Discrete Math: simplifying the proposition $\lnot [p \land (q \lor r) \land (\lnot p \lor \lnot q \lor r)]$ I want to simplify the below proposition
$\lnot [p \land (q \lor r) \land (\lnot p \lor \lnot q \lor r)]$
Taken from Discrete and Combinatorial Mathematics sec 2.2 ex. #6.

$\lnot[p \land (q \lor r) \land (\lnot p \lor \lnot q \lor r)]$
$\lnot[(p \land (q \lor r)) \land ((\lnot p \lor \lnot q) \lor r)] \quad $ //assoc (not sure if legal)
$\lnot(p \land (q \lor r)) \lor \lnot((\lnot p \lor \lnot q) \lor r) \quad$ //DM
$(\lnot p \lor \lnot(q \lor r)) \lor (\lnot(\lnot p \lor \lnot q) \land \lnot r) \quad$ //DM
$(\lnot p \lor (\lnot q \land \lnot r)) \lor ((p \land q) \land \lnot r) \quad$ //DM
$((\lnot p \land \lnot q) \lor (\lnot p \land \lnot r)) \lor ((p \land q) \land \lnot r) \quad$ //DIST

I'm not sure what to do at this point, maybe I should have temporary ignored the negation (I tried that before, and got nowhere).
 A: Yes, I would temporarily ignore negation.
$$\begin{array}{lcll}(p\land(q\lor r))\land(\lnot p\lor\lnot q\lor r) & \Leftrightarrow & (p\land(q\lor r)\land\lnot p)\lor(p\land(q\lor r)\land\lnot q)\lor(p\land(q\lor r)\land r)& \text{De Morgan's laws}\\ & \Leftrightarrow& \bot \lor (p\land\lnot q\land r) \lor (p\land r) & \text{Because: }(q\lor r)\land r\Leftrightarrow r\text{ and }(q\lor r)\land\lnot q\Leftrightarrow\lnot q\land r\\ & \Leftrightarrow& p\land((\lnot q\land r) \lor r)&\\ & \Leftrightarrow & p\land r \end{array}$$
and now you can add the final negation. The result should be, if I am right, $\lnot(p\land r)$, or $\lnot p\lor\lnot r$.
A: The association was legal, but you just choose the wrong literals.   You have a lone $p$ among the conjunction, so you should target the $\neg p$ inside the disjunction.   Use association and commutation.
$${\lnot\big(p \land (q \lor r) \land (\lnot p \lor \lnot q \lor r)\big)
\\\lnot\Big(p\land\big(\neg p\lor(\neg q\lor r)\big)\land (q \lor r)\Big) }$$
From here distribution will lead the way.
