theorem Th19:
  (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.(n+k)+-seq1.(n+k) by VALUED_1:8
      .= (seq^\k).n-seq1.(n+k) by NAT_1:def 3
      .= (seq^\k).n+-(seq1^\k).n by NAT_1:def 3
      .= (seq^\k).n+(-(seq1^\k)).n by VALUED_1:8
      .= ((seq^\k)-(seq1^\k)).n by VALUED_1:1;
  end;
  hence thesis;
end;
