# ListsWorking with Structured Data

# Pairs of Numbers

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

Here are two simple functions for extracting the first and
second components of a pair. The definitions also illustrate how
to do pattern matching on two-argument constructors.

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 *)

Since pairs are used quite a bit, 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.

The new notation can be used both in expressions and in
pattern matches (indeed, we've seen it already in the previous
chapter — this works because the pair notation is actually
provided as part of the standard library):

Compute (fst (3,5)).

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.

Let's try to prove a few simple facts about pairs.
If we state things in a particular (and slightly peculiar) way, we
can complete proofs with just reflexivity (and its built-in
simplification):

Theorem surjective_pairing' : ∀(n m : nat),

(n,m) = (fst (n,m), snd (n,m)).

Proof.

reflexivity. Qed.

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

Theorem surjective_pairing_stuck : ∀(p : natprod),

p = (fst p, snd p).

Proof.

simpl. (* Doesn't reduce anything! *)

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 : ∀(p : natprod),

p = (fst p, snd p).

Proof.

intros p. destruct p as [n m]. simpl. reflexivity. Qed.

Notice that, unlike its behavior with nats, destruct
doesn't generate an extra subgoal here. That's because natprods
can only be constructed in one way.

#### Exercise: 1 star (snd_fst_is_swap)

Theorem snd_fst_is_swap : ∀(p : natprod),

(snd p, fst p) = swap_pair p.

Proof.

(* FILL IN HERE *) Admitted.

(snd p, fst p) = swap_pair p.

Proof.

(* FILL IN HERE *) Admitted.

Theorem fst_swap_is_snd : ∀(p : natprod),

fst (swap_pair p) = snd p.

Proof.

(* FILL IN HERE *) Admitted.

fst (swap_pair p) = snd p.

Proof.

(* FILL IN HERE *) Admitted.

☐

# Lists of Numbers

*lists*of numbers like this: "A list is either the empty list or else a pair of a number and another list."

For example, here is a three-element list:

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 in case you are interested, here is roughly
what's going on. The right associativity annotation tells Coq
how to parenthesize expressions involving several 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,
(By the way, it's worth noting in passing that 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.

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]).
(at level 50, left associativity).

### 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.

Fixpoint app (l1 l2 : natlist) : natlist :=

match l1 with

| nil ⇒ l2

| h :: t ⇒ h :: (app t l2)

end.

Actually, app will be used a lot in some parts of what
follows, so it is 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].

Proof. reflexivity. Qed.

Example test_app2: nil ++ [4;5] = [4;5].

Proof. reflexivity. Qed.

Example test_app3: [1;2;3] ++ nil = [1;2;3].

Proof. 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"). Of course, the empty list has no first element, so 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.

Proof. reflexivity. Qed.

Example test_hd2: hd 0 [] = 0.

Proof. reflexivity. Qed.

Example test_tl: tl [1;2;3] = [2;3].

Proof. reflexivity. Qed.

### Exercises

#### Exercise: 2 stars, 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 :=

(* FILL IN HERE *) admit.

Example test_nonzeros:

nonzeros [0;1;0;2;3;0;0] = [1;2;3].

(* FILL IN HERE *) Admitted.

Fixpoint oddmembers (l:natlist) : natlist :=

(* FILL IN HERE *) admit.

Example test_oddmembers:

oddmembers [0;1;0;2;3;0;0] = [1;3].

(* FILL IN HERE *) Admitted.

Fixpoint countoddmembers (l:natlist) : nat :=

(* FILL IN HERE *) admit.

Example test_countoddmembers1:

countoddmembers [1;0;3;1;4;5] = 4.

(* FILL IN HERE *) Admitted.

Example test_countoddmembers2:

countoddmembers [0;2;4] = 0.

(* FILL IN HERE *) Admitted.

Example test_countoddmembers3:

countoddmembers nil = 0.

(* FILL IN HERE *) Admitted.

☐
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 requires
defining a new kind of pairs, but this is not the only way.)

#### Exercise: 3 stars, advanced (alternate)

Complete the definition of alternate, which "zips up" 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.Fixpoint alternate (l1 l2 : natlist) : natlist :=

(* FILL IN HERE *) admit.

Example test_alternate1:

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].

(* FILL IN HERE *) Admitted.

Example test_alternate3:

alternate [1;2;3] [4] = [1;4;2;3].

(* FILL IN HERE *) Admitted.

Example test_alternate4:

alternate [] [20;30] = [20;30].

(* FILL IN HERE *) Admitted.

☐

### Bags via Lists

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

Complete the following definitions for the functions count, sum, add, and member for bags.
All these proofs can be done just by reflexivity.

Example test_count1: 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.

(* 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, 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 :=

(* FILL IN HERE *) admit.

Example test_sum1: count 1 (sum [1;2;3] [1;4;1]) = 3.

