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Lists: Working with Structured Data

From LF Require Export Induction.
Module NatList.

(* ################################################################# *)

Pairs of Numbers

In an Inductive type definition, each constructor can take any number of arguments -- none (as with true and O), one (as with S), or more than one (as with nybble, and here):

Inductive natprod : Type :=
| pair (n1 n2 : nat).

This declaration can be read: "There is just one way to construct a pair of numbers: by applying the constructor pair to two arguments of type nat."

Check (pair 3 5).pair 3 5
     : natprod

Here are simple functions for extracting the first and second components of a pair.

Definition fst (p : natprod) : nat :=
  match p with
  | pair x y => x
  end.

Definition snd (p : natprod) : nat :=
  match p with
  | pair x y => y
  end.

Compute (fst (pair 3 5)).= 3
: nat

Since pairs will be used heavily, it is nice to be able to write them with the standard mathematical notation (x,y) instead of pair x y. We can tell Coq to allow this with a Notation declaration.

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

The new pair notation can be used both in expressions and in pattern matches.

Compute (fst (3,5)).= 3
: nat

Definition fst' (p : natprod) : nat :=
  match p with
  | (x,y) => x
  end.

Definition snd' (p : natprod) : nat :=
  match p with
  | (x,y) => y
  end.

Definition swap_pair (p : natprod) : natprod :=
  match p with
  | (x,y) => (y,x)
  end.

Note that pattern-matching on a pair (with parentheses: (x, y)) is not to be confused with the "multiple pattern" syntax (with no parentheses: x, y) that we have seen previously.

The above examples illustrate pattern matching on a pair with elements x and y, whereas minus below (taken from Basics) performs pattern matching on the values n and m.

Fixpoint minus (n m : nat) : nat := match n, m with | O , _ => O | S _ , O => n | S n', S m' => minus n' m' end.

The distinction is minor, but it is worth knowing that they are not the same. For instance, the following definitions are ill-formed:

Definition bad_fst (p : natprod) : nat := match p with | x, y => x end.

Definition bad_minus (n m : nat) : nat := match n, m with | (O , _ ) => O | (S _ , O ) => n | (S n', S m') => bad_minus n' m' end.

Let's try to prove a few simple facts about pairs.

If we state things in a slightly peculiar way, we can complete proofs with just reflexivity (and its built-in simplification):

Theorem surjective_pairing' : forall (n m : nat),
  (n,m) = (fst (n,m), snd (n,m)).forall n m : nat, (n, m) = (fst (n, m), snd (n, m))
Proof.forall n m : nat, (n, m) = (fst (n, m), snd (n, m))
  reflexivity.  Qed.

But reflexivity is not enough if we state the lemma in a more natural way:

Theorem surjective_pairing_stuck : forall (p : natprod),
  p = (fst p, snd p).forall p : natprod, p = (fst p, snd p)
Proof.forall p : natprod, p = (fst p, snd p)
  simpl. (* Doesn't reduce anything! *)forall p : natprod, p = (fst p, snd p)
Abort.

We have to expose the structure of p so that simpl can perform the pattern match in fst and snd. We can do this with destruct.

Theorem surjective_pairing : forall (p : natprod),
  p = (fst p, snd p).forall p : natprod, p = (fst p, snd p)
Proof.forall p : natprod, p = (fst p, snd p)
  intros p.p:natprod
p = (fst p, snd p)  destruct p as [n m].n, m:nat
(n, m) = (fst (n, m), snd (n, m))  simpl.n, m:nat
(n, m) = (n, m)  reflexivity.  Qed.

Notice that, unlike its behavior with nats, where it generates two subgoals, destruct generates just one subgoal here. That's because natprods can only be constructed in one way.

Exercise: 1 star, standard (snd_fst_is_swap)

Theorem snd_fst_is_swap : forall (p : natprod),
  (snd p, fst p) = swap_pair p.forall p : natprod, (snd p, fst p) = swap_pair p
Proof.forall p : natprod, (snd p, fst p) = swap_pair p
  (* FILL IN HERE *) Admitted.

☐

Exercise: 1 star, standard, optional (fst_swap_is_snd)

Theorem fst_swap_is_snd : forall (p : natprod),
  fst (swap_pair p) = snd p.forall p : natprod, fst (swap_pair p) = snd p
Proof.forall p : natprod, fst (swap_pair p) = snd p
  (* FILL IN HERE *) Admitted.

☐

(* ################################################################# *)

Lists of Numbers

Generalizing the definition of pairs, we can describe the type of lists of numbers like this: "A list is either the empty list or else a pair of a number and another list."

Inductive natlist : Type :=
  | nil
  | cons (n : nat) (l : natlist).

For example, here is a three-element list:

Definition mylist := cons 1 (cons 2 (cons 3 nil)).

As with pairs, it is more convenient to write lists in familiar programming notation. The following declarations allow us to use :: as an infix cons operator and square brackets as an "outfix" notation for constructing lists.

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

It is not necessary to understand the details of these declarations, but here is roughly what's going on in case you are interested. The right associativity annotation tells Coq how to parenthesize expressions involving multiple uses of :: so that, for example, the next three declarations mean exactly the same thing:

Definition mylist1 := 1 :: (2 :: (3 :: nil)).
Definition mylist2 := 1 :: 2 :: 3 :: nil.
Definition mylist3 := [1;2;3].

The at level 60 part tells Coq how to parenthesize expressions that involve both :: and some other infix operator. For example, since we defined + as infix notation for the plus function at level 50,

Notation "x + y" := (plus x y) (at level 50, left associativity).

the + operator will bind tighter than ::, so 1 + 2 :: [3] will be parsed, as we'd expect, as (1 + 2) :: [3] rather than 1 + (2 :: [3]).

