHl1'. reflexivity.
Qed.

(** Again, this Coq proof is not especially illuminating as a
    static written document -- it is easy to see what's going on if
    you are reading the proof in an interactive Coq session and you
    can see the current goal and context at each point, but this state
    is not visible in the written-down parts of the Coq proof.  So a
    natural-language proof -- one written for human readers -- will
    need to include more explicit signposts; in particular, it will
    help the reader stay oriented if we remind them exactly what the
    induction hypothesis is in the second case.  *)

(** _Theorem_: For all lists [l1], [l2], and [l3], 
   [(l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3)].

   _Proof_: By induction on [l1].

   - First, suppose [l1 = []].  We must show
[[
       ([] ++ l2) ++ l3 = [] ++ (l2 ++ l3),
]]
     which follows directly from the definition of [++].

   - Next, suppose [l1 = n::l1'], with
[[
       (l1' ++ l2) ++ l3 = l1' ++ (l2 ++ l3)
]]
     (the induction hypothesis). We must show
[[
       ((n :: l1') ++ l2) ++ l3 = (n :: l1') ++ (l2 ++ l3).
]]  
     By the definition of [++], this follows from
[[
       n :: ((l1' ++ l2) ++ l3) = n :: (l1' ++ (l2 ++ l3)),
]]
     which is immediate from the induction hypothesis.  []

  Here is an exercise to be worked together in class: *)

Theorem app_length : forall l1 l2 : natlist,
  length (l1 ++ l2) = (length l1) + (length l2).
Proof.
  (* WORKED IN CLASS *)
  intros l1 l2. induction l1 as [| n l1'].
  Case "l1 = nil".
    reflexivity.
  Case "l1 = cons".
    simpl. rewrite -> IHl1'. reflexivity.  Qed.

(** For a slightly more involved example of an inductive proof
    over lists, suppose we define a "cons on the right" function
    [snoc] like this... *)

Fixpoint snoc (l:natlist) (v:nat) : natlist := 
  match l with
  | nil    => [v]
  | h :: t => h :: (snoc t v)
  end.

(** ... and use it to define a list-reversing function [rev]
    like this: *)

Fixpoint rev (l:natlist) : natlist := 
  match l with
  | nil    => nil
  | h :: t => snoc (rev t) h
  end.

Example test_rev1:            rev [1,2,3] = [3,2,1].
Proof. reflexivity.  Qed.
Example test_rev2:            rev nil = nil.
Proof. reflexivity.  Qed.

(** Now let's prove some more list theorems using our newly
    defined [snoc] and [rev].  For something a little more challenging
    than the inductive proofs we've seen so far, let's prove that
    reversing a list does not change its length.  Our first attempt at
    this proof gets stuck in the successor case... *)

Theorem rev_length_firsttry : forall l : natlist,
  length (rev l) = length l.
Proof.
  intros l. induction l as [| n l'].
  Case "l = []".
    reflexivity.
  Case "l = n :: l'".
    simpl. (* Here we are stuck: the goal is an equality involving
              [snoc], but we don't have any equations in either the
              immediate context or the global environment that have
              anything to do with [snoc]! *)
Admitted.

(** So let's take the equation about [snoc] that would have
    enabled us to make progress and prove it as a separate lemma. *)

Theorem length_snoc : forall n : nat, forall l : natlist,
  length (snoc l n) = S (length l).
Proof.
  intros n l. induction l as [| n' l'].
  Case "l = nil".
    reflexivity.
  Case "l = cons n' l'".
    simpl. rewrite -> IHl'. reflexivity.  Qed. 

(** Now we can complete the original proof. *)

Theorem rev_length : forall l : natlist,
  length (rev l) = length l.
Proof.
  intros l. induction l as [| n l'].
  Case "l = nil".
    reflexivity.
  Case "l = cons".
    simpl. rewrite -> length_snoc. 
    rewrite -> IHl'. reflexivity.  Qed.

