First of all we have carefully to distinguish between
rit = rectified tesseract =
o3o3x4oand
rat = rectified triacontaditeron = rectified 5D crosspolytope =
o3x3o3o4o.
The latter one also can be given wrt. demipenteractic (hinnic) symmetry as
o3o3o *b3x3o,
where
hin = demipenteract (aka: hemipenteract) =
x3o3o *b3o3o,
while the former likewise can be given wrt. demitesseractic (hexic) symmetry as
x3o3x *b3o,
where
hex = demitesseract = hexadecachoron =
x3o3o4o =
o3o3o *b3x.
Thus the remark of
Mercejide would probably try to blend 2 identic members of hinnic symmetry,
when the 2 (assumed to be) non-identically decorated arms of the Dynkin symbol get reversed,
but the longer tail each would be kept in place: Then indeed the facets of that tail could be blendet out.
But such a blend supposedly would increase the total symmetry from demipenteractic (hinnic)
back to (full) penteractic (pentic) symmetry again. And I'd bet that Jonathan would count those there already...
--- rk