PolyPolymorphism and Higher-Order Functions

In this chapter we continue our development of basic concepts of functional programming. The critical new ideas are polymorphism (abstracting functions over the types of the data they manipulate) and higher-order functions (treating functions as data).

Require Export Lists.

Polymorphism

Polymorphic Lists

For the last couple of chapters, we've been working just with lists of numbers. Obviously, interesting programs also need to be able to manipulate lists with elements from other types — lists of strings, lists of booleans, lists of lists, etc. We could just define a new inductive datatype for each of these, for example...

Inductive boollist : Type :=
| bool_nil : boollist
| bool_cons : bool boollist boollist.

... but this would quickly become tedious, partly because we have to make up different constructor names for each datatype, but mostly because we would also need to define new versions of all our list manipulating functions (length, rev, etc.) for each new datatype definition.
To avoid all this repetition, Coq supports polymorphic inductive type definitions. For example, here is a polymorphic list datatype.

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

This is exactly like the definition of natlist from the previous chapter, except that the nat argument to the cons constructor has been replaced by an arbitrary type X, a binding for X has been added to the header, and the occurrences of natlist in the types of the constructors have been replaced by list X. (We can re-use the constructor names nil and cons because the earlier definition of natlist was inside of a Module definition that is now out of scope.)
What sort of thing is list itself? One good way to think about it is that list is a function from Types to Inductive definitions; or, to put it another way, list is a function from Types to Types. For any particular type X, the type list X is an Inductively defined set of lists whose elements are things of type X.
With this definition, when we use the constructors nil and cons to build lists, we need to tell Coq the type of the elements in the lists we are building — that is, nil and cons are now polymorphic constructors. Observe the types of these constructors:

Check nil.
(* ===> nil : forall X : Type, list X *)
Check cons.
(* ===> cons : forall X : Type, X -> list X -> list X *)

The " X" in these types can be read as an additional argument to the constructors that determines the expected types of the arguments that follow. When nil and cons are used, these arguments are supplied in the same way as the others. For example, the list containing 2 and 1 is written like this:

Check (cons nat 2 (cons nat 1 (nil nat))).

(We've gone back to writing nil and cons explicitly here because we haven't yet defined the [] and :: notations for the new version of lists. We'll do that in a bit.)
We can now go back and make polymorphic (or "generic") versions of all the list-processing functions that we wrote before. Here is length, for example:

Fixpoint length (X:Type) (l:list X) : nat :=
match l with
| nil ⇒ 0
| cons h tS (length X t)
end.

Note that the uses of nil and cons in match patterns do not require any type annotations: Coq already knows that the list l contains elements of type X, so there's no reason to include X in the pattern. (More precisely, the type X is a parameter of the whole definition of list, not of the individual constructors. We'll come back to this point later.)
As with nil and cons, we can use length by applying it first to a type and then to its list argument:

Example test_length1 :
length nat (cons nat 1 (cons nat 2 (nil nat))) = 2.
Proof. reflexivity. Qed.

To use our length with other kinds of lists, we simply instantiate it with an appropriate type parameter:

Example test_length2 :
length bool (cons bool true (nil bool)) = 1.
Proof. reflexivity. Qed.

Let's close this subsection by re-implementing a few other standard list functions on our new polymorphic lists:

Fixpoint app (X : Type) (l1 l2 : list X)
: (list X) :=
match l1 with
| nill2
| cons h tcons X h (app X t l2)
end.

Fixpoint snoc (X:Type) (l:list X) (v:X) : (list X) :=
match l with
| nilcons X v (nil X)
| cons h tcons X h (snoc X t v)
end.

Fixpoint rev (X:Type) (l:list X) : list X :=
match l with
| nilnil X
| cons h tsnoc X (rev X t) h
end.

Example test_rev1 :
rev nat (cons nat 1 (cons nat 2 (nil nat)))
= (cons nat 2 (cons nat 1 (nil nat))).
Proof. reflexivity. Qed.

Example test_rev2:
rev bool (nil bool) = nil bool.
Proof. reflexivity. Qed.

Module MumbleBaz.

Exercise: 2 stars (mumble_grumble)

Consider the following two inductively defined types.

Inductive mumble : Type :=
| a : mumble
| b : mumble nat mumble
| c : mumble.
Inductive grumble (X:Type) : Type :=
| d : mumble grumble X
| e : X grumble X.

