theorem
  p is one-to-one & x in rng p implies len(p - {x}) = len p - 1
proof
  assume that
A1: p is one-to-one and
A2: x in rng p;
  {x} c= rng p by A2,ZFMISC_1:31;
  then len(p - {x}) = len p - card{x} by A1,Th87;
  hence thesis by CARD_1:30;
end;
