> I tried the following code at the repl:
>
> (define-syntax foo
>   (syntax-rules ()
>     ((foo bar x)
>      (define-syntax bar
>        (syntax-rules (x)
>          ((bar c) 'matched)
>          ((bar z) 'unmatched))))))
>
> (foo bar c)
> (bar q)
>
> Chez, Chibi, Racket (R6RS), Larceny (R7R6, version from yesterday) all
> evaluate the last expression to matched.

> >  From R6RS 11.19:
> >
> >     A literal identifier matches an input subform if and only
> >     if the input subform is an identifier and either both its
> >     occurrence in the input expression and its occurrence in
> >     the list of literals have the same lexical binding, or the
> >     two identifiers have the same name and both have no lexical
> >     binding.
> >
> > R7RS 4.3.2:
> >
> >     An element in the input matches a literal identifier if and
> >     only if it is an identifier and either both its occurrence
> >     in the macro expression and its occurrence in the macro
> >     definition have the same lexical binding, or the two
> >     identifiers are the same and both have no lexical binding.
> >
> > The R6RS is more precise here, with the "have the same name"
> > explaining what the R7RS leaves open to more imaginative
> > interpretations by saying "are the same".
>
> Renamed identifiers are certainly not the same (e.g. are not
> free-identifier=?); by the very meaning of "rename", they also do not have
> the same name.

That's an overly simplistic view.  Tentative renaming of identifiers
(coloring, marking, whatever you want to call it) does not prevent
them from being matched by free-identifier=?, whose semantics is
described by R6RS 11.19 and even more explicitly by the R6RS Standard
Libraries document, section 12.5:

    The free-identifier=? procedure returns #t if and only if
    the two identifiers would resolve to the same binding if
    both were to appear in the output of a transformer outside
    of any bindings inserted by the transformer. (If neither
    of two like-named identifiers resolves to a binding, i.e.,
    both are unbound, they are considered to resolve to the same
    binding.) Operationally, two identifiers are considered
    equivalent by free-identifier=? if and only [if] the topmost
    matching substitution for each maps to the same binding
    (section 12.1) or the identifiers have the same name and
    no matching substitution.

The two R6RS documents are pretty clear about this.  Not so clear
is whether and when an identifier inserted via macro expansion is
to be considered bound (for the purposes of the paragraphs quoted)
because it ends up as a pattern variable in a use of syntax-rules
created by macro expansion.  That is subtle, because the question
of whether the inserted identifier becomes a pattern variable or
literal has to be determined by matching it against the literals,
which means you have to do some form of matching before you can do
the matching described for free-identifier=?.  This is probably
where I went wrong.

For that to make sense, you have to realize the word "hygienic"
is somewhat ill-defined in these corner cases.  It would be nice
if consistency between implementations were achieved entirely by
following a mathematically precise specification, but the truth
is that we end up tinkering with our systems until they pass the
test cases people care about.