(** [] *)

(** Coq generates induction principles for every datatype defined with
    [Inductive], including those that aren't recursive.  Although of
    course we don't need induction to prove properties of
    non-recursive datatypes, the idea of an induction principle still
    makes sense for them: it gives a way to prove that a property
    holds for all values of the type.

    These generated principles follow a similar pattern. If we define
    a type [t] with constructors [c1] ... [cn], Coq generates a
    theorem with this shape:

    t_ind : forall P : t -> Prop,
              ... case for c1 ... ->
              ... case for c2 ... -> ...
              ... case for cn ... ->
              forall n : t, P n

    The specific shape of each case depends on the arguments to the
    corresponding constructor.  Before trying to write down a general
    rule, let's look at some more examples. First, an example where
    the constructors take no arguments: *)

Inductive yesno : Type :=
  | yes : yesno
  | no : yesno.(** [] *)

(** Coq generates induction principles for every datatype defined with
    [Inductive], including those that aren't recursive.  Although of
    course we don't need induction to prove properties of
    non-recursive datatypes, the idea of an induction principle still
    makes sense for them: it gives a way to prove that a property
    holds for all values of the type.

    These generated principles follow a similar pattern. If we define
    a type [t] with constructors [c1] ... [cn], Coq generates a
    theorem with this shape:

    t_ind : forall P : t -> Prop,
              ... case for c1 ... ->
              ... case for c2 ... -> ...
              ... case for cn ... ->
              forall n : t, P n

    The specific shape of each case depends on the arguments to the
    corresponding constructor.  Before trying to write down a general
    rule, let's look at some more examples. First, an example where
    the constructors take no arguments: *)

Inductive yesno : Type :=
  | yes : yesno
  | no : yesno.(** [] *)

(** Coq generates induction principles for every datatype defined with
    [Inductive], including those that aren't recursive.  Although of
    course we don't need induction to prove properties of
    non-recursive datatypes, the idea of an induction principle still
    makes sense for them: it gives a way to prove that a property
    holds for all values of the type.

    These generated principles follow a similar pattern. If we define
    a type [t] with constructors [c1] ... [cn], Coq generates a
    theorem with this shape:

    t_ind : forall P : t -> Prop,
              ... case for c1 ... ->
              ... case for c2 ... -> ...
              ... case for cn ... ->
              forall n : t, P n

    The specific shape of each case depends on the arguments to the
    corresponding constructor.  Before trying to write down a general
    rule, let's look at some more examples. First, an example where
    the constructors take no arguments: *)

Inductive yesno : Type :=
  | yes : yesno
  | no : yesno.(* ===> yesno_ind : forall P : yesno -> Prop,
                      P yes  ->
                      P no  ->
                      forall y : yesno, P y *)

(** **** Exercise: 1 star, optional (rgb)  *)
(** Write out the induction principle that Coq will generate for the
    following datatype.  Write down your answer on paper or type it
    into a comment, and then compare it with what Coq prints. *)

Inductive rgb : Type :=
  | red : rgb
  | green : rgb
  | blue : rgb.(* ===> yesno_ind : forall P : yesno -> Prop,
                      P yes  ->
                      P no  ->
                      forall y : yesno, P y *)

(** **** Exercise: 1 star, optional (rgb)  *)
(** Write out the induction principle that Coq will generate for the
    following datatype.  Write down your answer on paper or type it
    into a comment, and then compare it with what Coq prints. *)

Inductive rgb : Type :=
  | red : rgb
  | green : rgb
  | blue : rgb.(* ===> yesno_ind : forall P : yesno -> Prop,
                      P yes  ->
                      P no  ->
                      forall y : yesno, P y *)

(** **** Exercise: 1 star, optional (rgb)  *)
(** Write out the induction principle that Coq will generate for the
    following datatype.  Write down your answer on paper or type it
    into a comment, and then compare it with what Coq prints. *)

Inductive rgb : Type :=
  | red : rgb
  | green : rgb
  | blue : rgb.(* ===> yesno_ind : forall P : yesno -> Prop,
                      P yes  ->
                      P no  ->
                      forall y : yesno, P y *)

