Re: [Haskell-cafe] A question about "monad laws"

Hi

...

(x >>= f) >>= g == x >>= (\v -> f v >>= g)

Or stated another way:

(x >>= f) >>= g == x >>= (f >>= g)

Which is totally wrong, woops. See this page for lots of details about the Monad Laws and quite a nice explanation of where you use them: http://www.haskell.org/haskellwiki/Monad_Laws Thanks Neil