theorem Th6:
  dom (g^) c= dom g & dom g /\ (dom g \ g"{0}) = dom g \ g"{0}
proof
  dom (g^) = dom g \ g"{0c} by Def2;
  hence dom (g^) c= dom g by XBOOLE_1:36;
  thus thesis by XBOOLE_1:28,36;
end;
