BasicsFunctional Programming in Coq
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########################## # PLEASE DO NOT DISTRIBUTE SOLUTIONS PUBLICLY # ##########################
(See the Preface for why.)
*)
Introduction
Data and Functions
Enumerated Types
Days of the Week
The new 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.
One point 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 next check that it
    works on some examples.  There are actually three different ways
    to do the examples 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 it 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: 
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 just now, 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."
 
    Third, we can ask Coq to extract, from our Definition, a
    program in another, more conventional, programming
    language (OCaml, Scheme, or Haskell) with a high-performance
    compiler.  This facility is very interesting, since it gives us a
    path 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; 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.
 - If you introduce a helper lemma that you end up being unable to prove, hence end it with Admitted, then make sure to also end the main theorem in which you use it with Admitted, not Qed. That will help you get partial credit, in case you use that main theorem to solve a later exercise.
 
coqc -Q . LF Basics.v
coqc -Q . LF BasicsTest.v
- If you submit multiple versions of the assignment, you may notice that they are given different names. This is fine: The most recent submission is the one that will be graded.
 - To hand in multiple files at the same time (if more than one chapter is assigned in the same week), you need to make a single submission with all the files at once using the button "Add another file" just above the comment box.
 
Booleans
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.
(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.  Whenever possible,
    we'll name our own definitions and theorems so that they exactly
    coincide with the ones in the standard library.) 
 
 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 infix 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.
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. 
 
 These examples are also an opportunity to introduce one more small
    feature of Coq's programming language: conditional expressions... 
Definition negb' (b:bool) : bool :=
if b then false
else true.
Definition andb' (b1:bool) (b2:bool) : bool :=
if b1 then b2
else false.
Definition orb' (b1:bool) (b2:bool) : bool :=
if b1 then true
else b2.
Coq's conditionals are exactly like those found in any other
    language, with one small generalization.  Since the bool type
    is not built in, Coq actually supports conditional expressions over
    any inductively defined type with exactly two constructors.  The
    guard is considered true if it evaluates to the first constructor
    in the Inductive definition and false if it evaluates to the
    second. 
 
 
    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, and make sure Coq accepts it.) The
    function should return true if either or both of its inputs are
    false. 
Exercise: 1 star, standard (nandb)
The command Admitted can be used as a placeholder for an incomplete proof. We 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.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.
☐
Exercise: 1 star, standard (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.
☐
Types
If the expression after Check is followed by a colon and a type,
    Coq will verify that the type of the expression matches the given
    type and halt with an error if not. 
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. 
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, each 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).
Let's look at this in a little more detail.
 
    An Inductive definition does two things:
 
 
    Constructor expressions are formed by applying a constructor
    to zero or more other constructors or constructor expressions,
    obeying the declared number and types of the constructor arguments.
    E.g.,
 
 In particular, the definitions of rgb and color say
    which constructor expressions belong to the sets rgb and
    color:
 
 
 We can define functions on colors using pattern matching just as
    we did for day and bool. 
-  It defines a set of new constructors. E.g., red,
      primary, true, false, monday, etc. are constructors.
 - It groups them into a new named type, like bool, rgb, or color.
 
- red
 - true
 - primary red
 - etc.
 
- red primary
 - true red
 - primary (primary red)
 - etc.
 
- red, green, and blue belong to the set rgb;
 - black and white belong to the set color;
 - if p is a constructor expression belonging to the set rgb, then primary p (pronounced "the constructor primary applied to the argument p") is a constructor expression belonging to the set color; and
 - constructor 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.
Since the primary constructor takes an argument, a pattern
    matching primary should include either a variable (as above --
    note that we can choose its name freely) 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.
The pattern "primary _" here is shorthand for "the constructor
    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.) 
Modules
Module Playground.
Definition b : rgb := blue.
End Playground.
Definition b : bool := true.
Check Playground.b : rgb.
Check b : bool.
A single constructor with multiple parameters can be used
    to create a tuple type. As an example, consider representing
    the four bits in a nybble (half a byte). We first define
    a datatype bit that resembles bool (using the
    constructors B0 and B1 for the two possible bit values)
    and then define the datatype nybble, which is essentially
    a tuple of four bits. 
Inductive bit : Type :=
| B0
| B1.
Inductive nybble : Type :=
| bits (b0 b1 b2 b3 : bit).
Check (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 B0.  We
    use 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 *)
End TuplePlayground.
Numbers
All the types we have defined so far -- both "enumerated
    types" such as day, bool, and bit and tuple types such as
    nybble built from them -- are finite.  The natural numbers, on
    the other hand, are an infinite set, so we'll need to use a
    slightly richer form of type declaration to represent them.
 
    There are many representations of numbers to choose from. We are
    most familiar with decimal notation (base 10), using the digits 0
    through 9, for example, to form the number 123.  You may have
    encountered hexadecimal notation (base 16), in which the same
    number is represented as 7B, or octal (base 8), where it is 173,
    or binary (base 2), where it is 1111011. Using an enumerated type
    to represent digits, we could use any of these as our
    representation natural numbers. Indeed, there are circumstances
    where each of these choices would be useful.
 
