# PolyPolymorphism and Higher-Order Functions

(* \$Date: 2013-01-04 20:13:58 -0500 (Fri, 04 Jan 2013) \$ *)

Require Export Lists.

# Polymorphism

## Polymorphic Lists

Up to this point, we've been working with lists of numbers. Of course, interesting programs also need to be able to manipulate lists whose elements are drawn 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like the definition of natlist from the previous chapter, except that the nat argument to the cons constructor has been replaced by an arbitrary type X, a binding for X has been added to the header, and the occurrences of natlist in the types of the constructors have been replaced by list X. (We can re-use the constructor names nil and cons because the earlier definition of natlist was inside of a Module definition that is now out of scope.)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 *)

(Side note on notation: In .v files, the "forall" quantifier is spelled out in letters. In the generated HTML files, is usually typeset as the usual mathematical "upside down A," but you'll see the spelled-out "forall" in a few places. This is just a quirk of typesetting: there is no difference in meaning.
The " X" in these types should 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 t => S (length X t)
end.

Note that the uses of nil and cons in match patterns do not require any type annotations: we already know that the list l contains elements of type X, so there's no reason to include X in the pattern. (More formally, 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
| nil => l2
| cons h t => cons X h (app X t l2)
end.

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

Fixpoint rev (X:Type) (l:list X) : list X :=
match l with
| nil => nil X
| cons h t => snoc 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.

### 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
| nil => l2
| cons h t => cons X h (app' X t l2)
end.

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

Check app'.
Check app.

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 t => S (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.

Implicit Arguments nil [[X]].
Implicit Arguments cons [[X]].
Implicit Arguments length [[X]].
Implicit Arguments app [[X]].
Implicit Arguments rev [[X]].
Implicit Arguments snoc [[X]].

(* 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 t => S (length'' t)
end.

(Note that we didn't even have to provide a type argument to the recursive call to length''.) We will use this style whenever possible, although we will continue to use use explicit Implicit 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, if we write:

(* 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].

### 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 bool 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.

Implicit 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.

Indeed, when no ambiguity results, we can even drop the enclosing parens:

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 simpl in (combine [1,2] [false,false,true,true]).
print?

#### Exercise: 2 stars, recommended (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
(* FILL IN HERE *)

Example test_split:
split (1,false),(2,false) = (1,2,false,false).
Proof. reflexivity.  Qed.
*)

(If you're reading the HTML version of this file, note that there's an unresolved typesetting problem in the example: several square brackets are missing. Refer to the .v file for the correct version.

## 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.

Implicit Arguments Some [[X]].
Implicit 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

#### Exercise: 2 stars, optional (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 :: t => if 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 n => n * n) 2 = 256.
Proof. reflexivity. Qed.

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

Example test_filter2':
filter (fun l => beq_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 which takes a list of natural numbers as input and keeps only those numbers which 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 x => false) [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.

#### Exercise: 3 stars, optional (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, recommended (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
| None => None
| Some x => Some (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
| nil => b
| h :: t => f 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 plus).
Eval simpl in (fold plus [1,2,3,4] 0).

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.

#### Exercise: 1 star, optional (fold_types_different)

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 rest of the chapter.

# Optional Material

This section needs more text!
Recall the definition of lists of booleans:
Inductive boollist : Type :=
boolnil  : boollist
boolcons : bool  boollist  boollist.
We saw how it could be generalized to "polymorphic lists" with elements of an arbitrary type X. Here's another way of generalizing it: an inductive family of "length-indexed" lists of booleans:

Inductive boolllist : nat Type :=
boollnil : boolllist O
| boollcons : n, bool boolllist n boolllist (S n).

Implicit Arguments boollcons [[n]].

Check (boollcons true (boollcons false (boollcons true boollnil))).

Fixpoint blapp {n1} (l1: boolllist n1)
{n2} (l2: boolllist n2)
: boolllist (n1 + n2) :=
match l1 with
| boollnil => l2
| boollcons _ h t => boollcons h (blapp t l2)
end.