(** For comparison, here are _informal_ proofs of these two theorems: 

    _Theorem_: For all numbers [n] and lists [l],
       [length (snoc l n) = S (length l)].
 
    _Proof_: By induction on [l].

    - First, suppose [l = []].  We must show
[[
        length (snoc [] n) = S (length []),
]]
      which follows directly from the definitions of
      [length] and [snoc].

    - Next, suppose [l = n'::l'], with
[[
        length (snoc l' n) = S (length l').
]]
      We must show
[[
        length (snoc (n' :: l') n) = S (length (n' :: l')).
]]
      By the definitions of [length] and [snoc], this
      follows from
[[
        S (length (snoc l' n)) = S (S (length l')),
]] 
      which is immediate from the induction hypothesis. [] *)
                        
(** _Theorem_: For all lists [l], [length (rev l) = length l].
    
    _Proof_: By induction on [l].  

      - First, suppose [l = []].  We must show
[[
          length (rev []) = length [],
]]
        which follows directly from the definitions of [length] 
        and [rev].
    
      - Next, suppose [l = n::l'], with
[[
          length (rev l') = length l'.
]]
        We must show
[[
          length (rev (n :: l')) = length (n :: l').
]]
        By the definition of [rev], this follows from
[[
          length (snoc (rev l') n) = S (length l')
]]
        which, by the previous lemma, is the same as
[[
          S (length (rev l')) = S (length l').
]]
        This is immediate from the induction hypothesis. [] *)

(** Obviously, the style of these proofs is rather longwinded
    and pedantic.  After the first few, we might find it easier to
    follow proofs that give a little less detail overall (since we can
    easily work them out in our own minds or on scratch paper if
    necessary) and just highlight the non-obvious steps.  In this more
    compressed style, the above proof might look more like this: *)

(** _Theorem_:
     For all lists [l], [length (rev l) = length l].

    _Proof_: First, observe that
[[
       length (snoc l n) = S (length l)
]]
     for any [l].  This follows by a straightforward induction on [l].
     The main property now follows by another straightforward
     induction on [l], using the observation together with the
     induction hypothesis in the case where [l = n'::l']. [] *)

(** Which style is preferable in a given situation depends on
    the sophistication of the expected audience and on how similar the
    proof at hand is to ones that the audience will already be
    familiar with.  The more pedantic style is a good default for
    present purposes. *)

(* ###################################################### *)
(** ** [SearchAbout] *)

(** We've seen that proofs can make use of other theorems we've
    already proved, using [rewrite], and later we will see other ways
    of reusing previous theorems.  But in order to refer to a theorem,
    we need to know its name, and remembering the names of all the
    theorems we might ever want to use can become quite difficult!  It
    is often hard even to remember what theorems have been proven,
    much less what they are named.

    Coq's [SearchAbout] command is quite helpful with this.  Typing
    [SearchAbout foo] will cause Coq to display a list of all theorems
    involving [foo].  For example, try uncommenting the following to
    see a list of theorems that we have proved about [rev]: *)

SearchAbout rev.

(** Keep [SearchAbout] in mind as you do the following exercises and
    throughout the rest of the course; it can save you a lot of time! *) 

(* ###################################################### *)
(** ** List Exercises, Part 1 *)

(** **** Exercise: 3 stars, recommended (list_exercises) *)
(** More practice with lists. *)

Theorem app_nil_end : forall l : natlist, 
  l ++ [] = l.   
Proof.
induction l.
 reflexivity.
 simpl. rewrite IHl. reflexivity.
Qed.

Example eg_rev_snoc : rev (snoc [1,2,3] 4) = (cons 4 (rev [1,2,3])).
reflexivity. Qed.

Lemma rev_snoc_is_cons_rev : forall n : nat, forall l : natlist,
 rev (snoc l n) = cons n (rev l).
Proof.
intro n.
induction l as [|m l'].
 reflexivity.
 simpl. rewrite IHl'. reflexivity.
Qed.

