I'm trying to get my head around the difference between these three interfaces, and when to use each
interface Foo (xs : List Nat) (n : Nat) bar where
...
interface Foo' (xs : List Nat) (bar : Nat -> Type) where
...
interface Foo'' (bar : List Nat -> Nat -> Type) where
...
I assume it's similar to Functor vs Show, but I'm a bit vague on that too. Any thoughts?
Context
I currently use Foo', and I'm using it in a type alias
Baz : Foo' xs bar => Type
Baz {bar} = forall m . Vect m Double -> bar m
but at usage site I need to specify xs twice: Baz {xs=[1]} {bar=MyFoo [1]} for some Foo' xs (MyFoo' xs) where, which I don't really understand because there's only one possible value that could work for xs in the application of Baz. I'm exploring the others to see if they work (no success yet), but I'm not very clear on what the API implications of each are.
Related
I'm trying to write a function which removes the zeros from a list. Here is the first try of my implementation.
Require Import Nat.
Require Import List.
Fixpoint shrink (x : list nat) list nat := (* Mistake! Needs a : list nat *)
match x with
| nil => nil
| h :: t => if h =? 0 then shrink t else (h :: shrink t)
end.
However, I get this error:
Error:
In environment
shrink : forall (x : Datatypes.list Datatypes.nat)
(list0 : ?T) (nat : ?T0#{list:=list0}),
Datatypes.list ?T1#{x:=x; list:=list0; x1:=x}
x : list nat
list : ?T
nat : ?T0
h : Datatypes.nat
t : Datatypes.list Datatypes.nat
The term "shrink t" has type
"forall (list0 : ?T#{x:=t}) (nat : ?T0#{x:=t; list:=list0}),
Datatypes.list ?T1#{x:=t; list:=list0; x1:=t}"
while it is expected to have type "Datatypes.list ?T1#{x1:=h :: t}".
The correct function implementation is this:
Fixpoint shrink (x : list nat) : list nat :=
match x with
| nil => nil
| h :: t => if h =? 0 then shrink t else (h :: shrink t)
end.
But how am I supposed to interpret the error, in particular (list0 : ?T#{x:=t})? In particular,
Why is it list0 instead of list?
What does ? mean before the T?
What does the #{ ... } mean after the T?
For your first point, this is just a printing thing but essentially local variables names are just hints for Coq and it may not respect them to avoid confusion. Here it's probably because of the list := list0 we see later.
Letters preceded by ? are existential variables or evars, used for unification. Here it doesn't know what the type of list0 should be because you did not give any, but it knows it has to be something. So for now it's ?T and it will try to figure out later what it should be by looking at its uses.
For nat it's the same, but it also knows it might depend on some list and in this case its value should be list0 so it does the instantiation ?T0{list:=list0}. This notation is for substitution in evars.
I invite you to look at the documentation of evars in the manual to know more.
I'm trying to remove all integers that are greater than 7 from a list as follows
filter (fun n => n > 7).
However I get the following error
The term "n > 7" has type "Prop" while it is expected to have type "bool".
I am new to Coq, how can I fix it?
The problem is that #List.filter nat expects a function of type nat -> bool but you supplied a function of type nat -> Prop. Here, #List.filter nat is List.filter applied to the type argument nat. One difference between bool and Prop is that bool is decidable while Prop is not: there are propositions P such that neither P nor ~P are known; one can always determine whether something is true or false.
In order to resolve this situation, you need to write a function of type nat -> bool that returns true when applied to an argument greater than 7 and false otherwise. You can take advantage of the fact that the standard library defines boolean comparison functions over the natural numbers. I would also suggest reading the first volume of Software Foundations to familiarize yourself with Coq. It is more accessible and easy-going than some other prominent introductions (it was used in a program verification course at my university that presupposed little functional programming experience).
Here is a minimal example using only the builtin list type and notations:
Require Import Coq.Lists.List.
Import ListNotations.
Fixpoint filterb {A : Type} (f : A -> bool) (xs : list A) : list A :=
match xs with
| [] => []
| x :: xs => if f x then x :: filterb f xs else filterb f xs
end.
Fixpoint ltb (n m : nat) : bool :=
match n, m with
| n , 0 => false
| 0 , S m => true
| S n, S m => ltb n m
end.
Eval compute in (filterb (fun n => ltb 7 n) [5;6;7;8;9]).
(* = [8;9] *)
I have a datatype that depends on a predicate P : a -> Type, meaning that some of its data constructors refer have an implicit P x as argument. I would like idris to be able to automatically infer this implicit. For this, I annotated the implicit with the keyword auto and I wrote a function isP : (x : a) -> Dec (P x) with a %hint annotation before the type declaration. Namely, something like:
module P
P : a -> Type
%hint
isP : (x : a) -> Dec (P x)
and in a separate file
module Main
import P
data Foo : Type where
Bar : (x : a) -> .{auto prf : P x} -> Foo
That said, I can't declare values of said Foo type because Idris claims the it can't infer prf.
