PermBasic Techniques for Comparisons and Permutations
 sort a sequence of numbers
 finite maps from numbers to (arbitrarytype) data
 finite maps from any ordered type to (arbitrarytype) data
 priority queues: finding/deleting the highest number in a set
 lessthan comparisons on natural numbers, and
 permutations (rearrangements of lists).
The Coq Standard Library
The LessThan Order on the Natural Numbers
And we write x <? y for the computation that returns true or
false depending on whether x is less than y:
Operation < is a reflection of <?, as discussed in
Logic and IndProp. The Nat module has a
theorem showing how they relate:
The Nat module contains a synonym for lt.
For unknown reasons, Nat does not define >=? or >?. So we
define them here:
Definition geb (n m : nat) := m <=? n.
Hint Unfold geb : core.
Infix ">=?" := geb (at level 70) : nat_scope.
Definition gtb (n m : nat) := m <? n.
Hint Unfold gtb : core.
Infix ">?" := gtb (at level 70) : nat_scope.
Hint Unfold geb : core.
Infix ">=?" := geb (at level 70) : nat_scope.
Definition gtb (n m : nat) := m <? n.
Hint Unfold gtb : core.
Infix ">?" := gtb (at level 70) : nat_scope.
The Lia Tactic
The hard way to prove this is by hand.
(* try to remember the name of the lemma about negation and ≤ *)
Search (¬ _ ≤ _ → _).
apply not_le in H_{0}.
(* > is defined using <. Let's convert all inequalities to <. *)
unfold gt in H_{0}.
unfold gt.
(* try to remember the name of the transitivity lemma about > *)
Search (_ < _ → _ < _ → _ < _).
apply Nat.lt_trans with j.
apply H.
apply Nat.lt_trans with (k3).
apply H_{0}.
(* Is k greater than k3? On the integers, sure. But we're working
with natural numbers, which truncate subtraction at zero. *)
Abort.
Theorem truncated_subtraction: ¬ (∀ k:nat, k > k  3).
Proof.
intros contra.
(* specialize applies a hypothesis to an argument *)
specialize (contra 0).
simpl in contra.
inversion contra.
Qed.
Search (¬ _ ≤ _ → _).
apply not_le in H_{0}.
(* > is defined using <. Let's convert all inequalities to <. *)
unfold gt in H_{0}.
unfold gt.
(* try to remember the name of the transitivity lemma about > *)
Search (_ < _ → _ < _ → _ < _).
apply Nat.lt_trans with j.
apply H.
apply Nat.lt_trans with (k3).
apply H_{0}.
(* Is k greater than k3? On the integers, sure. But we're working
with natural numbers, which truncate subtraction at zero. *)
Abort.
Theorem truncated_subtraction: ¬ (∀ k:nat, k > k  3).
Proof.
intros contra.
(* specialize applies a hypothesis to an argument *)
specialize (contra 0).
simpl in contra.
inversion contra.
Qed.
Since subtraction is truncated, does lia_example1 actually hold?
It does. Let's try again, the hard way, to find the proof.
Theorem lia_example1:
∀ i j k,
i < j →
¬ (k  3 ≤ j) →
k > i.
Proof. (* try again! *)
intros.
apply not_le in H_{0}.
unfold gt in H_{0}.
unfold gt.
(* try to remember the name ... *)
Search (_ < _ → _ ≤ _ → _ < _).
apply Nat.lt_le_trans with j.
apply H.
apply Nat.le_trans with (k3).
Search (_ < _ → _ ≤ _).
apply Nat.lt_le_incl.
auto.
apply Nat.le_sub_l.
Qed.
∀ i j k,
i < j →
¬ (k  3 ≤ j) →
k > i.
Proof. (* try again! *)
intros.
apply not_le in H_{0}.
unfold gt in H_{0}.
unfold gt.
(* try to remember the name ... *)
Search (_ < _ → _ ≤ _ → _ < _).
apply Nat.lt_le_trans with j.
apply H.
apply Nat.le_trans with (k3).
Search (_ < _ → _ ≤ _).
apply Nat.lt_le_incl.
auto.
apply Nat.le_sub_l.
Qed.
That was tedious. Here's a much easier way:
Lia is a decision procedure for integer linear arithemetic.