    The binary representation is valuable in computer hardware because
    the digits can be represented with just two distinct voltage
    levels, resulting in simple circuitry. Analogously, we wish here
    to choose a representation that makes proofs simpler.
 
    In fact, there is a representation of numbers that is even simpler
    than binary, namely unary (base 1), in which only a single digit
    is used (as one might do to count days in prison by scratching on
    the walls). To represent unary numbers with a Coq datatype, we use
    two constructors. The capital-letter O constructor represents
    zero.  When the S constructor is applied to the representation
    of the natural number n, the result is the representation of
    n+1, where S stands for "successor" (or "scratch" if one is in
    prison).  Here is the complete datatype definition. 
With this definition, 0 is represented by O, 1 by S O,
    2 by S (S O), and so on. 
 
 Informally, the clauses of the 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:
 
 
 These conditions are the precise force of the Inductive
    declaration.  They imply that the constructor expression O, the
    constructor expression S O, the constructor expression S (S
    O), the constructor expression S (S (S O)), and so on all
    belong to the set nat, while other constructor expressions, 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 (remember 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.
 
- the constructor expression O belongs to the set nat;
 - if n is a constructor expression belonging to the set nat, then S n is also a constructor expression belonging to the set nat; and
 - constructor expressions formed in these two ways are the only ones belonging to the set 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: 
The second branch can be read: "if n has the form S n'
    for some n', then return n'."  
 
 The following End command closes the current module,
    so nat will refer back to the type from the standard library.
    As mentioned earlier, it comes with special notation (as decimal
    numbers) unlike the above redefinition of nat. 
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 *)
The constructor S has the type nat → nat, just like functions
    such as pred and minustwo: 
These are all things that can be applied to a number to yield a
    number.  However, there is a fundamental difference between S
    and the other two: functions like pred and minustwo are
    defined by giving 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.)
 
    Now let's go on and define some more functions over numbers.
 
    For most interesting computations involving numbers, simple
    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.  Such functions are
    introduced with the keyword Fixpoint instead of Definition. 
We could define odd by a similar Fixpoint declaration, but
    here is a simpler way: 
Definition odd (n:nat) : bool :=
negb (even n).
Example test_odd1: odd 1 = true.
Proof. simpl. reflexivity. Qed.
Example test_odd2: odd 4 = false.
Proof. simpl. reflexivity. Qed.
(You may 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 discuss 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. 
The steps of simplification that Coq performs can be
    visualized as follows: 
(* plus 3 2
i.e. 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
i.e. 5 *)
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.
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, standard (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.
☐
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) : nat.
(The level, associativity, and nat_scope annotations
    control how these notations are treated by Coq's parser.  The
    details are not important for present purposes, but interested
    readers can refer to the "More on Notation" section at the end of
    this chapter.)
 
    Note that these declarations 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.
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.
We'll be using these (especially eqb) a lot, so 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.
We now have two symbols that look like equality: = and
    =?.  We'll have much more to say about the differences and
    similarities between them later. For now, the main thing to notice
    is that x = y is a logical claim -- a "proposition" -- that we
    can try to prove, while x =? y is an expression whose
    value (either true or false) we can compute. 
 
Exercise: 1 star, standard (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 want.)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.
☐
Proof by Simplification
(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. 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 the
    standard upside-down-A symbol.)
 
    This is a good place to mention that reflexivity is a bit more
    powerful than we have acknowledged. 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: 
Moreover, it will be useful 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. Note that we could have used
    another identifier instead of n in the intros clause, (though
    of course this might be confusing to human readers of the proof): 
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
    many 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.
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. 
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, standard (plus_id_exercise)
Remove "Admitted." and fill in the proof.(* SOOMER: KK: plus_id_exercise contains multiple hypotheses, and at
least one student was confused about this. Maybe we can talk about
→ being right-associative before it. *)
Theorem plus_id_exercise : ∀ n m o : nat,
n = m → m = o → n + m = m + o.
Proof.
(* FILL IN HERE *) Admitted.
☐
Check mult_n_O.
(* ===> forall n : nat, 0 = n * 0 *)
Check mult_n_Sm.
(* ===> forall n m : nat, n * m + n = n * S m *)
We can 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_n_0_m_0 : ∀ p q : nat,
(p × 0) + (q × 0) = 0.
Proof.
intros p q.
rewrite <- mult_n_O.
rewrite <- mult_n_O.
reflexivity. Qed.
Exercise: 1 star, standard (mult_n_1)
Use those two lemmas about multiplication that we just checked to prove the following theorem. Hint: recall that 1 is S O.Proof by Case Analysis
Theorem plus_1_neq_0_firsttry : ∀ n : nat,
(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 represents, 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.
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 the two
    generated subgoals.  The part of the proof script that comes after
    a bullet is the entire proof for the corresponding 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 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, improving
    readability. 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 -- e.g., 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 a single line.  Good style lies
    somewhere in the middle.  One reasonable guideline 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.
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.  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.
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) or any repetition
    of a bullet symbol (e.g. -- or ***) as a bullet.  We can also
    enclose sub-proofs in curly braces: 
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.
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. The choice of braces,
    bullets, or a combination of the two is purely a matter of
    taste. 
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.
Exercise: 2 stars, standard (andb_true_elim2)
Prove the following claim, marking cases (and subcases) with bullets when you use destruct. Hint: delay introducing the hypothesis until after you have an opportunity to simplify it.Theorem andb_true_elim2 : ∀ b c : bool,
andb b c = true → c = true.
Proof.
(* FILL IN HERE *) Admitted.
☐
intros x y. destruct y as [|y] eqn:E.
Theorem plus_1_neq_0' : ∀ n : nat,
(n + 1) =? 0 = false.
Proof.
intros [|n].
- reflexivity.
- reflexivity. Qed.
If there are no constructor arguments that need names, we can just
    write [] to get the case analysis. 
Theorem andb_commutative'' :
∀ b c, andb b c = andb c b.
Proof.
intros [] [].
- reflexivity.
- reflexivity.
- reflexivity.
- reflexivity.
Qed.
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.
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 product
    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, 42,
    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. 
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, standard, 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: 1 star, standard (identity_fn_applied_twice)
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.
☐
Exercise: 1 star, standard (negation_fn_applied_twice)
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 *)
(* Do not modify the following line: *)
Definition manual_grade_for_negation_fn_applied_twice : option (nat×string) := None.
(The last definition is used by the autograder.)  ☐ 
 