Which of the following are well-typed elements of grumble X for some type X?
• d (b a 5)
• d mumble (b a 5)
• d bool (b a 5)
• e bool true
• e mumble (b c 0)
• e bool (b c 0)
• c
(* FILL IN HERE *)

Exercise: 2 stars (baz_num_elts)

Consider the following inductive definition:

Inductive baz : Type :=
| x : baz baz
| y : baz bool baz.

How many elements does the type baz have? (* FILL IN HERE *)

End MumbleBaz.

Type Annotation Inference

Let's write the definition of app again, but this time we won't specify the types of any of the arguments. Will Coq still accept it?

Fixpoint app' X l1 l2 : list X :=
match l1 with
| nill2
| cons h tcons X h (app' X t l2)
end.

Indeed it will. Let's see what type Coq has assigned to app':

Check app'.
(* ===> forall X : Type, list X -> list X -> list X *)
Check app.
(* ===> forall X : Type, list X -> list X -> list X *)

It has exactly the same type type as app. Coq was able to use a process called type inference to deduce what the types of X, l1, and l2 must be, based on how they are used. For example, since X is used as an argument to cons, it must be a Type, since cons expects a Type as its first argument; matching l1 with nil and cons means it must be a list; and so on.
This powerful facility means we don't always have to write explicit type annotations everywhere, although explicit type annotations are still quite useful as documentation and sanity checks. You should try to find a balance in your own code between too many type annotations (so many that they clutter and distract) and too few (which forces readers to perform type inference in their heads in order to understand your code).

Type Argument Synthesis

Whenever we use a polymorphic function, we need to pass it one or more types in addition to its other arguments. For example, the recursive call in the body of the length function above must pass along the type X. But just like providing explicit type annotations everywhere, this is heavy and verbose. Since the second argument to length is a list of Xs, it seems entirely obvious that the first argument can only be X — why should we have to write it explicitly?
Fortunately, Coq permits us to avoid this kind of redundancy. In place of any type argument we can write the "implicit argument" _, which can be read as "Please figure out for yourself what type belongs here." More precisely, when Coq encounters a _, it will attempt to unify all locally available information — the type of the function being applied, the types of the other arguments, and the type expected by the context in which the application appears — to determine what concrete type should replace the _.
This may sound similar to type annotation inference — and, indeed, the two procedures rely on the same underlying mechanisms. Instead of simply omitting the types of some arguments to a function, like
app' X l1 l2 : list X :=
we can also replace the types with _, like
app' (X : _) (l1 l2 : _) : list X :=
which tells Coq to attempt to infer the missing information, just as with argument synthesis.
Using implicit arguments, the length function can be written like this:

Fixpoint length' (X:Type) (l:list X) : nat :=
match l with
| nil ⇒ 0
| cons h tS (length' _ t)
end.

In this instance, we don't save much by writing _ instead of X. But in many cases the difference can be significant. For example, suppose we want to write down a list containing the numbers 1, 2, and 3. Instead of writing this...

Definition list123 :=
cons nat 1 (cons nat 2 (cons nat 3 (nil nat))).

...we can use argument synthesis to write this:

Definition list123' := cons _ 1 (cons _ 2 (cons _ 3 (nil _))).

Implicit Arguments

If fact, we can go further. To avoid having to sprinkle _'s throughout our programs, we can tell Coq always to infer the type argument(s) of a given function. The Arguments directive specifies the name of the function or constructor, and then lists its argument names, with curly braces around any arguments to be treated as implicit.

Arguments nil {X}.
Arguments cons {X} _ _. (* use underscore for argument position that has no name *)
Arguments length {X} l.
Arguments app {X} l1 l2.
Arguments rev {X} l.
Arguments snoc {X} l v.

(* note: no _ arguments required... *)
Definition list123'' := cons 1 (cons 2 (cons 3 nil)).
Check (length list123'').

Alternatively, we can declare an argument to be implicit while defining the function itself, by surrounding the argument in curly braces. For example:

Fixpoint length'' {X:Type} (l:list X) : nat :=
match l with
| nil ⇒ 0
| cons h tS (length'' t)
end.

