Answer:
To prove that (AUB) NC CAU (BNC) and why z € (AUB) nC implies that a € AU (BNC), we can start by using the definition of set operations , specifically De Morgan's laws and set intersection.
First, let's rewrite the left-hand side of the equation using De Morgan's laws:
(AUB) NC = (A n C)' n (B n C)'
Next, we can expand (AUB) using the distributive law and simplify:
(A n C)' n (B n C)' = (A' n C') n (B' n C') = (A' n B') n C'
Now let's focus on the right-hand side of the equation:
CAU (BNC) = (C n A) U (C n B')
To prove the equivalence of the left and the right sides, we need to show that:
(A' n B') n C' = (C n A) U (C n B')
Let z € (AUB) nC, which means that z € AUB and z € C. This implies that z € A or z € B, and z € C.
If z € A and z € C, then z € A n C, which means that z € CA. Similarly, if z € B' and z € C, then z € B' n C, which means that z € C(BNC).
Therefore, z € CAU (BNC), which implies that (AUB) NC CAU (BNC).
Now to prove that z € (AUB) nC implies that a € AU (BNC):
Suppose that z € (AUB) nC. Then we know that z € C and z € A or z € B.
Without loss of generality , let's assume that z € A. This means that z € AU, which implies that a € AU for some a € A.
Now let's consider the case where z € BNC. This means that z € B' and z € C. If z € B' and a € A, then a € AU(BNC) since a can be in A or in B'(which means that it is not in B) and also in C. Thus, we have shown that z € (AUB) nC implies that a € AU(BNC).
Therefore, we have shown that (AUB) NC CAU (BNC) and why z €
Explanation: