Version 1 (modified by diatchki, 4 years ago) (diff) |
---|

Consistency of Functional Dependencies

The functional dependencies on a class restrict the instances that may be declared for a given class. The instances of a class:

- class C a b c | b -> c where f
- ...

are consistent with its functional dependency if the following invariant holds:

byFD(C,1): forall a1 a2 b c1 c2. (C a1 b c1, C a2 b c2) => (c1 ~ c2)

Please note that the question of FD-consistency is orthogonal to
instance coherence (i.e, uniqueness of evidence, overlapping instances,
etc.), and the decidability of type-checking of terms---for examples
of their independence, please see the `Examples` at the end of the document.

If we check that all instances in scope are consistent with their FDs,
then we can use the FD invariant `byFD` during type inference to
infer more precise types, report errors involving unsolvable contexts,
or accept programs that would be rejected without the invariant.

For example, if we have an instance:

I: instance F Int Int Char

and we have a constraint `C: F Int Int a`, then we can use the
FD-invariant to prove that `a` must be `Char`:

- byFD(C,1) (C,I)
- a ~ Char

Checking FD-Consistency

To ensure FD-consistency, before accepting an instance we need to check that it is compatible with all other instances that are already in scope. Note that we also need to perform the same check when combining imported instances. Consider adding a new instance to an FD-consistent set of instances:

I: instance P(as,bs) => C t1(as,bs) t2(as) t3(as,bs)

The notation `t(as)` states the variables in `t` are a subset of `as`.

- Check that
`I`is self-consistent (i.e., we can't use different instantiations of`I`to violate FD-consistency). Self consistency follows if we can prove the following theorem:

forall as bs cs. (P, P[cs/bs]) => t3[cs/bs] ~ t3[cs/bs]

- Check that
`I`is FD-consistent with all existing instances of the class. So, for each existing instance,`J`:

J: instance Q(xs) => C s1(xs) s2(xs) s3(xs)

we need to show that:

forall as bs xs. (P,Q,s2 ~ t2) => (s3 ~ t3)

Assuming no type-functions in instance heads, the equality assumption is equivalent to stating that

s2andt2may be unified, so another way to state our goal is:

forall as bs xs. (P[su], Q[su]) => (s3[su] ~ t3[su])

where

suis the most general unifier ofs2andt2.

Proving these two goals before accepting an instance is similar to the process of finding evidence for super-classes before accepting and instance. Also, note that while proving (2), it is not a problem if we find that we have assumed a contradiction: this simply means that the two instances can never be used at the same time, so the FD-consistency follows trivially.

Examples

- FD-consistency is orthogonal to instance coherence.

FD-consistent but not coherent:

instance C Int Int Int where f = definition_1 instance C Int Int Int where f = definition_2

Coherent but not FD-consistent:

instance C Int Int Char where ... instance C Char Int Bool where ...

- FD-consistency is orthogonal to termination of instances.

FD-consistent but "non-terminating":

instance C a b c => C a b c

Terminating but not FD-consistent:

instance C Int Int Char where ... instance C Char Int Bool where ...