theorem Th1:
  seq1=seq2-seq3 iff for n holds seq1.n=seq2.n-seq3.n
proof
  thus seq1=seq2-seq3 implies for n holds seq1.n=seq2.n-seq3.n
  proof
    assume
A1: seq1=seq2-seq3;
    now
      let n;
A2:   n in NAT by ORDINAL1:def 12;
      seq1.n=seq2.n+(-seq3).n by A1,VALUED_1:1,A2;
      then seq1.n=seq2.n+(-seq3.n) by VALUED_1:8;
      hence seq1.n=seq2.n-seq3.n;
    end;
    hence thesis;
  end;
  assume
A3: for n holds seq1.n=seq2.n-seq3.n;
  now
    let n be Element of NAT;
    thus seq1.n = seq2.n-seq3.n by A3
      .= seq2.n+-seq3.n
      .= seq2.n+(-seq3).n by VALUED_1:8
      .= (seq2+(-seq3)).n by VALUED_1:1;
  end;
  hence thesis by FUNCT_2:63;
end;