(Note that we didn't even have to provide a type argument to the recursive call to length''; indeed, it is invalid to provide one.) We will use this style whenever possible, although we will continue to use use explicit Argument declarations for Inductive constructors.
One small problem with declaring arguments Implicit is that, occasionally, Coq does not have enough local information to determine a type argument; in such cases, we need to tell Coq that we want to give the argument explicitly this time, even though we've globally declared it to be Implicit. For example, suppose we write this:

(* Definition mynil := nil.  *)

If we uncomment this definition, Coq will give us an error, because it doesn't know what type argument to supply to nil. We can help it by providing an explicit type declaration (so that Coq has more information available when it gets to the "application" of nil):

Definition mynil : list nat := nil.

Alternatively, we can force the implicit arguments to be explicit by prefixing the function name with @.

Check @nil.

Definition mynil' := @nil nat.

Using argument synthesis and implicit arguments, we can define convenient notation for lists, as before. Since we have made the constructor type arguments implicit, Coq will know to automatically infer these when we use the notations.

Notation "x :: y" := (cons x y)
(at level 60, right associativity).
Notation "[ ]" := nil.
Notation "[ x ; .. ; y ]" := (cons x .. (cons y []) ..).
Notation "x ++ y" := (app x y)
(at level 60, right associativity).

Now lists can be written just the way we'd hope:

Definition list123''' := [1; 2; 3].

Check ([3 + 4] ++ nil).

Exercises: Polymorphic Lists

Exercise: 2 stars, optional (poly_exercises)

Here are a few simple exercises, just like ones in the Lists chapter, for practice with polymorphism. Fill in the definitions and complete the proofs below.

Fixpoint repeat {X : Type} (n : X) (count : nat) : list X :=
(* FILL IN HERE *) admit.

Example test_repeat1:
repeat true 2 = cons true (cons true nil).
(* FILL IN HERE *) Admitted.

Theorem nil_app : X:Type, l:list X,
app [] l = l.
Proof.
(* FILL IN HERE *) Admitted.

Theorem rev_snoc : X : Type,
v : X,
s : list X,
rev (snoc s v) = v :: (rev s).
Proof.
(* FILL IN HERE *) Admitted.

Theorem rev_involutive : X : Type, l : list X,
rev (rev l) = l.
Proof.
(* FILL IN HERE *) Admitted.

Theorem snoc_with_append : X : Type,
l1 l2 : list X,
v : X,
snoc (l1 ++ l2) v = l1 ++ (snoc l2 v).
Proof.
(* FILL IN HERE *) Admitted.

Polymorphic Pairs

Following the same pattern, the type definition we gave in the last chapter for pairs of numbers can be generalized to polymorphic pairs (or products):

Inductive prod (X Y : Type) : Type :=
pair : X Y prod X Y.

Arguments pair {X} {Y} _ _.

As with lists, we make the type arguments implicit and define the familiar concrete notation.

Notation "( x , y )" := (pair x y).

We can also use the Notation mechanism to define the standard notation for pair types:

Notation "X × Y" := (prod X Y) : type_scope.

(The annotation : type_scope tells Coq that this abbreviation should be used when parsing types. This avoids a clash with the multiplication symbol.)
A note of caution: it is easy at first to get (x,y) and X×Y confused. Remember that (x,y) is a value built from two other values; X×Y is a type built from two other types. If x has type X and y has type Y, then (x,y) has type X×Y.
The first and second projection functions now look pretty much as they would in any functional programming language.

Definition fst {X Y : Type} (p : X × Y) : X :=
match p with (x,y) ⇒ x end.

Definition snd {X Y : Type} (p : X × Y) : Y :=
match p with (x,y) ⇒ y end.

The following function takes two lists and combines them into a list of pairs. In many functional programming languages, it is called zip. We call it combine for consistency with Coq's standard library. Note that the pair notation can be used both in expressions and in patterns...

Fixpoint combine {X Y : Type} (lx : list X) (ly : list Y)
: list (X×Y) :=
match (lx,ly) with
| ([],_) ⇒ []
| (_,[]) ⇒ []
| (x::tx, y::ty) ⇒ (x,y) :: (combine tx ty)
end.

Exercise: 1 star, optional (combine_checks)

• What is the type of combine (i.e., what does Check @combine print?)
• What does
Eval compute in (combine [1;2] [false;false;true;true]).
print?

Exercise: 2 stars (split)

The function split is the right inverse of combine: it takes a list of pairs and returns a pair of lists. In many functional programing languages, this function is called unzip.
Uncomment the material below and fill in the definition of split. Make sure it passes the given unit tests.

