theorem Th18:
  (seq+seq1)^\k = (seq^\k)+(seq1^\k)
proof
  now
    let n be Element of NAT;
A1:  n+k in NAT by ORDINAL1:def 12;
   thus ((seq+seq1)^\k).n = (seq+seq1).(n+k) by NAT_1:def 3
      .= seq.(n+k)+seq1.(n+k) by VALUED_1:1,A1
      .= (seq^\k).n+seq1.(n+k) by NAT_1:def 3
      .= (seq^\k).n+(seq1^\k).n by NAT_1:def 3
      .= ((seq^\k)+(seq1^\k)).n by VALUED_1:1;
  end;
  hence thesis;
end;