Theorem rev_involutive : forall l : natlist,
  rev (rev l) = l.
Proof.
induction l.
 reflexivity.
 simpl. (* rev n::l  =>  snoc (rev l) n *)
 set (l' := rev l). rewrite rev_snoc_is_cons_rev. rewrite<- IHl.
 reflexivity.
Qed.

Theorem snoc_append : forall (l:natlist) (n:nat),
  snoc l n = l ++ [n].
Proof.
intros l n. induction l as [|x xs].
 reflexivity.
 simpl. rewrite IHxs. reflexivity.
Qed. 

Theorem cons_app : forall x:nat, forall xs:natlist,
x :: xs = [x] ++ xs.
Proof.
intros x xs. reflexivity.
Qed.

Theorem cons_app_snoc : forall x z:nat, forall ys:natlist, cons x (snoc ys z) = [x] ++ ys ++ [z].
intros x z ys.
simpl. rewrite snoc_append. reflexivity.
Qed.

Theorem pivot__app_cons_is_ncons_app : forall y:nat, forall xs zs : natlist, 
xs ++ y :: zs = (snoc xs y) ++ zs.
intros y xs zs.
induction xs.
 reflexivity.
 simpl. rewrite IHxs. reflexivity.
Qed.

(** There is a short solution to the next exercise.  If you find
    yourself getting tangled up, step back and try to look for a
    simpler way. *)

Theorem app_ass4 : forall l1 l2 l3 l4 : natlist,
  l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4.
Proof.
 intros l1 l2 l3 l4.
 rewrite app_ass. set (x := l3 ++ l4).
 symmetry. apply app_ass. (* or rewrite app_ass. reflexivity. *)
Qed.

Lemma snoc_distributes_app : forall y:nat, forall xs ys : natlist,
snoc (xs ++ ys) y = xs ++ snoc ys y.
intros y xs ys.
induction ys as [|z zs].
 simpl. rewrite app_nil_end. rewrite snoc_append. reflexivity.
 simpl.
 rewrite snoc_append. rewrite cons_app_snoc. rewrite cons_app.
 rewrite app_ass. rewrite app_ass. reflexivity.
Qed.

Theorem distr_rev : forall xs ys : natlist,
  rev (xs ++ ys) = (rev ys) ++ (rev xs).
Proof.
intros xs ys.
induction xs as [|x xs].
 simpl. rewrite app_nil_end. reflexivity.
 simpl. rewrite IHxs. rewrite snoc_distributes_app. reflexivity.
Qed. 

(** An exercise about your implementation of [nonzeros]: *)

Lemma hoist_if : forall a b : Type, forall c:bool, forall f:a->b, forall x y:a,
(if c then f x else f y) = f (if c then x else y).
intros a b c f x y.
destruct c; reflexivity.
(* idiomatic coq would be to just inline when needed *)
Qed.

Lemma filter_distributes_append : forall f : nat -> bool, forall xs ys : natlist,
filter f (xs ++ ys) = filter f xs ++ filter f ys.
Proof.
intros f xs ys.
induction xs as [|x xs]. reflexivity.
simpl. rewrite IHxs.
(* idiomatic coq inline "destruct c; reflexivity." instead of the trivial "apply hoist_if"
set (c := f x).
destruct c; reflexivity.
Qed.
*)
rewrite cons_app. rewrite <- app_ass. 
set (g := fun list => list ++ filter f ys). 
 fold (g (if f x then x :: (filter f xs) else filter f xs)).
 fold (g ([x] ++ filter f xs)).
 fold (g (filter f xs)). simpl.
apply hoist_if.
Qed.

