The paragraph and example of eqv? on procedures with local state are
superfluous: it doesn't matter if a procedure has local state or not, as
the report has explicitly un-specified eqv? on procedures.  In this
sense, closures are not objects.

In 6.2.4, "A NaN always compares false to any number, including a NaN".
I thought the result of the discussion was that (eqv? +nan.0 +nan.0)
could be #t?  I didn't really follow it anyway.

Andy
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