Of course, these generalizions can be combined. Here's the length-indexed polymorphic version:

Inductive llist (X:Type) : nat Type :=
lnil : llist X O
| lcons : n, X llist X n llist X (S n).

Implicit Arguments lnil [[X]].
Implicit Arguments lcons [[X] [n]].

Check (lcons true (lcons false (lcons true lnil))).

Fixpoint lapp (X:Type)
{n1} (l1: llist X n1)
{n2} (l2: llist X n2)
: llist X (n1 + n2) :=
match l1 with
| lnil => l2
| lcons _ h t => lcons h (lapp X t l2)
end.

## The apply Tactic

We often encounter situations where the goal to be proved is exactly the same as some hypothesis in the context or some previously proved lemma.

Theorem silly1 : (n m o p : nat),
n = m
[n,o] = [n,p]
[n,o] = [m,p].
Proof.
intros n m o p eq1 eq2.
rewrite eq1.
(* At this point, we could finish with "rewrite eq2. reflexivity."
as we have done several times above. But we can achieve the
same effect in a single step by using the apply tactic instead: *)

apply eq2. Qed.

The apply tactic also works with conditional hypotheses and lemmas: if the statement being applied is an implication, then the premises of this implication will be added to the list of subgoals needing to be proved.

Theorem silly2 : (n m o p : nat),
n = m
((q r : nat), q = r [q,o] = [r,p])
[n,o] = [m,p].
Proof.
intros n m o p eq1 eq2.
apply eq2. apply eq1. Qed.

You may find it instructive to experiment with this proof and see if there is a way to complete it using just rewrite instead of apply.
Typically, when we use apply H, the statement H will begin with a binding some universal variables. When Coq matches the current goal against the conclusion of H, it will try to find appropriate values for these variables. For example, when we do apply eq2 in the following proof, the universal variable q in eq2 gets instantiated with n and r gets instantiated with m.

Theorem silly2a : (n m : nat),
(n,n) = (m,m)
((q r : nat), (q,q) = (r,r) [q] = [r])
[n] = [m].
Proof.
intros n m eq1 eq2.
apply eq2. apply eq1. Qed.

#### Exercise: 2 stars, optional (silly_ex)

Complete the following proof without using simpl.

Theorem silly_ex :
(n, evenb n = true oddb (S n) = true)
evenb 3 = true
oddb 4 = true.
Proof.
(* FILL IN HERE *) Admitted.
To use the apply tactic, the (conclusion of the) fact being applied must match the goal exactly — for example, apply will not work if the left and right sides of the equality are swapped.

Theorem silly3_firsttry : (n : nat),
true = beq_nat n 5
beq_nat (S (S n)) 7 = true.
Proof.
intros n H.
simpl.
(* Here we cannot use apply directly *)

In this case we can use the symmetry tactic, which switches the left and right sides of an equality in the goal.

Theorem silly3 : (n : nat),
true = beq_nat n 5
beq_nat (S (S n)) 7 = true.
Proof.
intros n H.
symmetry.
simpl. (* Actually, this simpl is unnecessary, since
apply will do a simpl step first. *)

apply H. Qed.

#### Exercise: 3 stars, recommended (apply_exercise1)

Theorem rev_exercise1 : (l l' : list nat),
l = rev l'
l' = rev l.
Proof.
(* Hint: you can use apply with previously defined lemmas, not
just hypotheses in the context.  Remember that SearchAbout is
your friend. *)

(* FILL IN HERE *) Admitted.

#### Exercise: 1 star (apply_rewrite)

Briefly explain the difference between the tactics apply and rewrite. Are there situations where both can usefully be applied?
(* FILL IN HERE *)

## 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. *)

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.

