reserve X,Y for set, x,y,z for object, i,j,n for natural number;

theorem Lem6:
  for f being one-to-one Function st X misses Y
  holds f.:X misses f.:Y
  proof
    let f be one-to-one Function;
    assume Z2: X misses Y;
    assume f.:X meets f.:Y;
    then consider x such that
A1: x in f.:X & x in f.:Y by XBOOLE_0:3;
    consider y such that
A2: y in dom f & y in X & x = f.y by A1,FUNCT_1:def 6;
    consider z being object such that
A3: z in dom f & z in Y & x = f.z by A1,FUNCT_1:def 6;
    y <> z by Z2,A2,A3,XBOOLE_0:3;
    hence contradiction by A3,A2,FUNCT_1:def 4;
  end;
