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Show that $(\lnot(p\lor(\lnot p\land q))$ is logically equivalent to $\lnot p \land \lnot q$.

I am wondering what I did is correct. Very new to learning simple logic.

$$(\lnot(p\lor(\lnot p\land q)) \equiv \lnot((p\lor\lnot p)\land(p\lor q)) \\ \equiv \lnot(\text{T} \land (p\lor q)) \\ \equiv\lnot(p \lor q) \\ \equiv \lnot p \land \lnot q \\ \square$$

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  • $\begingroup$ Seems good to me! $\endgroup$ – It'sNotALie. Jan 21 at 15:28
  • $\begingroup$ You can always build the truth table of both propositions and decide whether they match or not; if they match then, by definition, are equivalent. $\endgroup$ – manooooh Jan 21 at 15:40
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Only your last step is wrong: $$\neg(p \lor q) \equiv (\neg p \land \neg q)$$

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    $\begingroup$ Thank you! I fixed my mistake and it should be good now, haha! $\endgroup$ – DippyDog Jan 21 at 15:38
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    $\begingroup$ Yes indeed ! You may like this tool umsu.de/trees/… $\endgroup$ – SagarM Jan 21 at 15:39
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This is an alternative rount, and corrects for your misuse of DeMorgan's in the last step.

$$(\lnot(p\lor(\lnot p\land q))) \equiv \lnot p \land \lnot(\lnot p \land q)\tag{DeMorgan's}$$ $$\equiv \lnot p \land (p \lor \lnot q)\tag{DeMorgan's}$$ $$\equiv (\lnot p \land p) \lor (\lnot p \land \lnot q)\tag{Distributive Law}$$ $$\equiv F\lor (\lnot p \land \lnot q)$$ $$\equiv \lnot p \land \lnot q$$

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    $\begingroup$ Whoops! I made an issue. I fixed it in my solution. $\endgroup$ – DippyDog Jan 21 at 15:36
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    $\begingroup$ In the future, don't fix your post after other's provide corrections, because answerers are then chasing a moving target, and it render's answers meaningless to others. $\endgroup$ – amWhy Jan 21 at 15:39

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