(* I have not worked with any partners. *)
(** * Basics: Functional Programming in Coq *)

(* REMINDER:

          #####################################################
          ###  PLEASE DO NOT DISTRIBUTE SOLUTIONS PUBLICLY  ###
          #####################################################

   (See the [Preface] for why.)

*)

(* ################################################################# *)
(** * Introduction *)

(** The functional programming style is founded on simple, everyday
    mathematical intuition: If a procedure or method has no side
    effects, then (ignoring efficiency) all we need to understand
    about it is how it maps inputs to outputs -- that is, we can think
    of it as just a concrete method for computing a mathematical
    function.  This is one sense of the word "functional" in
    "functional programming."  The direct connection between programs
    and simple mathematical objects supports both formal correctness
    proofs and sound informal reasoning about program behavior.

    The other sense in which functional programming is "functional" is
    that it emphasizes the use of functions (or methods) as
    _first-class_ values -- i.e., values that can be passed as
    arguments to other functions, returned as results, included in
    data structures, etc.  The recognition that functions can be
    treated as data gives rise to a host of useful and powerful
    programming idioms.

    Other common features of functional languages include _algebraic
    data types_ and _pattern matching_, which make it easy to
    construct and manipulate rich data structures, and sophisticated
    _polymorphic type systems_ supporting abstraction and code reuse.
    Coq offers all of these features.

    The first half of this chapter introduces the most essential
    elements of Coq's functional programming language, called
    _Gallina_.  The second half introduces some basic _tactics_ that
    can be used to prove properties of Coq programs. *)

(* ################################################################# *)
(** * Enumerated Types *)

(** One notable aspect of Coq is that its set of built-in
    features is _extremely_ small.  For example, instead of providing
    the usual palette of atomic data types (booleans, integers,
    strings, etc.), Coq offers a powerful mechanism for defining new
    data types from scratch, with all these familiar types as
    instances.

    Naturally, the Coq distribution comes preloaded with an extensive
    standard library providing definitions of booleans, numbers, and
    many common data structures like lists and hash tables.  But there
    is nothing magic or primitive about these library definitions.  To
    illustrate this, we will explicitly recapitulate all the
    definitions we need in this course, rather than just getting them
    implicitly from the library. *)

(* ================================================================= *)
(** ** Days of the Week *)

(** To see how this definition mechanism works, let's start with
    a very simple example.  The following declaration tells Coq that
    we are defining a new set of data values -- a _type_. *)

Inductive day : Type :=
  | monday : day
  | tuesday : day
  | wednesday : day
  | thursday : day
  | friday : day
  | saturday : day
  | sunday : day.

(** The type is called [day], and its members are [monday],
    [tuesday], etc.  The second and following lines of the definition
    can be read "[monday] is a [day], [tuesday] is a [day], etc."

    Having defined [day], we can write functions that operate on
    days. *)

Definition next_weekday (d:day) : day :=
  match d with
  | monday    => tuesday
  | tuesday   => wednesday
  | wednesday => thursday
  | thursday  => friday
  | friday    => monday
  | saturday  => monday
  | sunday    => monday
  end.

(** One thing to note is that the argument and return types of
    this function are explicitly declared.  Like most functional
    programming languages, Coq can often figure out these types for
    itself when they are not given explicitly -- i.e., it can do _type
    inference_ -- but we'll generally include them to make reading
    easier. *)

(** Having defined a function, we should check that it works on
    some examples.  There are actually three different ways to do this
    in Coq.  First, we can use the command [Compute] to evaluate a
    compound expression involving [next_weekday]. *)

Compute (next_weekday friday).
(* ==> monday : day *)

Compute (next_weekday (next_weekday saturday)).
(* ==> tuesday : day *)

(** (We show Coq's responses in comments, but, if you have a
    computer handy, this would be an excellent moment to fire up the
    Coq interpreter under your favorite IDE -- either CoqIde or Proof
    General -- and try this for yourself.  Load this file, [Basics.v],
    from the book's Coq sources, find the above example, submit it to
    Coq, and observe the result.)

    Second, we can record what we _expect_ the result to be in the
    form of a Coq example: *)

Example test_next_weekday:
  (next_weekday (next_weekday saturday)) = tuesday.

(** This declaration does two things: it makes an
    assertion (that the second weekday after [saturday] is [tuesday]),
    and it gives the assertion a name that can be used to refer to it
    later.  Having made the assertion, we can also ask Coq to verify
    it, like this: *)

Proof. simpl. reflexivity.  Qed.

(** The details are not important for now (we'll come back to
    them in a bit), but essentially this can be read as "The assertion
    we've just made can be proved by observing that both sides of the
    equality evaluate to the same thing, after some simplification."

    Third, we can ask Coq to _extract_, from our [Definition], a
    program in some other, more conventional, programming
    language (OCaml, Scheme, or Haskell) with a high-performance
    compiler.  This facility is very interesting, since it gives us a
    way to go from proved-correct algorithms written in Gallina to
    efficient machine code.  (Of course, we are trusting the
    correctness of the OCaml/Haskell/Scheme compiler, and of Coq's
    extraction facility itself, but this is still a big step forward
    from the way most software is developed today.) Indeed, this is
    one of the main uses for which Coq was developed.  We'll come back
    to this topic in later chapters. *)

(* ================================================================= *)
(** ** Homework Submission Guidelines *)

(** If you are using Software Foundations in a course, your instructor
    may use automatic scripts to help grade your homework assignments.
    In order for these scripts to work correctly (so that you get full
    credit for your work!), please be careful to follow these rules:
      - The grading scripts work by extracting marked regions of the
        .v files that you submit.  It is therefore important that you
        do not alter the "markup" that delimits exercises: the
        Exercise header, the name of the exercise, the "empty square
        bracket" marker at the end, etc.  Please leave this markup
        exactly as you find it.
      - Do not delete exercises.  If you skip an exercise (e.g.,
        because it is marked Optional, or because you can't solve it),
        it is OK to leave a partial proof in your .v file, but in this
        case please make sure it ends with [Admitted] (not, for
        example [Abort]). *)

(* ================================================================= *)
(** ** Booleans *)

(** In a similar way, we can define the standard type [bool] of
    booleans, with members [true] and [false]. *)

Inductive bool : Type :=
  | true : bool
  | false : bool.

(** Although we are rolling our own booleans here for the sake
    of building up everything from scratch, Coq does, of course,
    provide a default implementation of the booleans, together with a
    multitude of useful functions and lemmas.  (Take a look at
    [Coq.Init.Datatypes] in the Coq library documentation if you're
    interested.)  Whenever possible, we'll name our own definitions
    and theorems so that they exactly coincide with the ones in the
    standard library.

    Functions over booleans can be defined in the same way as
    above: *)

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

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

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

(** The last two of these illustrate Coq's syntax for
    multi-argument function definitions.  The corresponding
    multi-argument application syntax is illustrated by the following
    "unit tests," which constitute a complete specification -- a truth
    table -- for the [orb] function: *)

Example test_orb1:  (orb true  false) = true.
Proof. simpl. reflexivity.  Qed.
Example test_orb2:  (orb false false) = false.
Proof. simpl. reflexivity.  Qed.
Example test_orb3:  (orb false true)  = true.
Proof. simpl. reflexivity.  Qed.
Example test_orb4:  (orb true  true)  = true.
Proof. simpl. reflexivity.  Qed.

(** We can also introduce some familiar syntax for the boolean
    operations we have just defined. The [Infix] command defines a new
    symbolic notation for an existing definition. *)

Infix "&&" := andb.
Infix "||" := orb.