Lemma nonzeros_length : forall l1 l2 : natlist,
  nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).
Proof.
intros l1 l2.
unfold nonzeros. apply filter_distributes_append.
   (* lazy beta in (fix nonzeros (l : natlist) : natlist := filter (fun x : nat => 0 \= x) l).*)
Qed.
(** [] *)

(*
Notation "x 'and' y" := (andb x y) (at level 66, right associativity) : bool_scope.
Check (true and true).
*)

(* ###################################################### *)
(** ** List Exercises, Part 2 *)

(** **** Exercise: 2 stars, recommended (list_design) *)
(** Design exercise: 
     - Write down a non-trivial theorem involving [cons]
       ([::]), [snoc], and [append] ([++]).  
     - Prove it.
*)
Print cons_app.
Print snoc_append.
Print cons_app_snoc.
Print pivot__app_cons_is_ncons_app.
(** [] *)

(** **** Exercise: 2 stars, optional (bag_proofs) *)
(** If you did the optional exercise about bags above, here are a
    couple of little theorems to prove about your definitions. *)

Theorem count_member_nonzero : forall (s : bag),
  1 ≤ (count 1 (1 :: s)) = true.
Proof.
induction s; reflexivity.
Qed.

(** The following lemma about [ble_nat] might help you in the next proof. *)

Theorem ble_n_Sn : forall n,
  ble_nat n (S n) = true.
Proof.
  intros n. induction n as [| n'].
  Case "0".  
    simpl.  reflexivity.
  Case "S n'".
    simpl.  rewrite IHn'.  reflexivity.  Qed.

Theorem subset_implies_smaller: forall A B : bag,
true= subset A B  ->  true= length B ≤ length A.
intros A B.
intro S.
(* complements *)
Admitted.

Theorem remove_decreases_count: forall (s : bag),
  (count 0 (remove_one 0 s)) ≤ (count 0 s) = true.
Proof.
induction s as [|x s].
 reflexivity.
 destruct x.
  simpl. rewrite ble_n_Sn. reflexivity.
  unfold remove_one. simpl. fold remove_one. apply IHs.
Qed.
(** [] *)

(** **** Exercise: 3 stars, optional (bag_count_sum) *)  
(** Write down an interesting theorem about bags involving the
    functions [count] and [sum], and prove it.

(* FILL IN HERE *)
[]
 *)


(* ###################################################### *)
(** * Options *)

(** Here is another type definition that is often useful in
    day-to-day programming: *)

Inductive maybe : Type :=
  | Some : nat -> maybe
  | None : maybe.  

Definition natoption := maybe.

(** One use of [natoption] is as a way of returning "error
    codes" from functions.  For example, suppose we want to write a
    function that returns the [n]th element of some list.  If we give
    it type [nat -> natlist -> nat], then we'll have to return some
    number when the list is too short! *)

Fixpoint index_bad (n:nat) (l:natlist) : nat :=
  match l with
  | nil => 42  (* arbitrary! *)
  | a :: l' => match beq_nat n O with 
               | true => a 
               | false => index_bad (pred n) l' 
               end
  end.

(** On the other hand, if we give it type [nat -> natlist ->
    natoption], then we can return [None] when the list is too short
    and [Some a] when the list has enough members and [a] appears at
    position [n]. *)

Fixpoint index (n:nat) (l:natlist) : natoption :=
  match l with
  | nil => None 
  | a :: l' => match beq_nat n O with 
               | true => Some a
               | false => index (pred n) l' 
               end
  end.

Example test_index1 :    index 0 [4,5,6,7]  = Some 4.
Proof. reflexivity.  Qed.
Example test_index2 :    index 3 [4,5,6,7]  = Some 7.
Proof. reflexivity.  Qed.
Example test_index3 :    index 10 [4,5,6,7] = None.
Proof. reflexivity.  Qed.

(** This example is also an opportunity to introduce one more
    small feature of Coq's programming language: conditional
    expressions... *)

Fixpoint index' (n:nat) (l:natlist) : natoption :=
  match l with
  | nil => None 
  | a :: l' => if beq_nat n O then Some a else index (pred n) l'
  end.

