reserve n,k,b for Nat, i for Integer;

theorem Th20:
  for n,k being Nat st k=10|^(2*n) - 1 holds 11 divides k
  proof
    defpred P[Nat] means ex k being Nat st k=10|^(2*$1)-1 & 11 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|^(2*k)-1 and
      A3: 11 divides l;
      consider m being Nat such that
      A4: l = 11 * m by A3,NAT_D:def 3;
      10|^(2*(k+1)) - 1 = 10|^(2*k+2) - 1
      .= (11*m+1)*10|^(1+1) - 1 by A2,A4,NEWTON:8
      .= (11*m+1)*(10|^1*10) - 1 by NEWTON:6
      .= (11*m+1)*(10*10) - 1
      .= 11*(m*100+9);
      hence P[k+1] by NAT_D:def 3;
    end;
    (10|^0) -1 = 1-1 by NEWTON:4
    .= 0; then
    A5: P[0] by NAT_D:6;
    for k being Nat holds P[k] from NAT_1:sch 2(A5,A1); then
    A6: ex l being Nat st l=10|^(2*n) - 1 & 11 divides l;
    assume k=10|^(2*n) - 1;
    hence thesis by A6;
  end;