(Expressions like "1 + 2 :: [3]" can be a little confusing when you read them in a .v file. The inner brackets, around 3, indicate a list, but the outer brackets, which are invisible in the HTML rendering, are there to instruct the "coqdoc" tool that the bracketed part should be displayed as Coq code rather than running text.)

The second and third Notation declarations above introduce the standard square-bracket notation for lists; the right-hand side of the third one illustrates Coq's syntax for declaring n-ary notations and translating them to nested sequences of binary constructors.

(* ----------------------------------------------------------------- *)

Repeat

A number of functions are useful for manipulating lists. For example, the repeat function takes a number n and a count and returns a list of length count where every element is n.

Fixpoint repeat (n count : nat) : natlist :=
  match count with
  | O => nil
  | S count' => n :: (repeat n count')
  end.

(* ----------------------------------------------------------------- *)

Length

The length function calculates the length of a list.

Fixpoint length (l:natlist) : nat :=
  match l with
  | nil => O
  | h :: t => S (length t)
  end.

(* ----------------------------------------------------------------- *)

Append

The app function concatenates (appends) two lists.

Fixpoint app (l1 l2 : natlist) : natlist :=
  match l1 with
  | nil    => l2
  | h :: t => h :: (app t l2)
  end.

Since app will be used extensively in what follows, it is again convenient to have an infix operator for it.

Notation "x ++ y" := (app x y)
                     (right associativity, at level 60).

Example test_app1:             [1;2;3] ++ [4;5] = [1;2;3;4;5].[1; 2; 3] ++ [4; 5] = [1; 2; 3; 4; 5]
Proof.[1; 2; 3] ++ [4; 5] = [1; 2; 3; 4; 5] reflexivity.  Qed.
Example test_app2:             nil ++ [4;5] = [4;5].[ ] ++ [4; 5] = [4; 5]
Proof.[ ] ++ [4; 5] = [4; 5] reflexivity.  Qed.
Example test_app3:             [1;2;3] ++ nil = [1;2;3].[1; 2; 3] ++ [ ] = [1; 2; 3]
Proof.[1; 2; 3] ++ [ ] = [1; 2; 3] reflexivity.  Qed.

(* ----------------------------------------------------------------- *)

Head (With Default) and Tail

Here are two smaller examples of programming with lists. The hd function returns the first element (the "head") of the list, while tl returns everything but the first element (the "tail"). Since the empty list has no first element, we must pass a default value to be returned in that case.

Definition hd (default:nat) (l:natlist) : nat :=
  match l with
  | nil => default
  | h :: t => h
  end.

Definition tl (l:natlist) : natlist :=
  match l with
  | nil => nil
  | h :: t => t
  end.

Example test_hd1:             hd 0 [1;2;3] = 1.hd 0 [1; 2; 3] = 1
Proof.hd 0 [1; 2; 3] = 1 reflexivity.  Qed.
Example test_hd2:             hd 0 [] = 0.hd 0 [ ] = 0
Proof.hd 0 [ ] = 0 reflexivity.  Qed.
Example test_tl:              tl [1;2;3] = [2;3].tl [1; 2; 3] = [2; 3]
Proof.tl [1; 2; 3] = [2; 3] reflexivity.  Qed.

(* ----------------------------------------------------------------- *)

Exercises

Exercise: 2 stars, standard, recommended (list_funs)

Complete the definitions of nonzeros, oddmembers, and countoddmembers below. Have a look at the tests to understand what these functions should do.

Fixpoint nonzeros (l:natlist) : natlist
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).nonzeros:natlist -> natlist
l:natlist
natlist Admitted.

Example test_nonzeros:
  nonzeros [0;1;0;2;3;0;0] = [1;2;3].nonzeros [0; 1; 0; 2; 3; 0; 0] = [1; 2; 3]
  (* FILL IN HERE *) Admitted.

Fixpoint oddmembers (l:natlist) : natlist
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).oddmembers:natlist -> natlist
l:natlist
natlist Admitted.

Example test_oddmembers:
  oddmembers [0;1;0;2;3;0;0] = [1;3].oddmembers [0; 1; 0; 2; 3; 0; 0] = [1; 3]
  (* FILL IN HERE *) Admitted.

Definition countoddmembers (l:natlist) : nat
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).l:natlist
nat Admitted.

Example test_countoddmembers1:
  countoddmembers [1;0;3;1;4;5] = 4.countoddmembers [1; 0; 3; 1; 4; 5] = 4
  (* FILL IN HERE *) Admitted.

Example test_countoddmembers2:
  countoddmembers [0;2;4] = 0.countoddmembers [0; 2; 4] = 0
  (* FILL IN HERE *) Admitted.

Example test_countoddmembers3:
  countoddmembers nil = 0.countoddmembers [ ] = 0
  (* FILL IN HERE *) Admitted.

☐

Exercise: 3 stars, advanced (alternate)

Complete the definition of alternate, which interleaves two lists into one, alternating between elements taken from the first list and elements from the second. See the tests below for more specific examples.

(Note: one natural and elegant way of writing alternate will fail to satisfy Coq's requirement that all Fixpoint definitions be "obviously terminating." If you find yourself in this rut, look for a slightly more verbose solution that considers elements of both lists at the same time. One possible solution involves defining a new kind of pairs, but this is not the only way.)

Fixpoint alternate (l1 l2 : natlist) : natlist
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).alternate:natlist -> natlist -> natlist
l1, l2:natlist
natlist Admitted.

