Coq QuickChick : Making propery Checkable, Decidable, Arbitrary (Gen) - coq

I am trying to test simple conjecture with QuickChick:
Conjecture lists_eq : forall (l : list string), l = l.
QuickChick lists_eq.
but I get this error:
Unable to satisfy the following constraints:
?arg_2 : "Checkable (forall l : list string, l = l)"
I know I need to make my property Instance of class Checkable, but how should I do it?
I made decidability on two lists (successfully)
Instance EqLists_Dec (x y : list string) : Dec (x = y).
I made Arbitrary for list (successfully)
Instance gen_list_string : Gen (list string)
But I couldn't make Property checkable, I know that bool type is Checkable in QC.
Here is the code, plese advice. Maybe I am doing it all in a wrong way.
From QuickChick Require Import QuickChick.
Require Import List ZArith. Import ListNotations.
Import QcNotation.
Set Warnings "-extraction-opaque-accessed,-extraction".
Require Import String. Local Open Scope string.
Definition genListString : G (list string) :=
elems_ ["Two" ; "Three"]
[
[ "One" ; "Nstasss" ; "SomeStr" ; "Yellow" ];
[ "111" ; "2222" ; "33333SomeStr" ; "44444Yellow" ]
].
(* Sample genListString. *)
Instance gen_list_string : Gen (list string) :=
{
arbitrary := genListString
}.
Notation "P '?'" :=
(match (#dec P _) with
| left _ => true
| right _ => false
end)
(at level 100).
Instance EqLists_Dec (x y : list string) : Dec (x = y).
Proof. dec_eq. Defined.
(* Problem in this function, couldn't make it work*)
Instance checkableDec `{P : Prop} `{Dec P} : Checkable P :=
{
checker p := if P? then ret ok else ret Failure
}.
Conjecture lists_eq : forall (l : list string), l = l.
QuickChick lists_eq.
(*
Unable to satisfy the following constraints:
?arg_2 : "Checkable (forall l : list string, l = l)"
*)

checkableDec already exists in QuickChick. You are actually missing a Shrink instance:
Instance Shrink_string : Shrink string :=
{| shrink x := [] |}.
To debug this you can Set Typeclasses Debug and then look at where the trace says no match for .... You have to be careful of false positives because there is backtracking, but with persistence and experience you'll get the hang of it.

Related

What does the term "?T#{x:=t}", or a type with a question mark and at sign surrounding it mean in Coq?

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.

