theorem
  g * h = h * g implies g * (h |^ i) = h |^ i * g
proof
  assume
A1: g * h = h * g;
  thus g * (h |^ i) = g |^ 1 * (h |^ i) by Th25
    .= h |^ i * (g |^ 1) by A1,Th38
    .= h |^ i * g by Th25;
end;
