The (x*y)^a formula seems valid for x, y = 0 also

[certik]

It's this formula: http://fungrim.org/entry/2090c3/, it seems it's valid even if x and/or y are 0.

[fredrik-johansson]

If you allow x = 0 or y = 0, you can get factors like 0^-1. It looks like it should still be correct if arithmetic with UnsignedInfinity and Undefined is defined consistently, but my test code can't check this (yet).

[certik]

I see, I forgot about that. Yes, I would not allow 1/0 at all. So it looks like the assumptions should be modified to something like this:

Assumptions(Or(     
    And(
        Element(x, SetMinus(CC, Set(0))),
        Element(y, SetMinus(CC, Set(0))),
        Element(a, CC)
    ),
    And(
        Element(x, CC),                  
        Element(y, CC),                  
        Element(a, OpenInterval(0, oo))                                 
    ),
))   

Maybe one can also generalize to Re(a) > 0 or something like that.