I'm trying to understand Agda's built in reflection mechanisms, so I decided to write a simple function that takes a string for an identifier,
a quoted type, and a quoted term, and simply defines a term of the given type with the given string identifier. Thus, I wrote the following:
testDefineName' : String → TC Type → TC Term → TC ⊤
testDefineName' s t a =
bindTC (freshName s)
(λ n → bindTC t
(λ t' → bindTC a
(λ a' → (bindTC (declareDef (arg (arg-info visible relevant) n) t')
(λ _ → defineFun n ((clause [] a') ∷ []))))))
unquoteDecl = (testDefineName' "test" (quoteTC ℕ) (quoteTC zero))
This type-checks, but, when I try to use "test" elsewhere I get a Not in scope: test error.
The documentation for unquoteDecl is kind of opaque. Appaently declarations should be of the form
unquoteDecl x_1 x_2 ... x_n = m
where the x_i are Names, and m has type TC \top, so maybe what I was trying to do isn't actually possible, but I still don't understand how this mechanism works: if m has to be of type TC ⊤, and thus cannot be a function of the names x_1 x_2 ... x_n, I don't see how it is possible to bring any new names into scope using unquoteDecl at all!
So, to sum up:
Is it possible to define a function like mine using Agda's reflection mechanism so that I can bring new names into scope using a String argument? What I want is something like:
<boilerplate code for unquoting> testDefineName' "test" (quoteTC ℕ) (quoteTC zero)
test' : ℕ
test' = test
to compile (i.e. I can actually use the new name, "test")
and if not, by what mechanism can m make use of the names x_1 ... x_n? Can m actually have a type like List Name → TC ⊤, contrary to the documentation?
Based on the way Ulf uses unquoteDecl, I have the impression that you need to list on the LHS the names which are going to extend the scope.
The problem with your setup is that you don't know the Name as you generate a fresh one from a String and have no way of getting hold of it AFAIK. I have the impression that freshName is only supposed to be used to generate internal names from inside a tactic.
If you make testDefineName' take a Name rather than a String then everything works out:
testDefineName' : Name → TC Type → TC Term → TC ⊤
testDefineName' n t a = bindTC t
$ λ t' → bindTC a
$ λ a' → bindTC (declareDef (arg (arg-info visible relevant) n) t')
$ λ _ → defineFun n ((clause [] a') ∷ [])
unquoteDecl test = testDefineName' test (quoteTC ℕ) (quoteTC zero)
Related
I wrote a function on the natural numbers that uses the operator _<?_ with the with-abstraction.
open import Data.Maybe
open import Data.Nat
open import Data.Nat.Properties
open import Relation.Binary.PropositionalEquality
open import Relation.Nullary
fun : ℕ → ℕ → Maybe ℕ
fun x y with x <? y
... | yes _ = nothing
... | no _ = just y
I would like to prove that if the result of computing with fun is nothing then the original two values (x and y) fulfill x < y.
So far all my attempts fall short to prove the property:
prop : ∀ (x y)
→ fun x y ≡ nothing
→ x < y
prop x y with fun x y
... | just _ = λ()
... | nothing = λ{refl → ?} -- from-yes (x <? y)}
-- This fails because the pattern matching is incomplete,
-- but it shouldn't. There are no other cases
prop' : ∀ (x y)
→ fun x y ≡ nothing
→ x < y
prop' x y with fun x y | x <? y
... | nothing | yes x<y = λ{refl → x<y}
... | just _ | no _ = λ()
--... | _ | _ = ?
In general, I've found that working with the with-abstraction is painful. It is probably due to the fact that with and | hide some magic in the background. I would like to understand what with and | really do, but the "Technical details" currently escape my understanding. Do you know where to look for to understand how to interpret them?
