theorem Th87:
  for A being finite set st p is one-to-one & A c= rng p holds
    len (p - A) = len p - card A
proof
  let A be finite set;
  assume that
A1: p is one-to-one and
A2: A c= rng p;
  A /\ rng p = A by A2,XBOOLE_1:28;
  hence thesis by A1,Th86;
end;
