Hi
Isn't _|_ = \x -> _|_?
_|_ `seq` () = _|_ (\x -> _|_) `seq` () = () Whether this is the fault of seq or not is your call... Thanks Neil _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe