theorem
  e = A \/ B implies A = e & B = e or A = e & B = {} or A = {} & B = e
proof
  assume
A1: e = A \/ B;
  then A c= e & B c= e by XBOOLE_1:7;
  then
  A = {} & B = e or A = e & B = {} or A = e & B = e or A = {} & B = {} by Th1;
  hence thesis by A1;
end;