Concrete solution
You need to case-split on the same element on which you case-split in your function:
prop : ∀ x y → fun x y ≡ nothing → x < y
prop x y _ with x <? y
... | yes p = p
In the older versions of Agda, you would have had to write the following:
prop-old : ∀ x y → fun x y ≡ nothing → x < y
prop-old x y _ with x <? y
prop-old _ _ refl | yes p = p
prop-old _ _ () | no _
But now you are able to completely omit a case when it leads to a direct contradiction, which is, in this case, that nothing and just smth can never be equal.
Detailed explanation
To understand how with works you first need to understand how definitional equality is used in Agda to reduce goals. Definitional equality binds a function call with its associated expression depending on the structure of its input. In Agda, this is easily seen by the use of the equal sign in the definition of the different cases of a function (although since Agda builds a tree of cases some definitional equalities might not hold in some cases, but let's forget this for now).
Let us consider the following definition of the addition over naturals:
_+_ : ℕ → ℕ → ℕ
zero + b = b
(suc a) + b = suc (a + b)
This definition provides two definitional equalities that bind zero + b with b and (suc a) + b with suc (a + b). The good thing with definitional equalities (as opposed to propositional equalities) is that Agda automatically uses them to reduce goals whenever possible. This means that, for instance, if in a further goal you have the element zero + p for any p then Agda will automatically reduce it to p.
To allow Agda to do such reduction, which is fundamental in most cases, Agda needs to know which of these two equalities can be exploited, which means a case-split on the first argument of this addition has to be made in any further proof about addition for a reduction to be possible. (Except for composite proofs based on other proofs which use such case-splits).
When using with you basically add additional definitional equalities depending on the structure of the additional element. This only makes sense, understanding that, that you need to case-split on said element when doing proofs about such a function, in order for Agda once again to be able to make use of these definitional equalities.
Let us take your example and apply this reasoning to it, first without the recent ability to omit impossible cases. You need to prove the following statement:
prop-old : ∀ x y → fun x y ≡ nothing → x < y
Introducing parameters in the context, you write the following line:
prop-old x y p = ?
Having written that line, you need to provide a proof of x < y with the elements in the context. x and y are just natural so you expect p to hold enough information for this result to be provable. But, in this case, p is just of type fun x y ≡ nothing which does not give you enough information. However, this type contains a call to function fun so there is hope ! Looking at the definition of fun, we can see that it yields two definitional equalities, which depend on the structure of x <? y. This means that adding this parameter to the proof by using with once more will allow Agda to make use of these equalities. This leads to the following code:
prop-old : ∀ x y → fun x y ≡ nothing → x < y
prop-old x y p with x <? y
prop-old _ _ p | yes q = ?
prop-old _ _ p | no q = ?
At that point, not only did Agda case-split on x <? y, but it also reduced the goal because it is able, in both cases, to use a specific definitional equality of fun. Let us take a closer look at both cases:
In the yes q case, p is now of type nothing ≡ nothing and q is of type x < y which is exactly what you want to prove, which means the goal is simply solved by:
prop-old _ _ p | yes q = q
I the no q case, something more interesting happens, which is somewhat harder to understand. After reduction, p is now of type just y ≡ nothing because Agda could use the second definitional equality of fun. Since _≡_ is a data type, it is possible to case-split on p which basically asks Agda: "Look at this data type and give me all the possible constructors for an element of type just y ≡ nothing". At first, Agda only finds one possible constructor, refl, but this constructor only builds an element of a type where both sides of the equality are the same, which is not the case here by definition because just and nothing are two distinct constructors from the same data type, Maybe. Agda then concludes that there are no possible constructors that could ever build an element of such type, hence this case is actually not possible, which leads to Agda replacing p with the empty pattern () and dismissing this case. This line is thus simply:
prop-old _ _ () | no _
In the more recent versions of Agda, as I explained earlier, some of these steps are done directly by Agda which allows us to directly omit impossible cases when the emptiness of a pattern can be deduced behind the curtain, which leads to the prettier:
prop : ∀ x y → fun x y ≡ nothing → x < y
prop x y _ with x <? y
... | yes p = p
But it is the same process, just done a bit more automatically. Hopefully, these elements will be of some use in your journey towards understanding Agda.
