(** * Lists: Working with Structured Data *)

Require Export Induction.
Require Export Coq.Init.Nat.
Module NatList.

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

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

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]." *)

Check (pair 3 5).

(** Here are simple functions for extracting the first and
    second components of a pair.  The definitions also illustrate how
    to do 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.

Compute (fst (pair 3 5)).
(* ===> 3 *)

(** 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).

(** The new pair notation can be used both in expressions and in
    pattern matches (indeed, we've actually seen this already in the
    [Basics] chapter, in the definition of the [minus] function --
    this works because the pair notation is also provided as part of
    the standard library): *)

Compute (fst (3,5)).

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 to prove a few simple facts about pairs.

    If we state things in a slightly peculiar way, we can complete
    proofs 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! *)
Abort.

(** 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]. *)

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

(** Notice that, unlike its behavior with [nat]s, [destruct]
    generates just one subgoal here.  That's because [natprod]s can
    only be constructed in one way. *)

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

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

(** Generalizing the definition of pairs, 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 mylist := cons 1 (cons 2 (cons 3 nil)).

(** As with pairs, it is more convenient to write lists in
    familiar programming notation.  The following 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 understand the details of 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 mylist1 := 1 :: (2 :: (3 :: nil)).
Definition mylist2 := 1 :: 2 :: 3 :: nil.
Definition mylist3 := [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])].

    (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, which are invisible in the HTML
    rendering, are there to instruct the "coqdoc" tool that the bracketed
    part should be displayed as Coq code rather than running text.)

    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. *)

(* ----------------------------------------------------------------- *)
(** *** Repeat *)

(** 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.

(* ----------------------------------------------------------------- *)
(** *** Length *)

(** 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.

(* ----------------------------------------------------------------- *)
(** *** Append *)

(** The [app] function concatenates (appends) 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.

(* ----------------------------------------------------------------- *)
(** *** Head (with default) and Tail *)

(** Here are two smaller examples of programming with lists.
    The [hd] function returns the first element (the "head") of the
    list, while [tl] returns everything but the first
    element (the "tail").
    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 tl (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_tl:              tl [1;2;3] = [2;3].
Proof. reflexivity.  Qed.

(* ----------------------------------------------------------------- *)
(** *** Exercises *)

(** **** Exercise: 2 stars, recommended (list_funs)  *)
(** Complete the definitions of [nonzeros], [oddmembers] and
    [countoddmembers] below. Have a look at the tests to understand
    what these functions should do. *)

Fixpoint nonzeros (l:natlist) : natlist :=
  match l with
  | nil => []
  | x :: xs => match x with
               | 0 => nonzeros xs
               | n => n :: (nonzeros xs)
                end
  end.

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

Definition isOdd (x : nat) : bool :=
  match x mod 2 with
    | 1 => true
    | _ => false
  end.

Fixpoint oddmembers (l:natlist) : natlist :=
  match l with
  | nil => []
  | x :: xs => match isOdd x with
                 | true => x :: oddmembers xs
                 | _ => oddmembers xs
               end
  end.

Example test_oddmembers:
  oddmembers [0;1;0;2;3;0;0] = [1;3].
Proof.
  simpl. reflexivity. Qed.

Fixpoint countoddmembers (l:natlist) : nat :=
  match l with
  | [] => 0
  | x :: xs => match isOdd x with
               | true => 1 + countoddmembers xs
               | false => countoddmembers xs
                end
end.

Example test_countoddmembers1:
  countoddmembers [1;0;3;1;4;5] = 4.
Proof.
  simpl. reflexivity. Qed.

Example test_countoddmembers2:
  countoddmembers [0;2;4] = 0.
Proof.
  reflexivity. Qed.

Example test_countoddmembers3:
  countoddmembers nil = 0.
Proof.
  reflexivity. Qed.

(** **** Exercise: 3 stars, advanced (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 and elegant 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.  (One possible solution requires
    defining a new kind of pairs, but this is not the only way.)  *)

Fixpoint alternate (l1 l2 : natlist) : natlist
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Example test_alternate1:
  alternate [1;2;3] [4;5;6] = [1;4;2;5;3;6].
  (* FILL IN HERE *) Admitted.

Example test_alternate2:
  alternate [1] [4;5;6] = [1;4;5;6].
  (* FILL IN HERE *) Admitted.

Example test_alternate3:
  alternate [1;2;3] [4] = [1;4;2;3].
  (* FILL IN HERE *) Admitted.

Example test_alternate4:
  alternate [] [20;30] = [20;30].
  (* FILL IN HERE *) Admitted.
(** [] *)

(* ----------------------------------------------------------------- *)
(** *** Bags via Lists *)

(** A [bag] (or [multiset]) is like a set, except that each element
    can appear multiple times rather than just once.  One possible
    implementation is to represent a bag of numbers as a list. *)

Definition bag := natlist.

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

Fixpoint count (v:nat) (s:bag) : nat :=
  match s with
  | [] => 0
  | x :: xs => if eqb v x then S (count v xs) else (count v xs)
  end.

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

Example test_count1:              count 1 [1;2;3;1;4;1] = 3.
Proof.
  simpl. reflexivity. Qed.
Example test_count2:              count 6 [1;2;3;1;4;1] = 0.
Proof.
  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 -- using max instead
    of sum -- 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.
Proof.
  reflexivity. Qed.

Definition add (v:nat) (s:bag) : bag :=
  v :: s.

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

Definition member (v:nat) (s:bag) : bool :=
  if ltb 0 (count v s) then true else false.

Example test_member1:             member 1 [1;4;1] = true.
Proof. reflexivity. Qed.

Example test_member2:             member 2 [1;4;1] = false.
Proof. reflexivity. Qed.

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

(** When [remove_one] is applied to a bag without the number to remove,
   it should return the same bag unchanged. *)

Fixpoint remove_one (v:nat) (s:bag) : bag
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Example test_remove_one1:
  count 5 (remove_one 5 [2;1;5;4;1]) = 0.
  (* FILL IN HERE *) Admitted.

Example test_remove_one2:
  count 5 (remove_one 5 [2;1;4;1]) = 0.
  (* FILL IN HERE *) Admitted.

Example test_remove_one3:
  count 4 (remove_one 5 [2;1;4;5;1;4]) = 2.
  (* FILL IN HERE *) Admitted.

Example test_remove_one4:
  count 5 (remove_one 5 [2;1;5;4;5;1;4]) = 1.
  (* FILL IN HERE *) Admitted.

Fixpoint remove_all (v:nat) (s:bag) : bag
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Example test_remove_all1:  count 5 (remove_all 5 [2;1;5;4;1]) = 0.
 (* FILL IN HERE *) Admitted.
Example test_remove_all2:  count 5 (remove_all 5 [2;1;4;1]) = 0.
 (* FILL IN HERE *) Admitted.
Example test_remove_all3:  count 4 (remove_all 5 [2;1;4;5;1;4]) = 2.
 (* FILL IN HERE *) Admitted.
Example test_remove_all4:  count 5 (remove_all 5 [2;1;5;4;5;1;4;5;1;4]) = 0.
 (* FILL IN HERE *) Admitted.