Is this because prf has type P x instead of Dec (P x) or is it because the %hint flag does not get imported?
In either case, how can I get Idris to use a Dec value to try to find an implicit?
The %hint flag is imported, as you guess, it's because Dec (P x) is different than P x.
There's a trick though, you can use this datatype:
data IsYes : prop -> Type where
SoTrue : IsYes (Yes prop)
(basically, you define a type that holds a Yes for a given property)
and then you can use default instead of auto to check if the property holds:
data Foo : Type where
Bar : (x : a) -> .{default SoTrue prf : IsYes (isP x)} -> Foo
Note: with this trick, you don't even need %hint anymore, you just check the outcome of isP at compile time.
Is there a way to use recursion with Coq's typeclasses? Like for e.g., in defining show for lists, if you want to call the show function for lists recursively, then you will have to use a fixpoint like so:
Require Import Strings.String.
Require Import Strings.Ascii.
Local Open Scope string_scope.
Class Show (A : Type) : Type :=
{
show : A -> string
}.
Section showNormal.
Instance showList {A : Type} `{Show A} : Show (list A) :=
{
show :=
fix lshow l :=
match l with
| nil => "[]"
| x :: xs => show x ++ " : " ++ lshow xs
end
}.
End showNormal.
Which is all well and good, but what if I want to define some helper function that I'll use for defining Show instances? Like I want to create a more DAZZLING show function called magicShow that prints stars around something...
Definition magicShow {A : Type} `{Show A} (a : A) : string :=
"** " ++ show a ++ " **".
Instance showMagicList {A : Type} `{Show A} : Show (list A) :=
{
show :=
fix lshow l :=
match l with
| nil => "[]"
| x :: xs => show x ++ " : " ++ magicShow xs
end
}.
However, in this case Coq can't find a show instance for the list xs to pass to magicShow:
Error:
Unable to satisfy the following constraints:
In environment:
A : Type
H : Show A
lshow : list A -> string
l : list A
x : A
xs : list A
?H : "Show (list A)"
Is there any way to do this in general? I.e., can you define a method for a typeclass using functions that rely upon the typeclass instance that you're defining?
No, there's no way to do this. This works in Haskell because arbitrary recursive bindings are allowed, and the language doesn't care about the order of bindings. Coq is more restrictive on both fronts. This makes sense if you think about what the desugaring looks like: the recursive call to show would refer to the currently-being-defined instance by name, but that binding isn't in scope yet. And you can't make the instance itself a fixpoint because you're recursing on the structure of a type, not on a value of an algebraic data type.
Your inline fixpoint works for show, but the problem gets thornier if your method implementations refer to each other, such as
newtype MyInteger = MyInteger Integer
instance Num MyInteger where
MyInteger m + MyInteger n = MyInteger $ m + n
negate (MyInteger m) = MyInteger $ negate m
m - n = m + negate n
-- other methods
Here, the calls to (+) and negate in the definition of (-) needs to refer to the definitions of (+) and negate above, but this also doesn't work in Coq. The only solution is to define all your methods separately, manually referencing each other, and then define the instance simply by setting each method to the one you defined above. For example,
Inductive MyInteger := Mk_MyInteger : Integer -> MyInteger.
Definition add__MyInteger (m n : MyInteger) : MyInteger :=
let 'Mk_MyInteger m' := m in
let 'Mk_MyInteger n' := n in
Mk_MyInteger (add m' n').
Definition negate__MyInteger (m : MyInteger) : MyInteger :=
let 'Mk_MyInteger m' := m in
Mk_MyInteger (negate m').
Definition sub__MyInteger (m n : MyInteger) : MyInteger :=
add__MyInteger m (negate__MyInteger n).
Instance Num__MyInteger : Num MyInteger := {|
add := add__MyInteger;
negate := negate__MyInteger;
sub := sub__MyInteger;
(* other methods *)
|}.
If you must do this, it can be simulated by explicitly using the constructor of the underlying Record (since "Typeclasses are Records", to quote from Software Foundations [1]), which can be instantiated using the function(s) being defined as a fixpoint. I'll post three examples and explain where this can be useful.
The example you posted could be solved like this (all code tested for Coq 8.10.1):
Require Import Strings.String.
Local Open Scope list_scope.
Local Open Scope string_scope.
Class Show (A : Type) : Type :=
{
show : A -> string
}.
Definition magicShow {A : Type} `{Show A} (a : A) : string :=
"** " ++ show a ++ " **".
Print Show.
(* Record Show (A : Type) : Type := Build_Show { show : A -> string }
*)
Check Build_Show.
(* Build_Show : forall A : Type, (A -> string) -> Show A *)
Check #magicShow.
(* #magicShow : forall A : Type, Show A -> A -> string *)
Instance showMagicList {A : Type} `{Show A} : Show (list A) :=
{
show :=
fix lshow l :=
match l with
| nil => "[]"
| x :: xs => show x ++ " : " ++ #magicShow _ (#Build_Show _ lshow) xs
end
}.