## Inversion

Recall the definition of natural numbers:
Inductive nat : Type :=
| O : nat
| S : nat  nat.
It is clear from this definition that every number has one of two forms: either it is the constructor O or it is built by applying the constructor S to another number. But there is more here than meets the eye: implicit in the definition (and in our informal understanding of how datatype declarations work in other programming languages) are two other facts:
• The constructor S is injective. That is, the only way we can have S n = S m is if n = m.
• The constructors O and S are disjoint. That is, O is not equal to S n for any n.
Similar principles apply to all inductively defined types: all constructors are injective, and the values built from distinct constructors are never equal. For lists, the cons constructor is injective and nil is different from every non-empty list. For booleans, true and false are unequal. (Since neither true nor false take any arguments, their injectivity is not an issue.)
Coq provides a tactic, called inversion, that allows us to exploit these principles in making proofs.
The inversion tactic is used like this. Suppose H is a hypothesis in the context (or a previously proven lemma) of the form
c a1 a2 ... an = d b1 b2 ... bm
for some constructors c and d and arguments a1 ... an and b1 ... bm. Then inversion H instructs Coq to "invert" this equality to extract the information it contains about these terms:
• If c and d are the same constructor, then we know, by the injectivity of this constructor, that a1 = b1, a2 = b2, etc.; inversion H adds these facts to the context, and tries to use them to rewrite the goal.
• If c and d are different constructors, then the hypothesis H is contradictory. That is, a false assumption has crept into the context, and this means that any goal whatsoever is provable! In this case, inversion H marks the current goal as completed and pops it off the goal stack.
The inversion tactic is probably easier to understand by seeing it in action than from general descriptions like the above. Below you will find example theorems that demonstrate the use of inversion and exercises to test your understanding.

Theorem eq_add_S : (n m : nat),
S n = S m
n = m.
Proof.
intros n m eq. inversion eq. reflexivity. Qed.

Theorem silly4 : (n m : nat),
[n] = [m]
n = m.
Proof.
intros n o eq. inversion eq. reflexivity. Qed.

As a convenience, the inversion tactic can also destruct equalities between complex values, binding multiple variables as it goes.

Theorem silly5 : (n m o : nat),
[n,m] = [o,o]
[n] = [m].
Proof.
intros n m o eq. inversion eq. reflexivity. Qed.

#### Exercise: 1 star (sillyex1)

Example sillyex1 : (X : Type) (x y z : X) (l j : list X),
x :: y :: l = z :: j
y :: l = x :: j
x = y.
Proof.
(* FILL IN HERE *) Admitted.

Theorem silly6 : (n : nat),
S n = O
2 + 2 = 5.
Proof.
intros n contra. inversion contra. Qed.

Theorem silly7 : (n m : nat),
false = true
[n] = [m].
Proof.
intros n m contra. inversion contra. Qed.

#### Exercise: 1 star (sillyex2)

Example sillyex2 : (X : Type) (x y z : X) (l j : list X),
x :: y :: l = []
y :: l = z :: j
x = z.
Proof.
(* FILL IN HERE *) Admitted.
While the injectivity of constructors allows us to reason (n m : nat), S n = S m n = m, the reverse direction of the implication, provable by standard equational reasoning, is a useful fact to record for cases we will see several times.

Lemma eq_remove_S : n m,
n = m S n = S m.
Proof. intros n m eq. rewrite eq. reflexivity. Qed.

Here's another illustration of inversion. This is a slightly roundabout way of stating a fact that we have already proved above. The extra equalities force us to do a little more equational reasoning and exercise some of the tactics we've seen recently.

Theorem length_snoc' : (X : Type) (v : X)
(l : list X) (n : nat),
length l = n
length (snoc l v) = S n.
Proof.
intros X v l. induction l as [| v' l'].
Case "l = []". intros n eq. rewrite eq. reflexivity.
Case "l = v' :: l'". intros n eq. simpl. destruct n as [| n'].
SCase "n = 0". inversion eq.
SCase "n = S n'".
apply eq_remove_S. apply IHl'. inversion eq. reflexivity. Qed.

## Varying the Induction Hypothesis

Here is a more realistic use of inversion to prove a property that is useful in many places later on...