Fixpoint subset (s1:bag) (s2:bag) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Example test_subset1:              subset [1;2] [2;1;4;1] = true.
 (* FILL IN HERE *) Admitted.
Example test_subset2:              subset [1;2;2] [2;1;4;1] = false.
 (* FILL IN HERE *) Admitted.

(* Do not modify the following line: *)
(* Definition manual_grade_for_bag_theorem : option (prod nat string) := None. *)
(* (** [] *) *)

(** **** Exercise: 3 stars, recommended (bag_theorem)  *)
(** Write down an interesting theorem [bag_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! *)

(* Theorem bag_theorem : forall s:bag , count 2 (sum (add 5 (add 2 (add 1 (add 3 s)))) (add 2 s)) = 2 + count 2 s. *)
(* Proof. *)
(*   intros s. *)
(*   simpl. *)
(* Qed. *)

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

(** 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
    "scrutinee" (the expression whose value is being "scrutinized" by
    the match) 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 (tl l).
Proof.
  intros l. destruct l as [| n l'].
  - (* l = nil *)
    reflexivity.
  - (* 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 proof scripts will not get you very far!
    It is important to work through the details of each one, using Coq
    and thinking about what each step achieves.  Otherwise it is more
    or less guaranteed that the exercises will make no sense when you
    get to them.  'Nuff said. *)

(* ================================================================= *)
(** ** Induction on Lists *)

(** Proofs by induction over datatypes like [natlist] are a
    little less familiar than standard natural number induction, but
    the idea is equally simple.  Each [Inductive] declaration defines
    a set of data values that can be built up using the declared
    constructors: a boolean can be either [true] or [false]; a number
    can be either [O] or [S] applied to another 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'], assuming that [P]
        is true for [l'].

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

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

(** Notice that, as when doing induction on natural numbers, the
    [as...] clause provided to the [induction] tactic gives a name to
    the induction hypothesis corresponding to the smaller list [l1']
    in the [cons] case. Once 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. *)

(** For comparison, here is an informal proof of the same theorem. *)

(** _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.  [] *)

(* ----------------------------------------------------------------- *)
(** *** Reversing a List *)

(** For a slightly more involved example of inductive proof over
    lists, suppose we use [app] to define a list-reversing function
    [rev]: *)

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

(* ----------------------------------------------------------------- *)
(** *** Properties of [rev] *)

(** Now let's prove some theorems about our newly defined [rev].
    For something a bit more challenging than what we've seen, let's
    prove that reversing a list does not change its length.  Our first
    attempt 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' IHl'].
  - (* l = [] *)
    reflexivity.
  - (* l = n :: l' *)
    (* This is the tricky case.  Let's begin as usual
       by simplifying. *)
    simpl.
    (* Now we seem to be stuck: the goal is an equality
       involving [++], but we don't have any useful equations
       in either the immediate context or in the global
       environment!  We can make a little progress by using
       the IH to rewrite the goal... *)
    rewrite <- IHl'.
    (* ... but now we can't go any further. *)
Abort.

(** So let's take the equation relating [++] and [length] that
    would have enabled us to make progress and state it as a separate
    lemma. *)

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' IHl1'].
  - (* l1 = nil *)
    reflexivity.
  - (* l1 = cons *)
    simpl. rewrite -> IHl1'. reflexivity.  Qed.

(** Note that, to make the lemma as general as possible, we
    quantify over _all_ [natlist]s, not just those that result from an
    application of [rev].  This should seem natural, because the truth
    of the goal clearly doesn't depend on the list having been
    reversed.  Moreover, it is easier to prove the more general
    property. *)

(** 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' IHl'].
  - (* l = nil *)
    reflexivity.
  - (* l = cons *)
    simpl. rewrite -> app_length, plus_comm.
    simpl. rewrite -> IHl'. reflexivity.  Qed.

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

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

    _Proof_: By induction on [l1].

    - First, suppose [l1 = []].  We must show

        length ([] ++ l2) = length [] + length l2,

      which follows directly from the definitions of
      [length] and [++].

    - Next, suppose [l1 = n::l1'], with

        length (l1' ++ l2) = length l1' + length l2.

      We must show

        length ((n::l1') ++ l2) = length (n::l1') + length l2).

      This follows directly from the definitions of [length] and [++]
      together with 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 ((rev l') ++ [n]) = S (length l')

        which, by the previous lemma, is the same as

          length (rev l') + length [n] = S (length l').

        This follows directly from the induction hypothesis and the
        definition of [length]. [] *)

(** The style of these proofs is rather longwinded and pedantic.
    After the first few, we might find it easier to follow proofs that
    give fewer details (which can easily work 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
    like this: *)

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

    _Proof_: First, observe that [length (l ++ [n]) = S (length l)]
     for any [l] (this follows by a straightforward induction on [l]).
     The main property again follows by 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 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 our
    present purposes. *)

(* ================================================================= *)
(** ** [Search] *)

(** We've seen that proofs can make use of other theorems we've
    already proved, e.g., using [rewrite].  But in order to refer to a
    theorem, we need to know its name!  Indeed, it is often hard even
    to remember what theorems have been proven, much less what they
    are called.

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

 (* Search rev. *)

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

    If you are using ProofGeneral, you can run [Search] with [C-c
    C-a C-a]. Pasting its response into your buffer can be
    accomplished with [C-c C-;]. *)

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

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

Theorem app_nil_r : forall l : natlist,
  l ++ [] = l.
Proof.
  intros l.
  induction l as [|l' IHl'].
  - reflexivity.
  - simpl. rewrite IHIHl'. reflexivity. Qed.

Theorem rev_app_distr: forall l1 l2 : natlist,
  rev (l1 ++ l2) = rev l2 ++ rev l1.
Proof.
  intros l1 l2.
  induction l1 as [|l' IHl'].
  - simpl. rewrite app_nil_r. reflexivity.
  - simpl. rewrite IHIHl'. rewrite app_assoc. reflexivity. Qed.

Theorem rev_involutive : forall l : natlist,
  rev (rev l) = l.
Proof.
  intros l.
  induction l as [|l' IHl'].
  - reflexivity.
  - simpl. rewrite rev_app_distr. rewrite IHIHl'. simpl. reflexivity. Qed.

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

Theorem app_assoc4 : forall l1 l2 l3 l4 : natlist,
  l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4.
Proof.
  intros l1 l2 l3 l4.
  repeat rewrite app_assoc. reflexivity. Qed.

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

Lemma nonzeros_app : forall l1 l2 : natlist,
  nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).
Proof.
  intros l1 l2.
  induction l1 as [|l' IHl'].
  - simpl. reflexivity.
  - induction l' as [|l'' IHl''].
    + simpl. rewrite IHIHl'. reflexivity.
    + simpl. rewrite IHIHl'. reflexivity. Qed.

(** **** Exercise: 2 stars (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 with
  | [] => if (length l2) =? 0 then true else false
  | x :: xs => match l2 with
               | [] => false
               | y :: ys => if eqb x y then beq_natlist xs ys else false
               end
  end.

Example test_beq_natlist1 :
  (beq_natlist nil nil = true).
Proof. reflexivity. Qed.