Example test_orb5:  false || false || true = true.
Proof. simpl. reflexivity. Qed.

(** _A note on notation_: In [.v] files, we use square brackets
    to delimit fragments of Coq code within comments; this convention,
    also used by the [coqdoc] documentation tool, keeps them visually
    separate from the surrounding text.  In the html version of the
    files, these pieces of text appear in a [different font].

    The command [Admitted] can be used as a placeholder for an
    incomplete proof.  We'll use it in exercises, to indicate the
    parts that we're leaving for you -- i.e., your job is to replace
    [Admitted]s with real proofs. *)

(** **** Exercise: 1 star (nandb)  *)
(** Remove "[Admitted.]" and complete the definition of the following
    function; then make sure that the [Example] assertions below can
    each be verified by Coq.  (Remove "[Admitted.]" and fill in each
    proof, following the model of the [orb] tests above.) The function
    should return [true] if either or both of its inputs are
    [false]. *)

Definition nandb (b1:bool) (b2:bool) : bool :=
  (negb (andb b1 b2)).

Example test_nandb1:               (nandb true false) = true.
Proof. simpl. reflexivity. Qed.
Example test_nandb2:               (nandb false false) = true.
Proof. simpl. reflexivity. Qed.
Example test_nandb3:               (nandb false true) = true.
Proof. simpl. reflexivity. Qed.
Example test_nandb4:               (nandb true true) = false.
Proof. simpl. reflexivity. Qed.
(** [] *)

(** **** Exercise: 1 star (andb3)  *)
(** Do the same for the [andb3] function below. This function should
    return [true] when all of its inputs are [true], and [false]
    otherwise. *)

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

Example test_andb31:                 (andb3 true true true) = true.
Proof. simpl. reflexivity. Qed.
Example test_andb32:                 (andb3 false true true) = false.
Proof. simpl. reflexivity. Qed.
Example test_andb33:                 (andb3 true false true) = false.
Proof. simpl. reflexivity. Qed.
Example test_andb34:                 (andb3 true true false) = false.
Proof. simpl. reflexivity. Qed.
  (** [] *)

(* ================================================================= *)
(** ** Function Types *)

(** Every expression in Coq has a type, describing what sort of
    thing it computes. The [Check] command asks Coq to print the type
    of an expression. *)

Check true.
(* ===> true : bool *)
Check (negb true).
(* ===> negb true : bool *)

(** Functions like [negb] itself are also data values, just like
    [true] and [false].  Their types are called _function types_, and
    they are written with arrows. *)

Check negb.
(* ===> negb : bool -> bool *)

(** The type of [negb], written [bool -> bool] and pronounced
    "[bool] arrow [bool]," can be read, "Given an input of type
    [bool], this function produces an output of type [bool]."
    Similarly, the type of [andb], written [bool -> bool -> bool], can
    be read, "Given two inputs, both of type [bool], this function
    produces an output of type [bool]." *)

(* ================================================================= *)
(** ** Modules *)

(** Coq provides a _module system_, to aid in organizing large
    developments.  In this course we won't need most of its features,
    but one is useful: If we enclose a collection of declarations
    between [Module X] and [End X] markers, then, in the remainder of
    the file after the [End], these definitions are referred to by
    names like [X.foo] instead of just [foo].  We will use this
    feature to introduce the definition of the type [nat] in an inner
    module so that it does not interfere with the one from the
    standard library (which we want to use in the rest because it
    comes with a tiny bit of convenient special notation).  *)

Module NatPlayground.

(* ================================================================= *)
(** ** Numbers *)

(** The types we have defined so far are examples of "enumerated
    types": their definitions explicitly enumerate a finite set of
    elements.  A more interesting way of defining a type is to give a
    collection of _inductive rules_ describing its elements.  For
    example, we can define (a unary representation of) the natural
    numbers as follows: *)

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

(** The clauses of this definition can be read:
      - [O] is a natural number (note that this is the letter "[O],"
        not the numeral "[0]").
      - [S] can be put in front of a natural number to yield another
        one -- if [n] is a natural number, then [S n] is too. *)

(** Let's look at this in a little more detail.

    Every inductively defined set ([day], [nat], [bool], etc.) is
    actually a set of _expressions_ built from _constructors_
    like [O], [S], [true], [false], [monday], etc.  The definition of
    [nat] says how expressions in the set [nat] can be built:

    - [O] and [S] are constructors;
    - the expression [O] belongs to the set [nat];
    - if [n] is an expression belonging to the set [nat], then [S n]
      is also an expression belonging to the set [nat]; and
    - expressions formed in these two ways are the only ones belonging
      to the set [nat]. *)

(** The same rules apply for our definitions of [day] and
    [bool]. (The annotations we used for their constructors are
    analogous to the one for the [O] constructor, indicating that they
    don't take any arguments.)

    The above conditions are the precise force of the [Inductive]
    declaration.  They imply that the expression [O], the expression
    [S O], the expression [S (S O)], the expression [S (S (S O))], and
    so on all belong to the set [nat], while other expressions built
    from data constructors, like [true], [andb true false], [S (S
    false)], and [O (O (O S))] do not.

    A critical point here is that what we've done so far is just to
    define a _representation_ of numbers: a way of writing them down.
    The names [O] and [S] are arbitrary, and at this point they have
    no special meaning -- they are just two different marks that we
    can use to write down numbers (together with a rule that says any
    [nat] will be written as some string of [S] marks followed by an
    [O]).  If we like, we can write essentially the same definition
    this way: *)

Inductive nat' : Type :=
  | stop : nat'
  | tick : nat' -> nat'.

(** The _interpretation_ of these marks comes from how we use them to
    compute. *)

(** We can do this by writing functions that pattern match on
    representations of natural numbers just as we did above with
    booleans and days -- for example, here is the predecessor
    function: *)

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

(** The second branch can be read: "if [n] has the form [S n']
    for some [n'], then return [n']."  *)

End NatPlayground.

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

(** Because natural numbers are such a pervasive form of data,
    Coq provides a tiny bit of built-in magic for parsing and printing
    them: ordinary arabic numerals can be used as an alternative to
    the "unary" notation defined by the constructors [S] and [O].  Coq
    prints numbers in arabic form by default: *)

Check (S (S (S (S O)))).
  (* ===> 4 : nat *)
Compute (minustwo 4).
  (* ===> 2 : nat *)

(** The constructor [S] has the type [nat -> nat], just like the
    functions [minustwo] and [pred]: *)

Check S.
Check pred.
Check minustwo.

(** These are all things that can be applied to a number to yield a
    number.  However, there is a fundamental difference between the
    first one and the other two: functions like [pred] and [minustwo]
    come with _computation rules_ -- e.g., the definition of [pred]
    says that [pred 2] can be simplified to [1] -- while the
    definition of [S] has no such behavior attached.  Although it is
    like a function in the sense that it can be applied to an
    argument, it does not _do_ anything at all!  It is just a way of
    writing down numbers.  (Think about standard arabic numerals: the
    numeral [1] is not a computation; it's a piece of data.  When we
    write [111] to mean the number one hundred and eleven, we are
    using [1], three times, to write down a concrete representation of
    a number.)

    For most function definitions over numbers, just pattern matching
    is not enough: we also need recursion.  For example, to check that
    a number [n] is even, we may need to recursively check whether
    [n-2] is even.  To write such functions, we use the keyword
    [Fixpoint]. *)

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

(** We can define [oddb] by a similar [Fixpoint] declaration, but here
    is a simpler definition: *)

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

Example test_oddb1:    oddb 1 = true.
Proof. simpl. reflexivity.  Qed.
Example test_oddb2:    oddb 4 = false.
Proof. simpl. reflexivity.  Qed.

