BasicsFunctional Programming in Coq
Introduction
Enumerated Types
Days of the Week
Inductive day : Type :=
| monday
| tuesday
| wednesday
| thursday
| friday
| saturday
| sunday.
| monday
| tuesday
| wednesday
| thursday
| friday
| saturday
| sunday.
The type is called day, and its members are monday,
    tuesday, 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.
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 *)
(* ==> 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.
(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
- 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).
- It is fine to use additional definitions (of helper functions, useful lemmas, etc.) in your solutions. You can put these between the exercise header and the theorem you are asked to prove.
       coqc -Q . LF Basics.v 
coqc -Q . LF BasicsTest.v 
    There is no need to hand in BasicsTest.v itself (or Preface.v).
coqc -Q . LF BasicsTest.v
Booleans
Inductive bool : Type :=
| true
| false.
| true
| false.
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.
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.
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 Notation command defines a new
    symbolic notation for an existing definition. 
Notation "x && y" := (andb x y).
Notation "x || y" := (orb x y).
Example test_orb5: false || false || true = true.
Proof. simpl. reflexivity. Qed.
Notation "x || y" := (orb x y).
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
    Admitteds 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. (I.e., 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
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Example test_nandb1: (nandb true false) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb2: (nandb false false) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb3: (nandb false true) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb4: (nandb true true) = false.
(* FILL IN HERE *) Admitted.
☐ 
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Example test_nandb1: (nandb true false) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb2: (nandb false false) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb3: (nandb false true) = true.
(* FILL IN HERE *) Admitted.
Example test_nandb4: (nandb true true) = false.
(* FILL IN HERE *) Admitted.
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
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Example test_andb31: (andb3 true true true) = true.
(* FILL IN HERE *) Admitted.
Example test_andb32: (andb3 false true true) = false.
(* FILL IN HERE *) Admitted.
Example test_andb33: (andb3 true false true) = false.
(* FILL IN HERE *) Admitted.
Example test_andb34: (andb3 true true false) = false.
(* FILL IN HERE *) Admitted.
☐ 
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Example test_andb31: (andb3 true true true) = true.
(* FILL IN HERE *) Admitted.
Example test_andb32: (andb3 false true true) = false.
(* FILL IN HERE *) Admitted.
Example test_andb33: (andb3 true false true) = false.
(* FILL IN HERE *) Admitted.
Example test_andb34: (andb3 true true false) = false.
(* FILL IN HERE *) Admitted.
Types
Check true.
(* ===> true : bool *)
Check (negb true).
(* ===> negb true : bool *)
(* ===> 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 *)
(* ===> 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." 
New Types from Old
Inductive rgb : Type :=
| red
| green
| blue.
Inductive color : Type :=
| black
| white
| primary (p : rgb).
| red
| green
| blue.
Inductive color : Type :=
| black
| white
| primary (p : rgb).
Let's look at this in a little more detail.
 
    Every inductively defined type (day, bool, rgb, color,
    etc.) contains a set of constructor expressions built from
    constructors like red, primary, true, false, monday,
    etc.  The definitions of rgb and color say how expressions in the
    sets rgb and color can be built:
 
 
 We can define functions on colors using pattern matching just as
    we have done for day and bool. 
- red, green, and blue are the constructors of rgb;
- black, white, and primary are the constructors of color;
- the expression red belongs to the set rgb, as do the expressions green and blue;
- the expressions black and white belong to the set color;
- if p is an expression belonging to the set rgb, then primary p (pronounced "the constructor primary applied to the argument p") is an expression belonging to the set color; and
- expressions formed in these ways are the only ones belonging to the sets rgb and color.
Definition monochrome (c : color) : bool :=
match c with
| black ⇒ true
| white ⇒ true
| primary p ⇒ false
end.
match c with
| black ⇒ true
| white ⇒ true
| primary p ⇒ false
end.
Since the primary constructor takes an argument, a pattern
    matching primary should include either a variable (as above) or
    a constant of appropriate type (as below). 
Definition isred (c : color) : bool :=
match c with
| black ⇒ false
| white ⇒ false
| primary red ⇒ true
| primary _ ⇒ false
end.
match c with
| black ⇒ false
| white ⇒ false
| primary red ⇒ true
| primary _ ⇒ false
end.
The pattern primary _ here is shorthand for "primary applied
    to any rgb constructor except red."  (The wildcard pattern _
    has the same effect as the dummy pattern variable p in the
    definition of monochrome.) 
Tuples
Inductive bit : Type :=
| B0
| B1.
Inductive nybble : Type :=
| bits (b0 b1 b2 b3 : bit).
Check (bits B1 B0 B1 B0).
(* ==> bits B1 B0 B1 B0 : nybble *)
| B0
| B1.
Inductive nybble : Type :=
| bits (b0 b1 b2 b3 : bit).
Check (bits B1 B0 B1 B0).
(* ==> bits B1 B0 B1 B0 : nybble *)
The bits constructor acts as a wrapper for its contents.
    Unwrapping can be done by pattern-matching, as in the all_zero
    function which tests a nybble to see if all its bits are O.
    Note that we are using underscore (_) as a wildcard pattern to
    avoid inventing variable names that will not be used.
Definition all_zero (nb : nybble) : bool :=
match nb with
| (bits B0 B0 B0 B0) ⇒ true
| (bits _ _ _ _) ⇒ false
end.
Compute (all_zero (bits B1 B0 B1 B0)).
(* ===> false : bool *)
Compute (all_zero (bits B0 B0 B0 B0)).
(* ===> true : bool *)
match nb with
| (bits B0 B0 B0 B0) ⇒ true
| (bits _ _ _ _) ⇒ false
end.
Compute (all_zero (bits B1 B0 B1 B0)).
(* ===> false : bool *)
Compute (all_zero (bits B0 B0 B0 B0)).
(* ===> true : bool *)
Modules
Module NatPlayground.
Numbers
Inductive nat : Type :=
| O
| S (n : nat).
| O
| S (n : nat).
With this definition, 0 is represented by O, 1 by S O,
    2 by S (S O), and so on. 
 
