What’s an example of a Monad which is an Alternative but not a MonadPlus?

In his answer to the question “Distinction between typeclasses MonadPlus, Alternative, and Monoid?”, Edward Kmett says that

Moreover, even if Applicative was a superclass of Monad, you’d wind up needing the MonadPlus class anyways, because obeying

empty <*> m = empty

isn’t strictly enough to prove that

empty >>= f = empty

So claiming that something is a MonadPlus is stronger than claiming it is Alternative.

It’s clear that any applicative functor which is not a monad is automatically an example of an Alternative which is not a MonadPlus, but Edward Kmett’s answer implies that there exists a monad which is an Alternative but not a MonadPlus: its empty and <|> would satisfy the Alternative laws,¹ but not the MonadPlus laws.² I can’t come up with an example of this by myself; does anybody know of one?

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¹ I wasn’t able to find a canonical reference for a set of Alternative laws, but I lay out what I believe them to be roughly halfway through my answer to the question “Confused by the meaning of the Alternative type class and its relationship to other type classes” (search for the phrase “right distributivity”). The four laws I believe ought to hold are:

1. Right distributivity (of <*>):  (f <|> g) <*> a = (f <*> a) <|> (g <*> a)
2. Right absorption (for <*>):  empty <*> a = empty
3. Left distributivity (of fmap):  f <$> (a <|> b) = (f <$> a) <|> (f <$> b)
4. Left absorption (for fmap):  f <$> empty = empty

I’d also happily accept being given a more useful set of Alternative laws.

² I know that there’s some ambiguity about what the MonadPlus laws are; I’m happy with an answer that uses left distribution or left catch, although I would weakly prefer the former.

The clue to your answer is in the HaskellWiki about MonadPlus you linked to:

Which rules? Martin & Gibbons choose Monoid, Left Zero, and Left Distribution. This makes [] a MonadPlus, but not Maybe or IO.

So according to your favoured choice, Maybe isn't a MonadPlus (although there's an instance, it doesn't satisfy left distribution). Let's prove it satisfies Alternative.

Maybe is an Alternative

1. Right distributivity (of <*>): (f <|> g) <*> a = (f <*> a) <|> (g <*> a)

Case 1: f=Nothing:

(Nothing <|> g) <*> a =                    (g) <*> a  -- left identity <|>
                      = Nothing         <|> (g <*> a) -- left identity <|>
                      = (Nothing <*> a) <|> (g <*> a) -- left failure <*>

Case 2: a=Nothing:

(f <|> g) <*> Nothing = Nothing                             -- right failure <*>
                      = Nothing <|> Nothing                 -- left identity <|>
                      = (f <*> Nothing) <|> (g <*> Nothing) -- right failure <*>

Case 3: f=Just h, a = Just x

(Just h <|> g) <*> Just x = Just h <*> Just x                      -- left bias <|>
                          = Just (h x)                             -- success <*>
                          = Just (h x) <|> (g <*> Just x)          -- left bias <|>
                          = (Just h <*> Just x) <|> (g <*> Just x) -- success <*>

1. Right absorption (for <*>): empty <*> a = empty

That's easy, because

Nothing <*> a = Nothing    -- left failure <*>

1. Left distributivity (of fmap): f <$> (a <|> b) = (f <$> a) <|> (f <$> b)

Case 1: a = Nothing

f <$> (Nothing <|> b) = f <$> b                        -- left identity <|>
                 = Nothing <|> (f <$> b)          -- left identity <|>
                 = (f <$> Nothing) <|> (f <$> b)  -- failure <$>

Case 2: a = Just x

f <$> (Just x <|> b) = f <$> Just x                 -- left bias <|>
                     = Just (f x)                   -- success <$>
                     = Just (f x) <|> (f <$> b)     -- left bias <|>
                     = (f <$> Just x) <|> (f <$> b) -- success <$>

1. Left absorption (for fmap): f <$> empty = empty

Another easy one:

f <$> Nothing = Nothing   -- failure <$>

Maybe isn't a MonadPlus

Let's prove the assertion that Maybe isn't a MonadPlus: We need to show that mplus a b >>= k = mplus (a >>= k) (b >>= k) doesn't always hold. The trick is, as ever, to use some binding to sneak very different values out:

a = Just False
b = Just True

k True = Just "Made it!"
k False = Nothing

Now

mplus (Just False) (Just True) >>= k = Just False >>= k
                                     = k False
                                     = Nothing

here I've used bind (>>=) to snatch failure (Nothing) from the jaws of victory because Just False looked like success.

mplus (Just False >>= k) (Just True >>= k) = mplus (k False) (k True)
                                           = mplus Nothing (Just "Made it!")
                                           = Just "Made it!"

Here the failure (k False) was calculated early, so got ignored and we "Made it!".

So, mplus a b >>= k = Nothing but mplus (a >>= k) (b >>= k) = Just "Made it!".

You can look at this as me using >>= to break the left-bias of mplus for Maybe.

Validity of my proofs:

Just in case you felt I hadn't done enough tedious deriving, I'll prove the identities I used:

Firstly

Nothing <|> c = c      -- left identity <|>
Just d <|> c = Just d  -- left bias <|>

which come from the instance declaration

instance Alternative Maybe where
    empty = Nothing
    Nothing <|> r = r
    l       <|> _ = l

Secondly

f <$> Nothing = Nothing    -- failure <$>
f <$> Just x = Just (f x)  -- success <$>

which just come from (<$>) = fmap and

instance  Functor Maybe  where
    fmap _ Nothing       = Nothing
    fmap f (Just a)      = Just (f a)

Thirdly, the other three take a bit more work:

Nothing <*> c = Nothing        -- left failure <*>
c <*> Nothing = Nothing        -- right failure <*>
Just f <*> Just x = Just (f x) -- success <*>

Which comes from the definitions

instance Applicative Maybe where
    pure = return
    (<*>) = ap

ap :: (Monad m) => m (a -> b) -> m a -> m b
ap =  liftM2 id

liftM2  :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
liftM2 f m1 m2          = do { x1 <- m1; x2 <- m2; return (f x1 x2) }

instance  Monad Maybe  where
    (Just x) >>= k      = k x
    Nothing  >>= _      = Nothing
    return              = Just

so

mf <*> mx = ap mf mx
          = liftM2 id mf mx
          = do { f <- mf; x <- mx; return (id f x) }
          = do { f <- mf; x <- mx; return (f x) }
          = do { f <- mf; x <- mx; Just (f x) }
          = mf >>= \f ->
            mx >>= \x ->
            Just (f x)

so if mf or mx are Nothing, the result is also Nothing, whereas if mf = Just f and mx = Just x, the result is Just (f x)