A: No.
Proof.
Use these definitions
(<&>) :: Functor f => f a -> (a -> b) -> f b
(<&>) = flip fmap
empty :: ZipList a
empty = ZipList []
nat :: ZipList Integer
nat = ZipList [0..]Use this lemma (proof skipped):
pure x = x <$ nat
join (nat <&> \i -> nat <&> \j -> f i j) = join (nat <$> \i -> f i i)-- They are free variables
x :: Type
f :: Integer -> Integer -> x
q :: Integer -> ZipList (ZipList x)
q i = nat <&> \j -> if j < i then empty else pure (f i j)
ex :: ZipList (ZipList (ZipList x))
ex = nat <&> q
join . join $ ex
= join . join $ nat <&> \i -> nat <&> \j -> if j < i then empty else pure (f i j)
= join $ nat <&> \i -> if i < i then empty else pure (f i i)
= join . fmap pure $ nat <&> \i -> f i i
= nat <&> \i -> f i i
-- It must be:
join . fmap join $ ex = nat <&> \i -> f i i
fmap join ex = nat <&> \i -> join (q i)
-- By parametricity, join (q i) must keep its i-th element (f i i)
-- Generally, for any m >= 0:
join . join $ nat <&> \i -> q (i - m)
= join . join $ nat <&> \i -> nat <&> \j -> if j < i - m then empty else pure (f (i - m) j)
= join $ nat <&> \i -> pure (f (i - m) i)
= nat <&> \i -> f (i - m) i
-- Thus it must be:
join . fmap join $ nat <&> \i -> q (i - m) = nat <&> \i -> f (i - m) i
-- By parametricity, join (q (i - m)) must not discard (f (i - m) i)
-- In other words, for every m >= 0, join (q i) must not discard the element (f i (i + m)).
-- No finite list can contain infinite number of elements, thus there should be
g :: Integer -> Integer -> x
-- Such that
join (q i) = nat <&> g i
-- The above implies
join . fmap join $ nat <&> \i -> q (i + 1)
= join $ nat <&> \i -> join (q (i + 1))
= join $ nat <&> \i -> nat <&> g (i + 1)
= nat <&> g (i + 1) i
-- Then it contradicts to the monad law:
join . join $ nat <&> \i -> q (i + 1)
= join . join $ nat <&> \i -> nat <&> \j -> if j < i + 1 then empty else pure (f (i + 1) j)
= join $ nat <&> \i -> if i < i + 1 then empty else pure (f (i + 1) j)
= join $ nat <&> \i -> empty
= empty