theorem Th103:
  X1 misses X2 or Y1 misses Y2 implies [:X1,Y1:] misses [:X2,Y2:]
proof
  assume
A1: X1 misses X2 or Y1 misses Y2;
  assume not thesis;
  then consider z such that
A2: z in [:X1,Y1:] /\ [:X2,Y2:] by XBOOLE_0:4;
  ex x,y st z=[x,y] & x in X1 /\ X2 & y in Y1 /\ Y2 by A2,Th84;
  hence contradiction by A1;
end;
