# Showing Logical Equivalence

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.

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

• Seems good to me! – It'sNotALie. Jan 21 at 15:28
• 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. – manooooh Jan 21 at 15:40

## 2 Answers

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

• Thank you! I fixed my mistake and it should be good now, haha! – DippyDog Jan 21 at 15:38
• Yes indeed ! You may like this tool umsu.de/trees/… – SagarM Jan 21 at 15:39

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

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

• Whoops! I made an issue. I fixed it in my solution. – DippyDog Jan 21 at 15:36
• 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. – amWhy Jan 21 at 15:39