(** Coq's conditionals are exactly like those found in any other
    language, with one small generalization.  Since the boolean type
    is not built in, Coq actually allows conditional expressions over
    _any_ inductively defined type with exactly two constructors.  The
    guard is considered true if it evaluates to the first constructor
    in the [Inductive] definition and false if it evaluates to the
    second. *)

(** The function below pulls the [nat] out of a [natoption], returning
    a supplied default in the [None] case. *)

Definition option_elim (o : natoption) (d : nat) : nat :=
  match o with
  | Some n' => n'
  | None => d
  end.

(** **** Exercise: 2 stars (hd_opt) *)
(** Using the same idea, fix the [hd] function from earlier so we don't
   have to pass a default element for the [nil] case.  *)

Definition hd_opt (l : natlist) : natoption :=
match l with
| nil => None
| x::_ => Some x
end.

Example test_hd_opt1 : hd_opt [] = None.
reflexivity. Qed.
 Example test_hd_opt2 : hd_opt [1] = Some 1.
reflexivity. Qed.

Example test_hd_opt3 : hd_opt [5,6] = Some 5.
reflexivity. Qed.

Example test_hd_opt4:             hd_opt [1,2,3] = Some 1.
reflexivity.  Qed.
Example test_hd_opt5:             hd_opt [] = None.
reflexivity.  Qed.

(** [] *)

(** **** Exercise: 2 stars, optional (option_elim_hd) *)
(** This exercise relates your new [hd_opt] to the old [hd]. *)

Theorem option_elim_hd : forall (l:natlist) (default:nat),
  hd default l = option_elim (hd_opt l) default.
Proof.
  intros l default.
  induction l as [|x xs]; reflexivity.
Qed.  

(** [] *)

(** **** Exercise: 2 stars, recommended (beq_natlist) *)
(** Fill in the definition of [beq_natlist], which compares
    lists of numbers for equality.  Prove that [beq_natlist l l]
    yields [true] for every list [l]. *)

Fixpoint beq_natlist (l1 l2 : natlist) : bool :=
match l1, l2 with
| nil, nil => true
| nil, _ => false
| _, nil => false                
| (x::xs),(y::ys) => if x==y then beq_natlist xs ys else false
end.

Example test_beq_natlist1 :   (beq_natlist nil nil = true).
reflexivity. Qed.
Example test_beq_natlist2 :   beq_natlist [1,2,3] [1,2,3] = true.
reflexivity. Qed.
Example test_beq_natlist3 :   beq_natlist [1,2,3] [1,2,4] = false.
reflexivity. Qed.

Theorem beq_natlist_refl : forall l:natlist,
  true = beq_natlist l l.
Proof.
induction l as [|x xs].
 reflexivity. simpl. rewrite<- beq_nat_refl. apply IHxs.
Qed.
 
 (** [] *)

(* ###################################################### *)
(** * The [apply] Tactic *)

(** We often encounter situations where the goal to be proved is
    exactly the same as some hypothesis in the context or some
    previously proved lemma. *)

Theorem silly1 : forall (n m o p : nat),
     n = m  ->
     [n,o] = [n,p] ->
     [n,o] = [m,p].
Proof.
  intros n m o p eq1 eq2.
  rewrite <- eq1.
  (* At this point, we could finish with "[rewrite -> eq2. reflexivity.]"
     as we have done several times above.  But we can achieve the same
     effect in a single step by using the [apply] tactic instead: *)
  apply eq2.  Qed.

(** The [apply] tactic also works with _conditional_ hypotheses
    and lemmas: if the statement being applied is an implication, then
    the premises of this implication will be added to the list of
    subgoals needing to be proved. *)

Theorem silly2 : forall (n m o p : nat),
     n = m  ->
     (forall (q r : nat), q = r -> [q,o] = [r,p]) ->
     [n,o] = [m,p].
Proof.
  intros n m o p eq1 eq2. 
  apply eq2. apply eq1.  Qed.

