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I need to prove that $ \neg \big( p \lor ( \neg p \land q ) \big) $ is logically equivalent to $ \neg p \land \neg q $ using the laws of propositional logic instead of a truth table. I can't figure out which rules work to transform it to $ \neg p \land \neg q $. I am taking a logic course and I want to practice my logical thinking so that I can be ready on test day. This is one I am practicing with and I tried to do De Morgan's, then associativity then idempotence laws, but I don't think this is correct. I also tried to do absorption and then De Morgan's, but I don't think that works either because of the negations and because I need it to be equivalent to $ \neg p \land \neg q $.

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    $\begingroup$ Please include all the intermediate results so that we can better locate where you ran into trouble, and provide better answers. $\endgroup$
    – Trebor
    Sep 7, 2021 at 22:57
  • $\begingroup$ Use in first step distribution and then De Morgan. Add done steps to question if/when you stack and you'll get help. $\endgroup$
    – zkutch
    Sep 7, 2021 at 23:35

2 Answers 2

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Using distibution we have: $$ p \lor ( \neg p \land q ) \equiv \big(p\lor (\neg p)\big)\land \big(p\lor q\big) \equiv 1 \land \big(p\lor q\big) \equiv p\lor q $$ Then De Morgan gives answer

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Using au(bnc)=(aub)n(auc) we can say that pu(\pnq)= (pu\p)n(puq)= tn(puq)= (puq), t being tautology. Now (puq)= \pn\q by de morgans law.

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  • $\begingroup$ Im not sure what you mean by "au(bnc)=(aub)n(auc)" and "pu(\pnq)= (pu\p)n(puq)= tn(puq)= (puq)" and "(puq)= \pn\q" $\endgroup$
    – Skytendo
    Sep 7, 2021 at 23:42
  • $\begingroup$ @Skytendo I suppose that $u$ means $\vee$ and $n$ means $\wedge$. $\endgroup$
    – user0102
    Sep 8, 2021 at 0:10

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