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

Question

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

Moreover, even if Ap

Answer 1

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)