Haskell 将衍生过孔与幻像类型一起使用

很抱歉我的复述太长了，但我没能把它缩短。以下代码在最后一行之前编译良好：

{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE MultiParamTypeClasses, FlexibleInstances, UndecidableInstances #-}
{-# LANGUAGE DerivingVia, DerivingStrategies, StandaloneDeriving #-}
{-# LANGUAGE ScopedTypeVariables #-}

module Repro where

import Prelude hiding ((+))

class (Additive a) where
    (+) :: a -> a -> a

data Vector2D u = Vector2D { 
    x :: u, 
    y :: u 
}

addVector2 :: (Additive a) => Vector2D a -> Vector2D a -> Vector2D a 
addVector2 Vector2D { x = x1, y = y1 } (Vector2D { x = x2, y = y2 }) = 
    Vector2D { x = x1 + x2, y = y1 + y2 }

instance (Additive a) => Additive (Vector2D a) where
    (+) = addVector2

newtype Phantom1 d a = Phantom1 (Vector2D a) --Axial 

deriving via (Vector2D a) instance forall d . (Additive a) => (Additive (Phantom1 d a))

data Via a b = Via a

class IsoEvidence a b where
    convertTo :: a -> b
    convertFrom :: b -> a

instance forall a b . (IsoEvidence a b, Additive b) => (Additive (Via a b)) where
    (Via x) + (Via y) = Via $ convertFrom $ (convertTo x :: b) + (convertTo y :: b)  

newtype Phantom2 d a = Phantom2 (Vector2D a) --Offset 

instance (IsoEvidence (Phantom1 d a) (Phantom2 d a)) where
    convertTo (Phantom1 x) = Phantom2 x
    convertFrom (Phantom2 x) = Phantom1 x

deriving via (Via (Phantom2 d a) (Phantom1 d a)) 
    instance (Additive a, IsoEvidence (Phantom1 d a) (Phantom2 d a)) => 
        (Additive (Phantom2 d a))

之后，我得到以下错误：

• 无法匹配类型

  Vector2D a的表示形式

  与

  Via（phantom2da）（phantom1da）

它似乎在说它不能强迫“Via&c”变成“vector2da”，这很奇怪，因为它实际上是一个两层深的新类型，而且效果很好

---

我做错了什么？

通过新类型派生，但您编写了

data Via a b = Via a   -- should be newtype

我是个白痴。谢谢你。