(** (You will notice if you step through these proofs that
    [simpl] actually has no effect on the goal -- all of the work is
    done by [reflexivity].  We'll see more about why that is shortly.)

    Naturally, we can also define multi-argument functions by
    recursion.  *)

Module NatPlayground2.

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

(** Adding three to two now gives us five, as we'd expect. *)

Compute (plus 3 2).

(** The simplification that Coq performs to reach this conclusion can
    be visualized as follows: *)

(*  [plus (S (S (S O))) (S (S O))]
==> [S (plus (S (S O)) (S (S O)))]
      by the second clause of the [match]
==> [S (S (plus (S O) (S (S O))))]
      by the second clause of the [match]
==> [S (S (S (plus O (S (S O)))))]
      by the second clause of the [match]
==> [S (S (S (S (S O))))]
      by the first clause of the [match]
*)

(** As a notational convenience, if two or more arguments have
    the same type, they can be written together.  In the following
    definition, [(n m : nat)] means just the same as if we had written
    [(n : nat) (m : nat)]. *)

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

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

(** You can match two expressions at once by putting a comma
    between them: *)

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

(** The _ in the first line is a _wildcard pattern_.  Writing _ in a
    pattern is the same as writing some variable that doesn't get used
    on the right-hand side.  This avoids the need to invent a variable
    name. *)

End NatPlayground2.

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

(** **** Exercise: 1 star (factorial)  *)
(** Recall the standard mathematical factorial function:

       factorial(0)  =  1
       factorial(n)  =  n * factorial(n-1)     (if n>0)

    Translate this into Coq. *)

Fixpoint factorial (n:nat) : nat :=
  match n with
    | O => S O
    | 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.
  (** [] *)

(** We can make numerical expressions a little easier to read and
    write by introducing _notations_ for addition, multiplication, and
    subtraction. *)

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).

(** (The [level], [associativity], and [nat_scope] annotations
    control how these notations are treated by Coq's parser.  The
    details are not important for our purposes, but interested readers
    can refer to the optional "More on Notation" section at the end of
    this chapter.)

    Note that these do not change the definitions we've already made:
    they are simply instructions to the Coq parser to accept [x + y]
    in place of [plus x y] and, conversely, to the Coq pretty-printer
    to display [plus x y] as [x + y]. *)

(** When we say that Coq comes with almost nothing built-in, we really
    mean it: even equality testing for numbers is a user-defined
    operation!  We now define a function [beq_nat], which tests
    [nat]ural numbers for [eq]uality, yielding a [b]oolean.  Note the
    use of nested [match]es (we could also have used a simultaneous
    match, as we did in [minus].) *)

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.

(** The [leb] function tests whether its first argument is less than or
  equal to its second argument, yielding a boolean. *)

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.

Example test_leb1:             (leb 2 2) = true.
Proof. simpl. reflexivity.  Qed.
Example test_leb2:             (leb 2 4) = true.
Proof. simpl. reflexivity.  Qed.
Example test_leb3:             (leb 4 2) = false.
Proof. simpl. reflexivity.  Qed.

(** **** Exercise: 1 star (blt_nat)  *)
(** The [blt_nat] function tests [nat]ural numbers for [l]ess-[t]han,
    yielding a [b]oolean.  Instead of making up a new [Fixpoint] for
    this one, define it in terms of a previously defined function. *)

Definition blt_nat (n m : nat) : bool :=
  match (leb n m) with
    | true => match (beq_nat n m) with
      | true => false
      | false => true
      end
    | false => false
  end.

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.
(** [] *)

(* ################################################################# *)
(** * Proof by Simplification *)

(** Now that we've defined a few datatypes and functions, let's
    turn to stating and proving properties of their behavior.
    Actually, we've already started doing this: each [Example] in the
    previous sections makes a precise claim about the behavior of some
    function on some particular inputs.  The proofs of these claims
    were always the same: use [simpl] to simplify both sides of the
    equation, then use [reflexivity] to check that both sides contain
    identical values.

    The same sort of "proof by simplification" can be used to prove
    more interesting properties as well.  For example, the fact that
    [0] is a "neutral element" for [+] on the left can be proved just
    by observing that [0 + n] reduces to [n] no matter what [n] is, a
    fact that can be read directly off the definition of [plus].*)

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

(** (You may notice that the above statement looks different in
    the [.v] file in your IDE than it does in the HTML rendition in
    your browser, if you are viewing both. In [.v] files, we write the
    [forall] universal quantifier using the reserved identifier
    "forall."  When the [.v] files are converted to HTML, this gets
    transformed into an upside-down-A symbol.)

    This is a good place to mention that [reflexivity] is a bit
    more powerful than we have admitted. In the examples we have seen,
    the calls to [simpl] were actually not needed, because
    [reflexivity] can perform some simplification automatically when
    checking that two sides are equal; [simpl] was just added so that
    we could see the intermediate state -- after simplification but
    before finishing the proof.  Here is a shorter proof of the
    theorem: *)

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

(** Moreover, it will be useful later to know that [reflexivity]
    does somewhat _more_ simplification than [simpl] does -- for
    example, it tries "unfolding" defined terms, replacing them with
    their right-hand sides.  The reason for this difference is that,
    if reflexivity succeeds, the whole goal is finished and we don't
    need to look at whatever expanded expressions [reflexivity] has
    created by all this simplification and unfolding; by contrast,
    [simpl] is used in situations where we may have to read and
    understand the new goal that it creates, so we would not want it
    blindly expanding definitions and leaving the goal in a messy
    state.

    The form of the theorem we just stated and its proof are almost
    exactly the same as the simpler examples we saw earlier; there are
    just a few differences.

    First, we've used the keyword [Theorem] instead of [Example].
    This difference is mostly a matter of style; the keywords
    [Example] and [Theorem] (and a few others, including [Lemma],
    [Fact], and [Remark]) mean pretty much the same thing to Coq.

    Second, we've added the quantifier [forall n:nat], so that our
    theorem talks about _all_ natural numbers [n].  Informally, to
    prove theorems of this form, we generally start by saying "Suppose
    [n] is some number..."  Formally, this is achieved in the proof by
    [intros n], which moves [n] from the quantifier in the goal to a
    _context_ of current assumptions.

    The keywords [intros], [simpl], and [reflexivity] are examples of
    _tactics_.  A tactic is a command that is used between [Proof] and
    [Qed] to guide the process of checking some claim we are making.
    We will see several more tactics in the rest of this chapter and
    yet more in future chapters.

    Other similar theorems can be proved with the same pattern. *)

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.

(** The [_l] suffix in the names of these theorems is
    pronounced "on the left." *)

(** It is worth stepping through these proofs to observe how the
    context and the goal change.  You may want to add calls to [simpl]
    before [reflexivity] to see the simplifications that Coq performs
    on the terms before checking that they are equal.

    Although simplification is powerful enough to prove some fairly
    general facts, there are many statements that cannot be handled by
    simplification alone.  For instance, we cannot use it to prove
    that [0] is also a neutral element for [+] _on the right_. *)

Theorem plus_n_O : forall n, n = n + 0.
Proof.
  intros n. simpl. (* Doesn't do anything! *)
(** (Can you explain why this happens?  Step through both proofs
    with Coq and notice how the goal and context change.)