Theorem beq_nat_eq_FAILED : n m,
true = beq_nat n m n = m.
Proof.
intros n m H. induction n as [| n'].
Case "n = 0".
destruct m as [| m'].
SCase "m = 0". reflexivity.
SCase "m = S m'". simpl in H. inversion H.
Case "n = S n'".
destruct m as [| m'].
SCase "m = 0". simpl in H. inversion H.
SCase "m = S m'".
apply eq_remove_S.
(* stuck here because the induction hypothesis
talks about an extremely specific m *)

The inductive proof above fails because we've set up things so that the induction hypothesis (in the second subgoal generated by the induction tactic) is
true = beq_nat n' m n' = m .
This hypothesis makes a statement about n' together with the particular natural number m — that is, the number m, which was introduced into the context by the intros at the top of the proof, is "held constant" in the induction hypothesis. This induction hypothesis is not strong enough to make the induction step of the proof go through.
If we set up the proof slightly differently by introducing just n into the context at the top, then we get an induction hypothesis that makes a stronger claim:
m : nat, true = beq_nat n' m n' = m
Setting up the induction hypothesis this way makes the proof of beq_nat_eq go through:

Theorem beq_nat_eq : n m,
true = beq_nat n m n = m.
Proof.
intros n. induction n as [| n'].
Case "n = 0".
intros m. destruct m as [| m'].
SCase "m = 0". reflexivity.
SCase "m = S m'". simpl. intros contra. inversion contra.
Case "n = S n'".
intros m. destruct m as [| m'].
SCase "m = 0". simpl. intros contra. inversion contra.
SCase "m = S m'". simpl. intros H.
apply eq_remove_S. apply IHn'. apply H. Qed.

Similar issues will come up in many of the proofs below. If you ever find yourself in a situation where the induction hypothesis is insufficient to establish the goal, consider going back and doing fewer intros to make the IH stronger.

#### Exercise: 2 stars (beq_nat_eq_informal)

Give an informal proof of beq_nat_eq.

(* FILL IN HERE *)

#### Exercise: 3 stars (beq_nat_eq')

We can also prove beq_nat_eq by induction on m, though we have to be a little careful about which order we introduce the variables, so that we get a general enough induction hypothesis — this is done for you below. Finish the following proof. To get maximum benefit from the exercise, try first to do it without looking back at the one above.

Theorem beq_nat_eq' : m n,
beq_nat n m = true n = m.
Proof.
intros m. induction m as [| m'].
(* FILL IN HERE *) Admitted.

### Practice Session

#### Exercise: 2 stars, optional (practice)

Some nontrivial but not-too-complicated proofs to work together in class, and some for you to work as exercises. Some of the exercises may involve applying lemmas from earlier lectures or homeworks.

Theorem beq_nat_0_l : n,
true = beq_nat 0 n 0 = n.
Proof.
(* FILL IN HERE *) Admitted.

Theorem beq_nat_0_r : n,
true = beq_nat 0 n 0 = n.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars (apply_exercise2)

In the following proof opening, notice that we don't introduce m before performing induction. This leaves it general, so that the IH doesn't specify a particular m, but lets us pick. Finish the proof.

Theorem beq_nat_sym : (n m : nat),
beq_nat n m = beq_nat m n.
Proof.
intros n. induction n as [| n'].
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars (beq_nat_sym_informal)

Provide an informal proof of this lemma that corresponds to your formal proof above:
Theorem: For any nats n m, beq_nat n m = beq_nat m n.
Proof: (* FILL IN HERE *)

## Using Tactics on Hypotheses

By default, most tactics work on the goal formula and leave the context unchanged. However, most tactics also have a variant that performs a similar operation on a statement in the context.
For example, the tactic simpl in H performs simplification in the hypothesis named H in the context.

Theorem S_inj : (n m : nat) (b : bool),
beq_nat (S n) (S m) = b
beq_nat n m = b.
Proof.
intros n m b H. simpl in H. apply H. Qed.

