No, S03 is probably just wrong there.  Junctions are scalar values, and
don't flatten in list context.  Maybe we need something like:

    for =all(@foo) {...}

to iterate the junction.

: Is the whole block run once with 1,2 and 3, or does the
: junction go into the block and autothread each operation?

I expect =all(@foo) would do the former, while all(@foo) would do
the latter, in which case you might as well have used "given" instead.

: for all(1,2,3) {
:    next if $_ < 2;  # testing 1 or all(1,2,3) ?
:    %got{$_} = 1;
: }
: say %got.perl;      # "(('2', 1), ('3', 1))" or "()" ?

Well, { 2 => 1, 3 => 1 } is the more likely notation.

: The "no coupling" in s03 suggests to me that the right
: answer is "(('2', 1), ('3', 1))", but I'm just guessing.

I think =all(@foo) should do what you expect there.  Without the =
it should return { 1 => 1, 2 => 1, 3 => 1 } since there's only one
loop iteration, and it is *not* true that all(1,2,3) < 2.  If you'd
said

    for any(1,2,3) {...}

then it would have done the "next", because 1 < 2.

I should say that I don't see that =all() is different from =any().
They each just produce a list in "random" order.  Though I suppose,
if we say that =one(1,2,3) should randomly pick one value, then
=any(1,2,3) should pick anywhere from 1 to 3 values.  And, of course,
=none(1,2,3) should return a list of all the things that aren't 1, 2,
or 3 in random order.  Maybe a lazy implementation will be beneficial
at that point.  :-)