The lia tactic was made available by importing Lia at the
beginning of the file. The tactic
works with Coq types Z and nat, and these operators: < = >
≤ ≥ +  ¬, as well as multiplication by small integer
literals (such as 0,1,2,3...), and some uses of ∨, ∧, and ↔.
Lia does not "understand" other operators. It treats
expressions such as f x y as variables. That is, it
can prove f x y > a × b → f x y + 3 ≥ a × b, in the same way it
would prove u > v → u + 3 ≥ v.
Theorem lia_example_3 : ∀ (f : nat → nat → nat) a b x y,
f x y > a × b → f x y + 3 ≥ a × b.
Proof.
intros. lia.
Qed.
f x y > a × b → f x y + 3 ≥ a × b.
Proof.
intros. lia.
Qed.
Swapping
Definition maybe_swap (al: list nat) : list nat :=
match al with
 a :: b :: ar ⇒ if a >? b then b :: a :: ar else a :: b :: ar
 _ ⇒ al
end.
Example maybe_swap_123:
maybe_swap [1; 2; 3] = [1; 2; 3].
Proof. reflexivity. Qed.
Example maybe_swap_321:
maybe_swap [3; 2; 1] = [2; 3; 1].
Proof. reflexivity. Qed.
match al with
 a :: b :: ar ⇒ if a >? b then b :: a :: ar else a :: b :: ar
 _ ⇒ al
end.
Example maybe_swap_123:
maybe_swap [1; 2; 3] = [1; 2; 3].
Proof. reflexivity. Qed.
Example maybe_swap_321:
maybe_swap [3; 2; 1] = [2; 3; 1].
Proof. reflexivity. Qed.
Applying maybe_swap twice should give the same result as applying it once.
That is, maybe_swap is idempotent.
Theorem maybe_swap_idempotent: ∀ al,
maybe_swap (maybe_swap al) = maybe_swap al.
Proof.
intros [  a [  b al]]; simpl; try reflexivity.
destruct (a >? b) eqn:H_{1}; simpl.
 destruct (b >? a) eqn:H_{2}; simpl.
+
maybe_swap (maybe_swap al) = maybe_swap al.
Proof.
intros [  a [  b al]]; simpl; try reflexivity.
destruct (a >? b) eqn:H_{1}; simpl.
 destruct (b >? a) eqn:H_{2}; simpl.
+
Now what? We have a contradiction in the hypotheses: it
cannot hold that a is less than b and b is less than
a. Unfortunately, lia cannot immediately show that
for us, because it reasons about comparisons in Prop not
bool.
Fail lia.
Abort.
Abort.
Of course we could finish the proof by reasoning directly about
inequalities in bool. But this situation is going to occur
repeatedly in our study of sorting.
Let's set up some machinery to enable using lia on boolean
tests.
The reflect type, defined in the standard library (and presented
in IndProp), relates a proposition to a Boolean. That is,
a value of type reflect P b contains a proof of P if b is
true, or a proof of ¬ P if b is false.
Reflection
Print reflect.
(*
Inductive reflect (P : Prop) : bool > Set :=
 ReflectT : P > reflect P true
 ReflectF : ~ P > reflect P false
*)
(*
Inductive reflect (P : Prop) : bool > Set :=
 ReflectT : P > reflect P true
 ReflectF : ~ P > reflect P false
*)
The standard library proves a theorem that says if P is provable
whenever b = true is provable, then P reflects b.
Using that theorem, we can quickly prove that the propositional
(in)equality operators are reflections of the Boolean
operators.
Lemma eqb_reflect : ∀ x y, reflect (x = y) (x =? y).
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.eqb_eq.
Qed.
Lemma ltb_reflect : ∀ x y, reflect (x < y) (x <? y).
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.ltb_lt.
Qed.
Lemma leb_reflect : ∀ x y, reflect (x ≤ y) (x <=? y).
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.leb_le.
Qed.
Lemma gtb_reflect : ∀ x y, reflect (x > y) (x >? y).
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.ltb_lt.
Qed.
Lemma geb_reflect : ∀ x y, reflect (x ≥ y) (x >=? y).
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.leb_le.
Qed.
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.eqb_eq.
Qed.
Lemma ltb_reflect : ∀ x y, reflect (x < y) (x <? y).
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.ltb_lt.
Qed.
Lemma leb_reflect : ∀ x y, reflect (x ≤ y) (x <=? y).