Example test_beq_natlist2 :
  beq_natlist [1;2;3] [1;2;3] = true.
Proof. reflexivity. Qed.

Example test_beq_natlist3 :
  beq_natlist [1;2;3] [1;2;4] = false.
Proof. reflexivity. Qed.

Theorem beq_natlist_refl : forall l:natlist,
  true = beq_natlist l l.
Proof.
  intros l.
  induction l as [|l' IHl'].
  - reflexivity.
  - simpl. rewrite <- IHIHl'. induction l'.
    + reflexivity.
    + simpl. rewrite <- IHl'0. reflexivity. Qed.

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

(** Here are a couple of little theorems to prove about your
    definitions about bags above. *)

(* Theorem count_member_nonzero : forall (s : bag), *)
(*   leb 1 (count 1 (1 :: s)) = true. *)
(* Proof. *)

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

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

(* (** **** Exercise: 3 stars, advanced (remove_does_not_increase_count)  *) *)
(* Theorem remove_does_not_increase_count: forall (s : bag), *)
(*   leb (count 0 (remove_one 0 s)) (count 0 s) = true. *)
(* Proof. *)
(*   (* FILL IN HERE *) Admitted. *)
(* (** [] *) *)

(** **** Exercise: 3 stars, optional (bag_count_sum)  *)
(** Write down an interesting theorem [bag_count_sum] about bags
    involving the functions [count] and [sum], and prove it using
    Coq.  (You may find that the difficulty of the proof depends on
    how you defined [count]!) *)
(* FILL IN HERE *)
(** [] *)

(** **** Exercise: 4 stars, advanced (rev_injective)  *)
(** Prove that the [rev] function is injective -- that is,

    forall (l1 l2 : natlist), rev l1 = rev l2 -> l1 = l2.

    (There is a hard way and an easy way to do this.) *)

(* FILL IN HERE *)

(* Do not modify the following line: *)
(* Definition manual_grade_for_rev_injective : option (prod nat string) := None. *)
(** [] *)

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

(** 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 choose some number to return when the list is
    too short... *)

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

(** This solution is not so good: If [nth_bad] returns [42], we
    can't tell whether that value actually appears on the input
    without further processing. A better alternative is to change the
    return type of [nth_bad] to include an error value as a possible
    outcome. We call this type [natoption]. *)

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

(** We can then change the above definition of [nth_bad] to
    return [None] when the list is too short and [Some a] when the
    list has enough members and [a] appears at position [n]. We call
    this new function [nth_error] to indicate that it may result in an
    error. *)

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

Example test_nth_error1 : nth_error [4;5;6;7] 0 = Some 4.
Proof. reflexivity. Qed.
Example test_nth_error2 : nth_error [4;5;6;7] 3 = Some 7.
Proof. reflexivity. Qed.
Example test_nth_error3 : nth_error [4;5;6;7] 9 = None.
Proof. reflexivity. Qed.

(** (In the HTML version, the boilerplate proofs of these
    examples are elided.  Click on a box if you want to see one.)

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

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

(** **** Exercise: 2 stars (hd_error)  *)
(** 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_error (l : natlist) : natoption :=
  match l with
  | [] => None
  | x :: xs => Some x
                    end.

Example test_hd_error1 : hd_error [] = None.
Proof. reflexivity. Qed.

Example test_hd_error2 : hd_error [1] = Some 1.
Proof. reflexivity. Qed.

Example test_hd_error3 : hd_error [5;6] = Some 5.
Proof. reflexivity. Qed.

(** **** Exercise: 1 star, optional (option_elim_hd)  *)
(** This exercise relates your new [hd_error] to the old [hd]. *)

Theorem option_elim_hd : forall (l:natlist) (default:nat),
  hd default l = option_elim default (hd_error l).
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

End NatList.

(* ################################################################# *)
(** * Partial Maps *)

(** As a final illustration of how data structures can be defined in
    Coq, here is a simple _partial map_ data type, analogous to the
    map or dictionary data structures found in most programming
    languages. *)

(** First, we define a new inductive datatype [id] to serve as the
    "keys" of our partial maps. *)

Inductive id : Type :=
  | Id : nat -> id.

(** Internally, an [id] is just a number.  Introducing a separate type
    by wrapping each nat with the tag [Id] makes definitions more
    readable and gives us the flexibility to change representations
    later if we wish. *)

(** We'll also need an equality test for [id]s: *)

Definition beq_id (x1 x2 : id) :=
  match x1, x2 with
  | Id n1, Id n2 => beq_nat n1 n2
  end.

(** **** Exercise: 1 star (beq_id_refl)  *)
Theorem beq_id_refl : forall x, true = beq_id x x.
Proof.
  destruct x as [id].
  - simpl. rewrite <- beq_nat_refl. reflexivity. Qed.

(** Now we define the type of partial maps: *)

Module PartialMap.
Export NatList.

Inductive partial_map : Type :=
  | empty  : partial_map
  | record : id -> nat -> partial_map -> partial_map.

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

(** The [update] function overrides the entry for a given key in a
    partial map by shadowing it with a new one (or simply adds a new
    entry if the given key is not already present). *)

Definition update (d : partial_map)
                  (x : id) (value : nat)
                  : partial_map :=
  record x value d.

(** Last, the [find] function searches a [partial_map] for a given
    key.  It returns [None] if the key was not found and [Some val] if
    the key was associated with [val]. If the same key is mapped to
    multiple values, [find] will return the first one it
    encounters. *)

Fixpoint find (x : id) (d : partial_map) : natoption :=
  match d with
  | empty         => None
  | record y v d' => if beq_id x y
                     then Some v
                     else find x d'
  end.

(** **** Exercise: 1 star (update_eq)  *)
Theorem update_eq :
  forall (d : partial_map) (x : id) (v: nat),
    find x (update d x v) = Some v.
Proof.
  intros d x v.
  simpl. rewrite <- beq_id_refl. reflexivity. Qed.

(** **** Exercise: 1 star (update_neq)  *)
Theorem update_neq :
  forall (d : partial_map) (x y : id) (o: nat),
    beq_id x y = false -> find x (update d y o) = find x d.
Proof.
  intros d x y o H.
  simpl. rewrite H. reflexivity. Qed.

End PartialMap.

(** **** Exercise: 2 stars (baz_num_elts)  *)
(** Consider the following inductive definition: *)

Inductive baz : Type :=
  | Baz1 : baz -> baz
  | Baz2 : baz -> bool -> baz.

(** How _many_ elements does the type [baz] have?
    (Explain your answer in words, preferrably English.) *)

(* FILL IN HERE *)

(* Do not modify the following line: *)
(** [] *)


(**************************************
  Finish reading, Finish exercise
**************************************)

Require Import Nat Bool.

(* ================================================================= *)
(** * Proof by Induction *)

Theorem plus_n_O: forall n: nat, n = n + 0.
Proof.
  induction n as [| n' IHn'].
  - (* n = 0 *) reflexivity.
  - (* n = S n' *) simpl. rewrite <- IHn'. reflexivity.
Qed.

