Skip to content
GitLab
Projects
Groups
Snippets
Help
Loading...
Help
What's new
7
Help
Support
Community forum
Keyboard shortcuts
?
Submit feedback
Contribute to GitLab
Sign in / Register
Toggle navigation
Open sidebar
George Pirlea
Iris
Commits
711bead3
Commit
711bead3
authored
Feb 23, 2017
by
Ralf Jung
Browse files
Options
Browse Files
Download
Email Patches
Plain Diff
Note: If f^2 is contractive, that doesn't imply that f is non-expansive
parent
6456f1f9
Changes
1
Hide whitespace changes
Inline
Side-by-side
Showing
1 changed file
with
24 additions
and
4 deletions
+24
-4
theories/algebra/ofe.v
theories/algebra/ofe.v
+24
-4
No files found.
theories/algebra/ofe.v
View file @
711bead3
...
...
@@ -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
.
...
...
Write
Preview
Markdown
is supported
0%
Try again
or
attach a new file
.
Attach a file
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Cancel
Please
register
or
sign in
to comment