Let's analyze the given expression and verify if it's equal to $(A \cup B)$.

Given Expression:

$(A \cap B') \cup (A \cap B) \cup (A' \cap B) \cup (A' \cap B')$

Simplification:

1. First, let's break down each part:

   • $A \cap B'$: Elements in $A$ but not in $B$.
   • $A \cap B$: Elements in both $A$ and $B$.
   • $A' \cap B$: Elements in $B$ but not in $A$.
   • $A' \cap B'$: Elements in neither $A$ nor $B$.

2. Union of the first two terms:

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

   Since $B' \cup B$ is the universal set, this simplifies to:

   $A \cap \text{Universal Set} = A$

3. Union of the last two terms:

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

   Similarly, $B \cup B'$ is the universal set, so this simplifies to:

   $A' \cap \text{Universal Set} = A'$

4. Union of all simplified terms:

   $A \cup A'$

   Since $A \cup A'$ also represents the universal set, the final expression covers all possible elements in the universal set.

Now compare this to the right-hand side:

$A \cup B$

Since the universal set contains all elements, and $A \cup B$ is a subset of the universal set, we find:

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

Conclusion:

The given expression simplifies to the universal set, which is equal to $A \cup B$. Therefore, the expression is indeed equal to $A \cup B$.

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Would you like further details or have any questions? Here are some related questions for further exploration:

1. How can we represent the complement of a set using Venn diagrams?
2. What are the properties of set unions and intersections?
3. How does De Morgan's Law relate to the complement of the intersection or union of sets?
4. What happens when we take the intersection of more than two sets?
5. Can we express set differences using union and intersection?

Tip: Remember that $A \cup A' = \text{Universal Set}$, which can be useful when simplifying set expressions.