theorem
  rng R misses rng S implies R misses S
proof
  assume
A1:rng R misses rng S;
  assume R meets S;
  then consider x being object such that
A2: x in R and
A3: x in S by XBOOLE_0:3;
   consider y,z such that
A4: x = [y,z] by A2,Def1;
   z in rng R & z in rng S by A2,A3,A4,XTUPLE_0:def 13;
   hence contradiction by A1,XBOOLE_0:3;
end;
