Commit 711bead3 by Ralf Jung

### Note: If f^2 is contractive, that doesn't imply that f is non-expansive

parent 6456f1f9
 ... ... @@ -298,12 +298,32 @@ Definition fixpointK `{Cofe A, Inhabited A} k (f : A → A) Section fixpointK. Local Set Default Proof Using "Type*". Context `{Cofe A, Inhabited A} (f : A → A) (k : nat). Context `{f_contractive : !Contractive (Nat.iter k f)}. (* TODO: Can we get rid of this assumption, derive it from contractivity? *) Context {f_ne : NonExpansive f}. Context {f_contractive : Contractive (Nat.iter k f)} {f_ne : NonExpansive f}. (* Note than f_ne is crucial here: there are functions f such that f^2 is contractive, but f is not non-expansive. Consider for example f: SPred → SPred (where SPred is "downclosed sets of natural numbers"). Define f (using informative excluded middle) as follows: f(N) = N (where N is the set of all natural numbers) f({0, ..., n}) = {0, ... n-1} if n is even (so n-1 is at least -1, in which case we return the empty set) f({0, ..., n}) = {0, ..., n+2} if n is odd In other words, if we consider elements of SPred as ordinals, then we decreaste odd finite ordinals by 1 and increase even finite ordinals by 2. f is not non-expansive: Consider f({0}) = ∅ and f({0,1}) = f({0,1,2,3}). The arguments are clearly 0-equal, but the results are not. Now consider g := f^2. We have g(N) = N g({0, ..., n}) = {0, ... n+1} if n is even g({0, ..., n}) = {0, ..., n+4} if n is odd g is contractive. All outputs contain 0, so they are all 0-equal. Now consider two n-equal inputs. We have to show that the outputs are n+1-equal. Either they both do not contain n in which case they have to be fully equal and hence so are the results. Or else they both contain n, so the results will both contain n+1, so the results are n+1-equal. *) Let f_proper : Proper ((≡) ==> (≡)) f := ne_proper f. Existing Instance f_proper. Local Existing Instance f_proper. Lemma fixpointK_unfold : fixpointK k f ≡ f (fixpointK k f). Proof. ... ...
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