theorem Th13:
  a1|^(m+1)-b1|^(m+1) = ((a1|^m+b1|^m)*(a1-b1) + (a1+b1)*(a1|^m-b1|^m))/2
  proof
    thus a1|^(m+1)-b1|^(m+1) =
    ((a1|^m+b1|^m)*(a1-b1) + (a1|^1+b1|^1)*(a1|^m-b1|^m))/2 by Th5
    .= ((a1|^m+b1|^m)*(a1-b1) + (a1+b1)*(a1|^m-b1|^m))/2;
  end;
