ListsWorking with Structured Data
Inductive natprod : Type :=
| pair (n1 n2 : nat).
Check (pair 3 5) : natprod.
| pair (n1 n2 : nat).
Check (pair 3 5) : natprod.
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 *)
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 *)
Notation "( x , y )" := (pair x y).
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.
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.
Theorem surjective_pairing' : ∀ (n m : nat),
(n,m) = (fst (n,m), snd (n,m)).
Proof.
reflexivity. Qed.
(n,m) = (fst (n,m), snd (n,m)).
Proof.
reflexivity. Qed.
Theorem surjective_pairing_stuck : ∀ (p : natprod),
p = (fst p, snd p).
Proof.
simpl. (* Doesn't reduce anything! *)
Abort.
p = (fst p, snd p).
Proof.
simpl. (* Doesn't reduce anything! *)
Abort.
Theorem surjective_pairing : ∀ (p : natprod),
p = (fst p, snd p).
Proof.
intros p. destruct p as [n m]. simpl. reflexivity. Qed.
p = (fst p, snd p).
Proof.
intros p. destruct p as [n m]. simpl. reflexivity. Qed.
Inductive natlist : Type :=
| nil
| cons (n : nat) (l : natlist).
Definition mylist := cons 1 (cons 2 (cons 3 nil)).
| nil
| cons (n : nat) (l : natlist).
Definition mylist := cons 1 (cons 2 (cons 3 nil)).
Notation "x :: l" := (cons x l)
(at level 60, right associativity).
Notation "[ ]" := nil.
Notation "[ x ; .. ; y ]" := (cons x .. (cons y nil) ..).
(at level 60, right associativity).
Notation "[ ]" := nil.
Notation "[ x ; .. ; y ]" := (cons x .. (cons y nil) ..).
Now these all mean exactly the same thing:
Definition mylist1 := 1 :: (2 :: (3 :: nil)).
Definition mylist2 := 1 :: 2 :: 3 :: nil.
Definition mylist3 := [1;2;3].
Definition mylist2 := 1 :: 2 :: 3 :: nil.
Definition mylist3 := [1;2;3].
Fixpoint repeat (n count : nat) : natlist :=
match count with
| O ⇒ nil
| S count' ⇒ n :: (repeat n count')
end.
match count with
| O ⇒ nil
| S count' ⇒ n :: (repeat n count')
end.
Fixpoint length (l:natlist) : nat :=
match l with
| nil ⇒ O
| h :: t ⇒ S (length t)
end.
match l with
| nil ⇒ O
| h :: t ⇒ S (length t)
end.
Fixpoint app (l1 l2 : natlist) : natlist :=
match l1 with
| nil ⇒ l2
| h :: t ⇒ h :: (app t l2)
end.
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.
match l1 with
| nil ⇒ l2
| h :: t ⇒ h :: (app t l2)
end.
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.
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.
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.
What does the following function do?
Fixpoint foo (n : nat) : natlist :=
match n with
| 0 ⇒ nil
| S n' ⇒ n :: (foo n')
end.
match n with
| 0 ⇒ nil
| S n' ⇒ n :: (foo n')
end.
(Submit any response when you have the answer.)
Theorem nil_app : ∀ l : natlist,
[] ++ l = l.
Proof. reflexivity. Qed.
[] ++ l = l.
Proof. reflexivity. Qed.
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.
pred (length l) = length (tl l).
Proof.
intros l. destruct l as [| n l'].
- (* l = nil *)
reflexivity.
- (* l = cons n l' *)
reflexivity. Qed.
Usually, though, interesting theorems about lists require
induction for their proofs. We'll see how to do this next.
Induction on Lists
Theorem app_assoc : ∀ l1 l2 l3 : natlist,
(l1 ++ l2) ++ l3 = l1 ++ (l2 ++ l3).
Proof.
intros l1 l2 l3. induction l1 as [| n l1' IHl1'].
- (* l1 = nil *)
reflexivity.
- (* l1 = cons n l1' *)
simpl. rewrite → IHl1'. reflexivity. Qed.
(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.
For comparison, here is an informal proof of the same theorem.
- 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
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.
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.
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' *)
simpl.
rewrite <- IHl'.
Abort.
length (rev l) = length l.
Proof.
intros l. induction l as [| n l' IHl'].
