Gradient helper prevents having a 'top()' mixin

[ianstormtaylor]

I get this error when trying to declare a mixing with a name of top()

TypeError: expected string, ident or literal, but got function:top()

and declaring a gradient using a string doesn't work either linear-gradient('top', ...

Not sure if there is a solution that keeps both of the functions around...

[tj]

we could probably also check for a function of the same name just in case, shouldn't hurt