theorem Th9:
  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 Th8
    .= ((a1|^m+b1|^m)*(a1+b1) + (a1-b1)*(a1|^m-b1|^m))/2;
  end;
