[r6rs-discuss] Why is (eqv? g g) unspecified when g is a procedure?

[Andy Wingo]

On Mon 16 Jul 2012 18:18, John Cowan <cowan at mercury.ccil.org> writes:

> (eqv? eqv? eqv?) => #t.

Since we are on the R6RS list, does R6RS guarantee this?

Andy
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