Reading this answer prompted me to try to construct, and then prove, the canonical form of polymorphic container functions. The construction was straightforward, but the proof stumps me. Below is a simplified-minimized version of how I tried to write the proof.
The simplified version is proving that sufficiently polymorphic functions, due to parametricity, can't change their behaviour only based on the choice of parameter. Let's say we have functions of two arguments, one of a fixed type and one parametric:
PolyFun : Set → Set _
PolyFun A = ∀ {X : Set} → A → X → A
the property I'd like to prove:
open import Relation.Binary.PropositionalEquality
parametricity : ∀ {A X Y} → (f : PolyFun A) → ∀ a x y → f {X} a x ≡ f {Y} a y
parametricity f a x y = {!!}
Are statements like that provable from inside Agda?
Nope, there's no parametricity principle accessible in Agda (yet! [1]).
However you can use these combinators to build the type of the corresponding free theorem and postulate it:
http://wiki.portal.chalmers.se/agda/pmwiki.php?n=Libraries.LightweightFreeTheorems
[1] http://www.cse.chalmers.se/~mouling/share/PresheafModelParametericTT.pdf
I'm working in a simple library containing definitions and properties about substitutions for simple types. I'm using the following encoding for types:
data Ty (n : Nat) : Set where
var : Fin n -> Ty n
con : Ty n
app : Ty n -> Ty n -> Ty n
so, since variables are represented as finite sets, substitutions are just vectors:
Subst : Nat -> Nat -> Set
Subst n m = Vec (Ty m) n
The complete development is in the following paste: http://lpaste.net/107751
Using this encoding, I have defined several lemmas about such substitutions, but I do not know how to define a theorem that specifies that substitutions are idempotent. I believe that I must use some property like weakening in order to express this, but I can't figure out how.
Could someone give any directions or clues?
Thanks in advance.
Substitutions that produce expressions over fresh variables are indeed idempotent. But in order to express this theorem, you have to consider your substitution Subst n m as one operating on the joint variable set Subst (n + m) (n + m). Here is a variant that uses arbitrary variable sets A and B instead of Fin n and Fin m.
open import Relation.Binary.PropositionalEquality using (_≡_; refl)
-- Disjoint union.
data _+_ (A B : Set) : Set where
left : A → A + B
right : B → A + B
-- A set of variable names can be any type.
Names = Set
-- Simple expressions over a set of names.
data Tm (A : Names) : Set where
var : A → Tm A
app : Tm A → Tm A → Tm A
-- Substitute all variables from set A by terms over a name set B.
Subst : Names → Names → Set
Subst A B = A → Tm B
subst : ∀{A B} → Subst A B → Tm A → Tm B
subst σ (var x) = σ x
subst σ (app t u) = app (subst σ t) (subst σ u)
-- Rename all variables from set A to names from set B.
Rename : Names → Names → Set
Rename A B = A → B
rename : ∀{A B} → Rename A B → Tm A → Tm B
rename σ (var x) = var (σ x)
rename σ (app t u) = app (rename σ t) (rename σ u)
-- In order to speak about idempotency of substitutions whose domain A
-- is disjoint from the variable set B used in the range, we have to promote
-- them to work on the union A+B of variable sets.
-- The promoted substitution is the identity on B,
-- and the original substitution on A, but with the result transferred from B to A + B.
promote : ∀{A B} → Subst A B → Subst (A + B) (A + B)
promote σ (left x) = rename right (σ x)
promote σ (right x) = var (right x)
module _ {A B : Set} (σ₀ : Subst A B) where
-- Now assume a promoted substitution.
