Since the natural numbers support a decidable total order, the injection nat_of_ascii (a : ascii) : nat induces a decidable total order on the type ascii. What would be a concise, idiomatic way of expressing this in Coq? (With or without type classes, modules, etc.)
Such process is fairly routine and will depend on the library you have chosen. For order.v, based on math-comp, the process is totally mechanical [in fact, we'll develop a general construction for types with an injection to total orders later in the post]:
From Coq Require Import Ascii String ssreflect ssrfun ssrbool.
From mathcomp Require Import eqtype choice ssrnat.
Require Import order.
Import Order.Syntax.
Import Order.Theory.
Lemma ascii_of_natK : cancel nat_of_ascii ascii_of_nat.
Proof. exact: ascii_nat_embedding. Qed.
(* Declares ascii to be a member of the eq class *)
Definition ascii_eqMixin := CanEqMixin ascii_of_natK.
Canonical ascii_eqType := EqType _ ascii_eqMixin.
(* Declares ascii to be a member of the choice class *)
Definition ascii_choiceMixin := CanChoiceMixin ascii_of_natK.
Canonical ascii_choiceType := ChoiceType _ ascii_choiceMixin.
(* Specific stuff for the order library *)
Definition ascii_display : unit. Proof. exact: tt. Qed.
Open Scope order_scope.
(* We use the order from nat *)
Definition lea x y := nat_of_ascii x <= nat_of_ascii y.
Definition lta x y := ~~ (lea y x).
Lemma lea_ltNeq x y : lta x y = (x != y) && (lea x y).
Proof.
rewrite /lta /lea leNgt negbK lt_neqAle.
by rewrite (inj_eq (can_inj ascii_of_natK)).
Qed.
Lemma lea_refl : reflexive lea.
Proof. by move=> x; apply: le_refl. Qed.
Lemma lea_trans : transitive lea.
Proof. by move=> x y z; apply: le_trans. Qed.
Lemma lea_anti : antisymmetric lea.
Proof. by move=> x y /le_anti /(can_inj ascii_of_natK). Qed.
Lemma lea_total : total lea.
Proof. by move=> x y; apply: le_total. Qed.
(* We can now declare ascii to belong to the order class. We must declare its
subclasses first. *)
Definition asciiPOrderMixin :=
POrderMixin lea_ltNeq lea_refl lea_anti lea_trans.
Canonical asciiPOrderType := POrderType ascii_display ascii asciiPOrderMixin.
Definition asciiLatticeMixin := Order.TotalLattice.Mixin lea_total.
Canonical asciiLatticeType := LatticeType ascii asciiLatticeMixin.
Canonical asciiOrderType := OrderType ascii lea_total.
Note that providing an order instance for ascii gives us access to a large theory of total orders, plus operators, etc..., however the definition of total itself is fairly simple:
"<= is total" == x <= y || y <= x
where <= is a "decidable relation" and we assume, of course, decidability of equality for the particular type. Concretely, for an arbitrary relation:
Definition total (T: Type) (r : T -> T -> bool) := forall x y, r x y || r y x.
so if T is and order, and satisfies total, you are done.
More generally, you can define a generic principle to build such types using injections:
Section InjOrder.
Context {display : unit}.
Local Notation orderType := (orderType display).
Variable (T : orderType) (U : eqType) (f : U -> T) (f_inj : injective f).
Open Scope order_scope.
Let le x y := f x <= f y.
Let lt x y := ~~ (f y <= f x).
Lemma CO_le_ltNeq x y: lt x y = (x != y) && (le x y).
Proof. by rewrite /lt /le leNgt negbK lt_neqAle (inj_eq f_inj). Qed.
Lemma CO_le_refl : reflexive le. Proof. by move=> x; apply: le_refl. Qed.
Lemma CO_le_trans : transitive le. Proof. by move=> x y z; apply: le_trans. Qed.
Lemma CO_le_anti : antisymmetric le. Proof. by move=> x y /le_anti /f_inj. Qed.
Definition InjOrderMixin : porderMixin U :=
POrderMixin CO_le_ltNeq CO_le_refl CO_le_anti CO_le_trans.
End InjOrder.
Then, the ascii instance gets rewritten as follows:
Definition ascii_display : unit. Proof. exact: tt. Qed.
Definition ascii_porderMixin := InjOrderMixin (can_inj ascii_of_natK).