    When stuck in the middle of a proof, we can use the [Abort]
    command to give up on it for the moment. *)

Abort.

(** The next chapter will introduce _induction_, a powerful
    technique that can be used for proving this goal.  For the moment,
    though, let's look at a few more simple tactics. *)

(* ################################################################# *)
(** * Proof by Rewriting *)

(** This theorem is a bit more interesting than the others we've
    seen: *)

Theorem plus_id_example : forall n m:nat,
  n = m ->
  n + n = m + m.

(** Instead of making a universal claim about all numbers [n] and [m],
    it talks about a more specialized property that only holds when [n
    = m].  The arrow symbol is pronounced "implies."

    As before, we need to be able to reason by assuming we are given such
    numbers [n] and [m].  We also need to assume the hypothesis
    [n = m]. The [intros] tactic will serve to move all three of these
    from the goal into assumptions in the current context.

    Since [n] and [m] are arbitrary numbers, we can't just use
    simplification to prove this theorem.  Instead, we prove it by
    observing that, if we are assuming [n = m], then we can replace
    [n] with [m] in the goal statement and obtain an equality with the
    same expression on both sides.  The tactic that tells Coq to
    perform this replacement is called [rewrite]. *)

Proof.
  (* move both quantifiers into the context: *)
  intros n m.
  (* move the hypothesis into the context: *)
  intros H.
  (* rewrite the goal using the hypothesis: *)
  rewrite -> H.
  reflexivity.  Qed.

(** The first line of the proof moves the universally quantified
    variables [n] and [m] into the context.  The second moves the
    hypothesis [n = m] into the context and gives it the name [H].
    The third tells Coq to rewrite the current goal ([n + n = m + m])
    by replacing the left side of the equality hypothesis [H] with the
    right side.

    (The arrow symbol in the [rewrite] has nothing to do with
    implication: it tells Coq to apply the rewrite from left to right.
    To rewrite from right to left, you can use [rewrite <-].  Try
    making this change in the above proof and see what difference it
    makes.) *)

(** **** Exercise: 1 star (plus_id_exercise)  *)
(** Remove "[Admitted.]" and fill in the proof. *)

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

(** The [Admitted] command tells Coq that we want to skip trying
    to prove this theorem and just accept it as a given.  This can be
    useful for developing longer proofs, since we can state subsidiary
    lemmas that we believe will be useful for making some larger
    argument, use [Admitted] to accept them on faith for the moment,
    and continue working on the main argument until we are sure it
    makes sense; then we can go back and fill in the proofs we
    skipped.  Be careful, though: every time you say [Admitted] you
    are leaving a door open for total nonsense to enter Coq's nice,
    rigorous, formally checked world! *)

(** We can also use the [rewrite] tactic with a previously proved
    theorem instead of a hypothesis from the context. If the statement
    of the previously proved theorem involves quantified variables,
    as in the example below, Coq tries to instantiate them
    by matching with the current goal. *)

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

(** **** Exercise: 2 stars (mult_S_1)  *)
Theorem mult_S_1 : forall n m : nat,
  m = S n ->
  m * (1 + n) = m * m.
Proof.
  intros n m.
  intros H.
  rewrite -> plus_1_l.
  rewrite <- H.
  reflexivity. Qed.

  (* (N.b. This proof can actually be completed without using [rewrite],
     but please do use [rewrite] for the sake of the exercise.) *)
(** [] *)

(* ################################################################# *)
(** * Proof by Case Analysis *)

(** Of course, not everything can be proved by simple
    calculation and rewriting: In general, unknown, hypothetical
    values (arbitrary numbers, booleans, lists, etc.) can block
    simplification.  For example, if we try to prove the following
    fact using the [simpl] tactic as above, we get stuck. *)

Theorem plus_1_neq_0_firsttry : forall n : nat,
  beq_nat (n + 1) 0 = false.
Proof.
  intros n.
  simpl.  (* does nothing! *)
Abort.

(** The reason for this is that the definitions of both
    [beq_nat] and [+] begin by performing a [match] on their first
    argument.  But here, the first argument to [+] is the unknown
    number [n] and the argument to [beq_nat] is the compound
    expression [n + 1]; neither can be simplified.

    To make progress, we need to consider the possible forms of [n]
    separately.  If [n] is [O], then we can calculate the final result
    of [beq_nat (n + 1) 0] and check that it is, indeed, [false].  And
    if [n = S n'] for some [n'], then, although we don't know exactly
    what number [n + 1] yields, we can calculate that, at least, it
    will begin with one [S], and this is enough to calculate that,
    again, [beq_nat (n + 1) 0] will yield [false].

    The tactic that tells Coq to consider, separately, the cases where
    [n = O] and where [n = S n'] is called [destruct]. *)

Theorem plus_1_neq_0 : forall n : nat,
  beq_nat (n + 1) 0 = false.
Proof.
  intros n. destruct n as [| n'].
  - reflexivity.
  - reflexivity.   Qed.

(** The [destruct] generates _two_ subgoals, which we must then
    prove, separately, in order to get Coq to accept the theorem. The
    annotation "[as [| n']]" is called an _intro pattern_.  It tells
    Coq what variable names to introduce in each subgoal.  In general,
    what goes between the square brackets is a _list of lists_ of
    names, separated by [|].  In this case, the first component is
    empty, since the [O] constructor is nullary (it doesn't have any
    arguments).  The second component gives a single name, [n'], since
    [S] is a unary constructor.

    The [-] signs on the second and third lines are called _bullets_,
    and they mark the parts of the proof that correspond to each
    generated subgoal.  The proof script that comes after a bullet is
    the entire proof for a subgoal.  In this example, each of the
    subgoals is easily proved by a single use of [reflexivity], which
    itself performs some simplification -- e.g., the first one
    simplifies [beq_nat (S n' + 1) 0] to [false] by first rewriting
    [(S n' + 1)] to [S (n' + 1)], then unfolding [beq_nat], and then
    simplifying the [match].

    Marking cases with bullets is entirely optional: if bullets are
    not present, Coq simply asks you to prove each subgoal in
    sequence, one at a time. But it is a good idea to use bullets.
    For one thing, they make the structure of a proof apparent, making
    it more readable. Also, bullets instruct Coq to ensure that a
    subgoal is complete before trying to verify the next one,
    preventing proofs for different subgoals from getting mixed
    up. These issues become especially important in large
    developments, where fragile proofs lead to long debugging
    sessions.

    There are no hard and fast rules for how proofs should be
    formatted in Coq -- in particular, where lines should be broken
    and how sections of the proof should be indented to indicate their
    nested structure.  However, if the places where multiple subgoals
    are generated are marked with explicit bullets at the beginning of
    lines, then the proof will be readable almost no matter what
    choices are made about other aspects of layout.

    This is also a good place to mention one other piece of somewhat
    obvious advice about line lengths.  Beginning Coq users sometimes
    tend to the extremes, either writing each tactic on its own line
    or writing entire proofs on one line.  Good style lies somewhere
    in the middle.  One reasonable convention is to limit yourself to
    80-character lines.

    The [destruct] tactic can be used with any inductively defined
    datatype.  For example, we use it next to prove that boolean
    negation is involutive -- i.e., that negation is its own
    inverse. *)

Theorem negb_involutive : forall b : bool,
  negb (negb b) = b.
Proof.
  intros b. destruct b.
  - reflexivity.
  - reflexivity.  Qed.

