theorem
  x in rng p & p is one-to-one implies rng(p |-- x) misses {x}
proof
  assume x in rng p & p is one-to-one;
  then not x in rng(p |-- x) by Th46;
  then for y being object st y in rng(p |-- x) holds not y in {x}
   by TARSKI:def 1;
  hence thesis by XBOOLE_0:3;
end;