 The clauses of this definition can be read:
 
 
 Again, let's look at this in a little more detail.  The definition
    of nat says how expressions in the set nat can be built:
 
 
 The same rules apply for our definitions of day, bool,
    color, etc.
 
    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: 
- 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.
- 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.
Inductive nat' : Type :=
| stop
| tick (foo : nat').
| stop
| tick (foo : 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.
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.
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 decimal numerals can be used as an alternative to
    the "unary" notation defined by the constructors S and O.  Coq
    prints numbers in decimal form by default: 
Check (S (S (S (S O)))).
(* ===> 4 : nat *)
Definition minustwo (n : nat) : nat :=
match n with
| O ⇒ O
| S O ⇒ O
| S (S n') ⇒ n'
end.
Compute (minustwo 4).
(* ===> 2 : nat *)
(* ===> 4 : nat *)
Definition minustwo (n : nat) : nat :=
match n with
| O ⇒ O
| S O ⇒ O
| S (S n') ⇒ n'
end.
Compute (minustwo 4).
(* ===> 2 : nat *)
The constructor S has the type nat → nat, just like
    pred and functions like minustwo: 
Check S.
Check pred.
Check minustwo.
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 decimal 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.
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.
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.
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
*)
==> 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.
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.
End NatPlayground2.
Fixpoint exp (base power : nat) : nat :=
match power with
| O ⇒ S O
| S p ⇒ mult base (exp base p)
end.
match n, m with
| O , _ ⇒ O
| S _ , O ⇒ n
| S n', S m' ⇒ minus n' m'
end.
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
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Example test_factorial1: (factorial 3) = 6.
(* FILL IN HERE *) Admitted.
Example test_factorial2: (factorial 5) = (mult 10 12).
(* FILL IN HERE *) Admitted.
☐ 
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Example test_factorial1: (factorial 3) = 6.
(* FILL IN HERE *) Admitted.
Example test_factorial2: (factorial 5) = (mult 10 12).
(* FILL IN HERE *) Admitted.
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).
(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 "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 is a user-defined operation!
 