Example test_alternate1:
  alternate [1;2;3] [4;5;6] = [1;4;2;5;3;6].alternate [1; 2; 3] [4; 5; 6] = [1; 4; 2; 5; 3; 6]
  (* FILL IN HERE *) Admitted.

Example test_alternate2:
  alternate [1] [4;5;6] = [1;4;5;6].alternate [1] [4; 5; 6] = [1; 4; 5; 6]
  (* FILL IN HERE *) Admitted.

Example test_alternate3:
  alternate [1;2;3] [4] = [1;4;2;3].alternate [1; 2; 3] [4] = [1; 4; 2; 3]
  (* FILL IN HERE *) Admitted.

Example test_alternate4:
  alternate [] [20;30] = [20;30].alternate [ ] [20; 30] = [20; 30]
  (* FILL IN HERE *) Admitted.

☐

(* ----------------------------------------------------------------- *)

Bags via Lists

A bag (or multiset) is like a set, except that each element can appear multiple times rather than just once. One possible representation for a bag of numbers is as a list.

Definition bag := natlist.

Exercise: 3 stars, standard, recommended (bag_functions)

Complete the following definitions for the functions count, sum, add, and member for bags.

Fixpoint count (v:nat) (s:bag) : nat
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).count:nat -> bag -> nat
v:nat
s:bag
nat Admitted.

All these proofs can be done just by reflexivity.

Example test_count1:              count 1 [1;2;3;1;4;1] = 3.count 1 [1; 2; 3; 1; 4; 1] = 3
 (* FILL IN HERE *) Admitted.
Example test_count2:              count 6 [1;2;3;1;4;1] = 0.count 6 [1; 2; 3; 1; 4; 1] = 0
 (* FILL IN HERE *) Admitted.

Multiset sum is similar to set union: sum a b contains all the elements of a and of b. (Mathematicians usually define union on multisets a little bit differently -- using max instead of sum -- which is why we don't use that name for this operation.) For sum we're giving you a header that does not give explicit names to the arguments. Moreover, it uses the keyword Definition instead of Fixpoint, so even if you had names for the arguments, you wouldn't be able to process them recursively. The point of stating the question this way is to encourage you to think about whether sum can be implemented in another way -- perhaps by using functions that have already been defined.

Definition sum : bag -> bag -> bag
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).bag -> bag -> bag Admitted.

Example test_sum1:              count 1 (sum [1;2;3] [1;4;1]) = 3.count 1 (sum [1; 2; 3] [1; 4; 1]) = 3
 (* FILL IN HERE *) Admitted.

Definition add (v:nat) (s:bag) : bag
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).v:nat
s:bag
bag Admitted.

Example test_add1:                count 1 (add 1 [1;4;1]) = 3.count 1 (add 1 [1; 4; 1]) = 3
 (* FILL IN HERE *) Admitted.
Example test_add2:                count 5 (add 1 [1;4;1]) = 0.count 5 (add 1 [1; 4; 1]) = 0
 (* FILL IN HERE *) Admitted.

Definition member (v:nat) (s:bag) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).v:nat
s:bag
bool Admitted.

Example test_member1:             member 1 [1;4;1] = true.member 1 [1; 4; 1] = true
 (* FILL IN HERE *) Admitted.

Example test_member2:             member 2 [1;4;1] = false.member 2 [1; 4; 1] = false
(* FILL IN HERE *) Admitted.

☐

Exercise: 3 stars, standard, optional (bag_more_functions)

Here are some more bag functions for you to practice with.

When remove_one is applied to a bag without the number to remove, it should return the same bag unchanged. (This exercise is optional, but students following the advanced track will need to fill in the definition of remove_one for a later exercise.)

Fixpoint remove_one (v:nat) (s:bag) : bag
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).remove_one:nat -> bag -> bag
v:nat
s:bag
bag Admitted.

Example test_remove_one1:
  count 5 (remove_one 5 [2;1;5;4;1]) = 0.count 5 (remove_one 5 [2; 1; 5; 4; 1]) = 0
  (* FILL IN HERE *) Admitted.

Example test_remove_one2:
  count 5 (remove_one 5 [2;1;4;1]) = 0.count 5 (remove_one 5 [2; 1; 4; 1]) = 0
  (* FILL IN HERE *) Admitted.

Example test_remove_one3:
  count 4 (remove_one 5 [2;1;4;5;1;4]) = 2.count 4 (remove_one 5 [2; 1; 4; 5; 1; 4]) = 2
  (* FILL IN HERE *) Admitted.

Example test_remove_one4:
  count 5 (remove_one 5 [2;1;5;4;5;1;4]) = 1.count 5 (remove_one 5 [2; 1; 5; 4; 5; 1; 4]) = 1
  (* FILL IN HERE *) Admitted.

Fixpoint remove_all (v:nat) (s:bag) : bag
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).remove_all:nat -> bag -> bag
v:nat
s:bag
bag Admitted.

Example test_remove_all1:  count 5 (remove_all 5 [2;1;5;4;1]) = 0.count 5 (remove_all 5 [2; 1; 5; 4; 1]) = 0
 (* FILL IN HERE *) Admitted.
Example test_remove_all2:  count 5 (remove_all 5 [2;1;4;1]) = 0.count 5 (remove_all 5 [2; 1; 4; 1]) = 0
 (* FILL IN HERE *) Admitted.
Example test_remove_all3:  count 4 (remove_all 5 [2;1;4;5;1;4]) = 2.count 4 (remove_all 5 [2; 1; 4; 5; 1; 4]) = 2
 (* FILL IN HERE *) Admitted.
Example test_remove_all4:  count 5 (remove_all 5 [2;1;5;4;5;1;4;5;1;4]) = 0.count 5 (remove_all 5 [2; 1; 5; 4; 5; 1; 4; 5; 1; 4]) =
0
 (* FILL IN HERE *) Admitted.

Fixpoint subset (s1:bag) (s2:bag) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).subset:bag -> bag -> bool
s1, s2:bag
bool Admitted.

Example test_subset1:              subset [1;2] [2;1;4;1] = true.subset [1; 2] [2; 1; 4; 1] = true
 (* FILL IN HERE *) Admitted.
Example test_subset2:              subset [1;2;2] [2;1;4;1] = false.subset [1; 2; 2] [2; 1; 4; 1] = false
 (* FILL IN HERE *) Admitted.

☐

Exercise: 2 stars, standard, recommended (bag_theorem)

Write down an interesting theorem bag_theorem about bags involving the functions count and add, and prove it. Note that, since this problem is somewhat open-ended, it's possible that you may come up with a theorem which is true, but whose proof requires techniques you haven't learned yet. Feel free to ask for help if you get stuck!

(*
Theorem bag_theorem : ...
Proof.
  ...
Qed.
*)

(* Do not modify the following line: *)
Definition manual_grade_for_bag_theorem : option (nat*string) := None.
(* Note to instructors: For silly technical reasons, in this
   file (only) you will need to write [Some (Datatypes.pair 3 ""%string)]
   rather than [Some (3,""%string)] to enter your grade and comments. 

    [] *)

(* ################################################################# *)

Reasoning About Lists

As for numbers, simple facts about list-processing functions can sometimes be proved entirely by simplification. For example, the simplification performed by reflexivity is enough for this theorem...

Theorem nil_app : forall l:natlist,
  [] ++ l = l.forall l : natlist, [ ] ++ l = l
Proof.forall l : natlist, [ ] ++ l = l reflexivity. Qed.

...because the [] is substituted into the "scrutinee" (the expression whose value is being "scrutinized" by the match) in the definition of app, allowing the match itself to be simplified.

Also, as with numbers, it is sometimes helpful to perform case analysis on the possible shapes (empty or non-empty) of an unknown list.

Theorem tl_length_pred : forall l:natlist,
  pred (length l) = length (tl l).forall l : natlist,
Nat.pred (length l) = length (tl l)
Proof.forall l : natlist,
Nat.pred (length l) = length (tl l)
  intros l.l:natlist
Nat.pred (length l) = length (tl l) destruct l as [| n l'].Nat.pred (length [ ]) = length (tl [ ])
n:nat
l':natlist
Nat.pred (length (n :: l')) = length (tl (n :: l'))
  - (* l = nil *)Nat.pred (length [ ]) = length (tl [ ])
    reflexivity.
  - (* l = cons n l' *)n:nat
l':natlist
Nat.pred (length (n :: l')) = length (tl (n :: l'))
    reflexivity.  Qed.

Here, the nil case works because we've chosen to define tl nil = nil. Notice that the as annotation on the destruct tactic here introduces two names, n and l', corresponding to the fact that the cons constructor for lists takes two arguments (the head and tail of the list it is constructing).

Usually, though, interesting theorems about lists require induction for their proofs.

(* ----------------------------------------------------------------- *)

Micro-Sermon

Simply reading proof scripts will not get you very far! It is important to step through the details of each one using Coq and think about what each step achieves. Otherwise it is more or less guaranteed that the exercises will make no sense when you get to them. 'Nuff said.

(* ================================================================= *)

Induction on Lists

Proofs by induction over datatypes like natlist are a little less familiar than standard natural number induction, but the idea is equally simple. Each Inductive declaration defines a set of data values that can be built up using the declared constructors: a boolean can be either true or false; a number can be either O or S applied to another number; a list can be either nil or cons applied to a number and a list.

Moreover, applications of the declared constructors to one another are the only possible shapes that elements of an inductively defined set can have, and this fact directly gives rise to a way of reasoning about inductively defined sets: a number is either O or else it is S applied to some smaller number; a list is either nil or else it is cons applied to some number and some smaller list; etc. So, if we have in mind some proposition P that mentions a list l and we want to argue that P holds for all lists, we can reason as follows:

First, show that P is true of l when l is nil.
Then show that P is true of l when l is cons n l' for some number n and some smaller list l', assuming that P is true for l'.

Since larger lists can only be built up from smaller ones, eventually reaching nil, these two arguments together establish the truth of P for all lists l. Here's a concrete example:

Theorem app_assoc : forall l1 l2 l3 : natlist,
  (l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).forall l1 l2 l3 : natlist,
(l1 ++ l2) ++ l3 = l1 ++ l2 ++ l3
Proof.forall l1 l2 l3 : natlist,
(l1 ++ l2) ++ l3 = l1 ++ l2 ++ l3
  intros l1 l2 l3.l1, l2, l3:natlist
(l1 ++ l2) ++ l3 = l1 ++ l2 ++ l3 induction l1 as [| n l1' IHl1'].l2, l3:natlist
([ ] ++ l2) ++ l3 = [ ] ++ l2 ++ l3
n:nat
l1', l2, l3:natlist
IHl1':(l1' ++ l2) ++ l3 = l1' ++ l2 ++ l3
((n :: l1') ++ l2) ++ l3 = (n :: l1') ++ l2 ++ l3
  - (* l1 = nil *)l2, l3:natlist
([ ] ++ l2) ++ l3 = [ ] ++ l2 ++ l3
    reflexivity.
  - (* l1 = cons n l1' *)n:nat
l1', l2, l3:natlist
IHl1':(l1' ++ l2) ++ l3 = l1' ++ l2 ++ l3
((n :: l1') ++ l2) ++ l3 = (n :: l1') ++ l2 ++ l3
    simpl.n:nat
l1', l2, l3:natlist
IHl1':(l1' ++ l2) ++ l3 = l1' ++ l2 ++ l3
n :: (l1' ++ l2) ++ l3 = n :: l1' ++ l2 ++ l3 rewrite -> IHl1'.n:nat
l1', l2, l3:natlist
IHl1':(l1' ++ l2) ++ l3 = l1' ++ l2 ++ l3
n :: l1' ++ l2 ++ l3 = n :: l1' ++ l2 ++ l3 reflexivity.  Qed.