Similarly, the tactic apply L in H matches some conditional statement L (of the form L1 L2, say) against a hypothesis H in the context. However, unlike ordinary apply (which rewrites a goal matching L2 into a subgoal L1), apply L in H matches H against L1 and, if successful, replaces it with L2.
In other words, apply L in H gives us a form of "forward reasoning" — from L1 L2 and a hypothesis matching L1, it gives us a hypothesis matching L2. By contrast, apply L is "backward reasoning" — it says that if we know L1L2 and we are trying to prove L2, it suffices to prove L1.
Here is a variant of a proof from above, using forward reasoning throughout instead of backward reasoning.

Theorem silly3' : (n : nat),
(beq_nat n 5 = true beq_nat (S (S n)) 7 = true)
true = beq_nat n 5
true = beq_nat (S (S n)) 7.
Proof.
intros n eq H.
symmetry in H. apply eq in H. symmetry in H.
apply H. Qed.

Forward reasoning starts from what is given (premises, previously proven theorems) and iteratively draws conclusions from them until the goal is reached. Backward reasoning starts from the goal, and iteratively reasons about what would imply the goal, until premises or previously proven theorems are reached. If you've seen informal proofs before (for example, in a math or computer science class), they probably used forward reasoning. In general, Coq tends to favor backward reasoning, but in some situations the forward style can be easier to use or to think about.

#### Exercise: 3 stars, recommended (plus_n_n_injective)

You can practice using the "in" variants in this exercise.

Theorem plus_n_n_injective : n m,
n + n = m + m
n = m.
Proof.
intros n. induction n as [| n'].
(* Hint: use the plus_n_Sm lemma *)
(* FILL IN HERE *) Admitted.

## Using destruct on Compound Expressions

We have seen many examples where the destruct tactic is used to perform case analysis of the value of some variable. But sometimes we need to reason by cases on the result of some expression. We can also do this with destruct.
Here are some examples:

Definition sillyfun (n : nat) : bool :=
if beq_nat n 3 then false
else if beq_nat n 5 then false
else false.

Theorem sillyfun_false : (n : nat),
sillyfun n = false.
Proof.
intros n. unfold sillyfun.
destruct (beq_nat n 3).
Case "beq_nat n 3 = true". reflexivity.
Case "beq_nat n 3 = false". destruct (beq_nat n 5).
SCase "beq_nat n 5 = true". reflexivity.
SCase "beq_nat n 5 = false". reflexivity. Qed.

After unfolding sillyfun in the above proof, we find that we are stuck on if (beq_nat n 3) then ... else .... Well, either n is equal to 3 or it isn't, so we use destruct (beq_nat n 3) to let us reason about the two cases.

Theorem override_shadow : {X:Type} x1 x2 k1 k2 (f : natX),
(override (override f k1 x2) k1 x1) k2 = (override f k1 x1) k2.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, recommended (combine_split)

(*
Theorem combine_split : forall X Y (l : list (X * Y)) l1 l2,
split l = (l1, l2) ->
combine l1 l2 = l.
Proof.
intros X Y l. induction l as | [x y] l'.
(* FILL IN HERE *) Admitted.
*)

#### Exercise: 3 stars, optional (split_combine)

Thought exercise: We have just proven that for all lists of pairs, combine is the inverse of split. How would you state the theorem showing that split is the inverse of combine?
Hint: what property do you need of l1 and l2 for split combine l1 l2 = (l1,l2) to be true?
State this theorem in Coq, and prove it. (Be sure to leave your induction hypothesis general by not doing intros on more things than necessary.)

(* FILL IN HERE *)

## The remember Tactic

(Note: the remember tactic is not strictly needed until a bit later, so if necessary this section can be skipped and returned to when needed.)
We have seen how the destruct tactic can be used to perform case analysis of the results of arbitrary computations. If e is an expression whose type is some inductively defined type T, then, for each constructor c of T, destruct e generates a subgoal in which all occurrences of e (in the goal and in the context) are replaced by c.
Sometimes, however, this substitution process loses information that we need in order to complete the proof. For example, suppose we define a function sillyfun1 like this:

Definition sillyfun1 (n : nat) : bool :=
if beq_nat n 3 then true
else if beq_nat n 5 then true
else false.

