theorem
  (seq1/"seq)*Ns = (seq1*Ns)/"(seq*Ns)
proof
  thus (seq1/"seq)*Ns = (seq1(#)(seq"))*Ns .= (seq1*Ns)(#)((seq")*Ns) by Th2
    .= (seq1*Ns)(#)((seq*Ns)") by Th5
    .= (seq1*Ns)/"(seq*Ns);
end;