(** **** Exercise: 1 star, optional (rgb)  *)
(** Write out the induction principle that Coq will generate for the
    following datatype.  Write down your answer on paper or type it
    into a comment, and then compare it with what Coq prints. *)

Inductive rgb : Type :=
  | red : rgb
  | green : rgb
  | blue : rgb.(** [] *)

(** Here's another example, this time with one of the constructors
    taking some arguments. *)

Inductive natlist : Type :=
  | nnil : natlist
  | ncons : nat -> natlist -> natlist.(** [] *)

(** Here's another example, this time with one of the constructors
    taking some arguments. *)

Inductive natlist : Type :=
  | nnil : natlist
  | ncons : nat -> natlist -> natlist.(** [] *)

(** Here's another example, this time with one of the constructors
    taking some arguments. *)

Inductive natlist : Type :=
  | nnil : natlist
  | ncons : nat -> natlist -> natlist.(* ===> (modulo a little variable renaming)
   natlist_ind :
      forall P : natlist -> Prop,
         P nnil  ->
         (forall (n : nat) (l : natlist),
            P l -> P (ncons n l)) ->
         forall n : natlist, P n *)

(** **** Exercise: 1 star, optional (natlist1)  *)
(** Suppose we had written the above definition a little
   differently: *)

Inductive natlist1 : Type :=
  | nnil1 : natlist1
  | nsnoc1 : natlist1 -> nat -> natlist1.(* ===> (modulo a little variable renaming)
   natlist_ind :
      forall P : natlist -> Prop,
         P nnil  ->
         (forall (n : nat) (l : natlist),
            P l -> P (ncons n l)) ->
         forall n : natlist, P n *)

(** **** Exercise: 1 star, optional (natlist1)  *)
(** Suppose we had written the above definition a little
   differently: *)

Inductive natlist1 : Type :=
  | nnil1 : natlist1
  | nsnoc1 : natlist1 -> nat -> natlist1.(* ===> (modulo a little variable renaming)
   natlist_ind :
      forall P : natlist -> Prop,
         P nnil  ->
         (forall (n : nat) (l : natlist),
            P l -> P (ncons n l)) ->
         forall n : natlist, P n *)

(** **** Exercise: 1 star, optional (natlist1)  *)
(** Suppose we had written the above definition a little
   differently: *)

Inductive natlist1 : Type :=
  | nnil1 : natlist1
  | nsnoc1 : natlist1 -> nat -> natlist1.(** Now what will the induction principle look like? *)
(** [] *)

(** From these examples, we can extract this general rule:

    - The type declaration gives several constructors; each
      corresponds to one clause of the induction principle.
    - Each constructor [c] takes argument types [a1] ... [an].
    - Each [ai] can be either [t] (the datatype we are defining) or
      some other type [s].
    - The corresponding case of the induction principle says:

        - "For all values [x1]...[xn] of types [a1]...[an], if [P]
          holds for each of the inductive arguments (each [xi] of type
          [t]), then [P] holds for [c x1 ... xn]".
*)

(** **** Exercise: 1 star, optional (byntree_ind)  *)
(** Write out the induction principle that Coq will generate for the
    following datatype.  (Again, write down your answer on paper or
    type it into a comment, and then compare it with what Coq
    prints.) *)

Inductive byntree : Type :=
 | bempty : byntree
 | bleaf  : yesno -> byntree
 | nbranch : yesno -> byntree -> byntree -> byntree.(** Now what will the induction principle look like? *)
(** [] *)

(** From these examples, we can extract this general rule:

    - The type declaration gives several constructors; each
      corresponds to one clause of the induction principle.
    - Each constructor [c] takes argument types [a1] ... [an].
    - Each [ai] can be either [t] (the datatype we are defining) or
      some other type [s].
    - The corresponding case of the induction principle says:

        - "For all values [x1]...[xn] of types [a1]...[an], if [P]
          holds for each of the inductive arguments (each [xi] of type
          [t]), then [P] holds for [c x1 ... xn]".
*)