(** ****  Exercise: 2 stars, recommended (basic_induction)  *)
Theorem mult_0_r: forall n: nat, n * 0 = 0.
Proof.
  induction n as [|n' H].
  - reflexivity.
  - simpl.
    rewrite -> H.
    reflexivity.
Qed.

Theorem plus_n_Sm: forall n m: nat, S (n + m) = n + (S m).
Proof.
  intros n m.
  induction n as [|n' H].
  - reflexivity.
  - simpl.
    rewrite -> H.
    reflexivity.
Qed.

Theorem plus_comm: forall n m: nat, n + m = m + n.
Proof.
  intros n m.
  induction n as [|n' H].
  - simpl.
    rewrite <- plus_n_O.
    reflexivity.
  - simpl.
    rewrite -> H.
    rewrite -> plus_n_Sm.
    reflexivity.
Qed.

Theorem plus_assoc: forall n m p: nat, n + (m + p) = (n + m) + p.
Proof.
  intros m n p.
  induction n as [|n' H].
  - simpl.
    rewrite <- plus_n_O.
    reflexivity.
  - simpl.
    rewrite -> plus_comm.
    simpl.
    rewrite -> plus_comm.
    rewrite -> H.
    rewrite <- plus_n_Sm.
    simpl.
    reflexivity.
Qed.

(** **** Exercise: 2 stars (double_plus)  *)
Fixpoint double (n:nat) :=
  match n with
  | O => O
  | S n' => S (S (double n'))
  end.

Lemma double_plus: forall n, double n = n + n.
Proof.
  intros n.
  induction n as [|n' H].
  - reflexivity.
  - simpl.
    rewrite H.
    rewrite plus_n_Sm.
    reflexivity.
Qed.

(** **** Exercise: 2 stars, optional (evenb_S)  *)
Theorem evenb_S: forall n: nat, even (S n) = negb (even n).
Proof.
  intros n.
  induction n as [|n H].
  - reflexivity.
  - rewrite H.
    rewrite negb_involutive.
    reflexivity.
Qed.

(** **** Exercise: 1 star (destruct_induction)  *)
(*
  destruct: without Hypothesis.
  induction: with Hypothesis.
*)

(* ================================================================= *)
(** * Formal vs. Informal Proof *)

(** **** Exercise: 2 stars, advanced, recommended (plus_comm_informal)  *)
(*
  Proof: (* not do the exercise *)
*)

(** **** Exercise: 2 stars, optional (beq_nat_refl_informal)  *)
(*
  Proof: (* not do the exercise *)
*)


Add LoadPath "../Basics".
Require Export NamingCases.

(** We start with a basic example of induction : proving that
    [n + 0 = n] *)
Theorem plus_0_r : forall n : nat, n + 0 = n.

Proof.
   intros n. induction n as [| n'].
   Case "n = 0".
      simpl. reflexivity.
   Case "n = S n".
      simpl. rewrite -> IHn' . reflexivity. Qed.

(** Another induction example. This time, we prove
    that n-n = 0 *)
Theorem minus_diag : forall n,
   minus n n = 0.

Proof.
   intros n. induction n as [| n'].
   Case "n = 0".
      simpl. reflexivity.
   Case "n = S n".
      simpl. rewrite -> IHn'. reflexivity. Qed.

(** EXERCISE [**]: Basic induction proofs. Prove the
    following lemmas using induction *)

Theorem mult_0_r : forall n : nat,
   n * 0 = 0.

Proof.
   intros n. induction n as [| n'].
   Case "n = 0".
      simpl. reflexivity.
   Case "n = S n".
      simpl. rewrite -> IHn'. reflexivity. Qed.

Theorem plus_n_Sm : forall n m : nat,
   S (n + m) = n + (S m).

Proof.
   intros n m. induction n as [| n'].
   Case "n = 0".
      simpl. reflexivity.
   Case "n = S n".
      simpl. rewrite -> IHn'. reflexivity. Qed.

Theorem plus_comm : forall n m : nat,
   n + m = m + n.

(** Addition is commutative *)
Proof.
   intros n m. induction n as [| n'].
   Case "n = 0".
      simpl. rewrite -> plus_0_r. reflexivity.
   Case "n = S n".
      simpl. rewrite <- plus_n_Sm. rewrite <- IHn'. reflexivity. Qed.

(** Addition is associative *)
Theorem plus_assoc : forall n m p : nat,
   n + (m + p) = (n + m) + p.

Proof.
   intros n m p. induction n as [| n'].
   Case "n = 0".
     simpl. reflexivity.
   Case "n = S n".
     simpl. rewrite <- IHn'. reflexivity. Qed.

(** EXERCISE [**]: Use induction to prove the given fact
    [double_plus.] *)
Fixpoint double (n : nat) :=
   match n with
   | O => O
   | S n' => S (S (double n'))
   end.

Lemma double_plus : forall n, double n = n + n.

Proof.
   intros n. induction n as [| n'].
   Case "n = 0".
      simpl. reflexivity.
   Case "n = S n'".
      simpl. rewrite -> IHn'. rewrite -> plus_n_Sm. reflexivity. Qed.

(** EXERCISE [*]: Discuss the differences between the induction and
    destruct strategies *)

(** Using the [destruct] tactic results in Coq specifically going to
    the data type definition, and generating cases for each member
    of the data type. Induction is a more powerful reasoning principle:
    prove a base case P(b) (when working over the natural numbers, this is
    usually proving your proposition for [b = 0]). Then, prove that, for
    all n in N, P(n) => P(n+1). The common analogy is dominoes falling.
    You've already proved P(b), then using the proven inductive impl.,
    you get P(b+1), P(b+2), and so on and so forth. *)


Module Playground1.
Inductive nat : Type :=
  | O : nat
  | S : nat -> nat.

Definition pred (n : nat) : nat :=
  match n with
    | O => O
    | S n' => n'
  end.
End Playground1.

Definition minustwo (n : nat) : nat :=
  match n with
    | O => O
    | S O => O
    | S (S n') => n'
  end.
Check (S (S (S (S O)))).
Eval simpl in (minustwo 4).

Check S.
Check pred.
Check minustwo.

Fixpoint evenb (n:nat) : bool :=
  match n with
  | O => true
  | S O => false
  | S (S n') => evenb n'
  end.

Definition oddb (n:nat) : bool := negb (evenb n).

Example test_oddb1: (oddb (S O)) = true.
Proof. simpl. reflexivity. Qed.
Example test_oddb2: (oddb (S (S (S (S O))))) = false.
Proof. simpl. reflexivity. Qed.

Module Playground2.

Fixpoint plus (n : nat) (m : nat) : nat :=
  match n with
    | O => m
    | S n' => S (plus n' m)
  end.
Eval simpl in (plus (S (S (S O))) (S (S O))).

Fixpoint mult (n m : nat) : nat :=
  match n with
    | O => O
    | S n' => plus m (mult n' m)
  end.
Eval simpl in (mult 5 6).
Fixpoint minus (n m:nat) : nat :=
  match n, m with
  | O , _ => O
  | S _ , O => n
  | S n', S m' => minus n' m'
  end.

End Playground2.

Fixpoint exp (base power : nat) : nat :=
  match power with
    | O => S O
    | S p => mult base (exp base p)
  end.