(** You may find it instructive to experiment with this proof
    and see if there is a way to complete it using just [rewrite]
    instead of [apply]. *)

(** Typically, when we use [apply H], the statement [H] will
    begin with a [forall] binding some _universal variables_.  When
    Coq matches the current goal against the conclusion of [H], it
    will try to find appropriate values for these variables.  For
    example, when we do [apply eq2] in the following proof, the
    universal variable [q] in [eq2] gets instantiated with [n] and [r]
    gets instantiated with [m]. *)

Theorem silly2a : forall (n m : nat),
     (n,n) = (m,m)  ->
     (forall (q r : nat), (q,q) = (r,r) -> [q] = [r]) ->
     [n] = [m].
Proof.
  intros n m eq1 eq2.
  apply eq2. apply eq1.  Qed.

(** **** Exercise: 2 stars, optional (silly_ex) *)
(** Complete the following proof without using [simpl]. *)

Theorem silly_ex : 
     (forall n, evenb n = true -> oddb (S n) = true) ->
     evenb 3 = true ->
     oddb 4 = true.
Proof.
intros A B.
apply A. apply B.
Qed.

(** To use the [apply] tactic, the (conclusion of the) fact
    being applied must match the goal _exactly_ -- for example, [apply]
    will not work if the left and right sides of the equality are
    swapped. *)

Theorem silly3_firsttry : forall (n : nat),
     true = beq_nat n 5  ->
     beq_nat (S (S n)) 7 = true.
Proof.
  intros n H.
  simpl.
  (* here we cannot use [apply] directly *)
Admitted.

(** In this case we can use the [symmetry] tactic, which
    switches the left and right sides of an equality in the goal. *)

Theorem silly3 : forall (n : nat),
     true = beq_nat n 5  ->
     beq_nat (S (S n)) 7 = true.
Proof.
  intros n H.
  symmetry.
  simpl. (* Actually, this [simpl] is unnecessary, since 
            [apply] will do a [simpl] step first. *)  
  apply H.  Qed.         


(** **** Exercise: 3 stars, recommended (apply_exercise1) *)
Theorem rev_exercise1 : forall (l l' : natlist),
     l = rev l' ->
     l' = rev l.
Proof.
intros l l'. intro A. rewrite A. 
symmetry. apply rev_involutive.
Qed.

(** [] *)


(** **** Exercise: 1 star (apply_rewrite) *)
(** Briefly explain the difference between the tactics [apply] and
    [rewrite].  Are there situations where both can usefully be
    applied?

  (* apply is like simpl then rewrite then reflexivity. when you must rewrite and do something apply doesnt do, and then you can apply. *)
*)
(** [] *)

(* ###################################################### *)
(** * Varying the Induction Hypothesis *)

(** One subtlety in these inductive proofs is worth noticing here.
    For example, look back at the proof of the [app_ass] theorem.  The
    induction hypothesis (in the second subgoal generated by the
    [induction] tactic) is

      [ (l1' ++ l2) ++ l3 = l1' ++ l2 ++ l3 ].

    (Note that, because we've defined [++] to be right associative,
    the expression on the right of the [=] is the same as writing [l1'
    ++ (l2 ++ l3)].)

    This hypothesis makes a statement about [l1'] together with the
    _particular_ lists [l2] and [l3].  The lists [l2] and [l3], which
    were introduced into the context by the [intros] at the top of the
    proof, are "held constant" in the induction hypothesis.  If we set
    up the proof slightly differently by introducing just [n] into the
    context at the top, then we get an induction hypothesis that makes
    a stronger claim:

     [ forall l2 l3,  (l1' ++ l2) ++ l3 = l1' ++ l2 ++ l3 ]

    Use Coq to see the difference for yourself.

