theorem
  r <> 0 & j <= i implies r|^(i-'j) = r|^i / r|^j
proof
  assume that
A1: r <> 0 and
A2: j <= i;
   reconsider rr=r as Real;
  thus r|^i / r|^j = (Product(i |-> rr)) / rr|^j by NEWTON:def 1
    .= (Product(i |-> rr)) / (Product(j |-> rr)) by NEWTON:def 1
    .= Product((i-'j) |-> rr) by A1,A2,Th8
    .= r|^(i-'j) by NEWTON:def 1;
end;
