Coq: Strong specification of haskell's Replicate function - haskell

Question

I'm having a bit of trouble understanding the difference between strong and weak specification in Coq. For instance, if I wanted to write the replicate function (given a number n and a value x, it creates a list of length n, with all elements equal to x) using the strong specification way, how would I be able to do that? Apparently I have to write an Inductive "version" of the function but how?
Definition in Haskell:
myReplicate :: Int -> a -> [a]
myReplicate 0 _ = []
myReplicate n x | n > 0 = x:myReplicate (n-1) x
| otherwise = []
Definition of weak specification:
To define these functions with a weak specification and then add companion lemmas.
For instance, we define a function f : A->B and we prove a statement of the form ∀ x:A, Rx (fx), where R is a relation coding the intended input/output behaviour of the function.
Definition of strong specification:
To give a strong specification of the function: the type of this function directly states that the input is a value x of type A and that the output is the combination of a value v of type B and a proof that v satisfies Rxv.
This kind of specification usually relies on dependent types.
EDIT: I heard back from my teacher and apparently I have to do something similar to this, but for the replicate case:
"For example, if we want to extract a function that computes the length of a list from its specification, we can define a relation RelLength which establishes a relation between the expected input and output and then prove it. Like this:
Inductive RelLength (A:Type) : nat -> list A -> Prop :=
| len_nil : RelLength 0 nil
| len_cons : forall l x n, RelLength n l -> RelLength (S n) (x::l) .
Theorem len_corr : forall (A:Type) (l:list A), {n | RelLength n l}.
Proof.
…
Qed.
Recursive Extraction len_corr.
The function used to prove must use the list “recursor” directly (that’s why fixpoint won’t show up - it’s hidden in list_rect).
So you don’t need to write the function itself, only the relation, because the function will be defined by the proof."
Knowing this, how can I apply it to the replicate function case?

Answer 1

Just for fun, here's what it would look like in Haskell, where everything dependent-ish is much more annoying. This code uses some very new GHC features, mostly to make the types more explicit, but it could be modified quite easily to work with older GHC versions.
{-# language GADTs, TypeFamilies, PolyKinds, DataKinds, ScopedTypeVariables,
TypeOperators, TypeApplications, StandaloneKindSignatures #-}
{-# OPTIONS_GHC -Wincomplete-patterns #-}
module RelRepl where
import Data.Kind (Type)
import Data.Type.Equality ((:~:)(..))
-- | Singletons (borrowed from the `singletons` package).
type Sing :: forall (k :: Type). k -> Type
type family Sing
type instance Sing #Nat = SNat
type instance Sing #[a] = SList #a
-- The version of Sing in the singletons package has many more instances;
-- in any case, more can be added anywhere as needed.
-- Natural numbers, used at the type level
data Nat = Z | S Nat
-- Singleton representations of natural numbers, used
-- at the term level.
data SNat :: Nat -> Type where
SZ :: SNat 'Z
SS :: SNat n -> SNat ('S n)
-- Singleton lists
data SList :: forall (a :: Type). [a] -> Type where
SNil :: SList '[]
SCons :: Sing a -> SList as -> SList (a ': as)
-- The relation representing the `replicate` function.
data RelRepl :: forall (a :: Type). Nat -> a -> [a] -> Type where
Repl_Z :: forall x. RelRepl 'Z x '[]
Repl_S :: forall n x l. RelRepl n x l -> RelRepl ('S n) x (x ': l)
-- Dependent pairs, because those aren't natively supported.
data DPair :: forall (a :: Type). (a -> Type) -> Type where
MkDPair :: forall {a :: Type} (x :: a) (p :: a -> Type).
Sing x -> p x -> DPair #a p
-- Proof that every natural number and value produce a list
-- satisfying the relation.
repl_corr :: forall {a :: Type} (n :: Nat) (x :: a).
SNat n -> Sing x -> DPair #[a] (RelRepl n x)
repl_corr SZ _x = MkDPair SNil Repl_Z
repl_corr (SS n) x
| MkDPair l pf <- repl_corr n x
= MkDPair (SCons x l) (Repl_S pf)
-- Here's a proof that the relation indeed specifies
-- a *unique* function.
replUnique :: forall {a :: Type} (n :: Nat) (x :: a) (xs :: [a]) (ys :: [a]).
RelRepl n x xs -> RelRepl n x ys -> xs :~: ys
replUnique Repl_Z Repl_Z = Refl
replUnique (Repl_S pf1) (Repl_S pf2)
| Refl <- replUnique pf1 pf2
= Refl

Answer 2

A possible specification would look like this :
Inductive RelReplicate (A : Type) (a : A) : nat -> (list A) -> Prop :=
| rep0 : RelReplicate A a 0 nil
| repS : …
I did the zero case, leaving you the successor case. Its conclusion should be something like RelReplicate A a (S n) (a :: l).
As in your example, you can then try and prove something like
Theorem replicate_corr : forall (A:Type) (a : A) (n : nat), {l | ReplicateRel A a n l}.
which should be easy by induction on n.
If you want to check that your function replicate_corr corresponds to what you had in mind, you can try it on a few examples, with
Eval compute in (proj1_sig (rep_corr nat 0 3)).
which evaluates the first argument (the one corresponding to the "real function" and not the proof) of rep_corr. To be able to do that, you should end your Theorem with Defined rather than Qed so that Coq can evaluate it.