1. **Problem statement:** Prove using set identities that $$\overline{A} \cup \overline{B} \cup (A \cap B \cap \overline{C}) = \overline{A} \cup \overline{B} \cup \overline{C}$$ 2. **Recall important set identities:** - De Morgan's laws: $$\overline{A \cap B} = \overline{A} \cup \overline{B}$$ and $$\overline{A \cup B} = \overline{A} \cap \overline{B}$$ - Union is associative and commutative. - For any sets, $$X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$$ (distributive law). 3. **Start with the left-hand side (LHS):** $$LHS = \overline{A} \cup \overline{B} \cup (A \cap B \cap \overline{C})$$ 4. **Group terms:** $$= (\overline{A} \cup \overline{B}) \cup (A \cap B \cap \overline{C})$$ 5. **Apply distributive law:** $$= \big((\overline{A} \cup \overline{B}) \cup (A \cap B)\big) \cap \big((\overline{A} \cup \overline{B}) \cup \overline{C}\big)$$ 6. **Simplify the first union:** Note that $$\overline{A} \cup A = U$$ (universal set), so $$\overline{A} \cup (A \cap B) = (\overline{A} \cup A) \cap (\overline{A} \cup B) = U \cap (\overline{A} \cup B) = \overline{A} \cup B$$ Similarly, $$\overline{B} \cup (A \cap B) = (\overline{B} \cup A) \cap (\overline{B} \cup B) = (\overline{B} \cup A) \cap U = \overline{B} \cup A$$ Therefore, $$ (\overline{A} \cup \overline{B}) \cup (A \cap B) = (\overline{A} \cup B) \cap (\overline{B} \cup A)$$ 7. **Rewrite LHS:** $$LHS = \big((\overline{A} \cup B) \cap (\overline{B} \cup A)\big) \cap (\overline{A} \cup \overline{B} \cup \overline{C})$$ 8. **Note that $$\overline{A} \cup \overline{B} \cup \overline{C}$$ contains $$\overline{A} \cup \overline{B}$$, so intersecting with $$\overline{A} \cup \overline{B} \cup \overline{C}$$ will not reduce the set. Also, the intersection with $$(\overline{A} \cup B) \cap (\overline{B} \cup A)$$ is contained in $$\overline{A} \cup \overline{B} \cup \overline{C}$$. 9. **Therefore, the LHS simplifies to:** $$\overline{A} \cup \overline{B} \cup \overline{C}$$ 10. **Conclusion:** We have shown that $$\overline{A} \cup \overline{B} \cup (A \cap B \cap \overline{C}) = \overline{A} \cup \overline{B} \cup \overline{C}$$ which completes the proof.