
theorem Th6:
  for n,k being Nat st k=10|^n - 1 holds 9 divides k
proof
  defpred P[Nat] means ex k being Nat st k=(10|^$1)-1 & 9 divides k;
  let n,k be Nat;
A1: now
    let k be Nat;
    assume P[k];
    then consider l being Nat such that
A2: l=(10|^k)-1 and
A3: 9 divides l;
    consider m being Nat such that
A4: l = 9 * m by A3,NAT_D:def 3;
A5: 9 divides 9*(m*10+1) by NAT_D:def 3;
    10|^(k+1)-1 = (9*m+1)*10 -1 by A2,A4,NEWTON:6
      .= 9*(m*10+1);
    hence P[k+1] by A5;
  end;
  (10|^0) -1 = 1-1 by NEWTON:4
    .= 0;
  then
A6: P[0] by NAT_D:6;
  for k being Nat holds P[k] from NAT_1:sch 2(A6,A1);
  then
A7: ex l being Nat st l=10|^n -1 & 9 divides l;
  assume k=10|^n - 1;
  hence thesis by A7;
end;