Fixpoint split
{X Y : Type} (l : list (X×Y))
: (list X) × (list Y) :=
(* FILL IN HERE *) admit.

Example test_split:
split [(1,false);(2,false)] = ([1;2],[false;false]).
Proof.
(* FILL IN HERE *) Admitted.

Polymorphic Options

One last polymorphic type for now: polymorphic options. The type declaration generalizes the one for natoption in the previous chapter:

Inductive option (X:Type) : Type :=
| Some : X option X
| None : option X.

Arguments Some {X} _.
Arguments None {X}.

We can now rewrite the index function so that it works with any type of lists.

Fixpoint index {X : Type} (n : nat)
(l : list X) : option X :=
match l with
| [] ⇒ None
| a :: l'if beq_nat n O then Some a else index (pred n) l'
end.

Example test_index1 : index 0 [4;5;6;7] = Some 4.
Proof. reflexivity. Qed.
Example test_index2 : index 1 [[1];[2]] = Some [2].
Proof. reflexivity. Qed.
Example test_index3 : index 2 [true] = None.
Proof. reflexivity. Qed.

Exercise: 1 star, optional (hd_opt_poly)

Complete the definition of a polymorphic version of the hd_opt function from the last chapter. Be sure that it passes the unit tests below.

Definition hd_opt {X : Type} (l : list X) : option X :=
(* FILL IN HERE *) admit.

Once again, to force the implicit arguments to be explicit, we can use @ before the name of the function.

Check @hd_opt.

Example test_hd_opt1 : hd_opt [1;2] = Some 1.
(* FILL IN HERE *) Admitted.
Example test_hd_opt2 : hd_opt [[1];[2]] = Some [1].
(* FILL IN HERE *) Admitted.

Functions as Data

Higher-Order Functions

Like many other modern programming languages — including all functional languages (ML, Haskell, Scheme, etc.) — Coq treats functions as first-class citizens, allowing functions to be passed as arguments to other functions, returned as results, stored in data structures, etc.
Functions that manipulate other functions are often called higher-order functions. Here's a simple one:

Definition doit3times {X:Type} (f:XX) (n:X) : X :=
f (f (f n)).

The argument f here is itself a function (from X to X); the body of doit3times applies f three times to some value n.

Check @doit3times.
(* ===> doit3times : forall X : Type, (X -> X) -> X -> X *)

Example test_doit3times: doit3times minustwo 9 = 3.
Proof. reflexivity. Qed.

Example test_doit3times': doit3times negb true = false.
Proof. reflexivity. Qed.

Partial Application

In fact, the multiple-argument functions we have already seen are also examples of passing functions as data. To see why, recall the type of plus.

Check plus.
(* ==> nat -> nat -> nat *)

Each in this expression is actually a binary operator on types. (This is the same as saying that Coq primitively supports only one-argument functions — do you see why?) This operator is right-associative, so the type of plus is really a shorthand for nat (nat nat) — i.e., it can be read as saying that "plus is a one-argument function that takes a nat and returns a one-argument function that takes another nat and returns a nat." In the examples above, we have always applied plus to both of its arguments at once, but if we like we can supply just the first. This is called partial application.

Definition plus3 := plus 3.
Check plus3.

Example test_plus3 : plus3 4 = 7.
Proof. reflexivity. Qed.
Example test_plus3' : doit3times plus3 0 = 9.
Proof. reflexivity. Qed.
Example test_plus3'' : doit3times (plus 3) 0 = 9.
Proof. reflexivity. Qed.

Digression: Currying

In Coq, a function f : A B C really has the type A (B C). That is, if you give f a value of type A, it will give you function f' : B C. If you then give f' a value of type B, it will return a value of type C. This allows for partial application, as in plus3. Processing a list of arguments with functions that return functions is called currying, in honor of the logician Haskell Curry.
Conversely, we can reinterpret the type A B C as (A × B) C. This is called uncurrying. With an uncurried binary function, both arguments must be given at once as a pair; there is no partial application.
We can define currying as follows:

Definition prod_curry {X Y Z : Type}
(f : X × Y Z) (x : X) (y : Y) : Z := f (x, y).

As an exercise, define its inverse, prod_uncurry. Then prove the theorems below to show that the two are inverses.

Definition prod_uncurry {X Y Z : Type}
(f : X Y Z) (p : X × Y) : Z :=
(* FILL IN HERE *) admit.

