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Prove that $$(A\cup B)^c=A^c \cap B^c$$.

Proof: Let $$x\in (A\cup B)^c$$. Then $$x\notin (A\cup B)$$ by the definition of complement, so $$x\notin A$$ and $$x\notin B$$ by the definition of union. From this we learn that $$x\in A^c$$ and $$x\in B^c$$ by the definition of complement. Therefore, $$x\in A^c \cap B^c$$ by the definition of intersection.

Conversely, let $$x\in A^c \cap B^c$$. Then $$x\in A^c$$ and $$x\in B^c$$ by the definition of intersection, so $$x\notin A$$ and $$x\notin B$$ by the definition of complement. Thus $$x\notin (A\cup B)$$ by the definition of union, which implies that $$x\in (A\cup B)^c$$ by the definition of complement.

We have proven that $$(A\cup B)^c\subseteq A^c \cap B^c$$ and $$A^c \cap B^c\subseteq(A\cup B)^c$$, therefore $$(A\cup B)^c=A^c \cap B^c$$.