Canonical asciiPOrderType := POrderType ascii_display ascii ascii_porderMixin.
Lemma lea_total : #total ascii (<=%O)%O.
Proof. by move=> x y; apply: le_total. Qed.
Definition asciiLatticeMixin := Order.TotalLattice.Mixin lea_total.
Canonical asciiLatticeType := LatticeType ascii asciiLatticeMixin.
Canonical asciiOrderType := OrderType ascii lea_total.
Related
Let's say we want to prove that weakening the upper bound of a Data.Fin doesn't change the value of the number. The intuitive way to state this is:
weakenEq : (num : Fin n) -> num = weaken num
Let's now generate the defini... Hold on! Let's think a bit about that statement. num and weaken num have different types. Can we state the equality in this case?
The documentation on = suggests we can try to, but we might want to use ~=~ instead. Well, anyway, let's go ahead and generate the definition and case-split, resulting in
weakenEq : (num : Fin n) -> num = weaken num
weakenEq FZ = ?weakenEq_rhs_1
weakenEq (FS x) = ?weakenEq_rhs_2
The goal in the weakenEq_rhs_1 hole is FZ = FZ, which still makes sense from the value point of view. So we optimistically replace the hole with Refl, but only to fail:
When checking right hand side of weakenEq with expected type
FZ = weaken FZ
Unifying k and S k would lead to infinite value
A somewhat cryptic error message, so we wonder if that's really related to the types being different.
Anyway, let's try again, but now with ~=~ instead of =. Unfortunately, the error is still the same.
So, how one would state and prove that weaken x doesn't change the value of x? Does it really make sense to? What should I do if that's a part of a larger proof where I might want to rewrite a Vect n (Fin k) with Vect n (Fin (S k)) that's obtained by mapping weaken over the original vector?
If you really want to prove that the value of the Fin n does not change after applying the weaken function, you would need to prove the equality of these values:
weakenEq: (num: Fin n) -> finToNat num = finToNat $ weaken num
weakenEq FZ = Refl
weakenEq (FS x) = cong $ weakenEq x
To your second problem/Markus' comment about map (Data.Fin.finToNat) v = map (Data.Fin.finToNat . Data.Fin.weaken) v:
vectorWeakenEq : (v: Vect n (Fin k)) ->
map Fin.finToNat v = map (Fin.finToNat . Fin.weaken) v
vectorWeakenEq [] = Refl
vectorWeakenEq (x :: xs) =
rewrite sym $ weakenEq x in
cong {f=(::) (finToNat x)} (vectorWeakenEq xs)
And to see why num = weaken num won't work, let's take a look at a counterexample:
getSize : Fin n -> Nat
getSize _ {n} = n
Now with x : Fin n, getSize x = n != (n + 1) = getSize (weaken x). This won't happen with functions which only depend on the constructors, like finToNat. So you have to constrain yourself to those and prove that they behave like that.
From what I have read, eq_rect and equality seem deeply interlinked. Weirdly, I'm not able to find a definition on the manual for it.
Where does it come from, and what does it state?
If you use Locate eq_rect you will find that eq_rect is located in Coq.Init.Logic, but if you look in that file there is no eq_rect in it. So, what's going on?
When you define an inductive type, Coq in many cases automatically generates 3 induction principles for you, appending _rect, _rec, _ind to the name of the type.
To understand what eq_rect means you need its type,
Check eq_rect.
here we go:
eq_rect
: forall (A : Type) (x : A) (P : A -> Type),
P x -> forall y : A, x = y -> P y
and you need to understand the notion of Leibniz's equality:
Leibniz characterized the notion of equality as follows:
Given any x and y, x = y if and only if, given any predicate P, P(x) if and only if P(y).
In this law, "P(x) if and only if P(y)" can be weakened to "P(x) if P(y)"; the modified law is equivalent to the original, since a statement that applies to "any x and y" applies just as well to "any y and x".
Speaking less formally, the above quotation says that if x and y are equal, their "behavior" for every predicate is the same.
To see more clearly that Leibniz's equality directly corresponds to eq_rect we can rearrange the order of parameters of eq_rect into the following equivalent formulation:
eq_rect_reorder
: forall (A : Type) (P : A -> Type) (x y : A),
x = y -> P x -> P y
I'm learning the Lean proof assistant. An exercise in https://leanprover.github.io/theorem_proving_in_lean/inductive_types.html is to define the predecessor function for the natural numbers. Can someone help me with that?