If you are trying to define several typeclass methods like this, it's tricky to instantiate the record constructor, but it can be done by treating the functions as if they were defined by mutual recursion (although there doesn't necessarily have to be any actual mutual recursion). Here's a contrived example where Show now has two methods. Notice that the typeclass instance is added to the context with an anonymous let-in binding. Evidently, this is enough to satisfy Coq's typeclass resolution mechanism.
Require Import Strings.String.
Local Open Scope list_scope.
Local Open Scope string_scope.
Class Show (A : Type) : Type :=
{
show1 : A -> string
; show2 : A -> string
}.
Definition magicShow1 {A : Type} `{Show A} (a : A) : string :=
"** " ++ show1 a ++ " **".
Definition magicShow2 {A : Type} `{Show A} (a : A) : string :=
"** " ++ show2 a ++ " **".
Fixpoint show1__list {A : Type} `{Show A} (l : list A) : string :=
let _ := (#Build_Show _ show1__list show2__list) in
match l with
| nil => "[]"
| x :: xs => show1 x ++ " : " ++ magicShow1 xs
end
with show2__list {A : Type} `{Show A} (l : list A) : string :=
let _ := (#Build_Show _ show1__list show2__list) in
match l with
| nil => "[]"
| x :: xs => show1 x ++ " : " ++ magicShow2 xs
end.
Instance showMagicList {A : Type} `{Show A} : Show (list A) :=
{
show1 := show1__list
; show2 := show2__list
}.
So why would you want to do this? A good example is when you are defining decidable equality on (rose) trees. In the middle of the definition, we have to recursively appeal to decidable equality of list (tree A). We would like to use the standard library helper function Coq.Classes.EquivDec.list_eqdec [2], which shows how to pass decidable equality on a type A to list A. Since list_eqdec requires a typeclass instance (the very one we are in the middle of defining), we have to use the same trick above:
Require Import Coq.Classes.EquivDec.
Require Import Coq.Program.Utils.
Set Implicit Arguments.
Generalizable Variables A.
Inductive tree (A : Type) : Type :=
| leaf : A -> tree A
| node : list (tree A) -> tree A.
Program Instance tree_eqdec `(eqa : EqDec A eq) : EqDec (tree A) eq :=
{ equiv_dec := fix tequiv t1 t2 :=
let _ := list_eqdec tequiv in
match t1, t2 with
| leaf a1, leaf a2 =>
if a1 == a2 then in_left else in_right
| node ts1, node ts2 =>
if ts1 == ts2 then in_left else in_right
| _, _ => in_right
end
}.
Solve Obligations with unfold not, equiv, complement in * ;
program_simpl ; intuition (discriminate || eauto).
Next Obligation.
destruct t1;
destruct t2;
( program_simpl || unfold complement, not, equiv in *; eauto ).
Qed.
Solve Obligations with split; (intros; try unfold complement, equiv ; program_simpl).
(*
No more obligations remaining
tree_eqdec is defined
*)
Commentary: There is no constructor for creating a record of type EqDec (since it only has one class method), so to convince Coq that list (tree A) has decidable equality, the invocation is simply list_eqdec tequiv. For the uninitiated, Program here is simply allowing for holes in the definition of the instance to be filled in later as Obligations, which is more convenient than writing the appropriate proofs inline.
I am fooling around with Coq. Specifically, I am trying to implement mergesort and then prove that it works.
My attempt at an implementation was:
Fixpoint sort ls :=
match ls with
| nil => nil
| cons x nil => cons x nil
| xs =>
let (left, right) := split xs nil nil
in merge (sort left) (sort right)
end.
The errors that I get as a result of this are:
Error:
Recursive definition of sort is ill-formed.
In environment
sort : list nat -> list nat
ls : list nat
x : nat
l : list nat
y : nat
l0 : list nat
left : list nat
right : list nat
Recursive call to sort has principal argument equal to "left" instead of
one of the following variables: "l" "l0".
Recursive definition is:
"fun ls : list nat =>
match ls with
| nil => nil
| x :: nil => x :: nil
| x :: _ :: _ =>
let (left, right) := split ls nil nil in merge (sort left) (sort right)
end".
My interpretation of these errors is that l and l0 are ls without its head, x, and ls without x and the element after x (which I guess it decided to call y?). It is mad that I did not recurse on one of these lists and instead recursed on a locally defined list.
Am I only allowed to recurse on things that come directly from the pattern match? If yes, this seems like a severe limitation. Are there ways around it? I am guessing that Coq can't tell that the function will terminate. Is there some way to prove to it that left and right are smaller than xs?
It turns out that the chapter of CPDT on General Recursion addresses just that particular issue:
http://adam.chlipala.net/cpdt/html/GeneralRec.html
Read the section called Well-founded recursion, it implements the merge sort using well-founded recursion to help Coq's termination checker be happy.
There may be other ways to solve that problem using either Function or Program Fixpoint, but I think reading about well-founded recursion will not hurt.