theorem
  seq is non-zero implies -seq1/"seq=(-seq1)/"seq & seq1/"(-seq)=-seq1/" seq
proof
  assume
A1: seq is non-zero;
  thus -seq1/"seq=(-1r)(#)(seq1/"seq) .=((-1r)(#)seq1)(#)(seq") by Th12
    .=(-seq1)/"seq;
  thus seq1/"(-seq)=seq1(#)((-1r)(#)seq") by A1,Th40
    .=(-1r)(#)(seq1(#)(seq")) by Th13
    .=-(seq1/"seq);
end;