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.leb_le.
Qed.
Lemma gtb_reflect : ∀ x y, reflect (x > y) (x >? y).
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.ltb_lt.
Qed.
Lemma geb_reflect : ∀ x y, reflect (x ≥ y) (x >=? y).
Proof.
intros x y. apply iff_reflect. symmetry.
apply Nat.leb_le.
Qed.
Here's an example of how you could use these lemmas. Suppose you
have this simple program, (if a <? 5 then a else 2), and you
want to prove that it evaluates to a number smaller than 6. You
can use ltb_reflect "by hand":
Example reflect_example1: ∀ a,
(if a <? 5 then a else 2) < 6.
Proof.
intros a.
(* The next two lines aren't strictly necessary, but they
help make it clear what destruct does. *)
assert (R: reflect (a < 5) (a <? 5)) by apply ltb_reflect.
remember (a <? 5) as guard.
destruct R as [HH] eqn:HR.
× (* ReflectT *) lia.
× (* ReflectF *) lia.
Qed.
(if a <? 5 then a else 2) < 6.
Proof.
intros a.
(* The next two lines aren't strictly necessary, but they
help make it clear what destruct does. *)
assert (R: reflect (a < 5) (a <? 5)) by apply ltb_reflect.
remember (a <? 5) as guard.
destruct R as [HH] eqn:HR.
× (* ReflectT *) lia.
× (* ReflectF *) lia.
Qed.
For the ReflectT constructor, the guard a <? 5 must be equal
to true. The if expression in the goal has already been
simplified to take advantage of that fact. Also, for ReflectT to
have been used, there must be evidence H that a < 5 holds.
From there, all that remains is to show a < 5 entails a < 6.
The lia tactic, which is capable of automatically proving some
theorems about inequalities, succeeds.
For the ReflectF constructor, the guard a <? 5 must be equal
to false. So the if expression simplifies to 2 < 6, which is
immediately provable by lia.
A less didactic version of the above proof wouldn't do the
assert and remember: we can directly skip to destruct.
Example reflect_example1': ∀ a,
(if a <? 5 then a else 2) < 6.
Proof.
intros a. destruct (ltb_reflect a 5); lia.
Qed.
(if a <? 5 then a else 2) < 6.
Proof.
intros a. destruct (ltb_reflect a 5); lia.
Qed.
But even that proof is a little unsatisfactory. The original expression,
a <? 5, is not perfectly apparent from the expression ltb_reflect a 5
that we pass to destruct.
It would be nice to be able to just say something like destruct
(a <? 5) and get the reflection "for free." That's what we'll
engineer, next.
We're now going to build a tactic that you'll want to use, but
you won't need to understand the details of how to build it
yourself.
Let's put several of these reflect lemmas into a Hint database.
We call it bdestruct, because we'll use it in our
booleandestruction tactic:
A Tactic for Boolean Destruction
Hint Resolve ltb_reflect leb_reflect gtb_reflect geb_reflect eqb_reflect : bdestruct.
Here is the tactic, the body of which you do not need to
understand. Invoking bdestruct on Boolean expression b does
the same kind of reasoning we did above: reflection and
destruction. It also attempts to simplify negations involving
inequalities in hypotheses.
Ltac bdestruct X :=
This tactic makes quick, easytoread work of our running example.
Example reflect_example2: ∀ a,
(if a <? 5 then a else 2) < 6.
Proof.
intros.
bdestruct (a <? 5); (* instead of: destruct (ltb_reflect a 5). *)
lia.
Qed.
(if a <? 5 then a else 2) < 6.
Proof.
intros.
bdestruct (a <? 5); (* instead of: destruct (ltb_reflect a 5). *)
lia.
Qed.
Finishing the maybe_swap Proof
Theorem maybe_swap_idempotent: ∀ al,
maybe_swap (maybe_swap al) = maybe_swap al.
Proof.
intros [  a [  b al]]; simpl; try reflexivity.
bdestruct (a >? b); simpl.
maybe_swap (maybe_swap al) = maybe_swap al.
Proof.
intros [  a [  b al]]; simpl; try reflexivity.
bdestruct (a >? b); simpl.
Note how a > b is a hypothesis, rather than a >? b = true.
lia can take care of the contradictory propositional inequalities.
