# PolyPolymorphism and Higher-Order Functions

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

Require Export Lists.

## Polymorphic Lists

*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constructor has been replaced by an arbitrary type X, a binding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natlist in the types of the constructors have been replaced by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
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

*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 *)

(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

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
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 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.### Type 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...

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

*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

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

*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.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)

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.

*)

(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

*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.

☐

## Higher-Order Functions

*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.

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

Definition doit3times {X:Type} (f:X→X) (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

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.

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

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

*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 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

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)

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.
(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:

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].

= [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.

# Optional Material

## Non-Uniform Inductive Families (GADTs)

*This section needs more text*!

Inductive boollist : Type :=

boolnil : boollist

| boolcons : bool → boollist → boollist.

boolnil : boollist

| boolcons : bool → boollist → boollist.

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

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.

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 *)

Admitted.

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.

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.

☐
(* FILL IN HERE *)

☐

#### Exercise: 1 star (apply_rewrite)

Briefly explain the difference between the tactics apply and rewrite. Are there situations where both can usefully be applied?☐

## 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.

## 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. 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.

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.

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

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 *)

Admitted.

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
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:

*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.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.

☐
Theorem: For any nats n m, beq_nat n m = beq_nat m n.
Proof:
(* FILL IN HERE *)

☐

#### Exercise: 3 stars (beq_nat_sym_informal)

Provide an informal proof of this lemma that corresponds to your formal proof above:☐

## 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.

☐

## 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 *)

☐

## 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 bool_fn_applied_thrice :

∀(f : bool → bool) (b : bool),

f (f (f b)) = f b.

Proof.

(* FILL IN HERE *) Admitted.

∀(f : bool → bool) (b : bool),

f (f (f b)) = f b.

Proof.

(* FILL IN HERE *) Admitted.

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 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.

☐

# 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
- 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

# 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.

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#### 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 *)

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Module MumbleBaz.

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?

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- 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

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

Consider the following inductive definition:Inductive baz : Type :=

| x : baz → baz

| y : baz → bool → baz.

How

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*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:
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 *)

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#### 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 *)☐