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Re: Reasons (not) to vote Motion 27: NO

Arnold Neumaier wrote:

Said that, I still don't think that 5.8.5 is correct. Consider:
   * xx and/or yy is Empty (so that the box (xx,yy) is empty).
   * f(x,y) = sqrt(x) + y
One gets:
   * After applying domain on the box (xx,yy), one gets (Empty,emp)
     for both components.


Oops, this is a typo in the text; according to the semantics in (Eval1),
upon which the proof of the FTDIA is based, domain should yield
(Empty,ein).


Then this should be OK for bound expressions (but not for free
expressions).


If one deletes the decoration ein (which was not in the original set of
decorations anyway, but was added because of prior demand), and uses emp
instead, things also work alright for free expressions.


In my perspective, the difficulty isn't with the ein decoration but rather
the containment order:
   con \supset def \supset dac \supset ein
   con \supset emp \supset ill
Note that emp is not comparable to ein.

With Dominique's information order:
   con <=_D def <=_D dac <=_D ein
   con <=_D ndf <=_D ein
decoration ndf is comparable to ein.

So in Vincencet's example the true decorated result of
   f([-2,-1],[Empty]) = (Empty,ein).
Evaluating on free expressions one gets
   sqrt([-2,-1]) + Empty = (Empty,ndf).
However the following comparison
   (Empty,ndf) <=_DI (Empty,ein)    <==>
           Empty \subseteq Empty  and    ndf <=_D ein
is true.

Nate