# PolyPolymorphism and Higher-Order Functions

# Polymorphism

## Polymorphic Lists

*could*just define a new inductive datatype for each of these, for example...

... 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.
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constructor has been replaced by an arbitrary type X, a binding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natlist in the types of the constructors have been replaced by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.)
So what, exactly, is list? One good way to think about it
is that list is a
With this definition, when we use the constructors nil and
cons to build lists, we need to specify what type of lists we
are building — that is, nil and cons are now

*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.*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 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:

(We are writing nil and cons explicitly here because we
haven't yet defined the [] and :: notations. 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.
As with nil and cons, we can use length by applying it first
to a type and then to its list argument:

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

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 Inference

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':

It has exactly the same type type as app. Coq was able to
use a procedure called
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 strike a balance in your own code
between too many type annotations (which serve only to clutter and
distract) and too few (which force the reader to perform type
inference in their head in order to understand your code).
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
This may sound similar to type inference — in fact, the two
procedures rely on the same underlying mechanisms. Instead of
simply omitting the types of some arguments to a function, like
Using implicit arguments, the length function can be written
like this:

*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.### Argument Synthesis

*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 _.
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.
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...

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

### Implicit Arguments

*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'').

We can also 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):

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

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:

### Exercises: Polymorphic Lists

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

Here are a few simple exercises, just like ones in Lists.v, 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 snoc_with_append : ∀ X : Type,

∀ l1 l2 : list X,

∀ v : X,

snoc (l1 ++ l2) v = l1 ++ (snoc l2 v).

Proof.

(* FILL IN HERE *) Admitted.

☐
Following the same pattern, the type definition we gave in
the last chapter for pairs of numbers can be generalized to

## Polymorphic Pairs

*polymorphic pairs*(or*products*):
As with lists, we make the type arguments implicit and define the
familiar concrete notation.

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

*types*:
(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
The first and second projection functions now look pretty
much as they would in any functional programming language.

*value*consisting of a pair of values; X*Y is a*type*consisting of a pair of types. If x has type X and y has type Y, then (x,y) has type X*Y.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 (combine_checks)

Try answering the following questions on paper and checking your answers in coq:- 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.(*

Fixpoint split

(* FILL IN HERE *)

Example test_split:

split (1,false),(2,false) = (1,2,false,false).

Proof. reflexivity. Qed.

*)

☐
One last polymorphic type for now:

## Polymorphic Options

*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.
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.

☐
Like many other modern programming languages — including,
of course, all
Functions that manipulate other functions are often called

# Functions as Data

## Higher-Order Functions

*functional languages*— 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.*higher-order*functions. Here's a simple one:
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

Check plus.

The → is actually a

*binary*operator on types; that is, Coq primitively supports only one-argument functions. Moreover, 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.

*uncurrying*. With an uncurried binary function, both arguments must be given at once as a pair; there is no partial application.

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

(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.

☐
Here is a useful higher-order function, which takes a list
of Xs and a

## Filter

*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: X→bool) (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 Lists.v.

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

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, optional (filter_even_gt7)

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, optional (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.
(X → bool) → list X → list X * list X

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.

Fixpoint map {X Y:Type} (f:X→Y) (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:

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:
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].

= [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

Fixpoint fold {X Y:Type} (f: X→Y→Y) (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

*arguments*. Now let's look at some examples involving

*returning*functions as the results of other functions.

Definition constfun {X: Type} (x: X) : nat→X :=

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: nat→X) (k:nat) (x:X) : nat→X:=

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.
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!)

# More About Coq

## The unfold Tactic

Theorem unfold_example_bad : ∀ m n,

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. *)

Admitted.

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:nat→X),

(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 : nat→X),

f k1 = x1 →

beq_nat k2 k1 = false →

(override f k2 x2) k1 = x1.

Proof.

(* FILL IN HERE *) Admitted.

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.
Recall the definition of natural numbers:
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
Then inversion H instructs Coq to "invert" this equality to
extract the information it contains about these terms:
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.

## Inversion

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:
| O : nat

| S : nat → nat.

- 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.

c a1 a2 ... an = d b1 b2 ... bm

for some constructors c and d and arguments a1 ... an and
b1 ... bm.
- 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.

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.

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.

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.

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.

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.

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

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.

(* 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.

☐
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.

### 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 n 0 → 0 = n.

Proof.

(* FILL IN HERE *) Admitted.

☐

Theorem double_injective : ∀ n m,

double n = double m →

n = m.

Proof.

intros n. induction n as [| n'].

(* WORKED IN CLASS *)

Case "n = 0". simpl. intros m eq. destruct m as [| m'].

SCase "m = 0". reflexivity.

SCase "m = S m'". inversion eq.

Case "n = S n'". intros m eq. destruct m as [| m'].

SCase "m = 0". inversion eq.

SCase "m = S m'".

apply eq_remove_S. apply IHn'. inversion eq. reflexivity. Qed.

## Using Tactics on Hypotheses

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 L1→L2 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.

☐
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
Here are some examples:

## Using destruct on Compound Expressions

*expression*. We can also do this with destruct.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.

#### Exercise: 1 star (override_shadow)

Theorem override_shadow : ∀ {X:Type} x1 x2 k1 k2 (f : nat→X),

(override (override f k1 x2) k1 x1) k2 = (override f k1 x1) k2.

Proof.

(* FILL IN HERE *) Admitted.

(override (override f k1 x2) k1 x1) k2 = (override f k1 x1) k2.

Proof.

(* FILL IN HERE *) Admitted.

(*

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.

*)

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.

*)

☐
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.)

#### 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?(* FILL IN HERE *)

☐
(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:

## The remember Tactic

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... *)

Admitted.

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
The remember tactic allows us to introduce such a variable.

*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.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.

Theorem override_same : ∀ {X:Type} x1 k1 k2 (f : nat→X),

f k1 = x1 →

(override f k1 x1) k2 = f k2.

Proof.

(* FILL IN HERE *) Admitted.

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

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

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 : nat→X),

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.

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 : nat→X),

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.

☐
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
Here are the ones we've seen:

# Review

*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.- 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
- 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

# Additional Exercises

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

(* FILL IN HERE *)

☐

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

☐

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

Consider the following inductive definition:
How

☐

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

#### 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:
forallb negb [false,false] = true

forallb evenb [0,2,4,5] = false

forallb (beq_nat 5) [] = true

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

existsb oddb [1,0,0,0,0,3] = true

existsb evenb [] = false

*nonrecursive*Definition, existsb', using forallb and negb.

(* 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 n) l'

end.

Write an informal proof of the following theorem:
match l with

| [] => None

| a :: l' => if beq_nat n O then Some a else index (pred n) l'

end.

∀ X n l, length l = n → @index X (S n) l = None.

(* FILL IN HERE *)☐