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We will prove one part of DeMorgan's Laws. You will prove the other in your homework. You can see that most of the proof is missing. Click on the boxes to reveal the missing expressions.

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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 \((A\cup B)^c=A^c \cap B^c\).