simplify the following using laws of propositional logic:
~[p --> ~(p ^ q)]
Hello, mathman11!
Simplify the following using laws of propositional logic:
. . $\displaystyle \sim \big[p \to \:\sim(p \wedge q)\big]$
$\displaystyle \begin{array}{cccccccccc}1. & \sim \big[p \to \:\sim(p \wedge q)\big] && 1. & \text{Given} \\ 2. & \sim\big[\sim p\: \vee \sim(p \wedge q)\big] && 2. & \text{Def. of Impl'n} \\ 3. & \sim\big[\sim p \vee (\sim p \:\vee \sim q)\big] && 3. & \text{DeMorgan} \\ 4. & \sim \big[(\sim p\:\vee \sim p)\:\vee \sim q\big] && 4 . & \text{Associative} \\ 5. & \sim\big(\sim p \:\vee \sim q\big) && 5. & \text{Identity prop.} \\ 6. & p \wedge q && 6. & \text{DeMorgan} \end{array}$