(** **** Exercise: 1 star, optional (byntree_ind)  *)
(** Write out the induction principle that Coq will generate for the
    following datatype.  (Again, write down your answer on paper or
    type it into a comment, and then compare it with what Coq
    prints.) *)

Inductive byntree : Type :=
 | bempty : byntree
 | bleaf  : yesno -> byntree
 | nbranch : yesno -> byntree -> byntree -> byntree.(** Now what will the induction principle look like? *)
(** [] *)

(** From these examples, we can extract this general rule:

    - The type declaration gives several constructors; each
      corresponds to one clause of the induction principle.
    - Each constructor [c] takes argument types [a1] ... [an].
    - Each [ai] can be either [t] (the datatype we are defining) or
      some other type [s].
    - The corresponding case of the induction principle says:

        - "For all values [x1]...[xn] of types [a1]...[an], if [P]
          holds for each of the inductive arguments (each [xi] of type
          [t]), then [P] holds for [c x1 ... xn]".
*)

(** **** Exercise: 1 star, optional (byntree_ind)  *)
(** Write out the induction principle that Coq will generate for the
    following datatype.  (Again, write down your answer on paper or
    type it into a comment, and then compare it with what Coq
    prints.) *)

Inductive byntree : Type :=
 | bempty : byntree
 | bleaf  : yesno -> byntree
 | nbranch : yesno -> byntree -> byntree -> byntree.(** Now what will the induction principle look like? *)
(** [] *)

(** From these examples, we can extract this general rule:

    - The type declaration gives several constructors; each
      corresponds to one clause of the induction principle.
    - Each constructor [c] takes argument types [a1] ... [an].
    - Each [ai] can be either [t] (the datatype we are defining) or
      some other type [s].
    - The corresponding case of the induction principle says:

        - "For all values [x1]...[xn] of types [a1]...[an], if [P]
          holds for each of the inductive arguments (each [xi] of type
          [t]), then [P] holds for [c x1 ... xn]".
*)

(** **** Exercise: 1 star, optional (byntree_ind)  *)
(** Write out the induction principle that Coq will generate for the
    following datatype.  (Again, write down your answer on paper or
    type it into a comment, and then compare it with what Coq
    prints.) *)

Inductive byntree : Type :=
 | bempty : byntree
 | bleaf  : yesno -> byntree
 | nbranch : yesno -> byntree -> byntree -> byntree.(** [] *)

(** **** Exercise: 1 star, optional (ex_set)  *)
(** Here is an induction principle for an inductively defined
    set.

      ExSet_ind :
         forall P : ExSet -> Prop,
             (forall b : bool, P (con1 b)) ->
             (forall (n : nat) (e : ExSet), P e -> P (con2 n e)) ->
             forall e : ExSet, P e

    Give an [Inductive] definition of [ExSet]: *)

Inductive ExSet : Type :=
  (* FILL IN HERE *)
.(** [] *)

(* ################################################################# *)
(** * Polymorphism *)

(** Next, what about polymorphic datatypes?

    The inductive definition of polymorphic lists

      Inductive list (X:Type) : Type :=
        | nil : list X
        | cons : X -> list X -> list X.

    is very similar to that of [natlist].  The main difference is
    that, here, the whole definition is _parameterized_ on a set [X]:
    that is, we are defining a _family_ of inductive types [list X],
    one for each [X].  (Note that, wherever [list] appears in the body
    of the declaration, it is always applied to the parameter [X].)
    The induction principle is likewise parameterized on [X]:

      list_ind :
        forall (X : Type) (P : list X -> Prop),
           P [] ->
           (forall (x : X) (l : list X), P l -> P (x :: l)) ->
           forall l : list X, P l

    Note that the _whole_ induction principle is parameterized on
    [X].  That is, [list_ind] can be thought of as a polymorphic
    function that, when applied to a type [X], gives us back an
    induction principle specialized to the type [list X]. *)

(** **** Exercise: 1 star, optional (tree)  *)
(** Write out the induction principle that Coq will generate for
   the following datatype.  Compare your answer with what Coq
   prints. *)

Inductive tree (X:Type) : Type :=
  | leaf : X -> tree X
  | node : tree X -> tree X -> tree X.(** [] *)

(* ################################################################# *)
(** * Polymorphism *)

(** Next, what about polymorphic datatypes?