Coq difficulties in defining constructors using coerced types

Definitions
I'm working on formalizing a typed lambda calculus in Coq, and to keep the notation manageable I've come to rely a lot on coercions. However, I've been running into some difficulties which seem odd.
Right now I'm trying to work with the following types:
type: A descriptor of an allowable type in the language (like function, Unit, etc...)
var: A variable type, defined as nat
VarSets: set of vars
Judgement: a var/my_type pair
ty_ctx: Lists of judgements.
ctx_join: Pairs of ty_ctx's describing disjoint sets of variables
The actual definitions are all given below, except for ctx_join which is given in the next block
(* Imports *)
Require Import lang_spec.
From Coq Require Import MSets.
Require Import List.
Import ListNotations.
Module VarSet := Make(Nat_as_OT).
Inductive Judgement : Type :=
| judge (v : var) (t : type)
.
Definition ty_ctx := (list Judgement).
Definition disj_vars (s1 s2 : VarSet.t) := VarSet.Empty (VarSet.inter s1 s2).
Often I'd like to make statements like "this var does not appear in the set of vars bound by ty_ctx", and to that end I've set up a bunch of coercions between these types below.
(* Functions to convert between the different types listed above *)
Fixpoint var_to_varset (v : var) : VarSet.t :=
VarSet.singleton v.
Coercion var_to_varset : var >-> VarSet.t.
Fixpoint bound_variables (g : ty_ctx) : VarSet.t :=
match g with
| nil => VarSet.empty
| cons (judge v _) g' =>VarSet.union (VarSet.singleton v) (bound_variables g')
end.
Coercion bound_variables : ty_ctx >-> VarSet.t.
Inductive ctx_join :=
| join_single (g : ty_ctx)
| join_double (g1 g2 : ty_ctx)
(disjoint_proof : disj_vars g1 g2)
.
Fixpoint coerce_ctx_join (dj : ctx_join) : ty_ctx :=
match dj with
| join_single g => g
| join_double g1 g2 _ => g1 ++ g2
end.
Coercion coerce_ctx_join : ctx_join >-> ty_ctx.
Fixpoint coerce_judgement_to_ty_ctx (j : Judgement) : ty_ctx :=
cons j nil.
Coercion coerce_judgement_to_ty_ctx : Judgement >-> ty_ctx.
You'll notice that the definition of ctx_join relies on coercing its arguments from ty_ctx to VarSet.
I've drawn up the conversion hierarchy just to make things clear
The Problem
I'd like to define an inductive type with the following constructor
Inductive expr_has_type : ty_ctx -> nat -> type -> Prop :=
(* General Expressions *)
| ty_var (g : ty_ctx) (x : var) (t : type) (xfree : disj_vars x g)
: expr_has_type (join_double (judge x t) g xfree) x t
.
The problem is that when I do, I get the following error:
Error:
In environment
expr_has_type : ty_ctx -> nat -> type -> Prop
g : ty_ctx
x : var
t : type
xfree : disj_vars x g
The term "xfree" has type "disj_vars x g" while it is expected to have type
"disj_vars (judge x t) g" (cannot unify "VarSet.In a (VarSet.inter (judge x t) g)" and
"VarSet.In a (VarSet.inter x g)").
However, if I change the type of xfree to disj_vars (VarSet.singleton x) g, then the definition works fine! This seems very odd, as disj_vars is defined only on VarSets, and so it seems like x should automatically be converted toVarSet.singleton x since that's how the coercion is set up.
Even weirder is the fact that if I don't set up the coercion from vars to varsets, then Coq correctly complains about applying dis_vars to a var instead of a VarSet. So the coercion is definitely doing something
Can someone explain to me why the first definition fails? Given the coercions I've set up, to me it like all the definitions above should be equivalent
Note
Changing the type of xfree to disj_vars (judge x t) g also fixes the error. This seems odd too, since to be able to apply disj_vars to j := (judge x t), it first needs to be coerced to a ty_ctx via cons j nil, then to a VarSet via bound_variables, which should produce a VarSet containing only x (which is equivalent to VarSet.singleton x?). So this coercion chain seems to go off without a hitch, while the other one fails even though it's simpler
If you use Set Printing Coercions., the error message will be much more informative about the problem:
The term "xfree" has type "disj_vars (var_to_varset x) (bound_variables g)"
while it is expected to have type
"disj_vars (bound_variables (coerce_judgement_to_ty_ctx (judge x t)))
(bound_variables g)"
The problem is that the coercion of x into a VarSet.t is equal to Var.singleton x, while the coercion in judge reduces to VarSet.union (VarSet.singleton x) VarSet.empty. While these two are propositionally equal, they are not judgmentally equal, so as far as Coq is concerned, the term it came up with is ill-typed.