(** Note that the [destruct] here has no [as] clause because
    none of the subcases of the [destruct] need to bind any variables,
    so there is no need to specify any names.  (We could also have
    written [as [|]], or [as []].)  In fact, we can omit the [as]
    clause from _any_ [destruct] and Coq will fill in variable names
    automatically.  This is generally considered bad style, since Coq
    often makes confusing choices of names when left to its own
    devices.

    It is sometimes useful to invoke [destruct] inside a subgoal,
    generating yet more proof obligations. In this case, we use
    different kinds of bullets to mark goals on different "levels."
    For example: *)

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

(** Each pair of calls to [reflexivity] corresponds to the
    subgoals that were generated after the execution of the [destruct c]
    line right above it. *)

(** Besides [-] and [+], we can use [*] (asterisk) as a third kind of
    bullet.  We can also enclose sub-proofs in curly braces, which is
    useful in case we ever encounter a proof that generates more than
    three levels of subgoals: *)

Theorem andb_commutative' : forall b c, andb b c = andb c b.
Proof.
  intros b c. destruct b.
  { destruct c.
    { reflexivity. }
    { reflexivity. } }
  { destruct c.
    { reflexivity. }
    { reflexivity. } }
Qed.

(** Since curly braces mark both the beginning and the end of a
    proof, they can be used for multiple subgoal levels, as this
    example shows. Furthermore, curly braces allow us to reuse the
    same bullet shapes at multiple levels in a proof: *)

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

(** Before closing the chapter, let's mention one final
    convenience.  As you may have noticed, many proofs perform case
    analysis on a variable right after introducing it:

       intros x y. destruct y as [|y].

    This pattern is so common that Coq provides a shorthand for it: we
    can perform case analysis on a variable when introducing it by
    using an intro pattern instead of a variable name. For instance,
    here is a shorter proof of the [plus_1_neq_0] theorem above. *)

Theorem plus_1_neq_0' : forall n : nat,
  beq_nat (n + 1) 0 = false.
Proof.
  intros [|n].
  - reflexivity.
  - reflexivity.  Qed.

(** If there are no arguments to name, we can just write [[]]. *)

Theorem andb_commutative'' :
  forall b c, andb b c = andb c b.
Proof.
  intros [] [].
  - reflexivity.
  - reflexivity.
  - reflexivity.
  - reflexivity.
Qed.

(** **** Exercise: 2 stars (andb_true_elim2)  *)
(** Prove the following claim, marking cases (and subcases) with
    bullets when you use [destruct]. *)

Theorem andb_true_elim2 : forall b c : bool,
  andb b c = true -> c = true.
Proof.
  intros [] [] [].
  - reflexivity.
  - reflexivity.
  - reflexivity.
  - reflexivity.
Qed.

(** **** Exercise: 1 star (zero_nbeq_plus_1)  *)
Theorem zero_nbeq_plus_1 : forall n : nat,
  beq_nat 0 (n + 1) = false.
Proof.
  intros [|n].
  - reflexivity.
  - reflexivity.
Qed.
(** [] *)

(* ================================================================= *)
(** ** More on Notation (Optional) *)

(** (In general, sections marked Optional are not needed to follow the
    rest of the book, except possibly other Optional sections.  On a
    first reading, you might want to skim these sections so that you
    know what's there for future reference.)

    Recall the notation definitions for infix plus and times: *)

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

(** For each notation symbol in Coq, we can specify its _precedence
    level_ and its _associativity_.  The precedence level [n] is
    specified by writing [at level n]; this helps Coq parse compound
    expressions.  The associativity setting helps to disambiguate
    expressions containing multiple occurrences of the same
    symbol. For example, the parameters specified above for [+] and
    [*] say that the expression [1+2*3*4] is shorthand for
    [(1+((2*3)*4))]. Coq uses precedence levels from 0 to 100, and
    _left_, _right_, or _no_ associativity.  We will see more examples
    of this later, e.g., in the [Lists]
    chapter.

    Each notation symbol is also associated with a _notation scope_.
    Coq tries to guess what scope is meant from context, so when it
    sees [S(O*O)] it guesses [nat_scope], but when it sees the
    cartesian product (tuple) type [bool*bool] (which we'll see in
    later chapters) it guesses [type_scope].  Occasionally, it is
    necessary to help it out with percent-notation by writing
    [(x*y)%nat], and sometimes in what Coq prints it will use [%nat]
    to indicate what scope a notation is in.

    Notation scopes also apply to numeral notation ([3], [4], [5],
    etc.), so you may sometimes see [0%nat], which means [O] (the
    natural number [0] that we're using in this chapter), or [0%Z],
    which means the Integer zero (which comes from a different part of
    the standard library).

    Pro tip: Coq's notation mechanism is not especially powerful.
    Don't expect too much from it! *)

(* ================================================================= *)
(** ** Fixpoints and Structural Recursion (Optional) *)

(** Here is a copy of the definition of addition: *)

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

(** When Coq checks this definition, it notes that [plus'] is
    "decreasing on 1st argument."  What this means is that we are
    performing a _structural recursion_ over the argument [n] -- i.e.,
    that we make recursive calls only on strictly smaller values of
    [n].  This implies that all calls to [plus'] will eventually
    terminate.  Coq demands that some argument of _every_ [Fixpoint]
    definition is "decreasing."

    This requirement is a fundamental feature of Coq's design: In
    particular, it guarantees that every function that can be defined
    in Coq will terminate on all inputs.  However, because Coq's
    "decreasing analysis" is not very sophisticated, it is sometimes
    necessary to write functions in slightly unnatural ways. *)

(** **** Exercise: 2 stars, optional (decreasing)  *)
(** To get a concrete sense of this, find a way to write a sensible
    [Fixpoint] definition (of a simple function on numbers, say) that
    _does_ terminate on all inputs, but that Coq will reject because
    of this restriction. *)

(** Fixpoint sensible (n : nat) (m : nat) : nat :=
  match n with
  | O => m
  | (S n') => match m with
    | (S m') => (sensible n' m')
    | O => (sensible m n)
    end
  end. *)
(** [] *)

(* ################################################################# *)
(** * More Exercises *)

(** **** Exercise: 2 stars (boolean_functions)  *)
(** Use the tactics you have learned so far to prove the following
    theorem about boolean functions. *)

Theorem identity_fn_applied_twice :
  forall (f : bool -> bool),
  (forall (x : bool), f x = x) ->
  forall (b : bool), f (f b) = b.
Proof.
  intros f. intros l. intros [].
  - rewrite l. rewrite l. reflexivity.
  - rewrite l. rewrite l. reflexivity.
Qed.

(** Now state and prove a theorem [negation_fn_applied_twice] similar
    to the previous one but where the second hypothesis says that the
    function [f] has the property that [f x = negb x].*)

Theorem negation_fn_applied_twice :
  forall (f : bool -> bool),
  (forall (x : bool), f x = x) ->
  forall (b : bool), f (f b) = b.
Proof.
  intros f. intros l. intros [].
  - rewrite l. rewrite l. reflexivity.
  - rewrite l. rewrite l. reflexivity.
Qed.

(** **** Exercise: 3 stars, optional (andb_eq_orb)  *)
(** Prove the following theorem.  (Hint: This one can be a bit tricky,
    depending on how you approach it.  You will probably need both
    [destruct] and [rewrite], but destructing everything in sight is
    not the best way.) *)

Lemma andb_negb : forall (b : bool), (andb b (negb b)) = false.
Proof.
  intros [].
  - reflexivity.
  - reflexivity.
Qed.

Lemma orb_negb : forall (b : bool), (orb b (negb b)) = true.
Proof.
  intros [].
  - reflexivity.
  - reflexivity.
Qed.

