Simple logic equivalence incorrect I am having some problems negating a rather simple logical statement. I am currently taking a introduction course, so please bear with me if my question is silly.
I am supposed to turn this:
$$ \lnot( p \lor q \lor (\lnot p \land \lnot q \land r)) $$
Into:
$$ \lnot p \land \lnot q \land \lnot r $$
But I guess I am already some mistakes early in my simplification:
\begin{gather}
    \lnot p \land \lnot q \land \lnot(\lnot p \land \lnot q \land r) \\
    \lnot p \land \lnot q \land ( p \lor q \lor \lnot r) \\
    \mbox{?? No clue what to do now}
\end{gather}
Can someone spot where my train is off track?
 A: Your work is fine up to the point you left off: So, good work!
Picking up where you left off, we have:

¬p ^ ¬q ^ (p v q v ¬r)

We can distribute the conjunction $(\lnot p \land \lnot q) \land...$ over the parentheses (a disjunction) $(p \lor q \lor \lnot r)$:
$$(\lnot p \land \lnot q \land p) \lor (\lnot p \land \lnot q \land q) \lor (\lnot p \land \lnot q \land \lnot r)$$
$$(\color{blue}{\bf (\lnot p \land p)}\land \lnot q) \lor (\lnot p \land \color{blue}{\bf (\lnot q \land q)}) \lor (\lnot p \land \lnot q \land \lnot r)$$
The first two terms form contradictions, and are hence false. You are left with $$\lnot p \land \lnot q \land \lnot r$$
To see how we can eliminate the first two terms in parentheses:
$$(\color{blue}{\bf (\lnot p \land p)}\land \lnot q) \lor (\lnot p \land \color{blue}{\bf (\lnot q \land q)}) \lor (\lnot p \land \lnot q \land \lnot r)$$
$$\equiv (\color{blue}{\bf F}\land \lnot q) \lor (\lnot p \land \color{blue}{\bf F}) \lor (\lnot p \land \lnot q \land \lnot r)$$
$$\equiv F \lor F \lor (\lnot p \land \lnot q \land \lnot r)$$
$$\equiv \lnot p \land \lnot q \land \lnot r$$
