2
$\begingroup$

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.

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

$\endgroup$
2
  • $\begingroup$ Seems good to me! $\endgroup$ Jan 21, 2021 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, 2021 at 15:40

2 Answers 2

7
$\begingroup$

Only your last step is wrong: $$\neg(p \lor q) \equiv (\neg p \land \neg q)$$

$\endgroup$
3
  • 1
    $\begingroup$ Thank you! I fixed my mistake and it should be good now, haha! $\endgroup$
    – DippyDog
    Jan 21, 2021 at 15:38
  • 1
    $\begingroup$ Yes indeed ! You may like this tool umsu.de/trees/… $\endgroup$
    – SagarM
    Jan 21, 2021 at 15:39
  • $\begingroup$ @DippyDog: That tool does not work!! Without looking at the code, I can't be sure why, but it fails on some of my FOL exercises, and the most likely reason is that the professor who wrote it did not really understand the correct tableaux method. (Many philosophy professors fall into this category, including in my university.) $\endgroup$
    – user21820
    Apr 5, 2022 at 20:53
4
$\begingroup$

This is an alternative rount, and corrects for your misuse of DeMorgan's in the last step.

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

$\endgroup$
2
  • 1
    $\begingroup$ Whoops! I made an issue. I fixed it in my solution. $\endgroup$
    – DippyDog
    Jan 21, 2021 at 15:36
  • 5
    $\begingroup$ 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, 2021 at 15:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .