Am Mi., 23. Sept. 2020 um 15:53 Uhr schrieb Jim Rees <xxxxxx@gmail.com>:
>
> On Wed, Sep 23, 2020 at 9:23 AM Marc Nieper-Wißkirchen <xxxxxx@gmail.com> wrote:
>>
>>
>> > And 2nd question -- w.r.t. to free-identifier=?, what's the name of an alias to an unbound identifier?   It's own name, or the name of the original source id?
>>
>> The name of an identifier is a property of the identifier, not of its
>> binding. In particular, syntax->datum won't return the name of the
>> other identifier.
>>
>> > ;; pre-conditions: x and y are not bound
>> >
>> > (define (is-x?  id) (free-identifier=? id #'x))
>> >
>> > (let ()
>> >    (alias x y)
>> >    (is-x? x))
>>
>> This should return `#f' (after adding (syntax ...) as you wrote in
>> your next email).
>
>
> These answers seem to conflict with each other.
>
> (syntax->datum #'x) is always x - good that tells me that there's no special object class for the result of an alias form -- it's just an identifier, possibly bound, possibly not, and now the new feature is that it can be "bound" to be un-bound -- and if so, it should have the same behavior as the same-named identifier that was never bound in the first place -- (or is this my confusion? is an un-binding distinct from a never-binding?)
>
> Why would is-x? return #f?     The outer x and the inner x are both un-bound, and both have the same name x.

My mistake. I should have paid more attention to what I had read
before I answered. I answered to what (is-x? #'y) would evaluate, not
to what (is-x? #'x) would...

So, yes, the latter would return #t because of the reasons you gave.
Sorry for the confusion.