Commit 711bead3 authored by Ralf Jung's avatar 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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