Exercise: 3 stars, standard, 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.
☐
Exercise: 3 stars, standard (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 B0 and B1 (representing 0s and 1s), terminated by a Z. For comparison, in the unary representation, a number is a sequence of S constructors terminated by an O.
        decimal            binary                           unary
           0                       Z                              O
           1                    B1 Z                            S O
           2                B0 (B1 Z)                        S (S O)
           3                B1 (B1 Z)                     S (S (S O))
           4            B0 (B0 (B1 Z))                 S (S (S (S O)))
           5            B1 (B0 (B1 Z))              S (S (S (S (S O))))
           6            B0 (B1 (B1 Z))           S (S (S (S (S (S O)))))
           7            B1 (B1 (B1 Z))        S (S (S (S (S (S (S O))))))
           8        B0 (B0 (B0 (B1 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. 
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.
The following "unit tests" of your increment and binary-to-unary
    functions should pass after you have defined those functions correctly.
    Of course, unit tests don't fully demonstrate the correctness of
    your functions!  We'll return to that thought at the end of the
    next chapter. 
Example test_bin_incr1 : (incr (B1 Z)) = B0 (B1 Z).
(* FILL IN HERE *) Admitted.
Example test_bin_incr2 : (incr (B0 (B1 Z))) = B1 (B1 Z).
(* FILL IN HERE *) Admitted.
Example test_bin_incr3 : (incr (B1 (B1 Z))) = B0 (B0 (B1 Z)).
(* FILL IN HERE *) Admitted.
Example test_bin_incr4 : bin_to_nat (B0 (B1 Z)) = 2.
(* FILL IN HERE *) Admitted.
Example test_bin_incr5 :
bin_to_nat (incr (B1 Z)) = 1 + bin_to_nat (B1 Z).
(* FILL IN HERE *) Admitted.
Example test_bin_incr6 :
bin_to_nat (incr (incr (B1 Z))) = 2 + bin_to_nat (B1 Z).
(* FILL IN HERE *) Admitted.
☐
Testing Your Solutions
coqc -Q . LF Basics.v
coqc -Q . LF BasicsTest.v
-  First will be all the output produced by Basics.v itself.  At
      the end of that you will see COQC BasicsTest.v.
 -  Second, for each required exercise, there is a report that tells
      you its point value (the number of stars or some fraction
      thereof if there are multiple parts to the exercise), whether
      its type is ok, and what assumptions it relies upon.
If the type is not ok, it means you proved the wrong thing: most likely, you accidentally modified the theorem statement while you were proving it. The autograder won't give you any points for that, so make sure to correct the theorem.The assumptions are any unproved theorems which your solution relies upon. "Closed under the global context" is a fancy way of saying "none": you have solved the exercise. (Hooray!) On the other hand, a list of axioms means you haven't fully solved the exercise. (But see below regarding "Allowed Axioms.") If the exercise name itself is in the list, that means you haven't solved it; probably you have Admitted it.
 -  Third, you will see the maximum number of points in standard and
      advanced versions of the assignment.  That number is based on
      the number of stars in the non-optional exercises.
 -  Fourth, you will see a list of "Allowed Axioms".  These are
      unproved theorems that your solution is permitted to depend
      upon.  You'll probably see something about
      functional_extensionality for this chapter; we'll cover what
      that means in a later chapter.
 - Finally, you will see a summary of whether you have solved each exercise. Note that summary does not include the critical information of whether the type is ok (that is, whether you accidentally changed the theorem statement): you have to look above for that information.
 
(* 2021-03-18 17:23 *)
