Re: [Haskell-cafe] Fwd: incorrect MonadPlus law "v >> mzero = mzero"?

Indeed this issue is not limited merely to multiple failure values. >>> runMaybeT $ lift (putStrLn "effect") >> mzero effect >>> runMaybeT mzero So you're right. This law is being violated -- Dan Burton On Mon, Feb 3, 2014 at 12:21 PM, Petr Pudlák wrote:

Since nobody replied so far, I'm also forwarding to cafe. Petr

---------- Forwarded message ---------- From: Petr Pudlák Date: 2014-01-29 Subject: incorrect MonadPlus law "v >> mzero = mzero"? To: "libraries@haskell.org"

Hi,

this law apparently fails for a MonadPlus instance that has more than one possible failure value. Consider:

runIdentity . runErrorT $ ((ErrorT . Identity $ Left "failure") >> mzero :: ErrorT String Identity ())

evaluates to `Left "failure"`, which isn't equal to ErrorT's mzero `Left ""`.

This isn't just the case of ErrorT, it fails for any MonadPlus with multiple failure values. For example

lift (tell "foo") >> mzero :: MaybeT (Writer String) ()

is again distinct from mzero.

Actually, no monad transformer with a MonadPlus instance can satisfy the law, because the first part in front of `>> mzero` can introduce side effects in the underlying monad.

I'm not sure what should be the proper solution. Perhaps to change the laws to:

return x >> mzero = mzero (v >> mzero) >>= f = (v >> mzero)`

That is, if an expression ends with `mzero`, it behaves like `mzero`.

Petr

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