    Here is a function eqb, which tests natural numbers for
    equality, yielding a boolean.  Note the use of nested
    matches (we could also have used a simultaneous match, as we did
    in minus.) 
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.
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.
Similarly, 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.
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.
Since we'll be using these (especially eqb) a lot, let's give
    them infix notations. 
Notation "x =? y" := (eqb x y) (at level 70) : nat_scope.
Notation "x <=? y" := (leb x y) (at level 70) : nat_scope.
Example test_leb3': (4 <=? 2) = false.
Proof. simpl. reflexivity. Qed.
Notation "x <=? y" := (leb x y) (at level 70) : nat_scope.
Example test_leb3': (4 <=? 2) = false.
Proof. simpl. reflexivity. Qed.
Exercise: 1 star (ltb)
The ltb function tests natural numbers for less-than, yielding a boolean. Instead of making up a new Fixpoint for this one, define it in terms of a previously defined function. (It can be done with just one previously defined function, but you can use two if you need to.)
Definition ltb (n m : nat) : bool
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Notation "x <? y" := (ltb x y) (at level 70) : nat_scope.
Example test_ltb1: (ltb 2 2) = false.
(* FILL IN HERE *) Admitted.
Example test_ltb2: (ltb 2 4) = true.
(* FILL IN HERE *) Admitted.
Example test_ltb3: (ltb 4 2) = false.
(* FILL IN HERE *) Admitted.
☐ 
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Notation "x <? y" := (ltb x y) (at level 70) : nat_scope.
Example test_ltb1: (ltb 2 2) = false.
(* FILL IN HERE *) Admitted.
Example test_ltb2: (ltb 2 4) = true.
(* FILL IN HERE *) Admitted.
Example test_ltb3: (ltb 4 2) = false.
(* FILL IN HERE *) Admitted.
Proof by Simplification
Theorem plus_O_n : ∀ n : nat, 0 + n = n.
Proof.
intros n. simpl. reflexivity. Qed.
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
    ∀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' : ∀ n : nat, 0 + n = n.
Proof.
intros n. reflexivity. Qed.
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 ∀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 : ∀ n:nat, 1 + n = S n.
Proof.
intros n. reflexivity. Qed.
Theorem mult_0_l : ∀ n:nat, 0 * n = 0.
Proof.
intros n. reflexivity. Qed.
Proof.
intros n. reflexivity. Qed.
Theorem mult_0_l : ∀ 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. 
Theorem plus_id_example : ∀ n m:nat,
n = m →
n + n = m + m.
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.
(* 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 : ∀ n m o : nat,
n = m → m = o → n + m = m + o.
Proof.
(* FILL IN HERE *) Admitted.
☐ 
n = m → m = o → n + m = m + o.
Proof.
(* FILL IN HERE *) Admitted.
Theorem mult_0_plus : ∀ n m : nat,
(0 + n) * m = n * m.
Proof.
intros n m.
rewrite → plus_O_n.
reflexivity. Qed.
(0 + n) * m = n * m.
Proof.
intros n m.
rewrite → plus_O_n.
reflexivity. Qed.
Theorem mult_S_1 : ∀ n m : nat,
m = S n →
m * (1 + n) = m * m.
Proof.
(* FILL IN HERE *) Admitted.
(* (N.b. This proof can actually be completed with tactics other than
rewrite, but please do use rewrite for the sake of the exercise.) *)
☐ 
m = S n →
m * (1 + n) = m * m.
Proof.
(* FILL IN HERE *) Admitted.
(* (N.b. This proof can actually be completed with tactics other than
rewrite, but please do use rewrite for the sake of the exercise.) *)
Proof by Case Analysis
Theorem plus_1_neq_0_firsttry : ∀ n : nat,
(n + 1) =? 0 = false.
Proof.
intros n.
simpl. (* does nothing! *)
Abort.
(n + 1) =? 0 = false.
Proof.
intros n.
simpl. (* does nothing! *)
Abort.
The reason for this is that the definitions of both
    eqb 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 eqb 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 (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, (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 : ∀ n : nat,
(n + 1) =? 0 = false.
Proof.
intros n. destruct n as [| n'] eqn:E.
- reflexivity.
- reflexivity. Qed.
(n + 1) =? 0 = false.
Proof.
intros n. destruct n as [| n'] eqn:E.
- 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.
 
    In each subgoal, Coq remembers the assumption about n that is
    relevant for this subgoal — either n = 0 or n = S n' for some
    n'.  The eqn:E annotation tells destruct to give the name E to
    this equation.  (Leaving off the eqn:E annotation causes Coq to
    elide these assumptions in the subgoals.  This slightly
    streamlines proofs where the assumptions are not explicitly used,
    but it is better practice to keep them for the sake of
    documentation, as they can help keep you oriented when working
    with the subgoals.)
 