Example test_mult1: (mult 3 3) = 9.
Proof. simpl. reflexivity. Qed.

Fixpoint factorial (n:nat) : nat :=
  match n with
    | 0 => 1
    | S p => mult n (factorial p)
  end.

Example test_factorial1: (factorial 3) = 6.
Proof. simpl. reflexivity. Qed.
Example test_factorial2: (factorial 5) = (mult 10 12).
Proof. simpl. reflexivity. Qed.

Notation "x + y" := (plus x y)
                       (at level 50, left associativity)
                       : nat_scope.
Notation "x - y" := (minus x y)
                       (at level 50, left associativity)
                       : nat_scope.
Notation "x * y" := (mult x y)
                       (at level 40, left associativity)
                       : nat_scope.

Check ((0 + 1) + 1).

Fixpoint beq_nat (n m : nat) : bool :=
  match n with
  | O => match m with
         | O => true
         | S m' => false
         end
  | S n' => match m with
            | O => false
            | S m' => beq_nat n' m'
            end
  end.

Fixpoint ble_nat (n m : nat) : bool :=
  match n with
  | O => true
  | S n' =>
      match m with
      | O => false
      | S m' => ble_nat n' m'
      end
  end.

Example test_ble_nat1: (ble_nat 2 2) = true.
Proof. simpl. reflexivity. Qed.
Example test_ble_nat2: (ble_nat 2 4) = true.
Proof. simpl. reflexivity. Qed.
Example test_ble_nat3: (ble_nat 4 2) = false.
Proof. simpl. reflexivity. Qed.

Definition blt_nat (n m : nat) : bool :=
  andb (ble_nat n m) (notb (beq_nat n m)).

Example test_blt_nat1: (blt_nat 2 2) = false.
Proof. simpl. reflexivity. Qed.
Example test_blt_nat2: (blt_nat 2 4) = true.
Proof. simpl. reflexivity. Qed.
Example test_blt_nat3: (blt_nat 4 2) = false.
Proof. simpl. reflexivity. Qed.


Require Import Arith.
Require Import Bool.

(**************)
(* Exercise 1 *)
(**************)

(* We first want to prove that our definition of add satisfies commutativity. *)

Fixpoint add n m :=
  match n with 0 => m | S p => add p (S m) end.

Lemma addnS : forall n m, add n (S m) = S (add n m).
  induction n.
  - intros m; simpl.
    auto.
  - intros m; simpl.
    apply IHn.
Qed.

(* Q1. Prove the following two theorems: beware that you will probably need
  addnS. *)

(* Already done during the lecture ? *)
Lemma addn0 : forall n, add n 0 = n.
Admitted.

Lemma add_comm : forall n m, add n m = add m n.
Admitted.

(* Q2. Now state and prove the associativity of this function. *)

(* Q3. Now state and prove a theorem that expresses that the add function
  returns the same result as the addition available in the loaded libraries
  (given by function plus) *)

(*********************)
(* Exercise 2: lists *)
(*********************)
Require Import List  Bool_nat.
Require Import Coq.omega.Omega.

(* From lecture 2 *)
Class Eq (A : Type) := cmp : A -> A -> bool.

Infix "==" := cmp (at level 70, no associativity).

Instance bool_Eq : Eq nat := beq_nat.
(* beq_nat comes from the Coq library. *)

Fixpoint multiplicity (n : nat)(l : list nat) : nat  :=
  match l with
    nil => 0%nat
  | a :: l' =>
    if n == a then S (multiplicity n l')
    else multiplicity n l'
  end.

Definition is_perm (l l'  : list nat)  :=
  forall n, multiplicity n l = multiplicity n l'.

(* Q4. Show the following theorem :  *)

Lemma multiplicity_app  (x : nat) (l1 l2 : list nat) :
  multiplicity x (l1 ++ l2) = multiplicity x l1 + multiplicity x l2.
Admitted.

(* Note : for Q5 and Q6, you will probably have an opportunity
  to use the omega tactic *)

(* Q5. State and prove a theorem that expresses that element counts are
  preserved by reverse. *)

(* Q6. Show the following theorem. *)

Lemma is_perm_transpose x y l1 l2 l3 :
  is_perm (l1 ++ x::l2 ++ y::l3) (l1 ++ y :: l2 ++ x :: l3).
Admitted.

(* Q7 :  Complete the following lemma using only a reasonning
  on the function rev_app defined in Lecture3.v *)
(* Excerpt from Lectuer3.v - Begin *)
Fixpoint rev_app (A : Type)(l1 l2 : list A) : list A :=
  match l1 with
  | nil => l2
  | a :: tl => rev_app A tl (a :: l2)
  end.

Lemma rev_app_nil : forall A (l1 : list A),
rev_app A l1 nil = rev l1.
Proof.
intros A l1.
assert (Htmp: forall l2, rev_app A l1 l2 = rev l1 ++ l2).
+ induction l1; intros l2.
  * simpl. auto.
  * simpl.
    rewrite app_assoc_reverse.
    simpl.
    apply IHl1.
+ rewrite Htmp.
  rewrite app_nil_r.
  auto.
Qed.
(* Excerpt from Lecture3.v - End *)

Lemma rev_rev_id : forall A (l:list A), rev (rev l) = l.
Proof.
  intros.
  rewrite <- rev_app_nil.
  rewrite <- rev_app_nil.
Admitted.

(* Q8 : What does this function do? *)
Fixpoint mask (A : Type)(m : list bool)(s : list A) {struct m} :=
  match m, s with
  | b :: m', x :: s' => if b then x :: mask A m' s' else mask A m' s'
  | _, _ => nil
  end.

Arguments mask [A] _ _.

(* Q9 Prove that : *)
Lemma mask_cat : forall A m1 (s1 : list A),
    length m1 = length s1 ->
  forall m2 s2, mask (m1 ++ m2) (s1 ++ s2) = mask m1 s1 ++ mask m2 s2.
Admitted.

(**************)
(* Exercise 3 *)
(**************)

(* Define an inductive type formula : Type that represents the *)
(*abstract language of propositional logic without variables:
L = L /\ L | L \/ L | ~L | L_true | L_false
*)

(* Define a function formula_holds of type (formula -> Prop computing the *)
(* Coq formula corresponding to an element of type formula *)

(* Define  a function isT_formula of type (formula -> bool) computing *)
(* the intended truth value of (f : formula) *)

(* prove that is (idT_formula f) evaluates to true, then its *)
(*corresponding Coq formula holds ie.:

Require Import Bool.
Lemma isT_formula_correct : forall f : formula,
   isT_formula f = true <-> formula_holds f.
*)

(**************)
(* Exercise 4 *)
(**************)

(* We use the inductive type defined in the lecture: *)

Inductive natBinTree : Set :=
| Leaf : natBinTree
| Node : nat -> natBinTree -> natBinTree -> natBinTree.

(* Define a function which sums all the values present in the tree.

Define a function is_zero_present : natBinTree -> bool, which tests whether
the value 0 is present or not in the tree.

Prove several simple statements about the fonctions tree_size
and tree_height seen in the lecture

Define a function called mirror that computes the mirror image of a tree.

