To prove the set-theoretic identity:

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

we will proceed by showing that if an element $x$ belongs to the left-hand side, then it must also belong to the right-hand side.

Proof:

1. Assume $x \in (A \cup C) \cap (B \cup C)$:

   This means that: $x \in (A \cup C) \quad \text{and} \quad x \in (B \cup C).$

2. Break down the implications of $x \in (A \cup C)$:

   $x \in A \quad \text{or} \quad x \in C.$

   Similarly, from $x \in (B \cup C)$, we have:

   $x \in B \quad \text{or} \quad x \in C.$

3. Consider two main cases:

   • Case 1: $x \in C$

     If $x \in C$, then clearly $x \in (A \cap B) \cup C$ because $x \in C$. Thus, $x$ satisfies the right-hand side of the inclusion.

   • Case 2: $x \notin C$

     If $x \notin C$, then from $x \in (A \cup C)$, it must be that $x \in A$. Similarly, from $x \in (B \cup C)$, it must be that $x \in B$. Therefore, $x \in A \cap B$.

     Since $x \in A \cap B$, it follows that $x \in (A \cap B) \cup C$.

4. Conclusion:

   In both cases (whether $x \in C$ or $x \notin C$), we have shown that $x \in (A \cap B) \cup C$. Therefore, every element of $(A \cup C) \cap (B \cup C)$ is also an element of $(A \cap B) \cup C$, proving the inclusion:

   $(A \cup C) \cap (B \cup C) \subseteq (A \cap B) \cup C.$

---

Would you like further clarification or details on this proof?

Here are 5 related questions to extend your understanding:

1. What is the difference between union and intersection in set theory?
2. Can you think of a counterexample where the reverse inclusion does not hold?
3. How would the proof change if $A$, $B$, and $C$ are disjoint sets?
4. What are the implications of this inclusion in terms of Venn diagrams?
5. How does the identity change if you replace union with symmetric difference?

Tip: When working with set identities, breaking down the sets using element-based proofs often makes things clearer.