    In the present case, the difference between the two proofs is
    minor, since the definition of the [++] function just examines its
    first argument and doesn't do anything interesting with its second
    argument.  But we'll soon come to situations where setting up the
    induction hypothesis one way or the other can make the difference
    between a proof working and failing. *)

(** **** Exercise: 2 stars, optional (app_ass') *)
(** Give an alternate proof of the associativity of [++] with a more
    general induction hypothesis.  Complete the following (leaving the
    first line unchanged). *)

Theorem app_ass' : forall l1 l2 l3 : natlist, 
  (l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).   
Proof.
intros l1. induction l1 as [ |x xs].
 reflexivity.
 simpl. intros l2 l3. (*?? is this ok? *)
 replace (x :: (xs ++ l2) ++ l3) with ([x] ++ ((xs ++ l2) ++ l3)).
 replace (x :: xs ++ l2 ++ l3) with ([x] ++ (xs ++ (l2 ++ l3))).
 rewrite IHxs. reflexivity.
 Case "replacement". apply cons_app.
 Case "replacement". apply cons_app.
Qed.

(** [] *)

(** **** Exercise: 3 stars (apply_exercise2) *)
(** Notice that we don't introduce m before performing induction.
    This leaves it general, so that the IH doesn't specify a
    particular m, but lets us pick. *)

Theorem beq_nat_sym : forall (n m : nat),
  beq_nat n m = beq_nat m n.
Proof.
intros n. induction n as [| n'].
 simpl. destruct m; reflexivity.
 destruct m.
  reflexivity.
  simpl. apply IHn'.
Qed. 

(** [] *)

(** **** Exercise: 3 stars, recommended (beq_nat_sym_informal) *)
(** Provide an informal proof of this lemma that corresponds
    to your formal proof above:

   Theorem: For any [nat]s [n] [m], [beq_nat n m = beq_nat m n].

   Proof:
   (* FILL IN HERE *)
[]
 *)

End NatList.

(* ###################################################### *)
(** * Exercise: Dictionaries *)

Module Dictionary.

Inductive dictionary : Type :=
  | empty  : dictionary 
  | record : nat -> nat -> dictionary -> dictionary. 

(** This declaration can be read: "There are two ways to construct a
    [dictionary]: either using the constructor [empty] to represent an
    empty dictionary, or by applying the constructor [record] to
    a key, a value, and an existing [dictionary] to construct a
    [dictionary] with an additional key to value mapping." *)

Definition insert (key value : nat) (d : dictionary) : dictionary :=
  (record key value d).

(** Below is a function [find] that searches a [dictionary] for a
    given key.  It evaluates evaluates to [None] if the key was not
    found and [Some val] if the key was mapped to [val] in the
    dictionary. If the same key is mapped to multiple values, [find]
    will return the first one it finds. *)

Fixpoint find (key : nat) (d : dictionary) : option nat := 
  match d with 
  | empty         => None
  | record k v d' => if (beq_nat key k) then (Some v) else (find key d')
  end.

(** **** Exercise: 1 star (dictionary_invariant1) *)
(* Complete the following proof. *)
Theorem dictionary_invariant1 : forall (d : dictionary) (k v: nat),
  (find k (insert k v d)) = Some v.
Proof.
destruct d; simpl; intro k; rewrite<- beq_nat_refl; reflexivity.
Qed.

(** [] *)

(** **** Exercise: 1 star (dictionary_invariant2) *)
(* Complete the following proof. *)
Theorem dictionary_invariant2 : forall (d : dictionary) (m n o: nat),
  (beq_nat m n) = false -> (find m d) = (find m (insert n o d)).
Proof.
induction d as [| k v d];
intros n m o; intro F; simpl; rewrite F; reflexivity. (* why cant i "apply" instead of "simpl. rewrite. reflexivity."? *)
Qed. 
(** [] *)

End Dictionary.

(** The following declaration puts [beq_nat_sym] into the
    top-level namespace, so that we can use it later without having to
    write [NatList.beq_nat_sym]. *)

Definition beq_nat_sym := NatList.beq_nat_sym.

(*
Inductive bool := true | false.
type inference on datatype(:Type) and constructors(:bool)
*)

(*
Open Scope list_scope.

Theorem test : forall p : NatList.natprod, 
p = pair (NatList.fst p) (NatList.snd p).
Proof.

Qed.
*)