Notice that, as when doing induction on natural numbers, the as... clause provided to the induction tactic gives a name to the induction hypothesis corresponding to the smaller list l1' in the cons case. Once again, this Coq proof is not especially illuminating as a static document -- it is easy to see what's going on if you are reading the proof in an interactive Coq session and you can see the current goal and context at each point, but this state is not visible in the written-down parts of the Coq proof. So a natural-language proof -- one written for human readers -- will need to include more explicit signposts; in particular, it will help the reader stay oriented if we remind them exactly what the induction hypothesis is in the second case.

For comparison, here is an informal proof of the same theorem.

Theorem: For all lists l1, l2, and l3, (l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).

Proof: By induction on l1.

First, suppose l1 = []. We must show

(☐ ++ l2) ++ l3 = ☐ ++ (l2 ++ l3),

which follows directly from the definition of ++.
Next, suppose l1 = n::l1', with

(l1' ++ l2) ++ l3 = l1' ++ (l2 ++ l3)

(the induction hypothesis). We must show

((n :: l1') ++ l2) ++ l3 = (n :: l1') ++ (l2 ++ l3).

By the definition of ++, this follows from

n :: ((l1' ++ l2) ++ l3) = n :: (l1' ++ (l2 ++ l3)),

which is immediate from the induction hypothesis. ☐

(* ----------------------------------------------------------------- *)

Reversing a List

For a slightly more involved example of inductive proof over lists, suppose we use app to define a list-reversing function rev:

Fixpoint rev (l:natlist) : natlist :=
  match l with
  | nil    => nil
  | h :: t => rev t ++ [h]
  end.

Example test_rev1:            rev [1;2;3] = [3;2;1].rev [1; 2; 3] = [3; 2; 1]
Proof.rev [1; 2; 3] = [3; 2; 1] reflexivity.  Qed.
Example test_rev2:            rev nil = nil.rev [ ] = [ ]
Proof.rev [ ] = [ ] reflexivity.  Qed.

(* ----------------------------------------------------------------- *)

Properties of rev

Now, for something a bit more challenging than the proofs we've seen so far, let's prove that reversing a list does not change its length. Our first attempt gets stuck in the successor case...

Theorem rev_length_firsttry : forall l : natlist,
  length (rev l) = length l.forall l : natlist, length (rev l) = length l
Proof.forall l : natlist, length (rev l) = length l
  intros l.l:natlist
length (rev l) = length l induction l as [| n l' IHl'].length (rev [ ]) = length [ ]
n:nat
l':natlist
IHl':length (rev l') = length l'
length (rev (n :: l')) = length (n :: l')
  - (* l = [] *)length (rev [ ]) = length [ ]
    reflexivity.
  - (* l = n :: l' *)n:nat
l':natlist
IHl':length (rev l') = length l'
length (rev (n :: l')) = length (n :: l')
    (* This is the tricky case.  Let's begin as usual
       by simplifying. *)
    simpl.n:nat
l':natlist
IHl':length (rev l') = length l'
length (rev l' ++ [n]) = S (length l')
    (* Now we seem to be stuck: the goal is an equality
       involving [++], but we don't have any useful equations
       in either the immediate context or in the global
       environment!  We can make a little progress by using
       the IH to rewrite the goal... *)
    rewrite <- IHl'.n:nat
l':natlist
IHl':length (rev l') = length l'
length (rev l' ++ [n]) = S (length (rev l'))
    (* ... but now we can't go any further. *)
Abort.

So let's take the equation relating ++ and length that would have enabled us to make progress and state it as a separate lemma.

Theorem app_length : forall l1 l2 : natlist,
  length (l1 ++ l2) = (length l1) + (length l2).forall l1 l2 : natlist,
length (l1 ++ l2) = length l1 + length l2
Proof.forall l1 l2 : natlist,
length (l1 ++ l2) = length l1 + length l2
  (* WORKED IN CLASS *)
  intros l1 l2.l1, l2:natlist
length (l1 ++ l2) = length l1 + length l2 induction l1 as [| n l1' IHl1'].l2:natlist
length ([ ] ++ l2) = length [ ] + length l2
n:nat
l1', l2:natlist
IHl1':length (l1' ++ l2) = length l1' + length l2
length ((n :: l1') ++ l2) =
length (n :: l1') + length l2
  - (* l1 = nil *)l2:natlist
length ([ ] ++ l2) = length [ ] + length l2
    reflexivity.
  - (* l1 = cons *)n:nat
l1', l2:natlist
IHl1':length (l1' ++ l2) = length l1' + length l2
length ((n :: l1') ++ l2) =
length (n :: l1') + length l2
    simpl.n:nat
l1', l2:natlist
IHl1':length (l1' ++ l2) = length l1' + length l2
S (length (l1' ++ l2)) = S (length l1' + length l2) rewrite -> IHl1'.n:nat
l1', l2:natlist
IHl1':length (l1' ++ l2) = length l1' + length l2
S (length l1' + length l2) =
S (length l1' + length l2) reflexivity.  Qed.

