theorem Th50:
  proj1 R = (R~).:B & proj2 R = R.:A
proof
  thus proj1 R = dom R .= rng (R~) by RELAT_1:20
    .= (R~).:B by RELSET_1:22;
  thus proj2 R = rng R .= R.:A by RELSET_1:22;
end;
