reserve
  a,b,c,d,e for Ordinal,
  m,n for Nat,
  f for Ordinal-Sequence,
  x for object;
reserve S,S1,S2 for Sequence;

theorem
  for n being Nat holds 0 |^|^ (2*n) = 1 & 0 |^|^ (2*n+1) = 0
  proof
    defpred P[Nat] means 0 |^|^ (2*$1) = 1 & 0 |^|^ (2*$1+1) = 0;
A1: P[0] by Th13,Th16;
A2: now
      let n be Nat; assume
A3:   P[n];
      thus P[n+1]
      proof
        thus
A4:     0 |^|^ (2*(n+1)) = 0 |^|^ Segm(2*n+1+1)
        .= 0 |^|^ succ Segm(2*n+1) by NAT_1:38
        .= exp(0, 0) by A3,Th14
        .= 1 by ORDINAL2:43;
        thus 0 |^|^ (2*(n+1)+1) = 0 |^|^ Segm(2*(n+1)+1)
        .= 0 |^|^ succ Segm(2*(n+1)) by NAT_1:38
        .= exp(0, 1) by A4,Th14 .= 0 by ORDINAL2:46;
      end;
    end;
    thus for n being Nat holds P[n] from NAT_1:sch 2(A1,A2);
  end;
