Yes, the given set equation is valid. Let's prove it step by step.

Theorem:

For any two sets $A$ and $B$, $(A \cup B) - (A \cap B) = (A - B) \cup (B - A).$

Proof:

Left-hand side (LHS):

$(A \cup B) - (A \cap B)$ This is the set difference between the union of $A$ and $B$, and the intersection of $A$ and $B$.

• $A \cup B$ represents all elements in $A$ or $B$ (or both).
• $A \cap B$ represents all elements common to both $A$ and $B$.

So, $(A \cup B) - (A \cap B)$ represents all elements in $A \cup B$ that are not in $A \cap B$. This means we are removing all elements that are common to both sets, leaving behind the elements that are in $A$ but not in $B$, and elements that are in $B$ but not in $A$.

Thus, $(A \cup B) - (A \cap B) = \{ x \in A \cup B \ | \ x \notin A \cap B \}.$

Right-hand side (RHS):

$(A - B) \cup (B - A)$ This is the union of the set differences $A - B$ and $B - A$.

• $A - B$ represents all elements in $A$ that are not in $B$.
• $B - A$ represents all elements in $B$ that are not in $A$.

Thus, the union of $A - B$ and $B - A$ gives us all elements that are in $A$ but not in $B$, or in $B$ but not in $A$. In other words, $(A - B) \cup (B - A) = \{ x \in A \ | \ x \notin B \} \cup \{ x \in B \ | \ x \notin A \}.$

Conclusion:

Both sides of the equation describe the set of elements that are either in $A$ or $B$, but not in both. Therefore, we have shown that $(A \cup B) - (A \cap B) = (A - B) \cup (B - A).$

The theorem is proven. ✅

Would you like any further details on this proof?

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Here are 5 related questions to explore further:

1. How would the equation change if the universal set $U$ is involved?
2. Can we extend this set identity to three sets $A$, $B$, and $C$?
3. What is the symmetric difference of two sets, and how does it relate to this identity?
4. How can this identity be visualized using a Venn diagram?
5. What are some applications of set operations in real-life problems?

Tip: Always verify set identities by considering examples with specific sets to gain better intuition.