1. **Problem statement:** Prove that the complement of the union of sets $A$ and $B$ equals the intersection of their complements, i.e., $$\overline{A \cup B} = \overline{A} \cap \overline{B}.$$\n\n2. **Relevant formula (De Morgan's Law):** For any subsets $A$ and $B$ of a universal set $U$, the complement of their union is the intersection of their complements: $$\overline{A \cup B} = \overline{A} \cap \overline{B}.$$\n\n3. **Proof using element-wise argument:**\n- Let $x \in U$.\n- Suppose $x \in \overline{A \cup B}$. By definition of complement, this means $x \notin A \cup B$.\n- Since $x \notin A \cup B$, $x$ is neither in $A$ nor in $B$. So, $x \notin A$ and $x \notin B$.\n- By definition of complement, $x \in \overline{A}$ and $x \in \overline{B}$.\n- Therefore, $x \in \overline{A} \cap \overline{B}$.\n\n4. **Conversely:**\n- Suppose $x \in \overline{A} \cap \overline{B}$. Then $x \in \overline{A}$ and $x \in \overline{B}$, meaning $x \notin A$ and $x \notin B$.\n- Hence, $x \notin A \cup B$.\n- So, $x \in \overline{A \cup B}$.\n\n5. **Conclusion:** Since every element of $\overline{A \cup B}$ is in $\overline{A} \cap \overline{B}$ and vice versa, the two sets are equal: $$\boxed{\overline{A \cup B} = \overline{A} \cap \overline{B}}.$$\n\nThis completes the proof using De Morgan's law and element-wise reasoning.