theorem Th3:
  a in H implies a |^ n in H
proof
  defpred P[Nat] means a |^ $1 in H;
  assume
A1: a in H;
A2: now
    let n;
A3: a |^ (n + 1) = a |^ n * a by GROUP_1:34;
    assume P[n];
    hence P[n+1] by A1,A3,GROUP_2:50;
  end;
  a |^ 0 = 1_G by GROUP_1:25;
  then
A4: P[0] by GROUP_2:46;
  for n holds P[n] from NAT_1:sch 2(A4,A2);
  hence thesis;
end;