And suppose that we want to convince Coq of the rather obvious observation that sillyfun1 n yields true only when n is odd. By analogy with the proofs we did with sillyfun above, it is natural to start the proof like this:

Theorem sillyfun1_odd_FAILED : (n : nat),
sillyfun1 n = true
oddb n = true.
Proof.
intros n eq. unfold sillyfun1 in eq.
destruct (beq_nat n 3).
(* stuck... *)

We get stuck at this point because the context does not contain enough information to prove the goal! The problem is that the substitution peformed by destruct is too brutal — it threw away every occurrence of beq_nat n 3, but we need to keep at least one of these because we need to be able to reason that since, in this branch of the case analysis, beq_nat n 3 = true, it must be that n = 3, from which it follows that n is odd.
What we would really like is not to use destruct directly on beq_nat n 3 and substitute away all occurrences of this expression, but rather to use destruct on something else that is equal to beq_nat n 3. For example, if we had a variable that we knew was equal to beq_nat n 3, we could destruct this variable instead.
The remember tactic allows us to introduce such a variable.

Theorem sillyfun1_odd : (n : nat),
sillyfun1 n = true
oddb n = true.
Proof.
intros n eq. unfold sillyfun1 in eq.
remember (beq_nat n 3) as e3.
(* At this point, the context has been enriched with a new
variable e3 and an assumption that e3 = beq_nat n 3.
Now if we do destruct e3... *)

destruct e3.
(* ... the variable e3 gets substituted away (it
disappears completely) and we are left with the same
state as at the point where we got stuck above, except
that the context still contains the extra equality
assumption -- now with true substituted for e3 --
which is exactly what we need to make progress. *)

Case "e3 = true". apply beq_nat_eq in Heqe3.
rewrite Heqe3. reflexivity.
Case "e3 = false".
(* When we come to the second equality test in the
body of the function we are reasoning about, we can
use remember again in the same way, allowing us
to finish the proof. *)

remember (beq_nat n 5) as e5. destruct e5.
SCase "e5 = true".
apply beq_nat_eq in Heqe5.
rewrite Heqe5. reflexivity.
SCase "e5 = false". inversion eq. Qed.

#### Exercise: 2 stars

Theorem bool_fn_applied_thrice :
(f : bool bool) (b : bool),
f (f (f b)) = f b.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars (override_same)

Theorem override_same : {X:Type} x1 k1 k2 (f : natX),
f k1 = x1
(override f k1 x1) k2 = f k2.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, optional (filter_exercise)

This one is a bit challenging. Be sure your initial intros go only up through the parameter on which you want to do induction!

Theorem filter_exercise : (X : Type) (test : X bool)
(x : X) (l lf : list X),
filter test l = x :: lf
test x = true.
Proof.
(* FILL IN HERE *) Admitted.

## The apply...with... Tactic

The following silly example uses two rewrites in a row to get from [a,b] to [e,f].

Example trans_eq_example : (a b c d e f : nat),
[a,b] = [c,d]
[c,d] = [e,f]
[a,b] = [e,f].
Proof.
intros a b c d e f eq1 eq2.
rewrite eq1. rewrite eq2. reflexivity. Qed.

Since this is a common pattern, we might abstract it out as a lemma recording once and for all the fact that equality is transitive.

Theorem trans_eq : {X:Type} (n m o : X),
n = m m = o n = o.
Proof.
intros X n m o eq1 eq2. rewrite eq1. rewrite eq2.
reflexivity. Qed.

Now, we should be able to use trans_eq to prove the above example. However, to do this we need a slight refinement of the apply tactic.