    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 second one
    simplifies (S n' + 1) =? 0 to false by first rewriting (S n'
    + 1) to S (n' + 1), then unfolding eqb, 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 : ∀ b : bool,
negb (negb b) = b.
Proof.
intros b. destruct b eqn:E.
- reflexivity.
- reflexivity. Qed.
negb (negb b) = b.
Proof.
intros b. destruct b eqn:E.
- 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 : ∀ 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.
Proof.
intros b c. destruct b eqn:Eb.
- destruct c eqn:Ec.
+ reflexivity.
+ reflexivity.
- destruct c eqn:Ec.
+ 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' : ∀ 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.
Proof.
intros b c. destruct b eqn:Eb.
{ destruct c eqn:Ec.
{ reflexivity. }
{ reflexivity. } }
{ destruct c eqn:Ec.
{ 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 :
∀ 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.
∀ 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.
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.  (You'll also note one downside of this shorthand: we lose
    the equation recording the assumption we are making in each
    subgoal, which we previously got from the eqn:E annotation.) 
Theorem plus_1_neq_0' : ∀ n : nat,
(n + 1) =? 0 = false.
Proof.
intros [|n].
- reflexivity.
- reflexivity. Qed.
(n + 1) =? 0 = false.
Proof.
intros [|n].
- reflexivity.
- reflexivity. Qed.
If there are no arguments to name, we can just write []. 
Theorem andb_commutative'' :
∀ b c, andb b c = andb c b.
Proof.
intros [] [].
- reflexivity.
- reflexivity.
- reflexivity.
- reflexivity.
Qed.
∀ 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 : ∀ b c : bool,
andb b c = true → c = true.
Proof.
(* FILL IN HERE *) Admitted.
☐ 
andb b c = true → c = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem zero_nbeq_plus_1 : ∀ n : nat,
0 =? (n + 1) = false.
Proof.
(* FILL IN HERE *) Admitted.
☐ 
0 =? (n + 1) = false.
Proof.
(* FILL IN HERE *) Admitted.
More on Notation (Optional)
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.
(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! 
Fixpoint plus' (n : nat) (m : nat) : nat :=
match n with
| O ⇒ m
| S n' ⇒ S (plus' n' m)
end.
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. (If you choose to turn in this optional exercise as part of a homework assignment, make sure you comment out your solution so that it doesn't cause Coq to reject the whole file!)
(* FILL IN HERE *)
☐ 
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 :
∀ (f : bool → bool),
(∀ (x : bool), f x = x) →
∀ (b : bool), f (f b) = b.
Proof.
(* FILL IN HERE *) Admitted.
∀ (f : bool → bool),
(∀ (x : bool), f x = x) →
∀ (b : bool), f (f b) = b.
Proof.
(* FILL IN HERE *) Admitted.
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.
(* FILL IN HERE *)
(* The Import statement on the next line tells Coq to use the
standard library String module. We'll use strings more in later
chapters, but for the moment we just need syntax for literal
strings for the grader comments. *)
From Coq Require Export String.
(* Do not modify the following line: *)
Definition manual_grade_for_negation_fn_applied_twice : option (prod nat string) := None.
☐ 
(* The Import statement on the next line tells Coq to use the
standard library String module. We'll use strings more in later
chapters, but for the moment we just need syntax for literal
strings for the grader comments. *)
From Coq Require Export String.
(* Do not modify the following line: *)
Definition manual_grade_for_negation_fn_applied_twice : option (prod nat string) := None.
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.)
Theorem andb_eq_orb :
∀ (b c : bool),
(andb b c = orb b c) →
b = c.
Proof.
(* FILL IN HERE *) Admitted.
☐ 
∀ (b c : bool),
(andb b c = orb b c) →
b = c.
Proof.
(* FILL IN HERE *) Admitted.
Exercise: 3 stars (binary)
We can generalize our unary representation of natural numbers to the more efficient binary representation by treating a binary number as a sequence of constructors A and B (representing 0s and 1s), terminated by a Z. For comparison, in the unary representation, a number is a sequence of Ss terminated by an O.
        decimal            binary                           unary
0 Z O
1 B Z S O
2 A (B Z) S (S O)
3 B (B Z) S (S (S O))
4 A (A (B Z)) S (S (S (S O)))
5 B (A (B Z)) S (S (S (S (S O))))
6 A (B (B Z)) S (S (S (S (S (S O)))))
7 B (B (B Z)) S (S (S (S (S (S (S O))))))
8 A (A (A (B Z))) S (S (S (S (S (S (S (S O))))))) 
    Note that the low-order bit is on the left and the high-order bit
    is on the right — the opposite of the way binary numbers are
    usually written.  This choice makes them easier to manipulate. 
0 Z O
1 B Z S O
2 A (B Z) S (S O)
3 B (B Z) S (S (S O))
4 A (A (B Z)) S (S (S (S O)))
5 B (A (B Z)) S (S (S (S (S O))))
6 A (B (B Z)) S (S (S (S (S (S O)))))
7 B (B (B Z)) S (S (S (S (S (S (S O))))))
8 A (A (A (B Z))) S (S (S (S (S (S (S (S O)))))))
Inductive bin : Type :=
| Z
| A (n : bin)
| B (n : bin).
| Z
| A (n : bin)
| B (n : bin).
(a) Complete the definitions below of an increment function incr
        for binary numbers, and a function bin_to_nat to convert
        binary numbers to unary numbers. 
Fixpoint incr (m:bin) : bin
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Fixpoint bin_to_nat (m:bin) : nat
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Fixpoint bin_to_nat (m:bin) : nat
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
   (b) 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. 
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_binary : option (prod nat string) := None.
☐ 
(* Do not modify the following line: *)
Definition manual_grade_for_binary : option (prod nat string) := None.
