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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
