simplify $((p\lor(r\lor q))\land\neg(\neg q\land\neg r)$ using logic laws

Question

I have to simplify this statement

$((p\lor (r\lor q))\land\neg(\neg q\land\neg r)$ as much as I can the answer is $q\lor r $

and I know the laws and the order in which they should be applied

The first law that should be used is de Morgans however I don't understand the steps and how would the expression look after the execution of each Help me understand this it's pretty confusing

Answer 1

The given statement has one more opening parenthesis than closing parenthesis .... but I'll go with:

$(p\lor (r\lor q))\land \neg(\neg q\land\neg r)$

OK, like you said, DeMorgan seems like a good first step:

$(p\lor (r\lor q))\land (\neg\neg q\lor \neg \neg r)$

Two double negations gives:

$(p\lor (r\lor q))\land(q\lor r)$

By Commutation:

$(p\lor ( q \lor r))\land(q \lor r)$

Absorption:

$q \lor r$

Unfortunately, some textbooks do not give you Absorption. If not, you can do:

$(p \lor (q \lor r)) \land (q \lor r)$

Identity:

$(p \lor (q \lor r)) \land (\bot \lor (q \lor r))$

Distribution:

$(p \land \bot) \lor (q \lor r)$

Annihilation:

$\bot \lor (q \lor r)$

Identity:

$q \lor r$

Answer 2

$((p\lor(r\lor q))\land\neg(\neg q\land\neg r)\equiv(p\lor(r\lor q))\land\neg(\neg(q\lor r))\equiv(p\lor(r\lor q))\land(q\lor r)\equiv q\lor r$

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explanation:

$(1)$DeMorgan: $\neg(q\lor r)\equiv\neg q\land\neg r$ (you have: $\neg q\land\neg r\equiv\neg(q\lor r)$

$(2)$ $\neg(\neg(q\lor r)\equiv q\lor r$

$(3)$ $(p\lor\underbrace{(q\lor r)}_{s})\land\underbrace{(q\lor r)}_{s}\equiv(p\lor s)\land s\equiv s\equiv q\lor r$