Andras Slemmer wrote:
The reason you need the typeclass constraint is not in the use of 'f' but rather in the use of propagate and what you pass in to it (foo1/foo2 here). If you leave away the typeclass constraint what you're left with is:
propagate' Bar -> (forall a. a->a) -> Bar
The implementation of this function is the same as before, however your use of propagate' is restricted: forall a. a->a is a very "strict" type, in fact the only inhabitant of this type is 'id' (and bottom, but disregard that here), which means the only way to call propagate is to pass in 'id'. Try it yourself!
Indeed I have tried, it works as you say.
Related note: there is a proof that in fact the only inhabitant of (forall a. a -> a) is 'id' and it is the consequence of the "parametricity" property. It is a very neat result I suggest you look it up!
Interesting. I have tried to google on the topic, but I find mainly research articles. For example: https://www.google.fr/search?client=ubuntu&channel=fs&q=haskell+%22parametricity+property%22&ie=utf-8&oe=utf-8&gws_rd=cr&ei=P_acUvboDse_0QX9mYDADQ#channel=fs&q=%22parametricity+property%22+haskell Are there textbooks where a proof of this fact could be found? I'm an autodidact (who also benefits from help of guys like you), I don't know what lectures on type theory at university level could look like. Thanks TP