[Haskell-beginners] Type classes and synonyms

Brent Yorgey byorgey at seas.upenn.edu
Sun Nov 22 12:25:36 EST 2009

On Sun, Nov 22, 2009 at 04:12:29PM +0000, pbrowne wrote:
> On Sun, Nov 22, 2009 at 03:32:07PM +0000, Brent Yorgey wrote:
> > A "free theorem" is a *property* which
> > is satisfied by any function with a particular type.  For example, any
> > function with the type
> > 
> >   foo :: [a] -> Maybe a
> > 
> > necessarily satisfies
> > 
> >   foo (map f xs) = fmap f (foo xs)
> A theorem is a statement which has been proved on the basis of
> previously established theorems or axiom. So, should the definition of
> the function not satisfy the signature? I may be confusing terminology
> here. I am coming of an OBJ2/Maude/CafeOBJ background to Haskell. In
> CafeOBJ a *property* of an operation could, for example, be
> associativity. I am having difficulty in adjusting to Haskells level of
> formality.

I'm not quite sure what you're asking here.  My point is just that a
"free theorem" will be of the form

  "Any function f of type T, *no matter how f is implemented*, will
  always satisfy the following property:

    blah blah f blah = blah f blah

This has nothing to do with whether or not there is only one possible
implementation of f that does not involve undefined, which is a
different phenomenon.

> >> In Haskell the *theorems for free* means roughly that the
> >> definition of a class, instance or a function follows from the supplied
> >> types because they are the only types that don’t have undefined argumens
> >>  or undefined return types.
> Question: Is my generalization of applying the "free theorem" concept to
> classes and instances correct?

Classes don't have types, so I'm not sure what you mean by including
classes.  Generalizing to instances makes sense; type class instances
are just composed of a list of functions.  However, again, this notion
of there sometimes being only one possible implementation of a
particular type has nothing to do with the concept of "free
theorems". (At least, it has not much to do with it that I am aware

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