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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.
$$(\lnot(p\lor(\lnot p\land q)) \equiv \lnot((p\lor\lnot p)\land(p\lor q)) \\ \equiv \lnot(\text{T} \land (p\lor q)) \\ \equiv\lnot(p \lor q) \\ \equiv \lnot p \land \lnot q \\ \square$$
Only your last step is wrong: $$\neg(p \lor q) \equiv (\neg p \land \neg q)$$
This is an alternative rount, and corrects for your misuse of DeMorgan's in the last step.
$$(\lnot(p\lor(\lnot p\land q))) \equiv \lnot p \land \lnot(\lnot p \land q)\tag{DeMorgan's}$$ $$\equiv \lnot p \land (p \lor \lnot q)\tag{DeMorgan's}$$ $$\equiv (\lnot p \land p) \lor (\lnot p \land \lnot q)\tag{Distributive Law}$$ $$\equiv F\lor (\lnot p \land \lnot q)$$ $$\equiv \lnot p \land \lnot q$$