σ = promote σ₀
-- A promoted substitution has no effect on terms with variables only in B.
lemma : ∀ t → subst σ (rename right t) ≡ rename right t
lemma (var x) = refl
lemma (app t u) rewrite lemma t | lemma u = refl
-- Hence, it is idempotent.
idempotency : ∀ x → subst σ (σ x) ≡ σ x
idempotency (right x) = refl
idempotency (left x) = lemma (σ₀ x)
Is there a way to conveniently use multiple instantiations of EqReasoning where the underlying Setoid is not necessarily semantic equality (i.e. ≡-Reasoning cannot be used)? The reason that ≡-Reasoning is convenient is that the type argument is implicit and it uniquely determines the Setoid in use by automatically selecting semantic equality. When using arbitrary Setoids there is no such unique selection. However a number of structures provide a canonical way to lift a Setoid:
Data.Maybe and Data.Covec have a setoid member.
Data.Vec.Equality.Equality provides enough definitions to write a canonical Setoid lifting for Vec as well. Interestingly there is also is a slightly different equality available at Relation.Binary.Vec.Pointwise, but it does not provide a direct lifting either albeit implementing all the necessary bits.
Data.List ships a Setoid lifting in Relation.Binary.List.Pointwise.
Data.Container also knows how to lift a Setoid.
By using any of these structures, one automatically gets to work with multiple Setoids even if one started out with one. When proofs use these structures (multiple of them in a single proof), it becomes difficult to write down the proof, because EqReasoning must be instantiated for all of them even though each particular Setoid is kind of obvious. This can be done by renaming begin_, _≈⟨_⟩_ and _∎, but I don't consider this renaming convenient.
Consider for example, a proof in the Setoid of Maybe where a sequence of arguments needs to be wrapped in Data.Maybe.Eq.just (think cong just) or a proof in an arbitrary Setoid that temporarily needs to wrap things in a just constructor exploiting its injectivity.
Normally, the only way Agda can pick something for you is when it is uniquely determined by the context. In the case of EqReasoning, there isn't usually enough information to pin down the Setoid, even worse actually: you could have two different Setoids over the same Carrier and _≈_ (consider for example two definitionally unequal proofs of transitivity in the isEquivalence field).
However, Agda does allow special form of implicit arguments, which can be filled as long as there is only one value of the desired type. These are known as instance arguments (think instance as in Haskell type class instances).
To demonstrate roughly how this works:
postulate
A : Set
a : A
Now, instance arguments are wrapped in double curly braces {{}}:
elem : {{x : A}} → A
elem {{x}} = x
If we decide to later use elem somewhere, Agda will check for any values of type A in scope and if there's only one of them, it'll fill that one in for {{x : A}}. If we added:
postulate
b : A
Agda will now complain that:
Resolve instance argument _x_7 : A. Candidates: [a : A, b : A]
Because Agda already allows you to perform computations on type level, instance arguments are deliberately limited in what they can do, namely Agda won't perform recursive search to fill them in. Consider for example:
eq : ... → IsEquivalence _≈_ → IsEquivalence (Eq _≈_)
where Eq is Data.Maybe.Eq mentioned in your question. When you then require Agda to fill in an instance argument of a type IsEquivalence (Eq _≈_), it won't try to find something of type IsEquivalence _≈_ and apply eq to it.
With that out of the way, let's take a look at what could work. However, bear in mind that all this stands on unification and as such you might need to push it in the right direction here and there (and if the types you are dealing with become complex, unification might require you to give it so many directions that it won't be worth it in the end).
Personally, I find instance arguments a bit fragile and I usually avoid them (and from a quick check, it would seem that so does the standard library), but your experience might vary.
