(** * Lists: Working with Structured Data *)
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 here: *)
Inductive natprod : Type :=
| pair : nat -> nat -> natprod.
(** 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]." *)
Check (pair 3 5).
(** 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. *)
Notation "( x , y )" := (pair x y).
(** 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' : 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! *)
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).
Proof.
intros p. destruct p as [n m]. simpl. reflexivity. Qed.
(** Notice that, unlike its behavior with [nat]s, [destruct]
doesn't generate an extra subgoal here. That's because [natprod]s
can only be constructed in one way. *)
(** **** Exercise: 1 star (snd_fst_is_swap) *)
Theorem snd_fst_is_swap : forall (p : natprod),
(snd p, fst p) = swap_pair p.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, optional (fst_swap_is_snd) *)
Theorem fst_swap_is_snd : forall (p : natprod),
fst (swap_pair p) = snd p.
Proof.
(* 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 : natlist
| cons : nat -> natlist -> 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 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,
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.
(** 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
(* 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.
Fixpoint 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 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.
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.) *)
Fixpoint alternate (l1 l2 : 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 *)
(** A [bag] (or [multiset]) is like a set, except that each element
can appear multiple times rather than just once. One possible
implementation is to represent a bag of numbers as a list. *)
Definition bag := natlist.
(** **** Exercise: 3 stars, 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_ . *) . 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, 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_ . *) . 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, 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. *)
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 (s1:bag) (s2: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: 3 stars, 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.
*)
(** [] *)
(* ################################################################# *)
(** * Reasoning About Lists *)
(** As with 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.
Proof. reflexivity. Qed.
(** ... 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 : forall 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 *)
(** Simply reading example proof scripts will not get you very far!
It is important to work through the details of each one, using Coq
and thinking 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).
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 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].
Proof. reflexivity. Qed.
Example test_rev2: rev nil = nil.
Proof. reflexivity. Qed.
(* ----------------------------------------------------------------- *)
(** *** Proofs about reverse *)
(** Now let's prove some theorems about our newly defined [rev].
For something a bit more challenging than what we've seen, 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.
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 : forall 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 _all_ [natlist]s, 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.
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:
_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 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. *)
(* ================================================================= *)
(** ** [SearchAbout] *)
(** 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 [SearchAbout] command is quite helpful with this. Typing
[SearchAbout foo] 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]: *)
(* 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-;]. *)
(* ================================================================= *)
(** ** List Exercises, Part 1 *)
(** **** Exercise: 3 stars (list_exercises) *)
(** More practice with lists: *)
Theorem app_nil_r : 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.
(** 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.
Proof.
(* 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).
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
(* REPLACE THIS LINE WITH := _your_definition_ . *) . Admitted.
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,
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 : forall (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 : forall 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: forall (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,
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 *)
(** [] *)
(* ################################################################# *)
(** * Options *)
(** Suppose we want to write a function that returns the [n]th
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 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]. *)
Inductive natoption : Type :=
| Some : nat -> natoption
| None : 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.
(** (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 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
_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 (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.
(** [] *)
(** **** Exercise: 1 star, 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).
Proof.
(* 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 : nat -> id.
(** 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 [id]s: *)
Definition beq_id x1 x2 :=
match x1, x2 with
| Id n1, Id n2 => beq_nat n1 n2
end.
(** **** Exercise: 1 star (beq_id_refl) *)
Theorem beq_id_refl : forall x, true = beq_id x x.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** 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). *)
Definition update (d : partial_map)
(key : id) (value : nat)
: partial_map :=
record key 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 (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.
(** **** Exercise: 1 star (update_eq) *)
Theorem update_eq :
forall (d : partial_map) (k : id) (v: nat),
find k (update d k v) = Some v.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star (update_neq) *)
Theorem update_neq :
forall (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.
(** [] *)
End PartialMap.
(** **** Exercise: 2 stars (baz_num_elts) *)
(** Consider the following inductive definition: *)
Inductive baz : Type :=
| Baz1 : baz -> baz
| Baz2 : baz -> bool -> baz.
(** How _many_ elements does the type [baz] have? (Answer in English
or the natural language of your choice.)
(* FILL IN HERE *)
*)
(** [] *)
(** $Date: 2016-07-11 21:31:32 -0400 (Mon, 11 Jul 2016) $ *)