Prove that a tree and its mirror image have the same height.

Prove that mirror is involutive (ie the mirror image of the mirror image
of the tree is this tree).

It is possible to navigate in a binary tree, given a tree t and
a path like "from the root, go to the left subtree, then
 go to the right subtree, then go to the left subtree, etc. "

Such a path can be easily represented by a list of directions. *)

Inductive direction : Set := L (* go left *) | R (* go right *).

(* Define in Coq a function get_label that takes a tree and some path,
and returns the label at which one arrives following the path
(if any) *)

Fixpoint get_label (path : list direction)(t:natBinTree): option nat:= None.
(* TO DO *)

(* Consider the following function :
*)

Fixpoint zero_present (t: natBinTree) : bool :=
match t with Leaf => false
           | Node n t1 t2 =>  beq_nat n 0 ||
                              zero_present t1 ||
                              zero_present t2
end.
(*
Prove that whenever zero_present t = true, then there exists
some path p such that get_label p t = Some 0

*)

(**************)
(* Exercise 5 *)
(**************)

(* Define the function
split : forall A B : Set, list A * B -> (list A) * (list B)

which transforms a list of pairs into a pair of lists
and the function
combine : forall A B : Set, list A * list B -> list (A * B)
which transforms a pair of lists into a list of pairs.

Write and prove two theorems relating the two functions.
*)


(** * Lists: Working with Structured Data *)

Require Export Induction.

Module NatList.

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

(** In an [Inductive] type definition, each constructor can take
    any number of arguments -- 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]." *)

(** We can construct an element of [natprod] like this: *)

Check (pair 3 5).

(** *** *)

(** Here are two simple function definitions for extracting the
    first and second components of a pair.  (The definitions also
    illustrate how to do 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.

Eval compute in (fst (pair 3 5)).
(* ===> 3 *)

(** *** *)

(** 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).

(** 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 compute in (fst (3,5)).

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.

(** Note that [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! *)
Abort.

(** *** *)
(** 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.

(** **** Exercise: 1 star (snd_fst_is_swap) *)
Theorem snd_fst_is_swap : forall (p : natprod),
  (snd p, fst p) = swap_pair p.
Proof.
  intros. destruct p as [n m]. simpl. 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.
  intros. destruct p as [n m]. simpl. 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 mylist := 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 mylist1 := 1 :: (2 :: (3 :: nil)).
Definition mylist2 := 1 :: 2 :: 3 :: nil.
Definition mylist3 := [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, which are invisible in the HTML rendering, are there to
   instruct the "coqdoc" tool that the bracketed part should be
   displayed as Coq code rather than running text.)

   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. *)

(** *** Repeat *)
(** 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.

(** *** Length *)
(** 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.

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

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

Definition tl (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_tl:              tl [1;2;3] = [2;3].
Proof. reflexivity.  Qed.

(** **** Exercise: 2 stars (list_funs) *)
(** Complete the definitions of [nonzeros], [oddmembers] and
    [countoddmembers] below. Have a look at the tests to understand
    what these functions should do. *)

Fixpoint nonzeros (l:natlist) : natlist :=
  match l with
  | nil => nil
  | h :: t => match h with
              | 0 => (nonzeros t)
              | _ => [h] ++ (nonzeros t)
              end
end.

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

Fixpoint oddmembers (l:natlist) : natlist :=
  match l with
  | nil => nil
  | h :: t => match (oddb h) with
                | true => [h] ++ (oddmembers t)
                | false => oddmembers t
              end
end.

Example test_oddmembers:            oddmembers [0;1;0;2;3;0;0] = [1;3].
  Proof. simpl. reflexivity.
Qed.

Fixpoint countoddmembers (l:natlist) : nat :=
  match (oddmembers l) with
  | nil => 0
  | _ => (length (oddmembers l))
end.

Example test_countoddmembers1:    countoddmembers [1;0;3;1;4;5] = 4.
  Proof. simpl. reflexivity.
Qed.
Example test_countoddmembers2:    countoddmembers [0;2;4] = 0.
  Proof. simpl. reflexivity. Qed.

Example test_countoddmembers3:    countoddmembers nil = 0.
 Proof. simpl. reflexivity. Qed.
(** [] *)

(** **** Exercise: 3 stars, advanced (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 and elegant 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.  (One possible solution requires
    defining a new kind of pairs, but this is not the only way.)  *)

Fixpoint alternate (l1 l2 : natlist) : natlist :=
  (* FILL IN HERE *) admit.

Example test_alternate1:        alternate [1;2;3] [4;5;6] = [1;4;2;5;3;6].
 (* FILL IN HERE *) Admitted.
Example test_alternate2:        alternate [1] [4;5;6] = [1;4;5;6].
 (* FILL IN HERE *) Admitted.
Example test_alternate3:        alternate [1;2;3] [4] = [1;4;2;3].
 (* FILL IN HERE *) Admitted.
Example test_alternate4:        alternate [] [20;30] = [20;30].
 (* FILL IN HERE *) Admitted.
(** [] *)

(* ###################################################### *)
(** ** 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. *)

Fixpoint count (v:nat) (s:bag) : nat :=
  match s with
  | nil => 0
  | h :: t => match (beq_nat h v) with
              | true  => 1 + (count v t)
              | false => 0 + (count v t)
              end
end.

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

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

Definition add (v:nat) (s:bag) : bag := sum [v] s.

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

Definition member (v:nat) (s:bag) : bool :=  ble_nat 1 (count v s).

Eval compute in (member 2 [1;4;1]).

Example test_member1:             member 1 [1;4;1] = true.
Proof. simpl. reflexivity. Qed.
Example test_member2:             member 2 [1;4;1] = false.
Proof. simpl. 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 :=
  match s with
  | nil => nil
  | h :: t => match (beq_nat h v) with
              | true => t
              | false => [h] ++ (remove_one v t)
              end
end.

Eval compute in (remove_one 5 [2;1;5;4;1]).

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

Fixpoint remove_all (v:nat) (s:bag) : bag :=
  match s with
  | nil => nil
  | h :: t => match (beq_nat h v) with
              | true => remove_all v t
              | false => [h] ++ (remove_all v t)
end
end.

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

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

Eval compute in (subset [1;2;2] [2;1;4;1]).

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

(** **** Exercise: 3 stars (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! *)

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

(* ###################################################### *)
(** * 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 (tl 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'], assuming 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_assoc : 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.  *)

(** *** Informal version *)

(** _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.  []
*)

(** *** Another example *)
(**
  Here is a similar example 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.

(** *** Reversing a list *)
(** 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.

(** *** Proofs about reverse *)
(** 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'".
    (* This is the tricky case.  Let's begin as usual
       by simplifying. *)
    simpl.
    (* Now we seem to be 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]!

       We can make a little progress by using the IH to
       rewrite the goal... *)
    rewrite <- IHl'.
    (* ... but now we can't go any further. *)
Abort.

(** 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.