(* FILL IN HERE *) Admitted.

Definition add (v:nat) (s:bag) : bag :=

(* FILL IN HERE *) admit.

Example test_add1: count 1 (add 1 [1;4;1]) = 3.

(* FILL IN HERE *) Admitted.

Example test_add2: count 5 (add 1 [1;4;1]) = 0.

(* FILL IN HERE *) Admitted.

Definition member (v:nat) (s:bag) : bool :=

(* FILL IN HERE *) admit.

Example test_member1: member 1 [1;4;1] = true.

(* FILL IN HERE *) Admitted.

Example test_member2: member 2 [1;4;1] = false.

(* FILL IN HERE *) Admitted.

☐

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

Here are some more bag functions for you to practice with.Fixpoint remove_one (v:nat) (s:bag) : bag :=

(* When remove_one is applied to a bag without the number to remove,

it should return the same bag unchanged. *)

(* FILL IN HERE *) admit.

Example test_remove_one1:

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.

(* FILL IN HERE *) Admitted.

Example test_remove_one3:

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.

(* FILL IN HERE *) Admitted.

Fixpoint remove_all (v:nat) (s:bag) : bag :=

(* FILL IN HERE *) admit.

Example test_remove_all1: 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.

(* FILL IN HERE *) Admitted.

Example test_remove_all3: 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.

(* FILL IN HERE *) Admitted.

Fixpoint subset (s

_{1}:bag) (s

_{2}:bag) : bool :=

(* FILL IN HERE *) admit.

Example test_subset1: subset [1;2] [2;1;4;1] = true.

(* FILL IN HERE *) Admitted.

Example test_subset2: subset [1;2;2] [2;1;4;1] = false.

(* FILL IN HERE *) Admitted.

☐

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

Write down an interesting theorem bag_theorem about bags involving the functions count and add, and prove it. For this, replace the admit command below with the statement of your theorem. 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!
☐

# Reasoning About Lists

... because the [] is substituted into the match
"scrutinee" 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 : ∀l:natlist,

pred (length l) = length (tl l).

Proof.

intros l. destruct l as [| n l'].

- (* l = nil *)

reflexivity.

- (* l = cons 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

## Induction on Lists

*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'.

Theorem app_assoc : ∀l1 l2 l3 : natlist,

(l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).

Proof.

intros l1 l2 l3. induction l1 as [| n l1' IHl1'].

- (* l1 = nil *)

reflexivity.

- (* l1 = cons n l1' *)

simpl. rewrite → IHl1'. 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 written 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 fromn :: ((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].

Proof. reflexivity. Qed.

Example test_rev2: rev nil = nil.

Proof. reflexivity. Qed.

### Proofs about reverse

Theorem rev_length_firsttry : ∀l : natlist,

length (rev l) = length l.

Proof.

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

- (* l = *)

reflexivity.

- (* l = n :: l' *)

(* This is the tricky case. Let's begin as usual

by simplifying. *)

simpl.

(* 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'.

(* ... 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 prove it as a separate
lemma.

Theorem app_length : ∀l1 l2 : natlist,

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

Proof.

(* WORKED IN CLASS *)

intros l1 l2. induction l1 as [| n l1' IHl1'].

- (* l1 = nil *)

reflexivity.

- (* l1 = cons *)

simpl. rewrite → IHl1'. reflexivity. Qed.

Note that, to make the lemma as general as possible, we
quantify over
Now we can complete the original proof.

*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.Theorem rev_length : ∀l : natlist,

length (rev l) = length l.

Proof.

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

- (* l = nil *)

reflexivity.

- (* l = cons *)

simpl. rewrite → app_length, plus_comm.

rewrite → IHl'. reflexivity. Qed.

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

*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 showlength ((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 showlength (rev (n :: l')) = length (n :: l').By the definition of rev, this follows fromlength ((rev l') ++ [n]) = S (length l')which, by the previous lemma, is the same aslength (rev l') + length [n] = S (length l').This follows directly from the induction hypothesis and the definition of length. ☐

*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'. ☐## SearchAbout

(* SearchAbout rev. *)

Keep SearchAbout 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 SearchAbout with C-c
C-a C-a. Pasting its response into your buffer can be
accomplished with C-c C-;.

Theorem app_nil_r : ∀l : natlist,

l ++ [] = l.

Proof.

(* FILL IN HERE *) Admitted.

Theorem rev_involutive : ∀l : natlist,

rev (rev l) = l.

Proof.

(* 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 : ∀l1 l2 l3 l4 : natlist,

l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4.

Proof.

(* FILL IN HERE *) Admitted.

An exercise about your implementation of nonzeros:

Lemma nonzeros_app : ∀l1 l2 : natlist,

nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 2 stars (beq_natlist)

Fill in the definition of beq_natlist, which compares lists of numbers for equality. Prove that beq_natlist l l yields true for every list l.Fixpoint beq_natlist (l1 l2 : natlist) : bool :=

(* FILL IN HERE *) admit.

