Hi,
On 22.01.2011, at 08:12, Sebastian Fischer wrote:
Also, Jan, I don't understand your comment about continuation
monads. Maybe I am a bit numb today.. What property do you mean do
continuation monads have or not?
I was wrong there. If there exist values x and y with x /= y and you
have a function f such that f x /= f y then we have f _|_ = _|_ (at
least if f is a sequential function). I thought this property might
fail if x and y are functions but I was totally wrong.
Therefore the laws
mzero >>= f = mzero
and
return x >>= f = f x
together with
mzero /= mplus m n
and
mzero /= mplus (m >>= f) (n >>= f)
for some m and n implies that we have _|_ >>= f = _|_ if >>= is
sequential.
Cheers, Jan
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