(**
    Note that we make the lemma as _general_ as possible: in particular,
    we quantify over _all_ [natlist]s, not just those that result
    from an application of [rev]. This should seem natural,
    because the truth of the goal clearly doesn't depend on
    the list having been reversed.  Moreover, it is much easier
    to prove the more general property.
*)

(** 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 fewer details (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! *)

(** Also, if you are using ProofGeneral, you can run [SearchAbout]
    with [C-c C-a C-a]. Pasting its response into your buffer can be
    accomplished with [C-c C-;]. *)

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

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

Theorem app_nil_end : forall l : natlist,
  l ++ [] = l.
Proof.
  intros. induction l as [| n l'].
  Case "[]".
  simpl. reflexivity.
  Case "l1 = cons".
  simpl. rewrite IHl'. reflexivity.
Qed.

Theorem rev_snoc : forall l : natlist, forall n : nat,
  rev (snoc l n) = n :: (rev l).
Proof.
  intros. induction l.
  Case "nil".
  simpl. reflexivity.
  Case "l = cons".
  simpl. rewrite IHl. simpl. reflexivity.
Qed.

Theorem rev_involutive : forall l : natlist,
  rev (rev l) = l.
Proof.
  intros. induction l.
  Case "nil".
  simpl. reflexivity.
  Case "cons".
  simpl. rewrite rev_snoc. rewrite IHl. 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_assoc4 : forall l1 l2 l3 l4 : natlist,
  l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4.
Proof.
  intros. rewrite app_assoc. rewrite app_assoc. reflexivity.
Qed.

Theorem snoc_append : forall (l:natlist) (n:nat),
  snoc l n = l ++ [n].
Proof.
  intros. induction l.
  Case "[]".
  simpl. reflexivity.
  Case "cons".
  simpl. rewrite IHl. reflexivity.
Qed.

Theorem distr_rev : forall l1 l2 : natlist,
  rev (l1 ++ l2) = (rev l2) ++ (rev l1).
Proof.
  intros. induction l1.
  Case "[]".
  simpl. rewrite app_nil_end. reflexivity.
  Case "l1 = cons".
  simpl.
  rewrite snoc_append.
  rewrite IHl1.
  rewrite snoc_append.
  rewrite app_assoc.
  reflexivity.
Qed.

Lemma nonzeros_app : forall l1 l2 : natlist,
  nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).
Proof.
  intros. induction l1.
  Case "[]".
  simpl. reflexivity.
  Case "cons".
    destruct n as [| n'].
    simpl. apply IHl1.
    simpl. rewrite IHl1. reflexivity.
Qed.
(** [] *)

(** **** Exercise: 2 stars (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 with
  | nil => match l2 with
           | nil => true
           | _   => false
           end
  | h :: t => match l2 with
              | nil => false
              | x :: xs => match (beq_nat h x) with
                           | true => beq_natlist t xs
                           | false => false
end
end
end.

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

Theorem beq_nat_refl : forall n:nat,
  true = beq_nat n n.
Proof.
  intros. induction n.
  simpl.  reflexivity.
  simpl. rewrite IHn. reflexivity.
Qed.

Theorem beq_natlist_refl : forall l:natlist,
   true = beq_natlist l l.
Proof.
   intros. induction l.
   simpl. reflexivity.
   simpl. rewrite <- beq_nat_refl. rewrite IHl. reflexivity.
Qed.
(** [] *)

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

(** **** Exercise: 2 stars (list_design) *)
(** Design exercise:
     - Write down a non-trivial theorem involving [cons]
       ([::]), [snoc], and [app] ([++]).
     - Prove it. *)

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

(** **** Exercise: 3 stars, advanced (bag_proofs) *)
(** Here are a couple of little theorems to prove about your
    definitions about bags earlier in the file. *)

Theorem count_member_nonzero : forall (s : bag),
  ble_nat 1 (count 1 (1 :: s)) = true.
Proof.
  (* FILL IN HERE *) Admitted.

(** 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 remove_decreases_count: forall (s : bag),
  ble_nat (count 0 (remove_one 0 s)) (count 0 s) = true.
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** 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 *)
(** [] *)

(** **** Exercise: 4 stars, advanced (rev_injective) *)
(** Prove that the [rev] function is injective, that is,

    forall (l1 l2 : natlist), rev l1 = rev l2 -> l1 = l2.

There is a hard way and an easy way to solve this exercise.
*)

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

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

(** 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]. *)

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

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 (d : nat) (o : natoption) : 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 :=
  (* FILL IN HERE *) admit.

Example test_hd_opt1 : hd_opt [] = None.
 (* FILL IN HERE *) Admitted.

Example test_hd_opt2 : hd_opt [1] = Some 1.
 (* FILL IN HERE *) Admitted.

Example test_hd_opt3 : hd_opt [5;6] = Some 5.
 (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 1 star, 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 default (hd_opt l).
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(* ###################################################### *)
(** * Dictionaries *)

(** As a final illustration of how fundamental data structures
    can be defined in Coq, here is the declaration of a simple
    [dictionary] data type, using numbers for both the keys and the
    values stored under these keys.  (That is, a dictionary represents
    a finite map from numbers to numbers.) *)

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).

(** Here 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) : natoption :=
  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.
  intros. induction k.
  simpl. reflexivity.
  simpl. 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.
  intros. induction d.
  simpl. rewrite H. reflexivity.
  simpl. rewrite H. reflexivity.
Qed.

End Dictionary.

End NatList.

(* $Date: 2014-01-28 13:19:45 -0500 (Tue, 28 Jan 2014) $ *)


Require Import Bool Arith List.

(* 6. Write the script that proves the following formula
   forall (P Q : nat -> Prop),
   forall (x y : nat), (forall z, P z -> Q z) -> x = y -> P x ->
     P x /\ Q y. *)

Lemma ex6 :
  forall (P Q : nat -> Prop),
   forall (x y : nat), (forall z, P z -> Q z) -> x = y -> P x -> P x /\ Q y.
Proof.
intros P Q.
intros x y.
intros z.

intros x_eq_y.
intros px.

split.

exact px.

rewrite  <- x_eq_y.

apply z.
exact px.

Qed.

(* 8. Write the script that proves the following formula
   forall P : nat -> Prop, (forall x, P x) ->
     exists y:nat, P y /\ y = 0 *)

Lemma ex8 :
  forall P : nat -> Prop, (forall x, P x) -> exists y:nat, P y /\ y = 0.
Proof.
intros P.
intros Px.

exists 0.

split.

apply Px.
reflexivity.
Qed.

(* 1. Write a predicate multiple_of type nat -> nat -> Prop,
   so that multiple a b expresses that a is a multiple of b
  (in other words, there exists a number k such that a = k * b). *)

Definition multiple_of (a b : nat) : Prop :=
 exists k,  a = k * b.

(* 2. Write a formula using natural numbers that expresses that when
   n is even (a multiple of 2) then n * n is even. *)

Definition even (n : nat) :=
  multiple_of n 2.

Check forall n, even n -> even (n * n).

(* 3. Write a formula using natural numbers that expresses that when
   a number n is a multiple of some k, then n * n is a multiple of k
   (you don't have to prove it yet). *)

Definition mul_sq (n k : nat) :=
 forall (n k : nat), (multiple_of n k) -> ( multiple_of (n * n) k).

Check forall n k, mul_sq n k.