(Thought exercise: before running these commands, can you calculate the types of prod_curry and prod_uncurry?)

Check @prod_curry.
Check @prod_uncurry.

Theorem uncurry_curry : (X Y Z : Type) (f : X Y Z) x y,
prod_curry (prod_uncurry f) x y = f x y.
Proof.
(* FILL IN HERE *) Admitted.

Theorem curry_uncurry : (X Y Z : Type)
(f : (X × Y) Z) (p : X × Y),
prod_uncurry (prod_curry f) p = f p.
Proof.
(* FILL IN HERE *) Admitted.

Filter

Here is a useful higher-order function, which takes a list of Xs and a predicate on X (a function from X to bool) and "filters" the list, returning a new list containing just those elements for which the predicate returns true.

Fixpoint filter {X:Type} (test: Xbool) (l:list X)
: (list X) :=
match l with
| [] ⇒ []
| h :: tif test h then h :: (filter test t)
else filter test t
end.

For example, if we apply filter to the predicate evenb and a list of numbers l, it returns a list containing just the even members of l.

Example test_filter1: filter evenb [1;2;3;4] = [2;4].
Proof. reflexivity. Qed.

Definition length_is_1 {X : Type} (l : list X) : bool :=
beq_nat (length l) 1.

Example test_filter2:
filter length_is_1
[ [1; 2]; [3]; [4]; [5;6;7]; []; [8] ]
= [ [3]; [4]; [8] ].
Proof. reflexivity. Qed.

We can use filter to give a concise version of the countoddmembers function from the Lists chapter.

Definition countoddmembers' (l:list nat) : nat :=
length (filter oddb l).

Example test_countoddmembers'1: countoddmembers' [1;0;3;1;4;5] = 4.
Proof. reflexivity. Qed.
Example test_countoddmembers'2: countoddmembers' [0;2;4] = 0.
Proof. reflexivity. Qed.
Example test_countoddmembers'3: countoddmembers' nil = 0.
Proof. reflexivity. Qed.

Anonymous Functions

It is a little annoying to be forced to define the function length_is_1 and give it a name just to be able to pass it as an argument to filter, since we will probably never use it again. Moreover, this is not an isolated example. When using higher-order functions, we often want to pass as arguments "one-off" functions that we will never use again; having to give each of these functions a name would be tedious.
Fortunately, there is a better way. It is also possible to construct a function "on the fly" without declaring it at the top level or giving it a name; this is analogous to the notation we've been using for writing down constant lists, natural numbers, and so on.

Example test_anon_fun':
doit3times (fun nn × n) 2 = 256.
Proof. reflexivity. Qed.

Here is the motivating example from before, rewritten to use an anonymous function.

Example test_filter2':
filter (fun lbeq_nat (length l) 1)
[ [1; 2]; [3]; [4]; [5;6;7]; []; [8] ]
= [ [3]; [4]; [8] ].
Proof. reflexivity. Qed.

Exercise: 2 stars (filter_even_gt7)

Use filter (instead of Fixpoint) to write a Coq function filter_even_gt7 that takes a list of natural numbers as input and returns a list of just those that are even and greater than 7.

Definition filter_even_gt7 (l : list nat) : list nat :=
(* FILL IN HERE *) admit.

Example test_filter_even_gt7_1 :
filter_even_gt7 [1;2;6;9;10;3;12;8] = [10;12;8].
(* FILL IN HERE *) Admitted.

Example test_filter_even_gt7_2 :
filter_even_gt7 [5;2;6;19;129] = [].
(* FILL IN HERE *) Admitted.

Exercise: 3 stars (partition)

Use filter to write a Coq function partition:
partition : X : Type,
(X  bool list X  list X × list X
Given a set X, a test function of type X bool and a list X, partition should return a pair of lists. The first member of the pair is the sublist of the original list containing the elements that satisfy the test, and the second is the sublist containing those that fail the test. The order of elements in the two sublists should be the same as their order in the original list.

Definition partition {X : Type} (test : X bool) (l : list X)
: list X × list X :=
(* FILL IN HERE *) admit.

Example test_partition1: partition oddb [1;2;3;4;5] = ([1;3;5], [2;4]).
(* FILL IN HERE *) Admitted.
Example test_partition2: partition (fun xfalse) [5;9;0] = ([], [5;9;0]).
(* FILL IN HERE *) Admitted.

Map

Another handy higher-order function is called map.