    The inductive definition of polymorphic lists

      Inductive list (X:Type) : Type :=
        | nil : list X
        | cons : X -> list X -> list X.

    is very similar to that of [natlist].  The main difference is
    that, here, the whole definition is _parameterized_ on a set [X]:
    that is, we are defining a _family_ of inductive types [list X],
    one for each [X].  (Note that, wherever [list] appears in the body
    of the declaration, it is always applied to the parameter [X].)
    The induction principle is likewise parameterized on [X]:

      list_ind :
        forall (X : Type) (P : list X -> Prop),
           P [] ->
           (forall (x : X) (l : list X), P l -> P (x :: l)) ->
           forall l : list X, P l

    Note that the _whole_ induction principle is parameterized on
    [X].  That is, [list_ind] can be thought of as a polymorphic
    function that, when applied to a type [X], gives us back an
    induction principle specialized to the type [list X]. *)

(** **** Exercise: 1 star, optional (tree)  *)
(** Write out the induction principle that Coq will generate for
   the following datatype.  Compare your answer with what Coq
   prints. *)

Inductive tree (X:Type) : Type :=
  | leaf : X -> tree X
  | node : tree X -> tree X -> tree X.(** [] *)

(* ################################################################# *)
(** * Polymorphism *)

(** Next, what about polymorphic datatypes?

    The inductive definition of polymorphic lists

      Inductive list (X:Type) : Type :=
        | nil : list X
        | cons : X -> list X -> list X.

    is very similar to that of [natlist].  The main difference is
    that, here, the whole definition is _parameterized_ on a set [X]:
    that is, we are defining a _family_ of inductive types [list X],
    one for each [X].  (Note that, wherever [list] appears in the body
    of the declaration, it is always applied to the parameter [X].)
    The induction principle is likewise parameterized on [X]:

      list_ind :
        forall (X : Type) (P : list X -> Prop),
           P [] ->
           (forall (x : X) (l : list X), P l -> P (x :: l)) ->
           forall l : list X, P l

    Note that the _whole_ induction principle is parameterized on
    [X].  That is, [list_ind] can be thought of as a polymorphic
    function that, when applied to a type [X], gives us back an
    induction principle specialized to the type [list X]. *)

(** **** Exercise: 1 star, optional (tree)  *)
(** Write out the induction principle that Coq will generate for
   the following datatype.  Compare your answer with what Coq
   prints. *)

Inductive tree (X:Type) : Type :=
  | leaf : X -> tree X
  | node : tree X -> tree X -> tree X.(** [] *)

(** **** Exercise: 1 star, optional (mytype)  *)
(** Find an inductive definition that gives rise to the
    following induction principle:

      mytype_ind :
        forall (X : Type) (P : mytype X -> Prop),
            (forall x : X, P (constr1 X x)) ->
            (forall n : nat, P (constr2 X n)) ->
            (forall m : mytype X, P m ->
               forall n : nat, P (constr3 X m n)) ->
            forall m : mytype X, P m
*) 
(** [] *)

(** **** Exercise: 1 star, optional (foo)  *)
(** Find an inductive definition that gives rise to the
    following induction principle:

      foo_ind :
        forall (X Y : Type) (P : foo X Y -> Prop),
             (forall x : X, P (bar X Y x)) ->
             (forall y : Y, P (baz X Y y)) ->
             (forall f1 : nat -> foo X Y,
               (forall n : nat, P (f1 n)) -> P (quux X Y f1)) ->
             forall f2 : foo X Y, P f2
*) 
(** [] *)

(** **** Exercise: 1 star, optional (foo')  *)
(** Consider the following inductive definition: *)