Note that, to make the lemma as general as possible, we quantify over all natlists, not just those that result from an application of rev. This should seem natural, because the truth of the goal clearly doesn't depend on the list having been reversed. Moreover, it is easier to prove the more general property.

Now we can complete the original proof.

Theorem rev_length : forall l : natlist,
  length (rev l) = length l.forall l : natlist, length (rev l) = length l
Proof.forall l : natlist, length (rev l) = length l
  intros l.l:natlist
length (rev l) = length l induction l as [| n l' IHl'].length (rev [ ]) = length [ ]
n:nat
l':natlist
IHl':length (rev l') = length l'
length (rev (n :: l')) = length (n :: l')
  - (* l = nil *)length (rev [ ]) = length [ ]
    reflexivity.
  - (* l = cons *)n:nat
l':natlist
IHl':length (rev l') = length l'
length (rev (n :: l')) = length (n :: l')
    simpl.n:nat
l':natlist
IHl':length (rev l') = length l'
length (rev l' ++ [n]) = S (length l') rewrite -> app_length, plus_comm.n:nat
l':natlist
IHl':length (rev l') = length l'
length [n] + length (rev l') = S (length l')
    simpl.n:nat
l':natlist
IHl':length (rev l') = length l'
S (length (rev l')) = S (length l') rewrite -> IHl'.n:nat
l':natlist
IHl':length (rev l') = length l'
S (length l') = S (length l') reflexivity.  Qed.

For comparison, here are informal proofs of these two theorems:

Theorem: For all lists l1 and l2, length (l1 ++ l2) = length l1 + length l2.

Proof: By induction on l1.

First, suppose l1 = []. We must show

length (☐ ++ l2) = length ☐ + length l2,

which follows directly from the definitions of length and ++.
Next, suppose l1 = n::l1', with

length (l1' ++ l2) = length l1' + length l2.

We must show

length ((n::l1') ++ l2) = length (n::l1') + length l2).

This follows directly from the definitions of length and ++ together with the induction hypothesis. ☐

Theorem: For all lists l, length (rev l) = length l.

Proof: By induction on l.

First, suppose l = []. We must show

length (rev ☐) = length ☐,

which follows directly from the definitions of length and rev.
Next, suppose l = n::l', with

length (rev l') = length l'.

We must show

length (rev (n :: l')) = length (n :: l').

By the definition of rev, this follows from

length ((rev l') ++ n) = S (length l')

which, by the previous lemma, is the same as

length (rev l') + length n = S (length l').

This follows directly from the induction hypothesis and the definition of length. ☐

The style of these proofs is rather longwinded and pedantic. After the first few, we might find it easier to follow proofs that give fewer details (which we can easily work out in our own minds or on scratch paper if necessary) and just highlight the non-obvious steps. In this more compressed style, the above proof might look like this:

Theorem: For all lists l, length (rev l) = length l.

Proof: First, observe that length (l ++ [n]) = S (length l) for any l (this follows by a straightforward induction on l). The main property again follows by induction on l, using the observation together with the induction hypothesis in the case where l = n'::l'. ☐

Which style is preferable in a given situation depends on the sophistication of the expected audience and how similar the proof at hand is to ones that the audience will already be familiar with. The more pedantic style is a good default for our present purposes.

(* ================================================================= *)

Search

We've seen that proofs can make use of other theorems we've already proved, e.g., using rewrite. But in order to refer to a theorem, we need to know its name! Indeed, it is often hard even to remember what theorems have been proven, much less what they are called.

Coq's Search command is quite helpful with this. Typing Search foo into your .v file and evaluating this line will cause Coq to display a list of all theorems involving foo. For example, try uncommenting the following line to see a list of theorems that we have proved about rev:

(*  Search rev. *)

Keep Search in mind as you do the following exercises and throughout the rest of the book; it can save you a lot of time!

If you are using ProofGeneral, you can run Search with C-c C-a C-a. Pasting its response into your buffer can be accomplished with C-c C-;.

(* ================================================================= *)

List Exercises, Part 1

Exercise: 3 stars, standard (list_exercises)

More practice with lists:

Theorem app_nil_r : forall l : natlist,
  l ++ [] = l.forall l : natlist, l ++ [ ] = l
Proof.forall l : natlist, l ++ [ ] = l
  (* FILL IN HERE *) Admitted.

Theorem rev_app_distr: forall l1 l2 : natlist,
  rev (l1 ++ l2) = rev l2 ++ rev l1.forall l1 l2 : natlist,
rev (l1 ++ l2) = rev l2 ++ rev l1
Proof.forall l1 l2 : natlist,
rev (l1 ++ l2) = rev l2 ++ rev l1
  (* FILL IN HERE *) Admitted.

Theorem rev_involutive : forall l : natlist,
  rev (rev l) = l.forall l : natlist, rev (rev l) = l
Proof.forall l : natlist, rev (rev l) = l
  (* FILL IN HERE *) Admitted.

There is a short solution to the next one. If you find yourself getting tangled up, step back and try to look for a simpler way.

Theorem app_assoc4 : forall l1 l2 l3 l4 : natlist,
  l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4.forall l1 l2 l3 l4 : natlist,
l1 ++ l2 ++ l3 ++ l4 = ((l1 ++ l2) ++ l3) ++ l4
Proof.forall l1 l2 l3 l4 : natlist,
l1 ++ l2 ++ l3 ++ l4 = ((l1 ++ l2) ++ l3) ++ l4
  (* FILL IN HERE *) Admitted.

