# Further simplifying this logical equivalence

I am new to the logic of compound statements, and I'm working on simplifying, but I'm having a tough time wrapping my head around how to reduce these statements.

I have the following, which thus far I think is accurate (it is supposed to simplify to $\neg p$):

$\neg(p \vee \neg q) \vee (\neg p \wedge \neg q) \equiv (\neg p \wedge q) \vee (\neg p \vee q)\ (De\ Morgan's\ Laws) \\$

However, I'm not sure where to go from here. I've put together a few options which I think make decent sense, but I honestly have no clue which one is correct, or what particular law is being demonstrated.

$\equiv \neg p \vee (\neg q \vee q)$

$\equiv \neg p \vee (\neg q \wedge q)$

$\equiv \neg p \wedge (\neg q \vee q)$

$\equiv \neg p \wedge (\neg q \wedge q)$

Hopefully one of these is right, or maybe none of them are. I'm just wondering if someone can clarify if I'm heading in remotely the right direction or not.

• It is not clear what are you asking .... $¬(p∨¬q)∨(¬p∧¬q) \equiv (¬p∧q)∨(¬p∧¬q)$ by De Morgan and by Distributivity $\equiv ¬p∧(q∨¬q) \equiv ¬p∧TRUE \equiv ¬p$. Jan 6 '16 at 21:01

First, note that, by De Morgan, $$\neg(p\lor \neg q) \equiv (\neg p \land q).\tag{*}$$ Therefore, \begin{align} \neg(p\lor \neg q) \lor (\neg p \land \neg q) &\equiv (\neg p \land q) \lor (\neg p \land \neg q) \tag{by *} \\ &\equiv \neg p \land (q \lor \neg q) \tag{\land distributes over \lor}\\ &\equiv \neg p. \end{align}