Anyways, here we go. I constructed a (totally nonsensical) example to demonstrate how to do it. Some boilerplate first:
open import Data.Maybe
open import Data.Nat
open import Relation.Binary
import Relation.Binary.EqReasoning as EqR
To make this example self-contained, I wrote some kind of Setoid with natural numbers as its carrier:
data _≡ℕ_ : ℕ → ℕ → Set where
z≡z : 0 ≡ℕ 0
s≡s : ∀ {m n} → m ≡ℕ n → suc m ≡ℕ suc n
ℕ-setoid : Setoid _ _
ℕ-setoid = record
{ _≈_ = _≡ℕ_
; isEquivalence = record
{ refl = refl
; sym = sym
; trans = trans
}
}
where
refl : Reflexive _≡ℕ_
refl {zero} = z≡z
refl {suc _} = s≡s refl
sym : Symmetric _≡ℕ_
sym z≡z = z≡z
sym (s≡s p) = s≡s (sym p)
trans : Transitive _≡ℕ_
trans z≡z q = q
trans (s≡s p) (s≡s q) = s≡s (trans p q)
Now, the EqReasoning module is parametrized over a Setoid, so usually you do something like this:
open EqR ℕ-setoid
However, we'd like to have the Setoid parameter to be implicit (instance) rather than explicit, so we define and open a dummy module:
open module Dummy {c ℓ} {{s : Setoid c ℓ}} = EqR s
And we can write this simple proof:
idʳ : ∀ n → n ≡ℕ (n + 0)
idʳ 0 = z≡z
idʳ (suc n) = begin
suc n ≈⟨ s≡s (idʳ n) ⟩
suc (n + 0) ∎
Notice that we never had to specify ℕ-setoid, the instance argument picked it up because it was the only type-correct value.
Now, let's spice it up a bit. We'll add Data.Maybe.setoid into the mix. Again, because instance arguments do not perform a recursive search, we'll have to define the setoid ourselves:
Maybeℕ-setoid = setoid ℕ-setoid
_≡M_ = Setoid._≈_ Maybeℕ-setoid
I'm going to postulate few stupid things just to demonstrate that Agda indeed picks correct setoids:
postulate
comm : ∀ m n → (m + n) ≡ℕ (n + m)
eq0 : ∀ n → n ≡ℕ 0
eq∅ : just 0 ≡M nothing
lem : ∀ n → just (n + 0) ≡M nothing
lem n = begin
just (n + 0) ≈⟨ just
(begin
n + 0 ≈⟨ comm n 0 ⟩
n ≈⟨ eq0 n ⟩
0 ∎
)⟩
just 0 ≈⟨ eq∅ ⟩
nothing ∎
I figured an alternative to the proposed solution using instance arguments that slightly bends the requirements, but fits my purpose. The major burden in the question was having to explicitly open EqReasoning multiple times and especially having to invent new names for the contained symbols. A slight improvement would be to pass the correct Setoid once per relation proof. In other words passing it to begin_ or _∎ somehow. Then we could make the Setoid implicit for all the other functions!
import Relation.Binary.EqReasoning as EqR
import Relation.Binary using (Setoid)
module ExplicitEqR where
infix 1 begin⟨_⟩_
infixr 2 _≈⟨_⟩_ _≡⟨_⟩_
infix 2 _∎
begin⟨_⟩_ : ∀ {c l} (X : Setoid c l) → {x y : Setoid.Carrier X} → EqR._IsRelatedTo_ X x y → Setoid._≈_ X x y
begin⟨_⟩_ X p = EqR.begin_ X p
_∎ : ∀ {c l} {X : Setoid c l} → (x : Setoid.Carrier X) → EqR._IsRelatedTo_ X x x
_∎ {X = X} = EqR._∎ X
_≈⟨_⟩_ : ∀ {c l} {X : Setoid c l} → (x : Setoid.Carrier X) → {y z : Setoid.Carrier X} → Setoid._≈_ X x y → EqR._IsRelatedTo_ X y z → EqR._IsRelatedTo_ X x z
_≈⟨_⟩_ {X = X} = EqR._≈⟨_⟩_ X
_≡⟨_⟩_ : ∀ {c l} {X : Setoid c l} → (x : Setoid.Carrier X) → {y z : Setoid.Carrier X} → x ≡ y → EqR._IsRelatedTo_ X y z → EqR._IsRelatedTo_ X x z
_≡⟨_⟩_ {X = X} = EqR._≡⟨_⟩_ X
Reusing the nice example from Vitus answer, we can write it:
lem : ∀ n → just (n + 0) ≡M nothing
lem n = begin⟨ Data.Maybe.setoid ℕ-setoid ⟩
just (n + 0) ≈⟨ just
(begin⟨ ℕ-setoid ⟩
n + 0 ≈⟨ comm n 0 ⟩
n ≈⟨ eq0 n ⟩
0 ∎
)⟩
just 0 ≈⟨ eq∅ ⟩
nothing ∎
where open ExplicitEqR
One still has to mention the Setoids in use, to avoid the use of instance arguments as presented by Vitus. However the technique makes it significantly more convenient.
