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

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

• The answer cannot be $p \land r$ : note that the given statement is completely 'symmetric' in terms of $q$ and $r$, and yet in the 'answer' you get a $q$ but not an $r$. So, I can immediately tell somethng is wrong here: either the given statement is not what you say it is, or the answer is not what you day it is (or both). i also note that the given statement misses a parenthesis ... my money would be on you not having written the given statement correctly: can you please check? Feb 5, 2020 at 21:21
• yup my mistake edited it Feb 5, 2020 at 21:26
• Same comment: the answer still contains $r$ but not $q$, but the given is 'symmetric' with regard to $r$ and $q$. Please check your given statement as well. Feb 5, 2020 at 21:28
• the statement is an exact copy directly from my discrete mathematics assignment so not there is nothing wrong with it and it doesn't miss a parenthesis. Feb 5, 2020 at 21:34
• Well, then I can understand you are confused because the answer must be wrong! Feb 5, 2020 at 21:36

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:

$$q \lor r$$

• this is not the statement , this is the statement ( p v(r v p)) v ~(~q ^ ~r) Feb 16, 2020 at 20:21
• @Somebody OK ... next time make sure you write down the right statement in the first place! Anyway, I'll change my Answer accordingly. Give me a sec Feb 16, 2020 at 21:05
• @Somebody Hey, it all makes sense now! Feb 16, 2020 at 21:28
• Yea it does, thank you for your answer. I appreciate the help 🥰 Feb 16, 2020 at 21:33

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

explanation:

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

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

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

• can you list the logic laws you used in the order you used them, please Feb 16, 2020 at 20:56
• I advise you to draw diagrams and analyze each law to get an intuitive insight. Feb 16, 2020 at 21:05
• okay, I understood it but what is the law in the 3rd step called? Feb 16, 2020 at 21:09
• @Somebody, I have no idea how it's called, we don't learn propositional calculus that way at the university. Just draw a diagram or write down the truth table and think. Feb 16, 2020 at 21:12
• you did help, thank you for your answer Feb 16, 2020 at 21:32