Hello, atwarwithmaths!

We use DeMorgan's Theorem just once.

Then we use a variety of other laws:

. .

On the right, we have:Using DeMorgans Theorem, show that:

\overline{A \cap B}) \;=\;(A \cup B) \:\cap\: (\overline{A} \cup \overline{B})\qquad\text{(DeMorgan)}" alt="(A \cup B) \:\cap \\overline{A \cap B}) \;=\;(A \cup B) \:\cap\: (\overline{A} \cup \overline{B})\qquad\text{(DeMorgan)}" />

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. . . . . . . . . . . . A \cap \overline{B}) \qquad\text{(Law 2)} " alt="= \;(B \cap \overline{A}) \;\cup\A \cap \overline{B}) \qquad\text{(Law 2)} " />

. . . . . . . . . . . . \overline{A} \cap B)\qquad\text{(Comm.)} " alt="= \;(A \cap \overline{B})\:\cup\\overline{A} \cap B)\qquad\text{(Comm.)} " />