Retrieving constraints from GADT to ensure exhaustion of pattern matching in Coq

Let's define two helper types:
Inductive AB : Set := A | B.
Inductive XY : Set := X | Y.
Then two other types that depend on XY and AB
Inductive Wrapped : AB -> XY -> Set :=
| W : forall (ab : AB) (xy : XY), Wrapped ab xy
| WW : forall (ab : AB), Wrapped ab (match ab with A => X | B => Y end)
.
Inductive Wrapper : XY -> Set :=
WrapW : forall (xy : XY), Wrapped A xy -> Wrapper xy.
Note the WW constructor – it can only be value of types Wrapped A X and Wrapped B Y.
Now I would like to pattern match on Wrapper Y:
Definition test (wr : Wrapper Y): nat :=
match wr with
| WrapW Y w =>
match w with
| W A Y => 27
end
end.
but I get error
Error: Non exhaustive pattern-matching: no clause found for pattern WW _
Why does it happen? Wrapper forces contained Wrapped to be A version, the type signature forces Y and WW constructor forbids being A and Y simultaneously. I don't understand why this case is being even considered, while I am forced to check it which seems to be impossible.
How to workaround this situation?
Let's simplify:
Inductive MyTy : Set -> Type :=
MkMyTy : forall (A : Set), A -> MyTy A.
Definition extract (m : MyTy nat) : nat :=
match m with MkMyTy _ x => S x end.
This fails:
The term "x" has type "S" while it is expected to have type "nat".
wat.
This is because I said
Inductive MyTy : Set -> Type
This made the first argument to MyTy an index of MyTy, as opposed to a parameter. An inductive type with a parameter may look like this:
Inductive list (A : Type) : Type :=
| nil : list A
| cons : A -> list A -> list A.
Parameters are named on the left of the :, and are not forall-d in the definition of each constructor. (They are still present in the constructors' types outside of the definition: cons : forall (A : Type), A -> list A -> list A.) If I make the Set a parameter of MyTy, then extract can be defined:
Inductive MyTy (A : Set) : Type :=
MkMyTy : A -> MyTy A.
Definition extract (m : MyTy nat) : nat :=
match m with MkMyTy _ x => S x end.
The reason for this is that, on the inside, a match ignores anything you know about the indices of the scrutinee from the outside. (Or, rather, the underlying match expression in Gallina ignores the indices. When you write a match in the source code, Coq tries to convert it into the primitive form while incorporating information from the indices, but it often fails.) The fact that m : MyTy nat in the first version of extract simply did not matter. Instead, the match gave me S : Set (the name was automatically chosen by Coq) and x : S, as per the constructor MkMyTy, with no mention of nat. Meanwhile, because MyTy has a parameter in the second version, I actually get x : nat. The _ is really a placeholder this time; it is mandatory to write it as _, because there's nothing to match, and you can Set Asymmetric Patterns to make it disappear.
The reason we distinguish between parameters and indices is because parameters have a lot of restrictions—most notably, if I is an inductive type with parameters, then the parameters must appear as variables in the return type of each constructor:
Inductive F (A : Set) : Set := MkF : list A -> F (list A).
(* ^--------^ BAD: must appear as F A *)
In your problem, we should make parameters where we can. E.g. the match wr with Wrap Y w => _ end bit is wrong, because the XY argument to Wrapper is an index, so the fact that wr : Wrapper Y is ignored; you would need to handle the Wrap X w case too. Coq hasn't gotten around to telling you that.
Inductive Wrapped (ab : AB) : XY -> Set :=
| W : forall (xy : XY), Wrapped ab xy
| WW : Wrapped ab (match ab with A => X | B => Y end).
Inductive Wrapper (xy : XY) : Set := WrapW : Wrapped A xy -> Wrapper xy.
And now your test compiles (almost):
Definition test (wr : Wrapper Y): nat :=
match wr with
| WrapW _ w => (* mandatory _ *)
match w with
| W _ Y => 27 (* mandatory _ *)
end
end.
because having the parameters gives Coq enough information for its match-elaboration to use information from Wrapped's index. If you issue Print test., you can see that there's a bit of hoop-jumping to pass information about the index Y through the primitive matchs which would otherwise ignore it. See the reference manual for more information.
The solution turned out to be simple but tricky:
Definition test (wr : Wrapper Y): nat.
refine (match wr with
| WrapW Y w =>
match w in Wrapped ab xy return ab = A -> xy = Y -> nat with
| W A Y => fun _ _ => 27
| _ => fun _ _ => _
end eq_refl eq_refl
end);
[ | |destruct a]; congruence.
Defined.
The issue was that Coq didn't infer some necessary invariants to realize that WW case is ridiculous. I had to explicitly give it a proof for it.
In this solution I changed match to return a function that takes two proofs and brings them to the context of our actual result:
ab is apparently A
xy is apparently Y
I have covered real cases ignoring these assumptions, and I deferred "bad" cases to be proven false later which turned to be trivial. I was forced to pass the eq_refls manually, but it worked and does not look that bad.

Recursive use of typeclass methods in Coq

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.

Pattern matching from a Record type

I have a record type, for example:
Record matrixInt : Type := mkMatrixInt {
const : vector nat dim;
args : vector (matrix dim dim) argCnt
}.
I have a pattern matching where it returns the type of matrixInt, I called it p for example: (where function_name p will return a type of matrixInt. I want to separate p into 2 fields: const and args, for example the draft code I want:
Definition my_function cons arg p :=
match function_name p with
| const => const + cons
| args => args + arg
end.
Could you please help me to write the pattern matching for p that returns 2 fields const; args ?
Thank you very much!
For the record (pun intended):
Record test :=
{ T : Type
; t : T
}.
(* The fields names are indeed accessor functions *)
Definition foo (x : test) : T x := t x.
(* you can destruct a record by matching against its data constructor *)
Definition bar (x : test) : T x :=
match x as _x return T _x with
| Build_test T' t' => t'
end.
(* You can even destruct a record with a let *)
Definition baz (x : test) : T x :=
let (T', t') as _x return T _x := x in t'.
Thank you I have found the answer:(const p) and (args p)