Lists: Products, Lists and Options
The preceding line imports all of our definitions from Basics.v. For
it to work, you need to compile Basics.v into Basics.vo. (This is
like making a .class file from a .java file, or a .o file from a
.c file.)
Here are two ways to compile your code:
In this file, we again use the Module feature to wrap all of the
definitions for pairs and lists of numbers in a module so that,
later, we can reuse the same names for improved (generic) versions
of the same operations.
- CoqIDE
- Command line
Pairs of numbers
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."
Here are some simple function definitions illustrating 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.
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 instruct 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 this already in the previous
chapter -- the comma is provided as part of the standard
library):
Eval simpl in (fst (3,4)).
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 and prove a few simple facts about pairs. If we
state the lemmas in a particular (and slightly peculiar) way, we
can prove them with just reflexivity (and evaluation):
Theorem surjective_pairing' : forall (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 : forall (p : natprod),
p = (fst p, snd p).
Proof.
simpl. (* Doesn't reduce anything! *)
Admitted.
We have to expose the structure of p so that simple can
perform the pattern match in fst and snd. We can do this with
destruct.
Notice that, unlike for nats, destruct doesn't generate an
extra subgoal here. That's because natprods can only be
constructed in one way.
Theorem surjective_pairing : forall (p : natprod),
p = (fst p, snd p).
Proof.
intros p. destruct p as (n,m). simpl. reflexivity. Qed.
Notice that Coq allows us to use the notation we introduced
for pairs in the "as..." pattern telling it what variables to
bind.
Exercise: 2 stars
Theorem snd_fst_is_swap : forall (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 : forall (p : natprod),
fst (swap_pair p) = snd p.
Proof.
(* FILL IN HERE *) Admitted.
fst (swap_pair p) = snd p.
Proof.
(* FILL IN HERE *) Admitted.
☐
Generalizing the definition of pairs a little, 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."
Lists of numbers
For example, here is a three-element list:
As with pairs, it is more convenient to write lists in
familiar mathematical notation. The following two 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 fully understand 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 l_123' := 1 :: (2 :: (3 :: nil)).
Definition l_123'' := 1 :: 2 :: 3 :: nil.
Definition l_123''' := [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 are there to instruct the "coqdoc" tool that the bracketed
part should be displayed as Coq code rather than running text.
These brackets don't appear in the generated HTML.)
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.
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.
Notation "x + y" := (plus x y)
(at level 50, left associativity).
+ will bind tighter than ::, so 1 + 2 :: [3] will be
parsed correctly as (1 + 2) :: [3] rather than 1 + (2 :: [3]).
(at level 50, left associativity).
Fixpoint repeat (n count : nat) : natlist :=
match count with
| O => nil
| S count' => n :: (repeat n count')
end.
The length function calculates the length of a list.
The app ("append") function concatenates two lists.
Fixpoint app (l1 l2 : natlist) : natlist :=
match l1 with
| nil => l2
| h :: t => h :: (app t l2)
end.
In fact, 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.
Here are two more small examples of programming with lists.
The hd function returns the first element (the "head") of the
list, while tl ("tail") returns everything but tqhe first
element. Of course, the empty list has no first element, so we
must make an arbitrary choice in that case.
Definition hd (l:natlist) : nat :=
match l with
| nil => 0 (* arbitrary! *)
| h :: t => h
end.
Definition tl (l:natlist) : natlist :=
match l with
| nil => nil
| h :: t => t
end.
Example test_hd: hd [1,2,3] = 1.
Proof. reflexivity. Qed.
Example test_tl: tl [1,2,3] = [2,3].
Proof. reflexivity. Qed.
Exercise: 1 star (list_funs)
Complete the definitions of nonzeros, oddmembers and countoddmembers below.
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.
(* 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.
☐
One natural way of writing alternate with a single
match..end construct 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 with two nested match...end constructs.
Note that each match must be terminated by an end.
Exercise: 2 stars (alternate)
Complete the definition of alternate.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.
☐
A bag (or multiset) is a set where each element can
appear any finite number of times. One reasonable implementation
of bags is to represent a bag of numbers as a list.
Bags via lists
Exercise: 3 stars (bag_functions)
As an exercise, complete the following definitions for the functions count, union, 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.
(* FILL IN HERE *) Admitted.
Example test_count2: count 6 [1,2,3,1,4,1] = 0.
(* FILL IN HERE *) Admitted.
For union 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 union can be implemented in another way --
perhaps by using functions that have already been defined.
