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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

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  • $\begingroup$ 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? $\endgroup$ – Bram28 Feb 5 '20 at 21:21
  • $\begingroup$ yup my mistake edited it $\endgroup$ – Somebody Feb 5 '20 at 21:26
  • $\begingroup$ 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. $\endgroup$ – Bram28 Feb 5 '20 at 21:28
  • $\begingroup$ 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. $\endgroup$ – Somebody Feb 5 '20 at 21:34
  • $\begingroup$ Well, then I can understand you are confused because the answer must be wrong! $\endgroup$ – Bram28 Feb 5 '20 at 21:36
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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:

$q \lor r$

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  • $\begingroup$ this is not the statement , this is the statement ( p v(r v p)) v ~(~q ^ ~r) $\endgroup$ – Somebody Feb 16 '20 at 20:21
  • $\begingroup$ @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 $\endgroup$ – Bram28 Feb 16 '20 at 21:05
  • $\begingroup$ @Somebody Hey, it all makes sense now! $\endgroup$ – Bram28 Feb 16 '20 at 21:28
  • $\begingroup$ Yea it does, thank you for your answer. I appreciate the help 🥰 $\endgroup$ – Somebody Feb 16 '20 at 21:33
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$((p\lor(r\lor q))\land\neg(\neg q\land\neg r)\equiv(p\lor(r\lor q))\land\neg(\neg(q\lor r))\equiv(p\lor(r\lor q))\land(q\lor r)\equiv q\lor r$


explanation:

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

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

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

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  • $\begingroup$ can you list the logic laws you used in the order you used them, please $\endgroup$ – Somebody Feb 16 '20 at 20:56
  • $\begingroup$ I advise you to draw diagrams and analyze each law to get an intuitive insight. $\endgroup$ – Invisible Feb 16 '20 at 21:05
  • $\begingroup$ okay, I understood it but what is the law in the 3rd step called? $\endgroup$ – Somebody Feb 16 '20 at 21:09
  • $\begingroup$ @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. $\endgroup$ – Invisible Feb 16 '20 at 21:12
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    $\begingroup$ you did help, thank you for your answer $\endgroup$ – Somebody Feb 16 '20 at 21:32

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