theorem
  (a |^ i) |^ b = (a |^ b) |^ i
proof
  per cases;
  suppose
    i >= 0;
    then i = |.i.| by ABSVALUE:def 1;
    hence thesis by Lm4;
  end;
  suppose
A1: i < 0;
    hence a |^ i |^ b = (a |^ |.i.|)" |^ b by GROUP_1:30
      .= (a |^ |.i.| |^ b)" by Th26
      .= ((a |^ b) |^ |.i.|)" by Lm4
      .= (a |^ b) |^ i by A1,GROUP_1:30;
  end;
end;
