# ListsWorking with Structured Data

# Pairs of Numbers

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

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 *)

Since pairs will be used heavily in what follows, 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.

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.

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, for example, the definition of minus in
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:

(* Can't match on a pair with multiple patterns: *)

Definition bad_fst (p : natprod) : nat :=

match p with

| x, y ⇒ x

end.

(* Can't match on multiple values with pair patterns: *)

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.
Now let's try to prove a few simple facts about pairs.
If we state properties of pairs in a slightly peculiar way, we can
sometimes complete their proofs with just reflexivity (and its
built-in simplification):

Fixpoint minus (n m : nat) : nat :=

match n, m with

| O , _ ⇒ O

| S _ , O ⇒ n

| S n', S m' ⇒ minus n' m'

end.

(* Can't match on a pair with multiple patterns: *)

Definition bad_fst (p : natprod) : nat :=

match p with

| x, y ⇒ x

end.

(* Can't match on multiple values with pair patterns: *)

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.

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.

Instead, we need 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, 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 : ∀ (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 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.
Next let's look at several functions for constructing and
manipulating lists. First, the repeat function takes a number
n and a count and returns a list of length count in which
every element is n.

Notation "x + y" := (plus x y)

(at level 50, left associativity).

### Repeat

Fixpoint repeat (n count : nat) : natlist :=

match count with

| O ⇒ nil

| S count' ⇒ n :: (repeat n count')

end.

Fixpoint app (l

_{1}l

_{2}: natlist) : natlist :=

match l

_{1}with

| nil ⇒ l

_{2}

| h :: t ⇒ h :: (app t l

_{2})

end.

Since app will be used extensively, 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].

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 and Tail

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_hd

_{1}: hd 0 [1;2;3] = 1.

Proof. reflexivity. Qed.

Example test_hd

_{2}: hd 0 [] = 0.

Proof. reflexivity. Qed.

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

Proof. reflexivity. Qed.

### Exercises

#### Exercise: 2 stars, standard, especially useful (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_ ." *). Admitted.

Example test_nonzeros:

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_ ." *). Admitted.

Example test_oddmembers:

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_ ." *). Admitted.

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.

☐

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

Complete the following 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.Fixpoint alternate (l

_{1}l

_{2}: natlist) : natlist

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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, standard, especially useful (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_ ." *). Admitted.

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 -- using max instead
of sum -- which is why we don't call this operation union.) 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 one or more functions that have already been defined.

Definition sum : bag → bag → bag

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Example test_sum1: 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_ ." *). Admitted.

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

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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, standard, optional (bag_more_functions)

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

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

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: 2 stars, standard, especially useful (add_inc_count)

Adding a value to a bag should increase the value's count by one. State that as a theorem and prove it.
(*

Theorem bag_theorem : ...

Proof.

...

Qed.

*)

(* Do not modify the following line: *)

Definition manual_grade_for_add_inc_count : option (nat×string) := None.

☐

Theorem bag_theorem : ...

Proof.

...

Qed.

*)

(* Do not modify the following line: *)

Definition manual_grade_for_add_inc_count : option (nat×string) := None.

☐

# Reasoning About Lists

...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 : ∀ 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. We'll see how to do this next.
(Micro-Sermon: As we get deeper into this material, 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

*only*possible shapes that elements of an inductively defined set can have.

*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 : ∀ l

_{1}l

_{2}l

_{3}: natlist,

(l

_{1}++ l

_{2}) ++ l

_{3}= l

_{1}++ (l

_{2}++ l

_{3}).

Proof.

intros l

_{1}l

_{2}l

_{3}. induction l

_{1}as [| n l

_{1}' IHl1'].

- (* l

_{1}= nil *)

reflexivity.

- (* l

_{1}= cons n l

_{1}' *)

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

_{1}' in the cons case.*Theorem*: For all lists l_{1}, l_{2}, and l_{3}, (l_{1}++ l_{2}) ++ l_{3}= l_{1}++ (l_{2}++ l_{3}).*Proof*: By induction on l_{1}.- First, suppose l
_{1}= []. We must show

([] ++ l_{2}) ++ l_{3}= [] ++ (l_{2}++ l_{3}), - Next, suppose l
_{1}= n::l_{1}', with

(l_{1}' ++ l_{2}) ++ l_{3}= l_{1}' ++ (l_{2}++ l_{3})

((n :: l_{1}') ++ l_{2}) ++ l_{3}= (n :: l_{1}') ++ (l_{2}++ l_{3}).

n :: ((l_{1}' ++ l_{2}) ++ l_{3}) = n :: (l_{1}' ++ (l_{2}++ l_{3})),

### Reversing a List

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.

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 : ∀ l : natlist,

length (rev l) = length l.

Proof.

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

- (* l = nil *)

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 at the point where we got
stuck and state it as a separate lemma.

Theorem app_length : ∀ l

_{1}l

_{2}: natlist,

length (l

_{1}++ l

_{2}) = (length l

_{1}) + (length l

_{2}).

Proof.

(* WORKED IN CLASS *)

intros l

_{1}l

_{2}. induction l

_{1}as [| n l

_{1}' IHl1'].

- (* l

_{1}= nil *)

reflexivity.

- (* l

_{1}= 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.

simpl. rewrite → IHl'. rewrite plus_comm.

reflexivity.

Qed.

For comparison, here are informal proofs of these two theorems:
The style of these proofs is rather longwinded and pedantic.
After reading a couple like this, 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:
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 they will already be familiar with.
The more pedantic style is a good default for our present
purposes.

*Theorem*: For all lists l_{1}and l_{2}, length (l_{1}++ l_{2}) = length l_{1}+ length l_{2}.*Proof*: By induction on l_{1}.- First, suppose l
_{1}= []. We must show

length ([] ++ l_{2}) = length [] + length l_{2}, - Next, suppose l
_{1}= n::l_{1}', with

length (l_{1}' ++ l_{2}) = length l_{1}' + length l_{2}.

length ((n::l_{1}') ++ l_{2}) = length (n::l_{1}') + length l_{2}.

*Theorem*: For all lists l, length (rev l) = length l.*Proof*: By induction on l.- First, suppose l = []. We must show

length (rev []) = length [], - Next, suppose l = n::l', with

length (rev l') = length l'.

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

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

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

*Theorem*: For all lists l, length (rev l) = length l.*Proof*: First, observe that length (l ++ [n]) = S (length l) for any l, 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'. ☐## Search

Or say you've forgotten the name of the theorem showing that plus
is commutative. You can use a pattern to search for all theorems
involving the equality of two additions.

You'll see a lot of results there, nearly all of them from the
standard library. To restrict the results, you can search inside
a particular module:

You can also make the search more precise by using variables in
the search pattern instead of wildcards:

The question mark in front of the variable is needed to indicate
that it is a variable in the search pattern, rather than a
variable that is expected to be in scope currently.
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!
Your IDE likely has its own functionality to help with searching.
For example, in ProofGeneral, you can run Search with C-c C-a
C-a, and paste its response into your buffer with C-c C-;.

Theorem app_nil_r : ∀ l : natlist,

l ++ [] = l.

Proof.

(* FILL IN HERE *) Admitted.

Theorem rev_app_distr: ∀ l

_{1}l

_{2}: natlist,

rev (l

_{1}++ l

_{2}) = rev l

_{2}++ rev l

_{1}.

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 : ∀ l

_{1}l

_{2}l

_{3}l

_{4}: natlist,

l

_{1}++ (l

_{2}++ (l

_{3}++ l

_{4})) = ((l

_{1}++ l

_{2}) ++ l

_{3}) ++ l

_{4}.

Proof.

(* FILL IN HERE *) Admitted.

An exercise about your implementation of nonzeros:

Lemma nonzeros_app : ∀ l

_{1}l

_{2}: natlist,

nonzeros (l

_{1}++ l

_{2}) = (nonzeros l

_{1}) ++ (nonzeros l

_{2}).

Proof.

(* 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 (l

_{1}l

_{2}: natlist) : bool

(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Example test_eqblist1 :

(eqblist nil nil = true).

(* FILL IN HERE *) Admitted.

Example test_eqblist2 :

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

(* FILL IN HERE *) Admitted.

Example test_eqblist3 :

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

(* FILL IN HERE *) Admitted.

Theorem eqblist_refl : ∀ l:natlist,

true = eqblist l l.

Proof.

(* FILL IN HERE *) Admitted.

☐

## List Exercises, Part 2

#### Exercise: 1 star, standard (count_member_nonzero)

Theorem count_member_nonzero : ∀ (s : bag),

1 <=? (count 1 (1 :: s)) = true.

Proof.

(* FILL IN HERE *) Admitted.

☐

1 <=? (count 1 (1 :: s)) = true.

Proof.

(* FILL IN HERE *) Admitted.

☐

Theorem leb_n_Sn : ∀ n,

n <=? (S n) = true.

Proof.

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

- (* 0 *)

simpl. reflexivity.

- (* S n' *)

simpl. rewrite IHn'. 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: ∀ (s : bag),

(count 0 (remove_one 0 s)) <=? (count 0 s) = true.

Proof.

(* FILL IN HERE *) Admitted.

☐

(count 0 (remove_one 0 s)) <=? (count 0 s) = true.

Proof.

(* 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 *)

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

Prove that the rev function is injective. There is a hard way and an easy way to do this.Theorem rev_injective : ∀ (l

_{1}l

_{2}: natlist),

rev l

_{1}= rev l

_{2}→ l

_{1}= l

_{2}.

Proof.

(* FILL IN HERE *) Admitted.

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# Options

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

match l with

| nil ⇒ 42

| a :: l' ⇒ match n with

| 0 ⇒ a

| S n' ⇒ nth_bad l' 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 n with

| O ⇒ Some a

| S n' ⇒ nth_error l' 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.

(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 bool type
is not built in, Coq actually supports 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, 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_ ." *). Admitted.

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.

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#### Exercise: 1 star, standard, 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.

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# 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 more flexibility.
We'll also need an equality test for ids:

Definition eqb_id (x

_{1}x

_{2}: id) :=

match x

_{1}, x

_{2}with

| Id n

_{1}, Id n

_{2}⇒ n

_{1}=? n

_{2}

end.

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 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).

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.

Theorem update_eq :

∀ (d : partial_map) (x : id) (v: nat),

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

Proof.

(* FILL IN HERE *) Admitted.

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∀ (d : partial_map) (x : id) (v: nat),

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

Proof.

(* FILL IN HERE *) Admitted.

☐

Theorem update_neq :

∀ (d : partial_map) (x y : id) (o: nat),

eqb_id x y = false → find x (update d y o) = find x d.

Proof.

(* FILL IN HERE *) Admitted.

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∀ (d : partial_map) (x y : id) (o: nat),

eqb_id x y = false → find x (update d y o) = find x d.

Proof.

(* FILL IN HERE *) Admitted.

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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.

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(* 2020-08-25 12:53 *)