An exercise about your implementation of nonzeros:

Lemma nonzeros_app : forall l1 l2 : natlist,
  nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).forall l1 l2 : natlist,
nonzeros (l1 ++ l2) = nonzeros l1 ++ nonzeros l2
Proof.forall l1 l2 : natlist,
nonzeros (l1 ++ l2) = nonzeros l1 ++ nonzeros l2
  (* FILL IN HERE *) Admitted.

☐

Exercise: 2 stars, standard (eqblist)

Fill in the definition of eqblist, which compares lists of numbers for equality. Prove that eqblist l l yields true for every list l.

Fixpoint eqblist (l1 l2 : natlist) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).eqblist:natlist -> natlist -> bool
l1, l2:natlist
bool Admitted.

Example test_eqblist1 :
  (eqblist nil nil = true).eqblist [ ] [ ] = true
 (* FILL IN HERE *) Admitted.

Example test_eqblist2 :
  eqblist [1;2;3] [1;2;3] = true.eqblist [1; 2; 3] [1; 2; 3] = true
(* FILL IN HERE *) Admitted.

Example test_eqblist3 :
  eqblist [1;2;3] [1;2;4] = false.eqblist [1; 2; 3] [1; 2; 4] = false
 (* FILL IN HERE *) Admitted.

Theorem eqblist_refl : forall l:natlist,
  true = eqblist l l.forall l : natlist, true = eqblist l l
Proof.forall l : natlist, true = eqblist l l
  (* FILL IN HERE *) Admitted.

☐

(* ================================================================= *)

List Exercises, Part 2

Here are a couple of little theorems to prove about your definitions about bags above.

Exercise: 1 star, standard (count_member_nonzero)

Theorem count_member_nonzero : forall (s : bag),
  1 <=? (count 1 (1 :: s)) = true.forall s : bag, (1 <=? count 1 (1 :: s)) = true
Proof.forall s : bag, (1 <=? count 1 (1 :: s)) = true
  (* FILL IN HERE *) Admitted.

☐

The following lemma about leb might help you in the next exercise.

Theorem leb_n_Sn : forall n,
  n <=? (S n) = true.forall n : nat, (n <=? S n) = true
Proof.forall n : nat, (n <=? S n) = true
  intros n.n:nat
(n <=? S n) = true induction n as [| n' IHn'].(0 <=? 1) = true
n':nat
IHn':(n' <=? S n') = true
(S n' <=? S (S n')) = true
  - (* 0 *)(0 <=? 1) = true
    simpl.true = true  reflexivity.
  - (* S n' *)n':nat
IHn':(n' <=? S n') = true
(S n' <=? S (S n')) = true
    simpl.n':nat
IHn':(n' <=? S n') = true
(n' <=? S n') = true  rewrite IHn'.n':nat
IHn':(n' <=? S n') = true
true = true  reflexivity.  Qed.

Before doing the next exercise, make sure you've filled in the definition of remove_one above.

Exercise: 3 stars, advanced (remove_does_not_increase_count)

Theorem remove_does_not_increase_count: forall (s : bag),
  (count 0 (remove_one 0 s)) <=? (count 0 s) = true.forall s : bag,
(count 0 (remove_one 0 s) <=? count 0 s) = true
Proof.forall s : bag,
(count 0 (remove_one 0 s) <=? count 0 s) = true
  (* FILL IN HERE *) Admitted.

☐

Exercise: 3 stars, standard, optional (bag_count_sum)

Write down an interesting theorem bag_count_sum about bags involving the functions count and sum, and prove it using Coq. (You may find that the difficulty of the proof depends on how you defined count!)

(* FILL IN HERE 

    [] *)

Exercise: 4 stars, advanced (rev_injective)

Prove that the rev function is injective -- that is,

forall (l1 l2 : natlist), rev l1 = rev l2 -> l1 = l2.

(There is a hard way and an easy way to do this.)

(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_rev_injective : option (nat*string) := None.

☐

(* ################################################################# *)

Options

Suppose we want to write a function that returns the nth element of some list. If we give it type nat → natlist → nat, then we'll have to choose some number to return when the list is too short...

Fixpoint nth_bad (l:natlist) (n:nat) : nat :=
  match l with
  | nil => 42  (* arbitrary! *)
  | a :: l' => match n =? O with
               | true => a
               | false => nth_bad l' (pred n)
               end
  end.

This solution is not so good: If nth_bad returns 42, we can't tell whether that value actually appears on the input without further processing. A better alternative is to change the return type of nth_bad to include an error value as a possible outcome. We call this type natoption.

Inductive natoption : Type :=
  | Some (n : nat)
  | None.

We can then change the above definition of nth_bad to return None when the list is too short and Some a when the list has enough members and a appears at position n. We call this new function nth_error to indicate that it may result in an error.

Fixpoint nth_error (l:natlist) (n:nat) : natoption :=
  match l with
  | nil => None
  | a :: l' => match n =? O with
               | true => Some a
               | false => nth_error l' (pred n)
               end
  end.

