[Haskell-cafe] Re: Non-termination of type-checking

Dan Doel dan.doel at gmail.com
Fri Jan 29 07:24:21 EST 2010

On Friday 29 January 2010 2:56:31 am oleg at okmij.org wrote:
> Here is a bit more simplified version of the example. The example has
> no value level recursion and no overt recursive types, and no impredicative
> polymorphism. The key is the observation, made earlier, that two types
> 	c (c ()) and R (c ())
> unify when c = R. Although the GADTs R c below is not recursive, when
> we instantiate c = R, it becomes recursive, with the negative
> occurrence. The trouble is imminent.
> We reach the conclusion that an instance of a non-recursive GADT
> can be a recursive type. GADT may harbor recursion, so to speak.
> The code below, when loaded into GHCi 6.10.4, diverges on
> type-checking. It type-checks when we comment-out absurd.
> {-# LANGUAGE GADTs, EmptyDataDecls #-}
> data False				-- No constructors
> data R c where				-- Not recursive
>     R :: (c (c ()) -> False) -> R (c ())
> -- instantiate c to R, so (c (c ())) and R (c ()) coincide
> -- and we obtain a recursive type
> --    mu R. (R (R ()) -> False) -> R (R ())
> cond_false :: R (R ()) -> False
> cond_false x@(R f) = f x
> absurd :: False
> absurd = cond_false (R cond_false)

This example is pretty weird in my view. For instance, consider:

  data S x

  type family T x :: *
  type family T ()     = R ()
  type family T (R ()) = S ()

  s'elim :: S x -> False
  s'elim x = undefined -- case x of {}

  what :: T (T ()) -> False
  what = s'elim

Now, we have T () ~ R (), and T (T ()) ~ S (R ()), so R (T ()) ~ R (R ()), no? 
And T :: * -> *, so it should be an acceptable argument to R. So, one would 
expect that I could write:

  r :: R (R ()) -- R (T ())
  r = R what

However, neither of those signatures for r goes through. With the second, I 
get a complaint that:

  Couldn't match expected type `S ()' against inferred type `()'
    Expected type: S (S ())
    Inferred type: T (T ())

though I'm not quite sure how it gets that error. With the R (R ()) choice, I 

  Couldn't match expected type `R (R ())'
         against inferred type `T (T ())'
    NB: `T' is a type function, and may not be injective

but the injectivity GHC seems to be relying on isn't F x ~ F y => x ~ y, but:

  f T ~ g T => f ~ g

or injectivity of ($ ()) in this case, so to speak. But this seems like a 
significantly weirder thing to take for granted than injectivity of type 
constructors. I suppose it's sufficient to limit c (in the constructor R's 
definition) to accept only data type constructors, but that's significantly 
more limiting than I'd expect to be the case (although, it does appear to be 
what GHC does).

Anyhow, I don't particularly find this 'scary', as Haskell/GHC isn't 
attempting to be an environment for theorem proving, and being able to encode 
(\x -> x x) (\x -> x x) through lack of logical consistency isn't giving up 
anything that hasn't already been given up multiple times (through general 
recursion, error and negative types at least). Non-termination of type 
checking is undesirable, of course.

-- Dan

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