{-
I am trying to apply model theoretic concepts to Haskell by considering type classes as theories and instances as models.
Then the declaration of a sub-class specifies a signature morphism from the superclass to the subclass.
In case below the theories (classes) are signature only (no default methods) so a signature morphisms can be considered as a theory morphisms.
The only purpose of the code below is to explore the concepts of model expansion[1] and model reduct [2] in Haskell
I am not trying to improve or rewrite the code itself for any particular application.
Neither am I trying to find OO style inheritance semantics in Haskell type classes.
Rather, I am wondering if the last instance of Worker (commented out) demonstrates that there is no model expansion in this case.
That is, for theory morphism the theory(Person) => theory(Worker) we do not get (model(Person) sub-model model(Worker)).

If there is no model expansion could it be because of the constructor discipline, which only allows variables, and constructors in the LHS argument patterns.

[1] http://www.informatik.uni-bremen.de/~cxl/papers/wadt04b.pdf
[2] http://en.wikipedia.org/wiki/Reduct

-}

constant::Int
constant = (1::Int)
fun1::Int -> Int
fun1 (constant::Int) = 8

class Person i n | i -> n where
 pid :: i
 name :: i -> n

-- There is a signature/theory morphism from Person to Worker
class Person i n => Worker i n s | i -> s where
 salary :: i -> s
  
-- model(Person)
instance Person Int [Char] where
 pid = (1::Int)
 name (1::Int)  = ("john"::[Char])


-- We can say that a model(Worker) can use model(Person).
instance Worker Int [Char] Int where
-- Hypothesis: pid on the RHS shows that a model(Person) is *available* in model(Person) (reduct)?
  salary i = if i == pid then 100 else 0
 
-- instance Worker Int [Char] Int where
-- Hypothesis: The model of Person cannot be expanded to a model of Worker(no model expansion)
-- pid below is not inherited from Person, it is just a local variable
--   salary pid = 100