Imho there are (at least) two ways of thinking about this. One
consistent way is the following:

Let c, d be "extended numbers" (reals, complexes, oo, zoo, etc).
Let f(x,y) be any function of two "ordinary numbers".

We define f(c,d) = e if for all expressions e1, e2 such that limit(e1)
= c, limit(e2) = d we have limit(f(e1, e2)) = e.
We define f(c,d) = nan if there exist expressions e1, e2, e3, e4 with
limit(e1) = c, limit(e2) = c, limit(e3) = d, limit(e4) = d and
limit(f(e1, e3)) != limit(f(e2, e4)).
[It is understood here that all limits are of the same kind, e.g. all
as x->0 in C or all es x->oo in R, or whatever.]

In particular then for f(x,y) = x - y, it follows that zoo-zoo = nan,
since e1=z, e2=z, e3=z+1, e4=z have the desired properties.

A slightly less ad-hoc way of thinking about this is using filters,
c.f. the wiki [1].

Actually I think that no matter what the "high-powered" definition is
going to be, the first property I described should be whenever the
result is not nan.

Obviously not more involved in sympy than you are, I'm just reporting
what I have caught up, hence take my words with a (big) grain of
salt :-).

[1] https://github.com/sympy/sympy/wiki/Infinities-and-Singularities

On 18 Apr., 17:33, Saptarshi Mandal <[email protected]> wrote:
> I am just having difficulty wrapping my head around zoo - zoo. Complex
> infinity should be unsigned. And even if you did write something like
> this should this be nan or zoo?
>
> Otoh, since zoo - zoo doesnt make sense, S.NaN should actually be the
> correct answer. Can someone more knowledgeable about this enlighten
> me?
>
> On Apr 18, 4:36 pm, "Alexey U. Gudchenko" <[email protected]> wrote:
>
> > 18.04.2011 08:20, smichr пишет:
>
> > > Does zoo absorb everything in an addition or multiplication (except
> > > maybe oo or another zoo)?
>
> > It is related with this old 
> > issue:http://code.google.com/p/sympy/issues/detail?id=360
>
> > And with this open pull 
> > request:https://github.com/sympy/sympy/pull/173/files
>
> > E.g. Those assertions have quit right behavior:
>
> > assert zoo == zoo
> > assert zoo != oo
> > assert 1*zoo == zoo
> > assert 2*zoo == zoo
> > assert 1 != zoo
> > assert zoo != 1
> > assert zoo != Symbol("x")**3
> > assert zoo + 1 == zoo + 1
> > assert zoo + 1 == zoo
> > assert 2 + zoo == zoo
> > assert 3*zoo + 2 == zoo
> > assert 1/zoo == 0
>
> > x = Symbol('x')
>
> > assert zoo + zoo == nan
> > assert zoo - zoo == nan
> > assert zoo + oo == nan
> > assert zoo - oo == nan
>
> > assert zoo + I == zoo
> > assert zoo * I == zoo
> > assert zoo + x + 2 + 3*I == zoo + x
> > assert zoo * x * 2 * 3*I * -oo == zoo * x
>
> > assert zoo * (1+I) == zoo
> > assert zoo * (4+I+x) != zoo
>
> > assert zoo * 0 == nan
> > assert zoo * nan == nan
> > assert 0 * zoo == nan
> > assert nan * zoo == nan
> > assert zoo + nan == nan
>
> > assert zoo**zoo == nan
> > assert zoo**0   == nan
> > assert zoo**2   == zoo
> > assert zoo**(-1) == 0
>
> > assert abs(zoo) == oo
>
> > If anyone disagree with those assertions please let us know.
>
> > --
> > Alexey U.
>
>

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