theorem Th33:
  A c= B implies P.(B \ A) = P.B - P.A
proof
  assume
A1: A c= B;
  A misses (B \ A) by XBOOLE_1:79;
  then P.A + P.(B \ A) = P.(A \/ (B \ A)) by Def8
    .= P.B by A1,XBOOLE_1:45;
  hence thesis;
end;