Definition union : bag -> bag -> bag :=
(* FILL IN HERE *) admit.
Example test_union1: count 1 (union [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.
(* FILL IN HERE *) admit.
Example test_union1: count 1 (union [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 extra (optional) bag functions for you to practice with.Fixpoint remove_one (v:nat) (s:bag) : bag :=
(* Note that when remove_one is applied (nonsensically) to the empty
bag, it's OK to return the empty bag. *)
(* 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]) = 0.
(* FILL IN HERE *) Admitted.
Fixpoint subset (s1:bag) (s2: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 (bag_theorem)
Write down an interesting 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!
(* FILL IN HERE *)
Reasoning about lists
... because the [] is substituted into the match position
in the definition of app, allowing the match itself to be
simplified.
Also like 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).
Proof.
intros l. destruct l as [| n l'].
Case "l = nil".
reflexivity.
Case "l = cons n l'".
reflexivity. Qed.
Here, the nil case works because we've chosen 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.
Proofs by induction over data types like natlist are
perhaps a little less familiar than standard natural number
induction, but the basic idea is equally simple. Each Inductive
declaration defines a set of data values that can be built up from
the declared constructors: a number can be either O or S
applied to a number; a boolean can be either true or false; 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:
Since larger lists can only be built up from smaller ones,
stopping eventually with nil, these two things together
establish the truth of P for all lists l.
Induction on lists
- 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', asssuming that P is true for l'.
Theorem ass_app : forall l1 l2 l3 : natlist,
l1 ++ (l2 ++ l3) = (l1 ++ l2) ++ l3.
Proof.
intros l1 l2 l3. induction l1 as [| n l1'].
Case "l1 = nil".
reflexivity.
Case "l1 = cons n l1'".
simpl. rewrite -> IHl1'. reflexivity. Qed.
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. A
natural-language proof needs to include more explicit signposts --
in particular, it helps the reader a lot to be reminded exactly
what the induction hypothesis is in the second case.
Theorem: For all l1, l2, and l3,
l1 ++ (l2 ++ l3) = (l1 ++ l2) ++ l3.
Proof: By induction on l1.
An exercise to be worked together:
- First, suppose l1 = []. We must show:
[] ++ (l2 ++ l3) = ([] ++ l2) ++ l3which 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) ++ l3By the definition of ++, this follows fromn :: (l1' ++ l2 ++ l3) = n :: ((l1' ++ l2) ++ l3)which is immediate from the induction hypothesis. ☐
Theorem app_length : forall l1 l2 : natlist,
length (l1 ++ l2) = plus (length l1) (length l2).
Proof.
(* WORKED IN CLASS *)
intros l1 l2. induction l1 as [| n l1'].
Case "l1 = nil".
reflexivity.
Case "l1 = cons".
simpl. rewrite -> IHl1'. reflexivity. Qed.
For a slightly more involved example of an inductive proof
over lists, suppose we define a "cons on the right" function
snoc like this...
Fixpoint snoc (l:natlist) (v:nat) : natlist :=
match l with
| nil => [v]
| h :: t => h :: (snoc t v)
end.
... and use it to define a list-reversing function rev
like this:
Fixpoint rev (l:natlist) : natlist :=
match l with
| nil => nil
| h :: t => snoc (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.
Now we prove some more list theorems using our newly defined
snoc and rev. Let's try something a little more intricate:
proving that reversing a list does not change its length. Our
first attempt at this proof gets stuck in the successor case...
Theorem rev_length_firsttry : forall l : natlist,
length (rev l) = length l.
Proof.
intros l. induction l as [| n l'].
Case "l = nil".
reflexivity.
Case "l = cons".
simpl. (* Here we get stuck: the goal is an equality involving
snoc, but we don't have any equations in either the
immediate context or the global environment that have
anything to do with snoc! *)
Admitted.
So let's take the equation about snoc that would have
enabled us to make progress and prove it as a separate lemma.
Theorem length_snoc : forall n : nat, forall l : natlist,
length (snoc l n) = S (length l).
Proof.
intros n l. induction l as [| n' l'].
Case "l = nil".
reflexivity.
Case "l = cons n' l'".
simpl. rewrite -> IHl'. reflexivity. Qed.
Now we can complete the original proof.
Theorem rev_length : forall l : natlist,
length (rev l) = length l.
Proof.
intros l. induction l as [| n l'].
Case "l = nil".
reflexivity.
Case "l = cons".
simpl. rewrite -> length_snoc.
rewrite -> IHl'. reflexivity. Qed.
