The image shows a set of exercises involving set theory identities. You are asked to prove several set relations, given that $A$ and $B$ are subsets of a universal set $U$. Let’s go through each of the problems listed:

4.1: $\overline{A \cup B} = \overline{A} \cap \overline{B}$

This is De Morgan's Law. According to De Morgan's laws: $\overline{A \cup B} = \overline{A} \cap \overline{B}$ This holds true as it shows that the complement of a union is equal to the intersection of the complements.

4.2: $\overline{A \cup B} = B \setminus A$

For this, we need to interpret the left-hand side ($\overline{A \cup B}$) and simplify it step by step:

• $\overline{A \cup B}$ represents everything that is not in $A \cup B$, i.e., elements not in either $A$ or $B$.
• $B \setminus A$ represents all the elements in $B$ but not in $A$.

By analyzing the two sides, we see that both sets represent the same elements—those in $B$ but not in $A$, so this is also true.

4.3: $A \cap \overline{A \cup B} = \emptyset$

Here, the idea is to prove that the intersection of $A$ and the complement of $A \cup B$ is the empty set.

• $A \cap \overline{A \cup B}$ represents the elements in $A$ that are not in $A \cup B$.
• However, all elements in $A$ are part of $A \cup B$, meaning no elements from $A$ can be in $\overline{A \cup B}$.

Thus, the intersection must be empty: $A \cap \overline{A \cup B} = \emptyset$.

4.4: $(A \cup B) \cap (A \cap \overline{B}) = B$

Let's break this into steps:

• $A \cup B$ represents all elements in $A$ or $B$.
• $A \cap \overline{B}$ represents the elements in $A$ that are not in $B$.
• Their intersection is the part of $A \cup B$ that also lies in $A \cap \overline{B}$.

Through set manipulations, this expression simplifies to $B$, as it essentially captures all elements in $B$ based on exclusions from $A$.

4.5: $\overline{A \cap B} \setminus A = \overline{A} \cup B$

• $\overline{A \cap B}$ is the complement of the intersection of $A$ and $B$, meaning all elements not in both $A$ and $B$.
• The set difference $\overline{A \cap B} \setminus A$ removes the elements of $A$ from $\overline{A \cap B}$.
• This simplifies to $\overline{A} \cup B$, since removing $A$ leaves elements that are in $B$ or not in $A$.

4.6: $\overline{B} \cup \left[(A \cap \overline{B}) \cap B \right] = U$

• The term $A \cap \overline{B}$ gives elements in $A$ but not in $B$.
• The intersection of $(A \cap \overline{B}) \cap B$ is empty since there are no elements in both $\overline{B}$ and $B$.
• Therefore, $\overline{B} \cup \emptyset = \overline{B}$, and $\overline{B} \cup B = U$, because $B$ and its complement together cover the universal set.

Would you like further details on any specific proof or clarification? Here are some related questions:

1. How does De Morgan's law apply in set theory?
2. Can the complement of a set always be expressed as a set difference?
3. How can we visually represent $A \cap \overline{A \cup B}$?
4. What is the significance of the universal set $U$ in these operations?
5. How do set operations like union and intersection relate to logical operators?

Tip: When working with set theory, Venn diagrams can be a useful tool to visualize the relationships between sets, especially when dealing with complements and intersections.