Theorem andb_eq_orb :
  forall (b c : bool),
  (andb b c = orb b c) ->
  b = c.
Proof.
  intros [] [].
  - reflexivity.
  - rewrite andb_negb. rewrite orb_negb. intros h. rewrite<-h. reflexivity.
  - rewrite andb_negb. rewrite orb_negb. intros h. rewrite<-h. reflexivity.
  - reflexivity.
Qed.
(** [] *)

(** **** Exercise: 3 stars (binary)  *)
(** Consider a different, more efficient representation of natural
    numbers using a binary rather than unary system.  That is, instead
    of saying that each natural number is either zero or the successor
    of a natural number, we can say that each binary number is either

      - zero,
      - twice a binary number, or
      - one more than twice a binary number.

    (a) First, write an inductive definition of the type [bin]
        corresponding to this description of binary numbers.

    (Hint: Recall that the definition of [nat] above,

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

    says nothing about what [O] and [S] "mean."  It just says "[O] is
    in the set called [nat], and if [n] is in the set then so is [S
    n]."  The interpretation of [O] as zero and [S] as successor/plus
    one comes from the way that we _use_ [nat] values, by writing
    functions to do things with them, proving things about them, and
    so on.  Your definition of [bin] should be correspondingly simple;
    it is the functions you will write next that will give it
    mathematical meaning.)

    (b) Next, write an increment function [incr] for binary numbers,
        and a function [bin_to_nat] to convert binary numbers to unary
        numbers.

    (c) Write five unit tests [test_bin_incr1], [test_bin_incr2], etc.
        for your increment and binary-to-unary functions.  (A "unit
        test" in Coq is a specific [Example] that can be proved with
        just [reflexivity], as we've done for several of our
        definitions.)  Notice that incrementing a binary number and
        then converting it to unary should yield the same result as
        first converting it to unary and then incrementing. *)

Inductive bin : Type :=
  | B : bin
  | T : bin -> bin
  | P : bin -> bin.

Fixpoint incr (b : bin) : bin :=
  match b with
    | B => P B
    | T b' => P b'
    | P b' => T (incr b')
  end.

Fixpoint bin_to_nat (b : bin) : nat :=
  match b with
    | B => 0
    | T b' => mult (bin_to_nat b') 2
    | P b' => S (mult (bin_to_nat b') 2)
  end.

Example test_bin_nat_convert1:  bin_to_nat B = 0.
Proof. simpl. reflexivity.  Qed.
Example test_bin_nat_convert2:  bin_to_nat (P (T (P B))) = 5.
Proof. simpl. reflexivity.  Qed.
Example test_bin_nat_convert3:  bin_to_nat (T (T (T (P B)))) = 8.
Proof. simpl. reflexivity.  Qed.
Example test_bin_incr1:         bin_to_nat (P (incr (P B))) = 5.
Proof. simpl. reflexivity.  Qed.
Example test_bin_incr2:         bin_to_nat (incr (P (incr (P B)))) =
  2 * ((bin_to_nat (P B)) + 1) + 1 + 1.
Proof. simpl. reflexivity.  Qed.
Example test_bin_incr3:         bin_to_nat (T (P (incr (P B)))) =
  bin_to_nat (T (incr (incr (incr (incr (incr (T B))))))).
Proof. simpl. reflexivity.  Qed.
(** [] *)

(** $Date: 2017-08-24 17:13:02 -0400 (Thu, 24 Aug 2017) $ *)


(** * Prop: Propositions and Evidence *)

Require Import Coq.Arith.Plus.
Require Import Cases.

(** In previous chapters, we have seen many examples of factual
    claims (_propositions_) and ways of presenting evidence of their
    truth (_proofs_).  In particular, we have worked extensively with
    _equality propositions_ of the form [e1 = e2], with
    implications ([P -> Q]), and with quantified propositions
    ([forall x, P]).

    In this chapter we take a deeper look at the way propositions are
    expressed in Coq and at the structure of the logical evidence that
    we construct when we carry out proofs.

    Some of the concepts in this chapter may seem a bit abstract on a
    first encounter.  We've included a _lot_ of exercises, most of
    which should be quite approachable even if you're still working on
    understanding the details of the text.  Try to work as many of
    them as you can, especially the one-starred exercises.

*)
(* ##################################################### *)
(** * Inductively Defined Propositions *)

(** This chapter will take us on a first tour of the
    propositional (logical) side of Coq.  As a running example, let's
    define a simple property of natural numbers -- we'll call it
    "[beautiful]." *)

(** Informally, a number is [beautiful] if it is [0], [3], [5], or the
    sum of two [beautiful] numbers.

    More pedantically, we can define [beautiful] numbers by giving four
    rules:

       - Rule [b_0]: The number [0] is [beautiful].
       - Rule [b_3]: The number [3] is [beautiful].
       - Rule [b_5]: The number [5] is [beautiful].
       - Rule [b_sum]: If [n] and [m] are both [beautiful], then so is
         their sum. *)

(** We will see many definitions like this one during the rest
    of the course, and for purposes of informal discussions, it is
    helpful to have a lightweight notation that makes them easy to
    read and write.  _Inference rules_ are one such notation: *)
(**
                              -----------                               (b_0)
                              beautiful 0

                              ------------                              (b_3)
                              beautiful 3

                              ------------                              (b_5)
                              beautiful 5

                       beautiful n     beautiful m
                       ---------------------------                      (b_sum)
                              beautiful (n+m)
*)

(** Each of the textual rules above is reformatted here as an
    inference rule; the intended reading is that, if the _premises_
    above the line all hold, then the _conclusion_ below the line
    follows.  For example, the rule [b_sum] says that, if [n] and [m]
    are both [beautiful] numbers, then it follows that [n+m] is
    [beautiful] too.  The rules with no premises above the line are
    called _axioms_.

    These rules _define_ the property [beautiful].  That is, if we
    want to convince someone that some particular number is [beautiful],
    our argument must be based on these rules.  For a simple example,
    suppose we claim that the number [5] is [beautiful].  To support
    this claim, we just need to point out that rule [b_5] says so.
    Or, if we want to claim that [8] is [beautiful], we can support our
    claim by first observing that [3] and [5] are both [beautiful] (by
    rules [b_3] and [b_5]) and then pointing out that their sum, [8],
    is therefore [beautiful] by rule [b_sum].  This argument can be
    expressed graphically with the following _proof tree_: *)
(**
         ----------- (b_3)   ----------- (b_5)
         beautiful 3         beautiful 5
         ------------------------------- (b_sum)
                   beautiful 8
    Of course, there are other ways of using these rules to argue that
    [8] is [beautiful], for instance:
         ----------- (b_5)   ----------- (b_3)
         beautiful 5         beautiful 3
         ------------------------------- (b_sum)
                   beautiful 8
*)

(** **** Exercise: 1 star (varieties_of_beauty) *)
(** How many different ways are there to show that [8] is [beautiful]? *)

(* FILL IN HERE *)
(* two "minimal" proofs and infinite proofs. 3+5, 5+3, 3+5+0, ... *)

(** In Coq, we can express the definition of [beautiful] as
    follows: *)

Inductive beautiful : nat -> Prop :=
  b_0   : beautiful 0
| b_3   : beautiful 3
| b_5   : beautiful 5
| b_sum : forall n m, beautiful n -> beautiful m -> beautiful (n+m).

(** The first line declares that [beautiful] is a proposition -- or,
    more formally, a family of propositions "indexed by" natural
    numbers.  (That is, for each number [n], the claim that "[n] is
    [beautiful]" is a proposition.)  Such a family of propositions is
    often called a _property_ of numbers.  Each of the remaining lines
    embodies one of the rules for [beautiful] numbers.