- (* l = nil *)
reflexivity.
- (* l = n :: l' *)
simpl.
rewrite <- IHl'.
Abort.
Theorem app_length : ∀ l1 l2 : natlist,
length (l1 ++ l2) = (length l1) + (length l2).
Proof.
(* WORK IN CLASS *) Admitted.
length (l1 ++ l2) = (length l1) + (length l2).
Proof.
(* WORK IN CLASS *) Admitted.
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 add_comm.
reflexivity.
Qed.
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 add_comm.
reflexivity.
Qed.
To prove the following theorem, which tactics will we need besides
intros, simpl, rewrite, and reflexivity? (1) none, (2) destruct,
(3) induction on n, (4) induction on l, or (5) can't be
done with the tactics we've seen.
Theorem foo1 : ∀ n :nat, ∀ l: natlist,
repeat n 0 = l → length l = 0.
Theorem foo1 : ∀ n :nat, ∀ l: natlist,
repeat n 0 = l → length l = 0.
What about the next one?
Theorem foo2 : ∀ n m:nat,
length (repeat n m) = m. Which tactics do we need besides intros, simpl, rewrite, and reflexivity? (1) none, (2) destruct, (3) induction on n, (4) induction on m, or (5) can't be done with the tactics we've seen.
Theorem foo2 : ∀ n m:nat,
length (repeat n m) = m. Which tactics do we need besides intros, simpl, rewrite, and reflexivity? (1) none, (2) destruct, (3) induction on n, (4) induction on m, or (5) can't be done with the tactics we've seen.
What about this one?
Theorem foo3: ∀ n :nat, ∀ l1 l2: natlist,
l1 ++ [n] = l2 → (1 <=? (length l2)) = true. Which tactics do we need besides intros, simpl, rewrite, and reflexivity? (1) none, (2) destruct, (3) induction on n, (4) induction on l1, (5) induction on l2, (6) can't be done with the tactics we've seen.
Theorem foo3: ∀ n :nat, ∀ l1 l2: natlist,
l1 ++ [n] = l2 → (1 <=? (length l2)) = true. Which tactics do we need besides intros, simpl, rewrite, and reflexivity? (1) none, (2) destruct, (3) induction on n, (4) induction on l1, (5) induction on l2, (6) can't be done with the tactics we've seen.
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.
match l with
| nil ⇒ 42
| a :: l' ⇒ match n with
| 0 ⇒ a
| S n' ⇒ nth_bad l' n'
end
end.
Inductive natoption : Type :=
| Some (n : nat)
| None.
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.
| Some (n : nat)
| None.
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.
End NatList.
Partial Maps
Inductive id : Type :=
| Id (n : nat).
| Id (n : nat).
Internally, an id is just a number. Introducing a separate type
by wrapping each nat with the tag Id makes definitions more
readable and gives us more flexibility.
We'll also need an equality test for ids:
Definition eqb_id (x1 x2 : id) :=
match x1, x2 with
| Id n1, Id n2 ⇒ n1 =? n2
end.
match x1, x2 with
| Id n1, Id n2 ⇒ n1 =? n2
end.
Inductive partial_map : Type :=
| empty
| record (i : id) (v : nat) (m : partial_map).
| empty
| record (i : id) (v : nat) (m : partial_map).
The update function overrides the entry for a given key in a partial map by shadowing it with a new one (or simply adds a new entry if the given key is not already present).
Definition update (d : partial_map)
(x : id) (value : nat)
: partial_map :=
record x value d.
(x : id) (value : nat)
: partial_map :=
record x value d.
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.
match d with
| empty ⇒ None
| record y v d' ⇒ if eqb_id x y
then Some v
else find x d'
end.
Is the following claim true or false?
Theorem quiz1 : ∀ (d : partial_map)
(x : id) (v: nat),
find x (update d x v) = Some v.
(x : id) (v: nat),
find x (update d x v) = Some v.
(1) True
(2) False
(3) Not sure
Is the following claim true or false?
Theorem quiz2 : ∀ (d : partial_map)
(x y : id) (o: nat),
eqb_id x y = false →
find x (update d y o) = find x d.
(x y : id) (o: nat),
eqb_id x y = false →
find x (update d y o) = find x d.
(1) True
(2) False
(3) Not sure