theorem Th1:
  a = h implies a |^ n = h |^ n
proof
  defpred P[Nat] means a |^ $1 = h |^ $1;
  assume
A1: a = h;
A2: now
    let n;
    assume
A3: P[n];
    a |^ (n + 1) = a |^ n * a by GROUP_1:34
      .= h |^ n * h by A1,A3,GROUP_2:43
      .= h |^ (n + 1) by GROUP_1:34;
    hence P[n+1];
  end;
  a |^ 0 = 1_G by GROUP_1:25
    .= 1_H by GROUP_2:44
    .= h |^ 0 by GROUP_1:25;
  then
A4: P[0];
  for n holds P[n] from NAT_1:sch 2(A4,A2);
  hence thesis;
end;
