theorem Th16:
  C c= BDD D implies D c= UBD C
proof
  assume
A1: C c= BDD D;
  assume
A2: not D c= UBD C;
  C misses D by A1,JORDAN1A:7,XBOOLE_1:63;
  then D c= BDD C by A2,JORDAN1K:19;
  hence contradiction by A1,Th15;
end;
