
theorem PAN:
  for a be Integer, n be Nat holds Parity (a|^n) = (Parity a)|^n
  proof
    let a be Integer, n be Nat;
    defpred P[Nat] means Parity (a|^$1) = (Parity a)|^$1;
    A1: P[0]
    proof
      Parity (1*a|^0) = 1*(Parity a)|^0;
      hence thesis;
    end;
    A2: for k be Nat holds P[k] implies P[k+1]
    proof
      let k be Nat;
      assume
      B1: Parity (a|^k) = (Parity a)|^k;
      Parity a|^(k+1) = Parity (a*a|^k) by NEWTON:6
     .= (Parity a)*(Parity a)|^k by B1,ILP;
      hence thesis by NEWTON:6;
    end;
    for x be Nat holds P[x] from NAT_1:sch 2(A1,A2);
    hence thesis;
  end;
