theorem Th23:
  not a in p implies not a|^k in p
  proof
    assume
A1: not a in p;
    not a|^k in p
    proof
      defpred P[Nat] means not a|^$1 in p;
A2:   P[1] by A1,BINOM:8;
A3:   for k holds P[k] implies P[k+1]
      proof
        let k;
        assume
A4:     P[k];
A5:     a|^(k+1) = (a|^k)*(a|^1) by BINOM:10;
        not (a|^1) in p by A1,BINOM:8;
        hence thesis by A4,A5,RING_1:def 1;
      end;
      for k holds P[k] from NAT_1:sch 10(A2,A3);
      hence thesis;
    end;
    hence thesis;
  end;
