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

theorem Th21:
  for n,k being Nat st k=10|^(2*n+1) + 1 holds 11 divides k
  proof
    defpred P[Nat] means ex k being Nat st k=10|^(2*$1+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)+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) + 1 = 10|^(2*k+1+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 INT_1:def 3;
  end;
  A5: P[0];
  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) + 1 & 11 divides l;
  assume k=10|^(2*n+1) + 1;
  hence thesis by A6;
end;
