Re: incorrect MonadPlus law "v >> mzero = mzero"?

On Wed, Feb 5, 2014 at 5:01 PM, wren ng thornton wrote:

On Wed, Feb 5, 2014 at 7:59 PM, wren ng thornton wrote:

...

The rules I'd expect MonadPlus to obey are:

mzero >>= f = mzero

(x `mplus` y) >> mzero = (x >> mzero) `mplus` (y >> mzero)

Er, sorry, that should've been:

(x `mplus` y) >> z = (x >> z) `mplus` (y >> z)

from which the above instance follows trivially.

I don't think this is correct. consider the original instance with x /= mzero. Then we have left side (x `mplus` y) >> mzero x >> mzero -- by def. of mplus right side (x >> mzero) `mplus` (y >> mzero) mzero'x `mplus` (y >> mzero) -- mzero'x is a zero with x's effects, this reduction comes from the monad laws (y >> mzero) -- by mplus Or concretely,

let x = lift (print "x!") :: MaybeT IO () let y = lift (print "y!") :: MaybeT IO () runMaybeT $ (x `mplus` y) >> mzero

"x!" Nothing

runMaybeT $ (x >> mzero) `mplus` (y >> mzero)

"x!" "y!" Nothing John