Example test_beq_natlist1 :

(beq_natlist nil nil = true).

(* FILL IN HERE *) Admitted.

Example test_beq_natlist2 :

beq_natlist [1;2;3] [1;2;3] = true.

(* FILL IN HERE *) Admitted.

Example test_beq_natlist3 :

beq_natlist [1;2;3] [1;2;4] = false.

(* FILL IN HERE *) Admitted.

Theorem beq_natlist_refl : ∀l:natlist,

true = beq_natlist l l.

Proof.

(* FILL IN HERE *) Admitted.

☐

## List Exercises, Part 2

#### Exercise: 3 stars, advanced (bag_proofs)

Here are a couple of little theorems to prove about your definitions about bags earlier in the file.Theorem count_member_nonzero : ∀(s : bag),

leb 1 (count 1 (1 :: s)) = true.

Proof.

(* FILL IN HERE *) Admitted.

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

Theorem ble_n_Sn : ∀n,

leb n (S n) = true.

Proof.

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

- (* 0 *)

simpl. reflexivity.

- (* S n' *)

simpl. rewrite IHn'. reflexivity. Qed.

Theorem remove_decreases_count: ∀(s : bag),

leb (count 0 (remove_one 0 s)) (count 0 s) = true.

Proof.

(* FILL IN HERE *) Admitted.

☐

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

Write down an interesting theorem bag_count_sum about bags involving the functions count and sum, and prove it.(* FILL IN HERE *)

☐

#### Exercise: 4 stars, advanced (rev_injective)

Prove that the rev function is injective — that is,
∀(l1 l2 : natlist), rev l1 = rev l2 → l1 = l2.

(There is a hard way and an easy way to do this.)
(* FILL IN HERE *)

☐

# Options

Fixpoint nth_bad (l:natlist) (n:nat) : nat :=

match l with

| nil ⇒ 42 (* arbitrary! *)

| a :: l' ⇒ match beq_nat 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.

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 beq_nat 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.

Proof. reflexivity. Qed.

Example test_nth_error2 : nth_error [4;5;6;7] 3 = Some 7.

Proof. reflexivity. Qed.

Example test_nth_error3 : nth_error [4;5;6;7] 9 = None.

Proof. reflexivity. Qed.

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 beq_nat 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 allows conditional expressions over
The function below pulls the nat out of a natoption, returning
a supplied default in the None case.

*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.Definition option_elim (d : nat) (o : natoption) : nat :=

match o with

| Some n' ⇒ n'

| None ⇒ d

end.

#### Exercise: 2 stars (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 :=

(* FILL IN HERE *) admit.

Example test_hd_error1 : hd_error [] = None.

(* FILL IN HERE *) Admitted.

Example test_hd_error2 : hd_error [1] = Some 1.

(* FILL IN HERE *) Admitted.

Example test_hd_error3 : hd_error [5;6] = Some 5.

(* FILL IN HERE *) Admitted.

☐

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

This exercise relates your new hd_error to the old hd.Theorem option_elim_hd : ∀(l:natlist) (default:nat),

hd default l = option_elim default (hd_error l).

Proof.

(* FILL IN HERE *) Admitted.

☐

# Partial Maps

*partial map*data type, analogous to the map or dictionary data structures found in most programming languages.

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 beq_id x

_{1}x

_{2}:=

match x

_{1}, x

_{2}with

| Id n

_{1}, Id n

_{2}⇒ beq_nat n

_{1}n

_{2}

end.

☐
Now we define the type of partial maps:

Module PartialMap.

Import NatList.

Inductive partial_map : Type :=

| empty : partial_map

| record : id → nat → partial_map → 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 (or adds a new entry if the given key is not already
present).

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 (key : id) (d : partial_map) : natoption :=

match d with

| empty ⇒ None

| record k v d' ⇒ if beq_id key k

then Some v

else find key d'

end.

Theorem update_eq :

∀(d : partial_map) (k : id) (v: nat),

find k (update d k v) = Some v.

Proof.

(* FILL IN HERE *) Admitted.

∀(d : partial_map) (k : id) (v: nat),

find k (update d k v) = Some v.

Proof.

(* FILL IN HERE *) Admitted.

Theorem update_neq :

∀(d : partial_map) (m n : id) (o: nat),

beq_id m n = false → find m (update d n o) = find m d.

Proof.

(* FILL IN HERE *) Admitted.

∀(d : partial_map) (m n : id) (o: nat),

beq_id m n = false → find m (update d n o) = find m d.

Proof.

(* FILL IN HERE *) Admitted.

☐

How
(* FILL IN HERE *)

☐

*many*elements does the type baz have?☐