[Iris-Club] Typos (, I think,) in the Iris 2.0 documentation
Ralf Jung
jung at mpi-sws.org
Tue Aug 16 09:45:55 CEST 2016
Hi,
> As far as I know, this is standard terminology/notation in category
> theory. Given a functor F : C1 -> C2, it is really two functions: F_0,
> the action on objects (mapping objects of C0 to objects of C1), and F_1,
> the action on arrows (mapping arrows of C0 to arrows of C1, and suitably
> compatible with F_0).
> So, in this case, we are saying that the function "\lambda f. F(f)",
> i.e. calling the functor F on arrows f of the category COFE, must be
> non-expansive.
>
> The idea behind these category-theoretic remarks is that if you know
> category theory, they should be helfpul; if not, just ignore them. :)
> Understanding them is certainly not required to just use Iris as a
> logic.
>
>
> Ah I see :-) May I ask which reference/textbook you are consulting to?
> Because it is different from the convention in the textbook I learned
> from (Mac Lane. Categories for the Working Mathematician. Springer GTM 5)..
I have no idea, I just know I saw this notation a few times while we did
our Category Theory seminar... Derek, do you remember?
> > - Page 9: in what sense "↑(S, T)" is the closure? I mean, is it minimal
> > w.r.t. the closedness? If so, is it obvious?
>
> It is adding exactly those elements that successors of elements in S.
> So I would say it is pretty obvious that this makes it the smallest set
> containing S that is closed under successors.
>
>
> I am concerned with the condition of the closedness that \mathcal{L}(s)
> # T.. Due to this restriction I believe the closure is in general not
> minimal w.r.t. the closedness.
Oh, right, we also added this well-formedness condition here because we
are lazy. However, considering T to be fixed, this condition is
monotone, i.e., if it rules out a set then it also rules out all larger
sets. How should this violate minimality?
> Another remark on the closedness is: the definition of s
> \stackrel{T}{\rightarrow} s' involves the existential quantification of
> T1 and T2,
Right, this matches exactly the definition in Coq.
> and I think there exists an alternative definition that does
> not involve that quantification:
> [(\mathcal{L}(s') # T) /\ s \rightarrow s' .]
> This is obtained by letting T1 and T2 be (\mathcal{L}(s') \setminus
> \mathcal{L}(s)) and (\mathcal{L}(s) \setminus \mathcal{L}(s')). I guess
> the definition in the documentation is more amenable to the proof, and
> for this reason you choose to define in that way. May I ask what is the
> reason?
Yeah, this looks equivalent. I honestly do not know any more why this
is defined the way it is; it is also possible that the definition
changed over time.
Kind regards,
Ralf
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