Example test_nth_error1 : nth_error [4;5;6;7] 0 = Some 4.nth_error [4; 5; 6; 7] 0 = Some 4
Proof.nth_error [4; 5; 6; 7] 0 = Some 4 reflexivity. Qed.
Example test_nth_error2 : nth_error [4;5;6;7] 3 = Some 7.nth_error [4; 5; 6; 7] 3 = Some 7
Proof.nth_error [4; 5; 6; 7] 3 = Some 7 reflexivity. Qed.
Example test_nth_error3 : nth_error [4;5;6;7] 9 = None.nth_error [4; 5; 6; 7] 9 = None
Proof.nth_error [4; 5; 6; 7] 9 = None reflexivity. Qed.

(In the HTML version, the boilerplate proofs of these examples are elided. Click on a box if you want to see one.)

This example is also an opportunity to introduce one more small feature of Coq's programming language: conditional expressions...

Fixpoint nth_error' (l:natlist) (n:nat) : natoption :=
  match l with
  | nil => None
  | a :: l' => if n =? O then Some a
               else nth_error' l' (pred n)
  end.

Coq's conditionals are exactly like those found in any other language, with one small generalization. Since the boolean type is not built in, Coq actually supports conditional expressions over any inductively defined type with exactly two constructors. The guard is considered true if it evaluates to the first constructor in the Inductive definition and false if it evaluates to the second.

The function below pulls the nat out of a natoption, returning a supplied default in the None case.

Definition option_elim (d : nat) (o : natoption) : nat :=
  match o with
  | Some n' => n'
  | None => d
  end.

Exercise: 2 stars, standard (hd_error)

Using the same idea, fix the hd function from earlier so we don't have to pass a default element for the nil case.

Definition hd_error (l : natlist) : natoption
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *).l:natlist
natoption Admitted.

Example test_hd_error1 : hd_error [] = None.hd_error [ ] = None
 (* FILL IN HERE *) Admitted.

Example test_hd_error2 : hd_error [1] = Some 1.hd_error [1] = Some 1
 (* FILL IN HERE *) Admitted.

Example test_hd_error3 : hd_error [5;6] = Some 5.hd_error [5; 6] = Some 5
 (* FILL IN HERE *) Admitted.

☐

Exercise: 1 star, standard, optional (option_elim_hd)

This exercise relates your new hd_error to the old hd.

Theorem option_elim_hd : forall (l:natlist) (default:nat),
  hd default l = option_elim default (hd_error l).forall (l : natlist) (default : nat),
hd default l = option_elim default (hd_error l)
Proof.forall (l : natlist) (default : nat),
hd default l = option_elim default (hd_error l)
  (* FILL IN HERE *) Admitted.

☐

End NatList.

(* ################################################################# *)

Partial Maps

As a final illustration of how data structures can be defined in Coq, here is a simple partial map data type, analogous to the map or dictionary data structures found in most programming languages.

First, we define a new inductive datatype id to serve as the "keys" of our partial maps.

Inductive id : Type :=
  | Id (n : nat).

Internally, an id is just a number. Introducing a separate type by wrapping each nat with the tag Id makes definitions more readable and gives us the flexibility to change representations later if we wish.

We'll also need an equality test for ids:

Definition eqb_id (x1 x2 : id) :=
  match x1, x2 with
  | Id n1, Id n2 => n1 =? n2
  end.

Exercise: 1 star, standard (eqb_id_refl)

Theorem eqb_id_refl : forall x, true = eqb_id x x.forall x : id, true = eqb_id x x
Proof.forall x : id, true = eqb_id x x
  (* FILL IN HERE *) Admitted.

☐

Now we define the type of partial maps:

Module PartialMap.
Export NatList.
  
Inductive partial_map : Type :=
  | empty
  | record (i : id) (v : nat) (m : partial_map).

This declaration can be read: "There are two ways to construct a partial_map: either using the constructor empty to represent an empty partial map, or by applying the constructor record to a key, a value, and an existing partial_map to construct a partial_map with an additional key-to-value mapping."

The update function overrides the entry for a given key in a partial map by shadowing it with a new one (or simply adds a new entry if the given key is not already present).

Definition update (d : partial_map)
                  (x : id) (value : nat)
                  : partial_map :=
  record x value d.

Last, the find function searches a partial_map for a given key. It returns None if the key was not found and Some val if the key was associated with val. If the same key is mapped to multiple values, find will return the first one it encounters.

Fixpoint find (x : id) (d : partial_map) : natoption :=
  match d with
  | empty         => None
  | record y v d' => if eqb_id x y
                     then Some v
                     else find x d'
  end.

Exercise: 1 star, standard (update_eq)

Theorem update_eq :
  forall (d : partial_map) (x : id) (v: nat),
    find x (update d x v) = Some v.forall (d : partial_map) (x : id) (v : nat),
find x (update d x v) = Some v
Proof.forall (d : partial_map) (x : id) (v : nat),
find x (update d x v) = Some v
 (* FILL IN HERE *) Admitted.

☐

Exercise: 1 star, standard (update_neq)

Theorem update_neq :
  forall (d : partial_map) (x y : id) (o: nat),
    eqb_id x y = false -> find x (update d y o) = find x d.forall (d : partial_map) (x y : id) (o : nat),
eqb_id x y = false -> find x (update d y o) = find x d
Proof.forall (d : partial_map) (x y : id) (o : nat),
eqb_id x y = false -> find x (update d y o) = find x d
 (* FILL IN HERE *) Admitted.

☐

End PartialMap.

Exercise: 2 stars, standard (baz_num_elts)

Consider the following inductive definition:

Inductive baz : Type :=
  | Baz1 (x : baz)
  | Baz2 (y : baz) (b : bool).

How many elements does the type baz have? (Explain in words, in a comment.)

(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_baz_num_elts : option (nat*string) := None.

☐

(* Wed Jan 9 12:02:44 EST 2019 *)