When proving theorems about a program that uses Boolean
comparisons, use bdestruct followed by lia, rather than
destruct followed by application of various theorems about
Boolean operators.
Permutations
Print Permutation.
(*
Inductive Permutation {A : Type} : list A > list A > Prop :=
 perm_nil : Permutation
 perm_skip : forall (x : A) (l l' : list A),
Permutation l l' >
Permutation (x :: l) (x :: l')
 perm_swap : forall (x y : A) (l : list A),
Permutation (y :: x :: l) (x :: y :: l)
 perm_trans : forall l l' l'' : list A,
Permutation l l' >
Permutation l' l'' >
Permutation l l''.
*)
(*
Inductive Permutation {A : Type} : list A > list A > Prop :=
 perm_nil : Permutation
 perm_skip : forall (x : A) (l l' : list A),
Permutation l l' >
Permutation (x :: l) (x :: l')
 perm_swap : forall (x y : A) (l : list A),
Permutation (y :: x :: l) (x :: y :: l)
 perm_trans : forall l l' l'' : list A,
Permutation l l' >
Permutation l' l'' >
Permutation l l''.
*)
You might wonder, "is that really the right definition?" And
indeed, it's important that we get a right definition, because
Permutation is going to be used in our specifications of
searching and sorting algorithms. If we have the wrong
specification, then all our proofs of "correctness" will be
useless.
It's not obvious that this is indeed the right specification of
permutations. (It happens to be, but that's not obvious.) To gain
confidence that we have the right specification, let's use it
prove some properties that permutations ought to have.
YOUR TASK: Add three more properties. Write them here:
Now, let's examine all the theorems in the Coq library about
permutations:
Exercise: 2 stars, standard (Permutation_properties)
Think of some desirable properties of the Permutation relation and write them down informally in English, or a mix of Coq and English. Here are four to get you started: 1. If Permutation al bl, then length al = length bl.
 2. If Permutation al bl, then Permutation bl al.
 3. [1;1] is NOT a permutation of [1;2].
 4. [1;2;3;4] IS a permutation of [3;4;2;1].
Which of the properties that you wrote down above have already
been proved as theorems by the Coq library developers? Answer
here:
(* Do not modify the following line: *)
Definition manual_grade_for_Permutation_properties : option (nat×string) := None.
☐
Definition manual_grade_for_Permutation_properties : option (nat×string) := None.
☐
Example butterfly: ∀ b u t e r f l y : nat,
Permutation ([b;u;t;t;e;r]++[f;l;y]) ([f;l;u;t;t;e;r]++[b;y]).
Proof.
intros.
Permutation ([b;u;t;t;e;r]++[f;l;y]) ([f;l;u;t;t;e;r]++[b;y]).
Proof.
intros.
Let's group [u;t;t;e;r] together on both sides. Tactic
change t with u replaces t with u. Terms t and u must
be convertible, here meaning that they evaluate to the same
term.
We don't actually need to know the list elements in
[u;t;t;e;r]. Let's forget about them and just remember them
as a variable named utter.
Likewise, let's group [f;l] and remember it as a variable.
Next, let's cancel fl from both sides. In order to do that,
we need to bring it to the beginning of each list. For the right
list, that follows easily from the associativity of ++.
But for the left list, we can't just use associativity.
Instead, we need to reason about permutations and use some
library theorems.
apply perm_trans with (fl ++ [y] ++ ([b] ++ utter)).
 replace (fl ++ [y] ++ [b] ++ utter) with ((fl ++ [y]) ++ [b] ++ utter).
+ apply Permutation_app_comm.
+ rewrite < app_assoc. reflexivity.

 replace (fl ++ [y] ++ [b] ++ utter) with ((fl ++ [y]) ++ [b] ++ utter).
+ apply Permutation_app_comm.
+ rewrite < app_assoc. reflexivity.

A library theorem will now help us cancel fl.
Next let's cancel utter.
apply perm_trans with (utter ++ [y] ++ [b]).
+ replace ([y] ++ [b] ++ utter) with (([y] ++ [b]) ++ utter).
× apply Permutation_app_comm.
× rewrite app_assoc. reflexivity.
+ apply Permutation_app_head.
+ replace ([y] ++ [b] ++ utter) with (([y] ++ [b]) ++ utter).
× apply Permutation_app_comm.
