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