theorem Th6:
  Product((n+1) |-> rr) = (Product(n|->rr))*rr
proof
  thus Product((n+1) |->rr) = (Product(n|->rr))*(Product(1|->rr))
     by RVSUM_1:104
    .= (Product(n|->rr))*rr by Th4;
end;
