theorem Th20:
  (seq")^\k = (seq^\k)"
proof
  now
    let n be Element of NAT;
    thus ((seq")^\k).n = (seq").(n+k) by NAT_1:def 3
      .= (seq.(n+k))" by VALUED_1:10
      .= ((seq^\k).n)" by NAT_1:def 3
      .= ((seq^\k)").n by VALUED_1:10;
  end;
  hence thesis;
end;
