On Thu, Jun 7, 2012 at 6:37 PM, [email protected]
<[email protected]> wrote:
> In [22]: f(y).atoms(f(x))
> Out[22]: set([f(y)])

I would say no since f(x) != f(y); I would expect f(y).atoms(f) to
return f(y), however (and it does).

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