Re: [Haskell-cafe] Puzzled by error with forall quantifier.

Iavor S. Diatchki wrote:

just skip the forall. "free" type variables are implicitly forall-quantified.

Of course! Thanks. The problem is that I actually wanted "exists". However as the type that exists is functionally dependent on other the parameters to the class I think I can deal with it using functional dependencies, as you suggest. Sorry for the elementary question. Cheers, Theo Norvell