Let's take a look at informal proofs of these two theorems:
Obviously, the style of these proofs is rather longwinded
and pedantic. After we've seen a few of them, we might begin to
find it easier to follow proofs that give a little less detail
overall (since we can easily work them 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
more like this:
Theorem:
For all lists l, length (rev l) = length l.
Proof: First, observe that
The main property now follows by another straightforward
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 on how similar the
proof at hand is to ones that the audience will already be
familiar with. The more pedantic style is usually a safe
fallback.
- Theorem: For all numbers n and lists l,
length (snoc l n) = S (length l).
- First, suppose l = []. We must show
length (snoc [] n) = S (length []),which follows directly from the definitions of length and snoc.
- Next, suppose l = n'::l', with
length (snoc l' n) = S (length l').We must showlength (snoc (n' :: l') n) = S (length (n' :: l'))By the definitions of length and snoc, this follows fromS (length (snoc l' n)) = S (S (length l')),which is immediate from the induction hypothesis. ☐
- First, suppose l = []. We must show
- Theorem: For all lists l, length (rev l) = length 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 (snoc (rev l') n) = S (length l')which, by the previous lemma, is the same asS (length (rev l')) = S (length l').This is immediate from the induction hypothesis. ☐
- First, suppose l = []. We must show
length (snoc l n) = S (length l)
for any l. This follows by a straightforward induction on l.
List exercises, Part 1
Exercise: 2 stars (list_exercises)
More practice with listsTheorem app_nil_end : forall l : natlist,
l ++ [] = l.
Proof.
(* FILL IN HERE *) Admitted.
Theorem rev_involutive : forall l : natlist,
rev (rev l) = l.
Proof.
(* FILL IN HERE *) Admitted.
Theorem distr_rev : forall l1 l2 : natlist,
rev (l1 ++ l2) = (rev l2) ++ (rev l1).
Proof.
(* FILL IN HERE *) Admitted.
There is a short solution to the next exercise. If you find
yourself getting tangled up, step back and try to look for a
simpler way...
Theorem ass_app4 : forall l1 l2 l3 l4 : natlist,
l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4.
Proof.
(* FILL IN HERE *) Admitted.
Theorem snoc_append : forall (l:natlist) (n:nat),
snoc l n = l ++ [n].
Proof.
(* FILL IN HERE *) Admitted.
l1 ++ (l2 ++ (l3 ++ l4)) = ((l1 ++ l2) ++ l3) ++ l4.
Proof.
(* FILL IN HERE *) Admitted.
Theorem snoc_append : forall (l:natlist) (n:nat),
snoc l n = l ++ [n].
Proof.
(* FILL IN HERE *) Admitted.
An exercise about your implementation of nonzeros.
Lemma nonzeros_length : forall l1 l2 : natlist,
nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).
Proof.
(* FILL IN HERE *) Admitted.
nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).
Proof.
(* FILL IN HERE *) Admitted.
☐
List exercises, part 2
Exercise: 2 stars (list_design)
Design exercise:- Write down a non-trivial theorem involving cons (::), snoc, and append (++).
- Prove it.
(* FILL IN HERE *)
☐
Exercise: 2 stars, optional (bag_proofs)
If you did the optional exercise about bags above, here are a couple of little theorems to prove about your definitions.
Theorem count_member_nonzero : forall (s : bag),
ble_nat 0 (count 1 (1 :: s)) = true.
Proof.
(* FILL IN HERE *) Admitted.
ble_nat 0 (count 1 (1 :: s)) = true.
Proof.
(* FILL IN HERE *) Admitted.
The following lemma about ble_nat might help you below.
Theorem ble_n_Sn : forall n,
ble_nat n (S n) = true.
Proof.
intros n. induction n as [| n'].
Case "0".
simpl. reflexivity.
Case "S n'".
simpl. rewrite IHn'. reflexivity. Qed.
Theorem remove_decreases_count: forall (s : bag),
ble_nat (count 0 (remove_one 0 s)) (count 0 s) = true.
Proof.
(* FILL IN HERE *) Admitted.
ble_nat n (S n) = true.
Proof.
intros n. induction n as [| n'].
Case "0".
simpl. reflexivity.
Case "S n'".
simpl. rewrite IHn'. reflexivity. Qed.
Theorem remove_decreases_count: forall (s : bag),
ble_nat (count 0 (remove_one 0 s)) (count 0 s) = true.