    We can use Coq's tactic scripting facility to assemble proofs that
    particular numbers are [beautiful].  *)

Theorem three_is_beautiful: beautiful 3.
Proof.
   (* This simply follows from the axiom [b_3]. *)
   apply b_3.
Qed.

Theorem eight_is_beautiful: beautiful 8.
Proof.
   (* First we use the rule [b_sum], telling Coq how to
      instantiate [n] and [m]. *)
   apply b_sum with (n:=3) (m:=5).
   (* appply (b_sum 3 5) *)
   (* To solve the subgoals generated by [b_sum], we must provide
      evidence of [beautiful 3] and [beautiful 5]. Fortunately we
      have axioms for both. *)
   apply b_3.
   apply b_5.
Qed.

(* ##################################################### *)
(** * Proof Objects *)

(** Look again at the formal definition of the [beautiful]
    property.  The opening keyword, [Inductive], has been used up to
    this point to declare new types of _data_, such as numbers and
    lists.  Does this interpretation also make sense for the Inductive
    definition of [beautiful]?  That is, can we view evidence of
    beauty as some kind of data structure? Yes, we can!

    The trick is to introduce an alternative pronunciation of "[:]".
    Instead of "has type," we can also say "is a proof of."  For
    example, the second line in the definition of [beautiful] declares
    that [b_0 : beautiful 0].  Instead of "[b_0] has type
    [beautiful 0]," we can say that "[b_0] is a proof of [beautiful 0]."
    Similarly for [b_3] and [b_5]. *)

(** This pun between types and propositions (between [:] as "has type"
    and [:] as "is a proof of" or "is evidence for") is called the
    _Curry-Howard correspondence_.  It proposes a deep connection
    between the world of logic and the world of computation.
<<
                 propositions  ~  types
                 proofs        ~  data values
>>
    Many useful insights follow from this connection.  To begin with, it
    gives us a natural interpretation of the type of [b_sum] constructor: *)

Check b_sum.
(* ===> b_sum : forall n m,
                  beautiful n ->
                  beautiful m ->
                  beautiful (n+m) *)

(** This can be read "[b_sum] is a constructor that takes four
    arguments -- two numbers, [n] and [m], and two values, of types
    [beautiful n] and [beautiful m] -- and yields evidence for the
    proposition [beautiful (n+m)]." *)

(** In view of this, we might wonder whether we can write an
    expression of type [beautiful 8] by applying [b_sum] to
    appropriate arguments.  Indeed, we can: *)

Check (b_sum 3 5 b_3 b_5).
Check (b_sum 3 5).
(* ===> beautiful (3 + 5) *)

(** The expression [b_sum 3 5 b_3 b_5] can be thought of as
    instantiating the parameterized constructor [b_sum] with the
    specific arguments [3] [5] and the corresponding proof objects for
    its premises [beautiful 3] and [beautiful 5] (Coq is smart enough
    to figure out that 3+5=8).  Alternatively, we can think of [b_sum]
    as a primitive "evidence constructor" that, when applied to two
    particular numbers, wants to be further applied to evidence that
    those two numbers are beautiful; its type,
[[
    forall n m, beautiful n -> beautiful m -> beautiful (n+m),
    expresses this functionality, in the same way that the polymorphic
    type [forall X, list X] in the previous chapter expressed the fact
    that the constructor [nil] can be thought of as a function from
    types to empty lists with elements of that type. *)

(** This gives us an alternative way to write the proof that [8] is
    beautiful: *)

Theorem eight_is_beautiful': beautiful 8.
Proof.
   exact (b_sum 3 5 b_3 b_5).
Qed.

(** Notice that we're using [apply] here in a new way: instead of just
    supplying the _name_ of a hypothesis or previously proved theorem
    whose type matches the current goal, we are supplying an
    _expression_ that directly builds evidence with the required
    type. *)

(* ##################################################### *)
(** ** Proof Scripts and Proof Objects *)

(** These proof objects lie at the core of how Coq operates.

    When Coq is following a proof script, what is happening internally
    is that it is gradually constructing a proof object -- a term
    whose type is the proposition being proved.  The tactics between
    the [Proof] command and the [Qed] instruct Coq how to build up a
    term of the required type.  To see this process in action, let's
    use the [Show Proof] command to display the current state of the
    proof tree at various points in the following tactic proof. *)

Theorem eight_is_beautiful'': beautiful 8.
Proof.
   Show Proof.
   apply b_sum with (n:=3) (m:=5).
   Show Proof.
   apply b_3.
   Show Proof.
   apply b_5.

   Show Proof.
Qed.

(** At any given moment, Coq has constructed a term with some
    "holes" (indicated by [?1], [?2], and so on), and it knows what
    type of evidence is needed at each hole.  In the [Show Proof]
    output, lines of the form [?1 -> beautiful n] record these
    requirements.  (The [->] here has nothing to do with either
    implication or function types -- it is just an unfortunate choice
    of concrete syntax for the output!)

    Each of the holes corresponds to a subgoal, and the proof is
    finished when there are no more subgoals.  At this point, the
    [Theorem] command gives a name to the evidence we've built and
    stores it in the global context. *)

(** Tactic proofs are useful and convenient, but they are not
    essential: in principle, we can always construct the required
    evidence by hand.  Indeed, we don't even need the [Theorem]
    command: we can instead use [Definition] to directly give a global
    name to a piece of evidence. *)

Definition eight_is_beautiful''' : beautiful 8 :=
  b_sum 3 5 b_3 b_5.

(** All these different ways of building the proof lead to exactly the
    same evidence being saved in the global environment. *)

Print eight_is_beautiful.
(* ===> eight_is_beautiful    = b_sum 3 5 b_3 b_5 : beautiful 8 *)
Print eight_is_beautiful'.
(* ===> eight_is_beautiful'   = b_sum 3 5 b_3 b_5 : beautiful 8 *)
Print eight_is_beautiful''.
(* ===> eight_is_beautiful''  = b_sum 3 5 b_3 b_5 : beautiful 8 *)
Print eight_is_beautiful'''.
(* ===> eight_is_beautiful''' = b_sum 3 5 b_3 b_5 : beautiful 8 *)

(** **** Exercise: 1 star (six_is_beautiful) *)
(** Give a tactic proof and a proof object showing that [6] is [beautiful]. *)

Theorem six_is_beautiful :
  beautiful 6.
Proof.
  apply b_sum with (n:=3) (m:=3); apply b_3.
Qed.

Definition six_is_beautiful' : beautiful 6 := (b_sum 3 3 b_3 b_3).

(* Digression: In the spirit of "everything is a proof object", consider how we
   defined the value "admit".  It has type T, for every single type T!
   That's impossible.  So we use Coq's Admitted tactic, which says,
   "I give up for now; just assume this can be proven." For instance,
   we can use Admitted to prove things that are blatantly False: *)
Theorem foo: False. Admitted.
(* We do the same thing for admit: *)
Definition admit {T: Type} : T.  Admitted.
(* If we then ask "what is the proof object for admit?", Coq responds *)
Print admit.
(* [*** [ admit : forall T : Type, T ]]
   which is its way of saying, "I don't know, but whatever it is,
   it has type forall T:Type, T."  In fact, Coq will say something
   similar for foo: *)
Print foo.
(* [*** foo : False ], or in other words "I don't know what this
   value is, but it has type False *)

(** [] *)

(** **** Exercise: 1 star (nine_is_beautiful) *)
(** Give a tactic proof and a proof object showing that [9] is [beautiful]. *)

Theorem nine_is_beautiful :
  beautiful 9.
Proof.
  (* FILL IN HERE *)
apply (b_sum 3 6 b_3 six_is_beautiful).
Qed.

