theorem
  x in rng p & p is one-to-one implies p -| x is one-to-one
proof
  assume x in rng p;
  then p = (p -| x) ^ <* x *> ^ (p |-- x) by Th51
    .= (p -| x) ^ (<* x *> ^ (p |-- x)) by FINSEQ_1:32;
  hence thesis by FINSEQ_3:91;
end;
