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.

  • $\begingroup$ It is not clear what are you asking .... $¬(p∨¬q)∨(¬p∧¬q) \equiv (¬p∧q)∨(¬p∧¬q)$ by De Morgan and by Distributivity $\equiv ¬p∧(q∨¬q) \equiv ¬p∧TRUE \equiv ¬p$. $\endgroup$ Jan 6, 2016 at 21:01

1 Answer 1


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


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