(* 4. Write a predicate odd of type nat -> Prop that characterize
   odd numbers like 3, 5, 37.  Avoid ~ (even ..). *)

Definition odd (n : nat) : Prop :=
  even (n + 1).

(* 9. Search the lemma proving associativity of multiplication *)

SearchAbout ( _ * _ * _ = _ *(_ * _) ).

(* 10. Write the script that proves that when n is a multiple of k,
   then n * n is also a multiple of k. *)

Lemma ex10 :
  forall (n k : nat), (multiple_of n k) -> ( multiple_of (n * n) k).
Proof.
unfold multiple_of.
intros n k.
intros k0.

(* 11. Search the lemmas proving the following properties:
   - distributivity of plus and mult
   - associativity of plus
   - 1 is a neutral elelemt for multiplication *)

SearchAbout ( 1 * _ = _ ).

(* 12. Show that the sum of two even numbers is even *)

Lemma ex12 : ...

(* 13. Write the script that proves that when n is odd,
   then n * n is also odd. *)

Lemma ex13 : ...


Inductive bool : Type :=
  | true
  | false.

Definition negb (b:bool) : bool :=
  match b with
  | false => true
  | true => false
  end.

Definition nandb (b1:bool) (b2:bool) : bool :=
  match b1 with
  | false => true
  | true => negb b2
  end.

Example test_nandb1:(nandb true false) = true.
Proof. simpl. reflexivity. Qed.

Example test_nandb2:(nandb false false) = true.
Proof. simpl. reflexivity. Qed.

Definition andb2 (b1:bool)(b2:bool) : bool :=
  match b1 with
  | false => false
  | true => b2
  end.

Notation "x && y" := (andb2 x y).

Definition andb3 (b1:bool)(b2:bool)(b3:bool) : bool :=
  match b1 with
  | false => false
  | true => andb2 b2 b3
  end.

Example test_andb31:(andb3 true true true) = true.
Proof. simpl. reflexivity. Qed.

Example test_andb32:(andb3 false true false) = false.
Proof. simpl. reflexivity. Qed.

Module NatPlayground.

Inductive nat : Type :=
  | O
  | S (n:nat).
End NatPlayground.

Fixpoint evenb (n:nat) : bool :=
  match n with
  | O => true
  | S O => false
  | S (S n') => evenb n'
  end.

Definition oddb (n:nat) : bool := negb (evenb n).

Module NatPlayground2.

Fixpoint plus (n m :nat) : nat :=
  match n with
  | O => m
  | S n' => S (plus n' m)
  end.

Fixpoint mult (n m : nat) : nat :=
  match n with
    | O => O
    | S n' => plus m (mult n' m)
  end.
Fixpoint minus (n m:nat) : nat :=
  match n, m with
  | O , _ => O
  | S _ , O => n
  | S n', S m' => minus n' m'
  end.
End NatPlayground2.
Compute (mult 3 2).

Fixpoint exp (base power : nat) : nat :=
  match power with
    | O => S O
    | S p => mult base (exp base p)
  end.

Notation "x + y" := (plus x y)
                       (at level 50, left associativity)
                       : nat_scope.
Notation "x - y" := (minus x y)
                       (at level 50, left associativity)
                       : nat_scope.
Notation "x * y" := (mult x y)
                       (at level 40, left associativity)
                       : nat_scope.

(*n的阶乘*)
Fixpoint factorial (n:nat) : nat :=
  match n with
  | O => S O
  | S n' => n * (factorial n')
  end.

Example test_factorial1 : (factorial 3) = 6.
Proof. simpl. reflexivity. Qed.

Fixpoint eqb (n m : nat) : bool :=
  match n with
  | O => match m with
         | O => true
         | S m' => false
         end
  | S n' => match m with
            | O => false
            | S m' => eqb n' m'
            end
  end.

Fixpoint leb (n m : nat) : bool :=
  match n with
  | O => true
  | S n' =>
      match m with
      | O => false
      | S m' => leb n' m'
      end
  end.

Fixpoint ltb (n m : nat) : bool :=
  match n with
  | O => match m with
          | S m' => true
          | O => false
          end
  | S n'=> match m with
          | O => false
          | S m' => ltb n' m'
          end
  end.

Notation "x =? y" := (eqb x y) (at level 70) : nat_scope.
Notation "x <=? y" := (leb x y) (at level 70) : nat_scope.
Notation "x <? y" := (ltb x y) (at level 70) : nat_scope.

Example test_ltb1: (ltb 2 2) = false.
Proof. simpl. reflexivity. Qed.

Example test_ltb3: (ltb 4 2) = false.
Proof. simpl. reflexivity. Qed.

Theorem plus_O_n : forall n : nat, 0 + n = n.
Proof.
  intros n. simpl. reflexivity. Qed.

Theorem plus_1_l : forall n:nat, 1 + n = S n.
Proof.
  intros n. reflexivity. Qed.

Theorem mult_0_l : forall n:nat, 0 * n = 0.
Proof.
  intros n. reflexivity. Qed.

Theorem plus_id_example : forall n m :nat,
 n = m -> n + n = m + m.
Proof. intros n m. intros H. rewrite <-H. reflexivity. Qed.

Theorem plus_id_exercise : forall n m o : nat,
  n = m -> m = o -> n + m = m + o.
Proof.
  intros n m o.
  intros H1.
  intros H2.
  rewrite H1, H2.
  reflexivity. Qed.

Theorem mult_0_plus : forall n m : nat,
  (0 + n) * m = n * m.
Proof.
  intros n m.
  rewrite -> plus_O_n.
  reflexivity. Qed.

Theorem mult_S_1 : forall n m : nat,
  m = S n -> m * (1+n) = m*m.
Proof.
  intros n m.
  intros H1.
  rewrite H1.
  reflexivity.
  Qed.

Theorem andb_commutative : forall b c, andb b c = andb c b.
Proof.
  intros b c. destruct b eqn:Eb.
  - destruct c eqn:Ec.
    + reflexivity.
    + reflexivity.
  - destruct c eqn:Ec.
    + reflexivity.
    + reflexivity.
Qed.

Theorem andb3_exchange :
  forall b c d, andb (andb b c) d = andb (andb b d) c.
Proof.
  intros b c d. destruct b eqn:Eb.
  - destruct c eqn:Ec.
    { destruct d eqn:Ed.
      - reflexivity.
      - reflexivity. }
    { destruct d eqn:Ed.
      - reflexivity.
      - reflexivity. }
  - destruct c eqn:Ec.
    { destruct d eqn:Ed.
      - reflexivity.
      - reflexivity. }
    { destruct d eqn:Ed.
      - reflexivity.
      - reflexivity. }
Qed.

Theorem andb_true_elim2 : forall b c : bool,
  andb2 b c = true -> c = true.
Proof.
  intros. destruct b eqn : Eb.
  - apply H.
  - destruct c eqn : Ec.
    + reflexivity.
    + rewrite<- H. simpl. reflexivity.
Qed.

Theorem zero_nbeq_plus_1 : forall n : nat,
  0 =? (n + 1) = false.
Proof.
  intros [|n'].
  - reflexivity.
  - reflexivity.
Qed.