This question comes from this answer in example of a functor that is Applicative but not a Monad: It is claimed that the
data PoE a = Empty | Pair a a deriving (
Since you are interested in how to do it systematically, here's how I found a counterexample with quickcheck:
{-# LANGUAGE DeriveFunctor #-}
import Control.Monad ((>=>))
import Test.QuickCheck
--
Defining an arbitrary instance to generate random PoE
s.
instance (Arbitrary a) => Arbitrary (PoE a) where
arbitrary = do
emptyq <- arbitrary
if emptyq
then return Empty
else Pair <$> arbitrary <*> arbitrary
And tests for the monad laws:
prop_right_id m = (m >>= return) == m
where
_types = (m :: PoE Int)
prop_left_id fun x = (return x >>= f) == f x
where
_types = fun :: Fun Int (PoE Int)
f = applyFun fun
prop_assoc fun gun hun x = (f >=> (g >=> h)) x == ((f >=> g) >=> h) x
where
_types = (fun :: Fun Int (PoE Int),
gun :: Fun Int (PoE Int),
hun :: Fun Int (PoE Int),
x :: Int)
f = applyFun fun
g = applyFun gun
h = applyFun hun
I don't get any failures for the identity laws, but prop_assoc
does generate a counterexample:
ghci> quickCheck prop_assoc
*** Failed! Falsifiable (after 7 tests and 36 shrinks):
{6->Pair 1 (-1), _->Empty}
{-1->Pair (-3) (-4), 1->Pair (-1) (-2), _->Empty}
{-3->Empty, _->Pair (-2) (-4)}
6
Not that it's terribly helpful for understanding why the failure occurs, it does give you a place to start. If we look carefully, we see that we are passing (-3)
and (-2)
to the third function; (-3)
maps to Empty
and (-2)
maps to a Pair
, so we can't defer to the laws of either of the two monads PoE
is composed of.