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 3 stars, optional (bag_count_union)
Write down an interesting theorem about bags involving the functions count and union, and prove it.
(* FILL IN HERE *)
One use of natoption is as a way of returning "error
codes" from functions. For example, 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 return some
number when the list is too short!
Fixpoint index_bad (n:nat) (l:natlist) : nat :=
match l with
| nil => 42 (* arbitrary! *)
| a :: l' => match beq_nat n O with
| true => a
| false => index_bad (pred n) l'
end
end.
On the other hand, if we give it type nat -> natlist ->
natoption, then we can return None when the list is too short
and Some a when the list has enough members and a appears at
position n.
Fixpoint index (n:nat) (l:natlist) : natoption :=
match l with
| nil => None
| a :: l' => match beq_nat n O with
| true => Some a
| false => index (pred n) l'
end
end.
Example test_index1 : index 0 [4,5,6,7] = Some 4.
Proof. reflexivity. Qed.
Example test_index2 : index 3 [4,5,6,7] = Some 7.
Proof. reflexivity. Qed.
Example test_index3 : index 10 [4,5,6,7] = None.
Proof. reflexivity. Qed.
This example is also an opportunity to introduce one more
small feature of Coq's programming language: conditional
expressions.
Fixpoint index' (n:nat) (l:natlist) : natoption :=
match l with
| nil => None
| a :: l' => if beq_nat n O then Some a else index (pred n) l'
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
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. This function pulls the nat out of a natoption, returning
a supplied default in the None case.
Definition option_elim (o : natoption) (d : nat) : nat :=
match o with
| Some n' => n'
| None => d
end.
Exercise: 2 stars
Using the same idea, fix the hd function from earlier so we don't have to return an arbitrary element in the nil case.
Definition hd_opt (l : natlist) : natoption :=
(* FILL IN HERE *) admit.
Example test_hd_opt1 : hd_opt [] = None.
(* FILL IN HERE *) Admitted.
Example test_hd_opt2 : hd_opt [1] = Some 1.
(* FILL IN HERE *) Admitted.
Example test_hd_opt3 : hd_opt [5,6] = Some 5.
(* FILL IN HERE *) Admitted.
(* FILL IN HERE *) admit.
Example test_hd_opt1 : hd_opt [] = None.
(* FILL IN HERE *) Admitted.
Example test_hd_opt2 : hd_opt [1] = Some 1.
(* FILL IN HERE *) Admitted.
Example test_hd_opt3 : hd_opt [5,6] = Some 5.
(* FILL IN HERE *) Admitted.
Theorem option_elim_hd : forall l:natlist,
hd l = option_elim (hd_opt l) 0.
Proof.
(* FILL IN HERE *) Admitted.
hd l = option_elim (hd_opt l) 0.
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 : forall l:natlist,
beq_natlist l l = true.
Proof.
(* FILL IN HERE *) Admitted.
(* 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 : forall l:natlist,
beq_natlist l l = true.
Proof.
(* FILL IN HERE *) Admitted.
☐
We often encounter situations where the goal to be proved is
exactly the same as some hypothesis in the context or some
previously proved lemma.
The apply tactic
Theorem silly1 : forall (n m o p : nat),
n = m ->
[n,o] = [n,p] ->
[n,o] = [m,p].
Proof.
intros n m o p eq1 eq2.
rewrite <- eq1.
(* At this point, we could finish with rewrite -> eq2. reflexivity. as we have done several times above. But we
can achieve the same effect in a single step by using the apply
tactic instead: *)
apply eq2. Qed.
The apply tactic also works with conditional hypotheses
and lemmas: if the statement being applied is an implication, then
the premises of this implication will be added to the list of
subgoals needing to be proved.
Theorem silly2 : forall (n m o p : nat),
n = m ->
(forall (q r : nat), q = r -> [q,o] = [r,p]) ->
[n,o] = [m,p].
Proof.
intros n m o p eq1 eq2.
apply eq2. apply eq1. Qed.
You may find it instructive to experiment with this proof
and see if there is a way to complete it using just rewrite
instead of apply.
Typically, when we use apply H, the statement H will
begin with a forall binding some "universal variables." When
Coq matches the current goal against the conclusion of H, it
will try to find appropriate values for these variables. For
example, when we do apply eq2 in the following proof, the
universal variable q in eq2 gets instantiated with n and r
gets instantiated with m.
Theorem silly2a : forall (n m : nat),
(n,n) = (m,m) ->
(forall (q r : nat), (q,q) = (r,r) -> [q] = [r]) ->
[n] = [m].