Fixpoint map {X Y:Type} (f:XY) (l:list X)
: (list Y) :=
match l with
| [] ⇒ []
| h :: t ⇒ (f h) :: (map f t)
end.

It takes a function f and a list l = [n1, n2, n3, ...] and returns the list [f n1, f n2, f n3,...] , where f has been applied to each element of l in turn. For example:

Example test_map1: map (plus 3) [2;0;2] = [5;3;5].
Proof. reflexivity. Qed.

The element types of the input and output lists need not be the same (map takes two type arguments, X and Y). This version of map can thus be applied to a list of numbers and a function from numbers to booleans to yield a list of booleans:

Example test_map2: map oddb [2;1;2;5] = [false;true;false;true].
Proof. reflexivity. Qed.

It can even be applied to a list of numbers and a function from numbers to lists of booleans to yield a list of lists of booleans:

Example test_map3:
map (fun n ⇒ [evenb n;oddb n]) [2;1;2;5]
= [[true;false];[false;true];[true;false];[false;true]].
Proof. reflexivity. Qed.

Map for options

Exercise: 3 stars (map_rev)

Show that map and rev commute. You may need to define an auxiliary lemma.

Theorem map_rev : (X Y : Type) (f : X Y) (l : list X),
map f (rev l) = rev (map f l).
Proof.
(* FILL IN HERE *) Admitted.

Exercise: 2 stars (flat_map)

The function map maps a list X to a list Y using a function of type X Y. We can define a similar function, flat_map, which maps a list X to a list Y using a function f of type X list Y. Your definition should work by 'flattening' the results of f, like so:
flat_map (fun n ⇒ [n;n+1;n+2]) [1;5;10]
= [1; 2; 3; 5; 6; 7; 10; 11; 12].

Fixpoint flat_map {X Y:Type} (f:X list Y) (l:list X)
: (list Y) :=
(* FILL IN HERE *) admit.

Example test_flat_map1:
flat_map (fun n ⇒ [n;n;n]) [1;5;4]
= [1; 1; 1; 5; 5; 5; 4; 4; 4].
(* FILL IN HERE *) Admitted.
Lists are not the only inductive type that we can write a map function for. Here is the definition of map for the option type:

Definition option_map {X Y : Type} (f : X Y) (xo : option X)
: option Y :=
match xo with
| NoneNone
| Some xSome (f x)
end.

Exercise: 2 stars, optional (implicit_args)

The definitions and uses of filter and map use implicit arguments in many places. Replace the curly braces around the implicit arguments with parentheses, and then fill in explicit type parameters where necessary and use Coq to check that you've done so correctly. (This exercise is not to be turned in; it is probably easiest to do it on a copy of this file that you can throw away afterwards.)

Fold

An even more powerful higher-order function is called fold. This function is the inspiration for the "reduce" operation that lies at the heart of Google's map/reduce distributed programming framework.

Fixpoint fold {X Y:Type} (f: XYY) (l:list X) (b:Y) : Y :=
match l with
| nilb
| h :: tf h (fold f t b)
end.

Intuitively, the behavior of the fold operation is to insert a given binary operator f between every pair of elements in a given list. For example, fold plus [1;2;3;4] intuitively means 1+2+3+4. To make this precise, we also need a "starting element" that serves as the initial second input to f. So, for example,
fold plus [1;2;3;4] 0
yields
1 + (2 + (3 + (4 + 0))).
Here are some more examples:

Check (fold andb).
(* ===> fold andb : list bool -> bool -> bool *)

Example fold_example1 : fold mult [1;2;3;4] 1 = 24.
Proof. reflexivity. Qed.

Example fold_example2 : fold andb [true;true;false;true] true = false.
Proof. reflexivity. Qed.

Example fold_example3 : fold app [[1];[];[2;3];[4]] [] = [1;2;3;4].
Proof. reflexivity. Qed.

Observe that the type of fold is parameterized by two type variables, X and Y, and the parameter f is a binary operator that takes an X and a Y and returns a Y. Can you think of a situation where it would be useful for X and Y to be different?

Functions For Constructing Functions

Most of the higher-order functions we have talked about so far take functions as arguments. Now let's look at some examples involving returning functions as the results of other functions.
To begin, here is a function that takes a value x (drawn from some type X) and returns a function from nat to X that yields x whenever it is called, ignoring its nat argument.

