 Hello and welcome to the session. Let us understand the following problem today. Given a non-empty set x, let's start the function from p of x cross p of x to p of x. We defined as a star b is equal to a minus b union b minus a for all a b belongs to p of x. Sure that the empty set phi is the identity element for the operation star and all elements a of p of x are invertible with a inverse is equal to a. Solution p of x is the power set of x. The x is a non-empty set. The set is the set of all subsets of x. The operation on p of x union b minus a for all a b belongs to p of x. Let us name it as 1. We have to show that the identity element for star, for star on p of x. We know phi is the identity element in p of x is equal to a which is equal to phi star a for all a belongs to p of x. But any a belongs to p of x, we have equal to a minus phi union equation 1 which is equal to phi dash union phi intersection a dash because phi is equal to a intersection phi dash which is equal to a intersection union because phi dash is equal to u which is the universal set. It is equal to a union phi which is equal to a. Now, a is equal to phi minus a union a minus phi which is equal to phi intersection a dash union a intersection phi dash which is equal to phi union a intersection universal set u which is equal to phi union a which is equal to a. Thus, a star phi is equal to phi star a which is equal to a for all a belongs to p of x. Therefore, phi is the identity element in p of x for the binary operation star on p of x. We have to show that elements of a of p of x invertible versus equal to a. Now, a belongs to p of x, we have equal to a minus a union a minus a which is equal to phi union phi which is equal to phi is invertible versus equal to a. I hope you understood the problem. Bye and have a nice day.