Example trans_eq_example' : (a b c d e f : nat),
[a,b] = [c,d]
[c,d] = [e,f]
[a,b] = [e,f].
Proof.
intros a b c d e f eq1 eq2.
(* If we simply tell Coq apply trans_eq at this point,
it can tell (by matching the goal against the
conclusion of the lemma) that it should instantiate X
with [nat]n with [a,b], and o with [e,f].
However, the matching process doesn't determine an
instantiation for m: we have to supply one explicitly
by adding with (m:=[c,d]) to the invocation of
apply. *)

apply trans_eq with (m:=[c,d]). apply eq1. apply eq2. Qed.

Actually, we usually don't have to include the name m in the with clause; Coq is often smart enough to figure out which instantiation we're giving. We could instead write: apply trans_eq with c,d.

#### Exercise: 3 stars, recommended (apply_exercises)

Example trans_eq_exercise : (n m o p : nat),
m = (minustwo o)
(n + p) = m
(n + p) = (minustwo o).
Proof.
(* FILL IN HERE *) Admitted.

Theorem beq_nat_trans : n m p,
true = beq_nat n m
true = beq_nat m p
true = beq_nat n p.
Proof.
(* FILL IN HERE *) Admitted.

Theorem override_permute : {X:Type} x1 x2 k1 k2 k3 (f : natX),
false = beq_nat k2 k1
(override (override f k2 x2) k1 x1) k3 = (override (override f k1 x1) k2 x2) k3.
Proof.
(* FILL IN HERE *) Admitted.

# Review

We've now seen a bunch of Coq's fundamental tactics — enough to do pretty much everything we'll want for a while. We'll introduce one or two more as we go along through the next few lectures, and later in the course we'll introduce some more powerful automation tactics that make Coq do more of the low-level work in many cases. But basically we've got what we need to get work done.
Here are the ones we've seen:
• intros: move hypotheses/variables from goal to context
• reflexivity: finish the proof (when the goal looks like e = e)
• apply: prove goal using a hypothesis, lemma, or constructor
• apply... in H: apply a hypothesis, lemma, or constructor to a hypothesis in the context (forward reasoning)
• apply... with...: explicitly specify values for variables that cannot be determined by pattern matching
• simpl: simplify computations in the goal
• simpl in H: ... or a hypothesis
• rewrite: use an equality hypothesis (or lemma) to rewrite the goal
• rewrite ... in H: ... or a hypothesis
• symmetry: changes a goal of the form t=u into u=t
• symmetry in H: changes a hypothesis of the form t=u into u=t
• unfold: replace a defined constant by its right-hand side in the goal
• unfold... in H: ... or a hypothesis
• destruct... as...: case analysis on values of inductively defined types
• induction... as...: induction on values of inductively defined types
• inversion: reason by injectivity and distinctness of constructors
• remember (e) as x: give a name (x) to an expression (e) so that we can destruct x without "losing" e
• assert (e) as H: introduce a "local lemma" e and call it H

#### Exercise: 2 stars, optional (fold_length)

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

Definition fold_length {X : Type} (l : list X) : nat :=
fold (fun _ n => S 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, recommended (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 *)

Module MumbleBaz.

#### Exercise: 2 stars, optional (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, optional (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.

#### Exercise: 4 stars, recommended (forall_exists_challenge)

Challenge problem: Define two recursive Fixpoints, forallb and existsb. The first checks whether every element in a list satisfies a given predicate:
forallb oddb [1,3,5,7,9] = true

forallb negb [false,false] = true

forallb evenb [0,2,4,5] = false

forallb (beq_nat 5) [] = true
The function existsb checks whether there exists an element in the list that satisfies a given predicate:
existsb (beq_nat 5) [0,2,3,6] = false

existsb (andb true) [true,true,false] = true

existsb oddb [1,0,0,0,0,3] = true

existsb evenb [] = false
Next, create a nonrecursive Definition, existsb', using forallb and negb.
Prove that existsb' and existsb have the same behavior.

(* FILL IN HERE *)

#### Exercise: 2 stars, optional (index_informal)

Recall the definition of the index function:
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 nl'
end.
Write an informal proof of the following theorem:
X n llength l = n  @index X (S nl = None.
(* FILL IN HERE *)