You are probably familiar with pattern-matching from Lean or some functional programming language, so here is a solution that uses this mechanism:
open nat
definition pred : ℕ → ℕ
| zero := zero
| (succ n) := n
Another way of doing this is using a recursor like so:
def pred (n : ℕ) : ℕ :=
nat.rec_on n 0 (λ p _, p)
Here, 0 is what we return if the argument is zero and (λ p _, p) is an anonymous function that takes two arguments: the predecessor (p) of n and the result of recursive call pred p. The anonymous function ignores the second argument and returns the predecessor.
I'm going through Terry Tao's real analysis textbook, which builds up fundamental mathematics from the natural numbers up. By formalizing as many of the proofs as possible, I hope to familiarize myself with both Idris and dependent types.
I have defined the following datatype:
data GE: Nat -> Nat -> Type where
Ge : (n: Nat) -> (m: Nat) -> GE n (n + m)
to represent the proposition that one natural number is greater than or equal to another.
I'm currently struggling to prove reflexivity of this relation, i.e. to construct the proof with signature
geRefl : GE n n
My first attempt was to simply try geRefl {n} = Ge n Z, but this has type Ge n (add n Z). To get this to unify with the desired signature, we obviously have to perform some kind of rewrite, presumably involving the lemma
plusZeroRightNeutral : (left : Nat) -> left + fromInteger 0 = left
My best attempt is the following:
geRefl : GE n n
geRefl {n} = x
where x : GE n (n + Z)
x = rewrite plusZeroRightNeutral n in Ge n Z
but this does not typecheck.
Can you please give a correct proof of this theorem, and explain the reasoning behind it?
The first problem is superficial: you are trying to apply the rewriting at the wrong place. If you have x : GE n (n + Z), then you'll have to rewrite its type if you want to use it as the definition of geRefl : GE n n, so you'd have to write
geRefl : GE n n
geRefl {n} = rewrite plusZeroRightNeutral n in x
But that still won't work. The real problem is you only want to rewrite a part of the type GE n n: if you just rewrite it using n + 0 = n, you'd get GE (n + 0) (n + 0), which you still can't prove using Ge n 0 : GE n (n + 0).
What you need to do is use the fact that if a = b, then x : GE n a can be turned into x' : GE n b. This is exactly what the replace function in the standard library provides:
replace : (x = y) -> P x -> P y
Using this, by setting P = GE n, and using Ge n 0 : GE n (n + 0), we can write geRefl as
geRefl {n} = replace {P = GE n} (plusZeroRightNeutral n) (Ge n 0)
(note that Idris is perfectly able to infer the implicit parameter P, so it works without that, but I think in this case it makes it clearer what is happening)
I have a simple lemma:
Lemma map2_comm: forall A (f:A->A->B) n (a b:t A n),
(forall x y, (f x y) = (f y x)) -> map2 f a b = map2 f b a.
which I was able to prove using standard equality (≡). Now I am need to prove the similar lemma using setoid equality (using CoRN MathClasses). I am new to this library and type classes in general and having difficulty doing so. My first attempt is:
Lemma map2_setoid_comm `{Equiv B} `{Equiv (t B n)} `{Commutative B A}:
forall (a b: t A n),
map2 f a b = map2 f b a.
Proof.
intros.
induction n.
dep_destruct a.
dep_destruct b.
simpl.
(here '=' is 'equiv'). After 'simpl' the goal is "(nil B)=(nil B)" or "[]=[]" using VectorNotations. Normally I would finish it using 'reflexivity' tactics but it gives me:
Tactic failure: The relation equiv is not a declared reflexive relation. Maybe you need to require the Setoid library.
I guess I need somehow to define reflexivity for vector types, but I am not sure how to do that. Please advise.
First of all the lemma definition needs to be adjusted to:
Lemma map2_setoid_comm : forall `{CO:Commutative B A f} `{SB: !Setoid B} ,
forall n:nat, Commutative (map2 f (n:=n)).
To be able to use reflexivity:
Definition vec_equiv `{Equiv A} {n}: relation (vector A n) := Vforall2 (n:=n) equiv.
Instance vec_Equiv `{Equiv A} {n}: Equiv (vector A n) := vec_equiv.