Proving a union by using the laws of algebra of sets and a new operator So I am given a new opeator A * B = complement of (A intersect B).  I apologize for not being able to post this with the correct syntax.
Using this new operator, I am to prove, algebraically, that the following is true:
(A * A) * (B * B) = A U B
I've gotten this far:

2nd part:

 A: You are almost there! Remember De Morgan's Law: 
$${\overline {A\cap B}}={\overline {A}}\cup {\overline {B}}$$
I will use two different notations for complement so you can see it better in the next line. Continuing from what you had,
$$\overline{A} * \overline{B} = \overline{(\overline{A} \cap \overline{B})} = \overline{A^c} \cup \overline{B^c}$$
Note that the complement of the complement of a set is just that set ($\overline{A^c} = A$) and hence you have:
$$(A * A) * (B * B) = A \cup B$$
Let me know if you need further explanation!
A: The exercice mainly aims at testing your knowledge of (1) idempotency of the intersection operation and of  (2) De Morgan's law for sets. 
Note : here I use  " X' "  to denote the complement of a set X and the symbol " X ''  " to denote the complement of the complement of X 
x belongs to (A*A) * (B * B) 
is equivalent to 
(1) x belongs to [ (A * A) Inter (B * B)]'
(2) x belongs to [ (A Inter A)' Inter (B Inter B)' ] ' 
(3) x belongs to [      A '     Inter      B'      ]' 
(4) x belongs to        A ''      Union    B '' 
(5) x belongs to        A         Union    B 
Reasons 
For (1) : definition of the * operator 
For (2) : again, definition of the * operator
For (3) : idempotency of the Intersection operation ( for all X, X Inter X = X) 
For (4) : De Morgan's law for sets ( The complement of X Inter Y is X' Union Y') 
For (5) : law of involution ( the complement of the complement of X is X itself) 