Proof.
intros n m eq1 eq2.
apply eq2. apply eq1. Qed.
Theorem silly_ex :
(forall n, evenb n = true -> oddb (S n) = true) ->
evenb 3 = true ->
oddb 4 = true.
Proof.
(* FILL IN HERE *) Admitted.
(forall n, evenb n = true -> oddb (S n) = true) ->
evenb 3 = true ->
oddb 4 = true.
Proof.
(* FILL IN HERE *) Admitted.
☐
To use the apply tactic, the (conclusion of the) fact
being applied must match the goal EXACTLY -- for example, apply
will not work if the left and right sides of the equality are
swapped.
Theorem silly3_firsttry : forall (n : nat),
true = beq_nat n 5 ->
beq_nat (S (S n)) 7 = true.
Proof.
intros n H.
simpl.
(* here we cannot use apply directly *)
Admitted.
In this case we can use the symmetry tactic, which
switches the left and right sides of an equality in the goal.
Theorem silly3 : forall (n : nat),
true = beq_nat n 5 ->
beq_nat (S (S n)) 7 = true.
Proof.
intros n H.
symmetry.
simpl. (* Actually, this line is unnecessary, since *)
apply H. (* apply will do a simpl step first. *) Qed.
Theorem rev_exercise1 : forall (l l' : natlist),
l = rev l' ->
l' = rev l.
Proof.
(* Hint: you can use apply with previously defined lemmas, not
just hypotheses in the context. *)
(* FILL IN HERE *) Admitted.
l = rev l' ->
l' = rev l.
Proof.
(* Hint: you can use apply with previously defined lemmas, not
just hypotheses in the context. *)
(* FILL IN HERE *) Admitted.
The next exercise is a little tricky. The first line of the proof
is provided, because it uses an idea we haven't seen before.
Notice that we don't introduce m before performing induction.
This leaves it general, so that the IH doesn't specify a
particular m, but lets us pick. We'll talk more about this below,
and later in the course in more depth.
Theorem beq_nat_sym : forall (n m : nat), forall (b : bool),
beq_nat n m = b -> beq_nat m n = b.
Proof.
intros n. induction n as [| n'].
(* FILL IN HERE *) Admitted.
☐
Theorem: For any nats n m and bool b, if
beq_nat n m = b then beq_nat m n = b.
Proof:
(* FILL IN HERE *)
☐ ☐
(* FILL IN HERE *)
☐
One subtlety in these inductive proofs is worth noticing here.
For example, look back at the proof of the ass_app theorem. The
induction hypothesis (in the second subgoal generated by the
induction tactic) is
l1' ++ l2 ++ l3 = (l1' ++ l2) ++ l3 .
(Note that, because we've defined ++ to be right associative,
the expression on the left of the = is the same as writing l1'
++ (l2 ++ l3).)
This hypothesis makes a statement about l1' together with the
particular lists l2 and l3. The lists l2 and l3, which
were introduced into the context by the intros at the top of the
proof, are "held constant" in the induction hypothesis. If we set
up the proof slightly differently by introducing just n into the
context at the top, then we get an induction hypothesis that makes
a stronger claim:
forall l2 l3, l1' ++ l2 ++ l3 = (l1' ++ l2) ++ l3
(Use Coq to see the difference for yourself.)
In the present case, the difference between the two proofs is
minor, since the definition of the ++ function just examines its
first argument and doesn't do anything interesting with its second
argument. But we'll soon come to situations where setting up the
induction hypothesis one way or the other can make the difference
between a proof working and failing.
Exercise: 3 stars (beq_nat_sym_informal)
Provide an informal proof of this lemma:☐ ☐
Exercise: 1 star (apply_rewrite)
Briefly explain the difference between the tactics apply and rewrite. Are there situations where either one can be applied?☐
Varying the Induction Hypothesis
Exercise: 2 stars
Give an alternate proof of the associativity of ++ with a more general induction hypothesis. Complete the following (leaving the first line unchanged).
Theorem ass_app' : forall l1 l2 l3 : natlist,
l1 ++ (l2 ++ l3) = (l1 ++ l2) ++ l3.
Proof.
intros l1. induction l1 as [ | n l1'].
(* FILL IN HERE *) Admitted.
l1 ++ (l2 ++ l3) = (l1 ++ l2) ++ l3.
Proof.
intros l1. induction l1 as [ | n l1'].
(* FILL IN HERE *) Admitted.
☐