Definition nine_is_beautiful' : beautiful 9 := (b_sum 3 6 b_3 six_is_beautiful).
  (* FILL IN HERE *)
(** [] *)

(* ##################################################### *)
(** ** Implications and Functions *)

(** In Coq's computational universe (where we've mostly been living
    until this chapter), there are two sorts of values with arrows in
    their types: _constructors_ introduced by [Inductive]-ly defined
    data types, and _functions_.

    Similarly, in Coq's logical universe, there are two ways of giving
    evidence for an implication: constructors introduced by
    [Inductive]-ly defined propositions, and... functions!

    For example, consider this statement: *)

Theorem b_plus3: forall (n:nat) (bn:beautiful n), beautiful (3+n).
Proof.
  intros.
  apply b_sum with (n:=3) (m:=n). exact b_3. exact bn.
  Show Proof.

(* fun (n : nat) (bn : beautiful n) => b_sum 3 n b_3 bn *)
Qed.

(** What is the proof object corresponding to [b_plus3]?

    We're looking for an expression whose _type_ is [forall n,
    beautiful n -> beautiful (3+n)] -- that is, a _function_ that
    takes two arguments (one number and a piece of evidence) and
    returns a piece of evidence!  Here it is: *)

Definition b_plus3' : forall n, beautiful n -> beautiful (3+n) :=
  fun n => fun H : beautiful n =>
    b_sum 3 n b_3 H.

Check b_plus3'.
(* ===> b_plus3' : forall n, beautiful n -> beautiful (3+n) *)

(** Recall that [fun n => blah] means "the function that, given [n],
    yields [blah]."  Another equivalent way to write this definition is: *)

Definition b_plus3'' (n : nat) (H : beautiful n) : beautiful (3+n) :=
  b_sum 3 n b_3 H.

Check b_plus3''.
(* ===> b_plus3'' : forall n, beautiful n -> beautiful (3+n) *)

(** **** Exercise: 2 stars (b_times2) *)
Theorem b_times2: forall n, beautiful n -> beautiful (2*n).
Proof.
  intros. simpl. apply b_sum. exact H. rewrite plus_0_r. exact H.
Qed.

Theorem b_times2': forall n, beautiful n -> beautiful (2*n).
Proof.
  intros. simpl. apply b_sum. exact H. apply b_sum. exact H. exact b_0.
Qed.

Print b_times2.
Print b_times2'.

(** **** Exercise: 3 stars, optional (b_times2') *)
(** Write a proof object corresponding to [b_times2] above *)

Definition b_times2'': forall n, beautiful n -> beautiful (2*n) :=
  (* FILL IN HERE *)
  fun (n:nat) (b_n : beautiful n) => b_sum n (n+0) b_n (b_sum n 0 b_n b_0).

(** **** Exercise: 2 stars (b_timesm) *)
Theorem b_timesm: forall n m, beautiful n -> beautiful (m*n).
Proof.
   (* FILL IN HERE *)

(* 3*n = 2*n + 1*n  <-  b_2_mult and b_sum.
 2+z * n = 2*n + z*n  <-  b_times2'' and H. *)
intros x y H. induction H.
 Case "x=0". rewrite Mult.mult_0_r. apply b_0.
 Case "x=3". (* 3*y = sum of 3s *) induction y as [|y].
  apply b_0.
  simpl. apply (b_sum 3 (y*3) b_3 IHy).
 Case "x=5". induction y as [|y].
  apply b_0.
  simpl. apply (b_sum 5 (y*5) b_5 IHy).
 Case "x=n+m".
  rewrite Mult.mult_plus_distr_l.
  apply (b_sum (y*n) (y*m) IHbeautiful1 IHbeautiful2).
Qed.

(** [] *)

(* ####################################################### *)
(** ** Induction Over Proof Objects *)

(** Since we use the keyword [Induction] to define primitive
    propositions together with their evidence, we might wonder whether
    there are some sort of induction principles associated with these
    definitions.  Indeed there are, and in this section we'll take a
    look at how they can be used.  *)

(** Besides _constructing_ evidence that numbers are beautiful, we can
    also _reason about_ such evidence. *)

(** The fact that we introduced [beautiful] with an [Inductive]
    declaration tells us not only that the constructors [b_0], [b_3],
    [b_5] and [b_sum] are ways to build evidence, but also that these
    two constructors are the _only_ ways to build evidence that
    numbers are beautiful. *)

(** In other words, if someone gives us evidence [E] for the assertion
    [beautiful n], then we know that [E] must have one of four shapes:

      - [E] is [b_0] (and [n] is [O]),
      - [E] is [b_3] (and [n] is [3]),
      - [E] is [b_5] (and [n] is [5]), or
      - [E] is [b_sum n1 n2 E1 E2] (and [n] is [n1+n2], where [E1] is
        evidence that [n1] is beautiful and [E2] is evidence that [n2]
        is beautiful). *)

(** This gives rise to an _induction principle_ for proofs -- i.e., we
    can use the [induction] tactic that we have already seen for
    reasoning about inductively defined _data_ to reason about
    inductively defined _evidence_.

    To illustrate this, let's define another property of numbers: *)

Inductive gorgeous : nat -> Prop :=
  g_0 : gorgeous 0
| g_plus3 : forall n, gorgeous n -> gorgeous (3+n)
| g_plus5 : forall n, gorgeous n -> gorgeous (5+n).

(** **** Exercise: 1 star (gorgeous_tree) *)
(** Write out the definition of [gorgeous] numbers using inference rule
    notation.

(* FILL IN HERE *)

----------
gorgeous 0

gorgeous n
----------
gorgeous (3+n)

gorgeous n
----------
gorgeous (5+n)

[]
*)

(** It seems intuitively obvious that, although [gorgeous] and
    [beautiful] are presented using slightly different rules, they are
    actually the same property in the sense that they are true of the
    same numbers.  Indeed, we can prove this. *)

Theorem gorgeous__beautiful : forall n,
  gorgeous n -> beautiful n.
Proof.
   intros n H.
   induction H as [|n'|n'].
   Case "g_0".
       apply b_0.
   Case "g_plus3".
       apply b_sum. apply b_3.
       apply IHgorgeous.
   Case "g_plus5".
       apply b_sum. apply b_5. apply IHgorgeous.
Qed.

(* Exercise: 2 stars (recommended)
   Prove the reverse direction. *)
Theorem gorgeous_sum: forall n m, gorgeous n -> gorgeous m -> gorgeous (n + m).
Proof.
  (* FILL IN HERE *)
intros n m N. induction N as [|n|n].
   Case "0". intro H. apply H.
   Case "+ 3". intro H. apply (g_plus3 (n+m) (IHN H)). (* modus ponens *)
   Case "+ 5". intro H. apply (g_plus5 (n+m) (IHN H)).
Qed.

Theorem beautiful__gorgeous : forall n,
  beautiful n -> gorgeous n.
Proof.
  (* FILL IN HERE *)
intros n H. induction H.
 Case "0". apply g_0.
 Case "3". apply (g_plus3 0 g_0).
 Case "5". apply (g_plus5 0 g_0).
 Case "n+m". apply (gorgeous_sum n m IHbeautiful1 IHbeautiful2).
Qed.