We are given that $X$ and $Y$ are subsets of a universal set $U$. We need to simplify two expressions involving set operations: intersection ($\cap$), union ($\cup$), and complement ($'$). The expressions are: i. $X \cap (X' \cup Y)$ ii. $[(X \cap Y)' \cap (X' \cup Y)]'$

1. Problem Description

We are given that $X$ and $Y$ are subsets of a universal set $U$. We need to simplify two expressions involving set operations: intersection ($\cap$), union ($\cup$), and complement ($'$). The expressions are:

i. $X \cap (X' \cup Y)$

ii. $[(X \cap Y)' \cap (X' \cup Y)]'$

2. Solution Steps

i. Simplifying $X \cap (X' \cup Y)$:

We can use the distributive property of intersection over union:

$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

So, $X \cap (X' \cup Y) = (X \cap X') \cup (X \cap Y)$

We know that $X \cap X' = \emptyset$, where $\emptyset$ is the empty set.

Thus, $X \cap (X' \cup Y) = \emptyset \cup (X \cap Y)$

Since the union of the empty set with any set is the set itself, we have:

$\emptyset \cup (X \cap Y) = X \cap Y$

Therefore, $X \cap (X' \cup Y) = X \cap Y$

ii. Simplifying $[(X \cap Y)' \cap (X' \cup Y)]'$:

We can use De Morgan's Law, which states that $(A \cap B)' = A' \cup B'$. Thus, $(X \cap Y)' = X' \cup Y'$.

Now we have $[(X' \cup Y') \cap (X' \cup Y)]'$.

We can factor out $X'$, since $X'$ appears in both terms of the intersection:

$[(X' \cup Y') \cap (X' \cup Y)]' = [X' \cup (Y' \cap Y)]'$

Since $Y' \cap Y = \emptyset$, the empty set, we have:

$[X' \cup \emptyset]' = [X']'$

The complement of a complement is the original set:

$[X']' = X$

Therefore, $[(X \cap Y)' \cap (X' \cup Y)]' = X$.

3. Final Answer

i. $X \cap Y$

ii. $X$