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$\begingroup$

What is the opposite of this expression?

$p \land ( q \lor r )$

Please suggest any theorem as a starting point.

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    Yes. I do mean "negation".2012-01-04

1 Answers 1

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Keep in mind that $\lnot (A\land B) = (\lnot A)\lor (\lnot B)$ and $\lnot (A\lor B) = (\lnot A)\land (\lnot B)$. Using this, you can figure out $\lnot (p \land ( q \lor r ))$ as follows: $\begin{eqnarray*}\lnot(p \land ( q \lor r )) &=& (\lnot p)\lor (\lnot(q\lor r))\\ &=& (\lnot p)\lor ((\lnot q)\land (\lnot r))\\\end{eqnarray*}$ which you can distribute out if you wish.