theorem
  seq is non-zero implies (seq1/"seq)(#)seq=seq1
proof
  assume
A1: seq is non-zero;
  now
    let n;
A2: seq.n<>0c by A1,Th4;
    thus ((seq1/"seq)(#)seq).n=((seq1(#)seq").n)*seq.n by VALUED_1:5
      .=(seq1.n)*(seq".n)*seq.n by VALUED_1:5
      .=(seq1.n)*(seq.n)"*seq.n by VALUED_1:10
      .=(seq1.n)*((seq.n)"*seq.n)
      .=(seq1.n)*1r by A2,XCMPLX_0:def 7
      .=seq1.n;
  end;
  hence thesis by FUNCT_2:63;
end;
