What are the relations between Alternative, MonadPlus(LeftCatch) and MonadPlus(LeftDistributive)?

Question

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

Assume $m$ is a monad. What are the relations betweem $m$ being an Alternative, a MonadPlusCatch and a MonadPlusDistr? For each of the six possible pairs, I'd like to have either a proof that one implies another, or a counter-example that it doesn't.

(I'm using

• MonadPlusCatch to distinguish a MonadPlus that satisfies the Left-Catch rule:

  mplus (return a) b = return a
  

• MonadPlusDistr to distinguish a MonadPlus that satifies Left-Distribution rule:

  mplus a b >>= k = mplus (a >>= k) (b >>= k)
  

see MonadPlus on HaskellWiki.)

---

My current knowledge + intuition is that:

1. MonadPlusDist $\rightarrow$ Alternative - likely true - it seems straightforward, I believe I have sketch of a proof, I'll check it and if it's correct, I'll post it AndrewC answered this part.
2. Alternative $\rightarrow$ MonadPlusDist - false - as AndrewC showed in his answer: Maybe is an Alternative, but it's known it's not MonadPlusDist (it's MonadPlusCatch).

3. MonadPlusCatch $\rightarrow$ Alternative - likely false - I believe that MaybeT (Either e) (or basically anything MaybeT m') should serve as a counterexample. The reason is that

   ((pure x) <|> g) <*> a =    -- LeftCatch
       (pure x) <*> a
   -- which in general cannot be equal to
   ((pure x) <*> a) <|> (g <*> a)
   

   again I'll check and post. (Interestingly, for just Maybe it's provable, because we can analyze if a is Just something or Nothing - see the aforementioned AndrewC's answer.)

4. Alternative $\rightarrow$ MonadPlusCatch - likely false - if we prove that MonadPlusDist $\rightarrow$ Alternative then [] will server as a counter-example. (Or we could explicitly prove Alternative laws for [].)
5. MonadPlusDist $\rightarrow$ MonadPlusCatch - false - [] is a known counter-example.
6. MonadPlusCatch $\rightarrow$ MonadPlusDist - false - Maybe is a known counter-example.

Answer 1

MonadPlusDist $\rightarrow$ Alternative is true.

Corollary: Alternative $\rightarrow$ MonadPlusCatch is false

(because as Petr Pudlák pointed out, [] is a counterexample - it doesn't satisfy MonadPlusCatch but does satisfy MonadPlusDist, hence Applicative)

Assumed: MonadPlusDist Laws

-- (mplus,mzero) is a monoid
mzero >>= k = mzero`                             -- left identity >>=
(a `mplus` b) >>= k  =  (a >>=k) `mplus` (b>>=k) -- left dist mplus

To prove: Alternative Laws

-- ((<|>),empty) is a monoid
(f <|> g) <*> a = (f <*> a) <|> (g <*> a) -- right dist <*>
empty <*> a = empty                       -- left identity <*>
f <$> (a <|> b) = (f <$> a) <|> (f <$> b) -- left dist <$>
f <$> empty = empty                       -- empty fmap

<*> expansion lemma
Assume we use the standard derivation of an applicative from a monad, namely (<*>) = ap and pure = return. Then

mf <*> mx = mf >>= \f -> mx >>= \x -> return (f x)

because

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

<$> expansion lemma
Assume we use the standard derivation of a functor from a monad, namely (<$>) = liftM. Then

f <$> mx = mx >>= return . f

because

f <$> mx = liftM f mx                    -- premise
         = do { x <- mx; return (f x) }  -- def(liftM)
         = mx >>= \x -> return (f x)     -- desugaring
         = mx >>= \x -> (return.f) x     -- def((.))
         = mx >>= return.f               -- eta-reduction 

Proof

Assume (<+>,m0) satisfy the MonadPlus laws. Trivially then it's a monoid.

Right Dist <*>

I'll prove

(mf <+> mg) <*> ma = (mf <*> ma) <+> (mg <*> ma) -- right dist <*>

because it's easier on the notation.

(mf <+> mg) <*> ma = (mf <+> mg) >>= \forg -> mx >>= \x -> return (forg x) -- <*> expansion
                   =     (mf >>= \f_g -> mx >>= \x -> return (f_g x))
                     <+> (mg >>= \f_g -> mx >>= \x -> return (f_g x))      -- left dist mplus
                   = (mf <*> mx) <+> (mg <*> mx)                           -- <*> expansion

Left Identity <*>

mzero <*> mx = mzero >>= \f -> mx >>= \x -> return (f x) -- <*> expansion
             = mzero                                     -- left identity >>=

as required.

Left Dist <$>

f <$> (a <|> b) = (f <$> a) <|> (f <$> b) -- left dist <$>

f <$> (a <+> b) = (a <+> b) >>= return . f              -- <$> expansion
                = (a >>= return.f) <+> (b >>= return.f) -- left dist mplus
                = (f <$> a) <+> (f <$> b)               -- <$> expansion

empty fmap

f <$> mzero = mzero >>= return.f   -- <$> expansion
            = mzero                -- left identity >>=

as required

Answer 2

A counter-example for MonadPlusCatch $\rightarrow$ Alternative

Indeed it's MaybeT Either:

{-# LANGUAGE FlexibleInstances #-}
import Control.Applicative
import Control.Monad
import Control.Monad.Trans.Maybe

instance (Show a, Show b) => Show (MaybeT (Either b) a) where
    showsPrec _ (MaybeT x) = shows x

main = print $
    let
        x = id :: Int -> Int
        g = MaybeT (Left "something")
        a = MaybeT (Right Nothing)
    -- print the left/right side of the left distribution law of Applicative:
    in ( ((return x) `mplus` g) `ap` a
       , ((return x) `ap` a) `mplus` (g `ap` a)
       )

The output is

(Right Nothing, Left "something")

which means that MaybeT Either fails the left distribution law of Applicative.

---

The reason is that

(return x `mplus` g) `ap` a

ignores g (due to LeftCatch) and evaluates just to

return x `ap` a

but this is different from what the other side evaluates to:

g `ap` a