Definition constfun {X: Type} (x: X) : natX :=
fun (k:nat) ⇒ x.

Definition ftrue := constfun true.

Example constfun_example1 : ftrue 0 = true.
Proof. reflexivity. Qed.

Example constfun_example2 : (constfun 5) 99 = 5.
Proof. reflexivity. Qed.

Similarly, but a bit more interestingly, here is a function that takes a function f from numbers to some type X, a number k, and a value x, and constructs a function that behaves exactly like f except that, when called with the argument k, it returns x.

Definition override {X: Type} (f: natX) (k:nat) (x:X) : natX:=
fun (k':nat) ⇒ if beq_nat k k' then x else f k'.

For example, we can apply override twice to obtain a function from numbers to booleans that returns false on 1 and 3 and returns true on all other arguments.

Definition fmostlytrue := override (override ftrue 1 false) 3 false.

Example override_example1 : fmostlytrue 0 = true.
Proof. reflexivity. Qed.

Example override_example2 : fmostlytrue 1 = false.
Proof. reflexivity. Qed.

Example override_example3 : fmostlytrue 2 = true.
Proof. reflexivity. Qed.

Example override_example4 : fmostlytrue 3 = false.
Proof. reflexivity. Qed.

Exercise: 1 star (override_example)

Before starting to work on the following proof, make sure you understand exactly what the theorem is saying and can paraphrase it in your own words. The proof itself is straightforward.

Theorem override_example : (b:bool),
(override (constfun b) 3 true) 2 = b.
Proof.
(* FILL IN HERE *) Admitted.
We'll use function overriding heavily in parts of the rest of the course, and we will end up needing to know quite a bit about its properties. To prove these properties, though, we need to know about a few more of Coq's tactics; developing these is the main topic of the next chapter. For now, though, let's introduce just one very useful tactic that will also help us with proving properties of some of the other functions we have introduced in this chapter.

The unfold Tactic

Sometimes, a proof will get stuck because Coq doesn't automatically expand a function call into its definition. (This is a feature, not a bug: if Coq automatically expanded everything possible, our proof goals would quickly become enormous — hard to read and slow for Coq to manipulate!)

3 + n = m
plus3 n + 1 = m + 1.
Proof.
intros m n H.
(* At this point, we'd like to do rewrite H, since
plus3 n is definitionally equal to 3 + n.  However,
Coq doesn't automatically expand plus3 n to its
definition. *)

Abort.

The unfold tactic can be used to explicitly replace a defined name by the right-hand side of its definition.

Theorem unfold_example : m n,
3 + n = m
plus3 n + 1 = m + 1.
Proof.
intros m n H.
unfold plus3.
rewrite H.
reflexivity. Qed.

Now we can prove a first property of override: If we override a function at some argument k and then look up k, we get back the overridden value.

Theorem override_eq : {X:Type} x k (f:natX),
(override f k x) k = x.
Proof.
intros X x k f.
unfold override.
rewrite beq_nat_refl.
reflexivity. Qed.

This proof was straightforward, but note that it requires unfold to expand the definition of override.

Exercise: 2 stars (override_neq)

Theorem override_neq : (X:Type) x1 x2 k1 k2 (f : natX),
f k1 = x1
beq_nat k2 k1 = false
(override f k2 x2) k1 = x1.
Proof.
(* FILL IN HERE *) Admitted.
As the inverse of unfold, Coq also provides a tactic fold, which can be used to "unexpand" a definition. It is used much less often.

Exercise: 2 stars (fold_length)

Many common functions on lists can be implemented in terms of fold. For example, here is an alternative definition of length:

Definition fold_length {X : Type} (l : list X) : nat :=
fold (fun _ nS n) l 0.

Example test_fold_length1 : fold_length [4;7;0] = 3.
Proof. reflexivity. Qed.

Prove the correctness of fold_length.

Theorem fold_length_correct : X (l : list X),
fold_length l = length l.
(* FILL IN HERE *) Admitted.

Exercise: 3 stars (fold_map)

We can also define map in terms of fold. Finish fold_map below.

Definition fold_map {X Y:Type} (f : X Y) (l : list X) : list Y :=
(* FILL IN HERE *) admit.

Write down a theorem in Coq stating that fold_map is correct, and prove it.

(* FILL IN HERE *)

(* \$Date: 2014-08-25 13:26:26 -0400 (Mon, 25 Aug 2014) \$ *)