Let's focus on that for a sec:
(define ((A)) 1) is the same as (define (A) (lambda () 1));;
defines procedure "(A)"
I wonder if you meant >>defines procedure "((A))"<< instead.
Assuming that, if "((A))" is just a name of the procedure, then
"A" and "(A)". Should not evaluate at all. Apparently above line
defines procedure "A" actually.
I know this is not that useful although very interesting in terms of
how it was implemented and how that (define ...) form interprets the
first argument.
A.
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