× rewrite app_assoc. reflexivity.
+ apply Permutation_app_head.
Finally we're left with just y and b.
That example illustrates a general method for proving permutations
involving cons :: and append ++:
 Identify some portion appearing in both sides.
 Bring that portion to the front on each side using lemmas such as Permutation_app_comm and perm_swap, with generous use of perm_trans.
 Use Permutation_app_head to cancel an appended head. You can also use perm_skip to cancel a single element.
Exercise: 3 stars, standard (permut_example)
Use the permutation rules in the library to prove the following theorem. The following Check commands are a hint about useful lemmas. You don't need all of them, and depending on your approach you will find some lemmas to be more useful than others. Use Search to find others, if you like.
Check perm_skip.
Check perm_trans.
Check Permutation_refl.
Check Permutation_app_comm.
Check app_assoc.
Check app_nil_r.
Check app_comm_cons.
Example permut_example: ∀ (a b: list nat),
Permutation (5 :: 6 :: a ++ b) ((5 :: b) ++ (6 :: a ++ [])).
Proof.
(* FILL IN HERE *) Admitted.
☐
Check perm_trans.
Check Permutation_refl.
Check Permutation_app_comm.
Check app_assoc.
Check app_nil_r.
Check app_comm_cons.
Example permut_example: ∀ (a b: list nat),
Permutation (5 :: 6 :: a ++ b) ((5 :: b) ++ (6 :: a ++ [])).
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 2 stars, standard (not_a_permutation)
Prove that [1;1] is not a permutation of [1;2]. Hints are given as Check commands.
Check Permutation_cons_inv.
Check Permutation_length_1_inv.
Example not_a_permutation:
¬ Permutation [1;1] [1;2].
Proof.
(* FILL IN HERE *) Admitted.
☐
Check Permutation_length_1_inv.
Example not_a_permutation:
¬ Permutation [1;1] [1;2].
Proof.
(* FILL IN HERE *) Admitted.
☐
Correctness of maybe_swap
Theorem maybe_swap_perm: ∀ al,
Permutation al (maybe_swap al).
Proof.
(* WORKED IN CLASS *)
unfold maybe_swap.
destruct al as [  a [  b al]].
 simpl. apply perm_nil.
 apply Permutation_refl.
 bdestruct (a >? b).
+ apply perm_swap.
+ apply Permutation_refl.
Qed.
Permutation al (maybe_swap al).
Proof.
(* WORKED IN CLASS *)
unfold maybe_swap.
destruct al as [  a [  b al]].
 simpl. apply perm_nil.
 apply Permutation_refl.
 bdestruct (a >? b).
+ apply perm_swap.
+ apply Permutation_refl.
Qed.
And, we can prove that maybe_swap permutes elements such that
the first is less than or equal to the second.
Definition first_le_second (al: list nat) : Prop :=
match al with
 a :: b :: _ ⇒ a ≤ b
 _ ⇒ True
end.
Theorem maybe_swap_correct: ∀ al,
Permutation al (maybe_swap al)
∧ first_le_second (maybe_swap al).
Proof.
intros. split.
 apply maybe_swap_perm.
 (* WORKED IN CLASS *)
unfold maybe_swap.
destruct al as [  a [  b al]]; simpl; auto.
bdestruct (a >? b); simpl; lia.
Qed.
match al with
 a :: b :: _ ⇒ a ≤ b
 _ ⇒ True
end.
Theorem maybe_swap_correct: ∀ al,
Permutation al (maybe_swap al)
∧ first_le_second (maybe_swap al).
Proof.
intros. split.
 apply maybe_swap_perm.
 (* WORKED IN CLASS *)
unfold maybe_swap.
destruct al as [  a [  b al]]; simpl; auto.
bdestruct (a >? b); simpl; lia.
Qed.
An Inversion Tactic
Ltac inv H := inversion H; clear H; subst.
Exercise: 3 stars, standard (Forall_perm)
Theorem Forall_perm: ∀ {A} (f: A → Prop) al bl,
Permutation al bl →
Forall f al → Forall f bl.
Proof.
(* FILL IN HERE *) Admitted.
☐
Permutation al bl →
Forall f al → Forall f bl.
Proof.
(* FILL IN HERE *) Admitted.
☐
Summary: Comparisons and Permutations
(* 20230823 11:34 *)