第一种方法对我很有效，尝试给他们明确的类签名

{-# Language DerivingStrategies       #-}
{-# Language PolyKinds                #-}
{-# Language StandaloneKindSignatures #-}

import Data.Kind

data DdManager = DdManager deriving stock (Eq, Show)
data DdNode    = DdNode    deriving stock (Eq, Show)
data Agent     = Agent     deriving stock (Eq, Show)
data Prp       = Prp       deriving stock (Eq, Show)

type    Dd :: k -> j -> i -> Type
newtype Dd x y z = ToDd DdNode
  deriving stock (Eq, Show)

type KnowStruct :: k -> j -> i -> Type
data KnowStruct a b c =
  KnS DdManager [Prp] (Dd a b c) [(Agent,[Prp])]
  deriving stock (Eq, Show)

是否也可以定义KnowStruct，而不使用所包含的Dd类型的上下文/参数规范的任何要求？
这是可能的，并且被称为存在量化，因为类型变量不出现在返回类型中。
由于a，b，c都是存在量化的，所以当你比较两个KnowStruct时，你是在比较Dd a1 b1 c1和Dd a2 b2 c2与完全不同类型的变量。

{-# Language ExistentialQuantification #-}
{-# Language StandaloneDeriving        #-}

import Data.Kind

data DdManager = DdManager deriving stock (Eq, Show)
data DdNode    = DdNode    deriving stock (Eq, Show)
data Agent     = Agent     deriving stock (Eq, Show)
data Prp       = Prp       deriving stock (Eq, Show)

-- I specialize this to some kind (`Type'). Otherwise the
-- kind would have to be an argument to `KnowStruct` below.
type    Dd :: Type -> Type -> Type -> Type
newtype Dd x y z = ToDd DdNode
  deriving stock (Eq, Show)

type KnowStruct :: Type
data KnowStruct = forall a b c.
  KnS DdManager [Prp] (Dd a b c) [(Agent,[Prp])]

deriving stock
  instance Show KnowStruct

我更喜欢（等效的）GADT语法

{-# Language GADTs #-}

type KnowStruct :: Type
data KnowStruct where
  KnS :: DdManager -> [Prp] -> Dd a b c -> [(Agent,[Prp])] -> KnownStruct