Here's what I understand about Relation.Binary.PropositionalEquality.TrustMe.trustMe: it seems to take an arbitrary x and y, and:
if x and y are genuinely equal, it becomes refl
if they are not, it behaves like postulate lie : x ≡ y.
Now, in the latter case it can easily make Agda inconsistent, but this in itself is not so much a problem: it just means that any proof using trustMe is a proof by appeal to authority. Moreover, though you can use such things to write coerce : {A B : Set} -> A -> B, it turns out to be the case that coerce {ℕ} {Bool} 0 doesn't reduce (at least, not according to C-c C-n), so it's really not analogous to, say, Haskell's semantic-stomping unsafeCoerce.
So what do I have to fear from trustMe? On the other hand, is there ever a reason to use it outside of implementing primitives?
Indeed, attempting to pattern match on trustMe which does not evaluate to refl results in a stuck term. Perhaps it is enlightening to see (part of) the code that defines the primitive operation behind trustMe, primTrustMe:
(u', v') <- normalise (u, v)
if (u' == v') then redReturn (refl $ unArg u) else
return (NoReduction $ map notReduced [a, t, u, v])
Here, u and v represent the terms x and y, respectively. The rest of the code can be found in the module Agda.TypeChecking.Primitive.
So yes, if x and y are not definitionally equal, then primTrustMe (and by extension trustMe) behaves as a postulate in the sense that evaluation simply gets stuck. However, there's one crucial difference when compiling Agda down to Haskell. Taking a look at the module Agda.Compiler.MAlonzo.Primitives, we find this code:
("primTrustMe" , Right <$> do
refl <- primRefl
flip runReaderT 0 $
term $ lam "a" (lam "A" (lam "x" (lam "y" refl))))
This looks suspicious: it always returns refl no matter what x and y are. Let's have a test module:
module DontTrustMe where
open import Data.Nat
open import Data.String
open import Function
open import IO
open import Relation.Binary.PropositionalEquality
open import Relation.Binary.PropositionalEquality.TrustMe
postulate
trustMe′ : ∀ {a} {A : Set a} {x y : A} → x ≡ y
transport : ℕ → String
transport = subst id (trustMe {x = ℕ} {y = String})
main = run ∘ putStrLn $ transport 42
Using trustMe inside transport, compiling the module (C-c C-x C-c) and running the resulting executable, we get... you guessed it right - a segfault.
If we instead use the postulate, we end up with:
DontTrustMe.exe: MAlonzo Runtime Error:
postulate evaluated: DontTrustMe.trustMe′
If you do not intend to compile your programs (at least using MAlonzo) then inconsistency should be your only worry (on the other hand, if you only typecheck your programs then inconsistency usually is kind of a big deal).
There are two use cases I can think of at the moment, first is (as you've said) for implementing primitives. The standard library uses trustMe in three places: in implementation of decidable equality for Names (Reflection module), Strings (Data.String module) and Chars (Data.Char module).
The second one is much like the first one, except that you provide the data type and the equality function yourself and then use trustMe to skip the proving and just use the equality function to define a decidable equality. Something like:
open import Data.Bool
open import Relation.Binary
open import Relation.Binary.PropositionalEquality
open import Relation.Nullary
data X : Set where
a b : X
eq : X → X → Bool
eq a a = true
eq b b = true
eq _ _ = false
dec-eq : Decidable {A = X} _≡_
dec-eq x y with eq x y
... | true = yes trustMe
... | false = no whatever
where postulate whatever : _
However, if you screw up eq, the compiler cannot save you.