On Wed, Oct 17, 2007 at 06:49:24PM -0700, Dan Weston wrote:
It would seem that this induction on SN requires strictness at every stage. Or am I missing something?
Yes you are missing something. Whether it exists in my message is less certain. First, note that the elements of SN[a] are lambda-terms, like (\x -> x) (\x -> x). Second, strong normalization means that *every* reduction sequence terminates. E.g. that term is strongly normalizing, since the only possible reduction leads to \x -> x, from which no further reductions apply. Third, by inhabit...when passed, I mean that if TERM1 inhabits SN[a -> b], and TERM2 inhabits SN[a], then (TERM1) (TERM2) inhabits SN[b]. No mention of evaluation required. Is it clear now? Stefan
Stefan O'Rear wrote:
2) the function must halt for all defined arguments
fix :: forall p . (p -> p) -> p fix f = let x = f x in x consider: foo :: ((a -> a) -> a) -> a foo x = x id foo is a valid proof of a true theorem, but does not halt for the defined argument 'fix'.
On Wed, Oct 17, 2007 at 03:06:33PM -0700, Dan Weston wrote: - An effective approach for handling this was found long ago in the context of prooving the strong normalization of the simply typed lambda calculus. Define a category of terms SN[a] for each type a by recursion on a. SN[a], for atomic a, consists of terms that halt. SN[a -> b] consists of functions that inhabit SN[b] when passed an argument in SN[a]. With that definition, prooving that every term of the STLC of type a inhabits SN[a], and thus halts, becomess a trivial induction on the syntax of terms. Stefan