Am Sonntag, den 06.12.2009, 11:33 -0800 schrieb Richard Fateman:
> Maybe it will not work if inf is embedded, e.g.
>
> 0 * (3+inf).
>
> since (intersection '($inf ...) '(0 ((mplus) 3 $inf)) is nit..
I have tried to implement the algorithm as general as possible. Nested
expressions are not a problem:
(%i10) 0*(3+inf);
(%o10) und
(%i11) 0*(3+inf)^2;
(%o11) und
The point is that the code only looks at the toplevel of the args,
because the routine can expect that nested infinities are already
simplified. I have done it this way, to avoid to look every time into
the nested lists of an expression.
Dieter Kaiser