我有下一个haskell代码成功运行：
type Relation a = [[a]]
isAEquivalentRelation :: Eq a => Relation a -> Bool
isAEquivalentRelation x = isASimetricRelation x && isAReflexiveRelation x && isATransitiveRelation x
isASimetricRelation :: Eq a => Relation a -> Bool
isASimetricRelation [] = True
isASimetricRelation (x:xs) = (elem (reverse x) (x:xs)) && (isASimetricRelation (filter (/= y) xs)) where y = (reverse x)
isAReflexiveRelation :: Eq a => Relation a -> Bool
isAReflexiveRelation [] = True
isAReflexiveRelation x = isAReflexiveRelationRec (filter (\ a -> a !! 0 == a !! 1) x) x
isAReflexiveRelationRec :: Eq a => Relation a -> Relation a -> Bool
isAReflexiveRelationRec [] _ = False
isAReflexiveRelationRec _ [] = True
isAReflexiveRelationRec pr (x:xs) = or ( map (== [x!!0,x!!0]) pr) && or ( map (== [x!!1,x!!1]) pr) && isAReflexiveRelationRec pr xs
isATransitiveRelation :: Eq a => Relation a -> Bool
isATransitiveRelation [] = True
isATransitiveRelation x = isATransitiveRelationRec x x
isATransitiveRelationRec :: Eq a => Relation a -> Relation a -> Bool
isATransitiveRelationRec _ [] = True
isATransitiveRelationRec original (x:xs) = and ( map (\ a -> elem a original) (map (\ a -> [x !! 0, a !! 1]) (filter (\ a -> x !! 1 == a !! 0) original)) ) && isATransitiveRelationRec original xs
areEqual :: Eq a => Relation a -> Relation a -> Bool
areEqual x y = subset x y && subset y x
subset :: Eq a => Relation a -> Relation a -> Bool
subset [] _ = True
subset _ [] = False
subset x y = and (map (\ a -> elem a y) x)
问题是我需要使用我实现的areEqual方法重新定义Relation类型的==但是我无法实现这一点。我已经尝试了很多关于类，实例和数据的东西但是我被困了，我如何声明一个通用的Relation类型并重新定义operator ==以匹配areEqual方法？