theorem
  (seq ^\k)^\m = (seq ^\m)^\k
proof
  now
    let n be Element of NAT;
    thus ((seq ^\k)^\m).n = (seq ^\k).(n + m) by NAT_1:def 3
      .= seq.(n + m + k) by NAT_1:def 3
      .= seq.(n + k + m)
      .= (seq ^\m).(n + k) by NAT_1:def 3
      .= ((seq ^\m)^\k).n by NAT_1:def 3;
  end;
  hence thesis by FUNCT_2:63;
end;