Inductive foo' (X:Type) : Type :=
  | C1 : list X -> foo' X -> foo' X
  | C2 : foo' X.(** [] *)

(** **** Exercise: 1 star, optional (mytype)  *)
(** Find an inductive definition that gives rise to the
    following induction principle:

      mytype_ind :
        forall (X : Type) (P : mytype X -> Prop),
            (forall x : X, P (constr1 X x)) ->
            (forall n : nat, P (constr2 X n)) ->
            (forall m : mytype X, P m ->
               forall n : nat, P (constr3 X m n)) ->
            forall m : mytype X, P m
*) 
(** [] *)

(** **** Exercise: 1 star, optional (foo)  *)
(** Find an inductive definition that gives rise to the
    following induction principle:

      foo_ind :
        forall (X Y : Type) (P : foo X Y -> Prop),
             (forall x : X, P (bar X Y x)) ->
             (forall y : Y, P (baz X Y y)) ->
             (forall f1 : nat -> foo X Y,
               (forall n : nat, P (f1 n)) -> P (quux X Y f1)) ->
             forall f2 : foo X Y, P f2
*) 
(** [] *)

(** **** Exercise: 1 star, optional (foo')  *)
(** Consider the following inductive definition: *)

Inductive foo' (X:Type) : Type :=
  | C1 : list X -> foo' X -> foo' X
  | C2 : foo' X.(** [] *)

(** **** Exercise: 1 star, optional (mytype)  *)
(** Find an inductive definition that gives rise to the
    following induction principle:

      mytype_ind :
        forall (X : Type) (P : mytype X -> Prop),
            (forall x : X, P (constr1 X x)) ->
            (forall n : nat, P (constr2 X n)) ->
            (forall m : mytype X, P m ->
               forall n : nat, P (constr3 X m n)) ->
            forall m : mytype X, P m
*) 
(** [] *)

(** **** Exercise: 1 star, optional (foo)  *)
(** Find an inductive definition that gives rise to the
    following induction principle:

      foo_ind :
        forall (X Y : Type) (P : foo X Y -> Prop),
             (forall x : X, P (bar X Y x)) ->
             (forall y : Y, P (baz X Y y)) ->
             (forall f1 : nat -> foo X Y,
               (forall n : nat, P (f1 n)) -> P (quux X Y f1)) ->
             forall f2 : foo X Y, P f2
*) 
(** [] *)

(** **** Exercise: 1 star, optional (foo')  *)
(** Consider the following inductive definition: *)

Inductive foo' (X:Type) : Type :=
  | C1 : list X -> foo' X -> foo' X
  | C2 : foo' X.(** What induction principle will Coq generate for [foo']?  Fill
   in the blanks, then check your answer with Coq.)

     foo'_ind :
        forall (X : Type) (P : foo' X -> Prop),
              (forall (l : list X) (f : foo' X),
                    _______________________ ->
                    _______________________   ) ->
             ___________________________________________ ->
             forall f : foo' X, ________________________
*)

(** [] *)

(* ################################################################# *)
(** * Induction Hypotheses *)

(** Where does the phrase "induction hypothesis" fit into this story?

    The induction principle for numbers

       forall P : nat -> Prop,
            P 0  ->
            (forall n : nat, P n -> P (S n))  ->
            forall n : nat, P n

   is a generic statement that holds for all propositions
   [P] (or rather, strictly speaking, for all families of
   propositions [P] indexed by a number [n]).  Each time we
   use this principle, we are choosing [P] to be a particular
   expression of type [nat->Prop].

   We can make proofs by induction more explicit by giving
   this expression a name.  For example, instead of stating
   the theorem [mult_0_r] as "[forall n, n * 0 = 0]," we can
   write it as "[forall n, P_m0r n]", where [P_m0r] is defined
   as... *)

Definition P_m0r (n:nat) : Prop :=
  n * 0 = 0.(** ... or equivalently: *)

Definition P_m0r' : nat->Prop :=
  fun n => n * 0 = 0.

Theorem plus_assoc' : forall n m p : nat, n + (m + p) = (n + m) + p.