> And at that point, all the power you get from saying things like there's only one possible implementation of a function f :: a -> a immediately go out of the window, because you don't have a value of type a, you have a value of type SomeConcreteTypeThatIKnowThingsAbout.

Not if your logic is primarily in terms of domain types, invariants encoded in those domain types as is necessary and economical.

> And at that point, all the power you get from saying things like there's only one possible implementation of a function f :: a -> a immediately go out of the window, because you don't have a value of type a, you have a value of type SomeConcreteTypeThatIKnowThingsAbout.

Can you give an example? I can only see this being the case if your logic is done at the concrete type level.