Disjunction Form I'm a little stuck on finding an equivalent disjunctive form to 
$P \rightarrow (Q \rightarrow \neg P \wedge R) $. I have only gotten 
$\neg P \vee Q (\neg P \vee Q) \wedge (\neg P \vee R) $. But I know that is not correct. Once I get the disjunction form, I know how to get the normal form, but for some reason I'm having trouble with the truth equivalences.
Thanks
 A: Let's start from the beginning, since in your work, you're missing a connective between $Q$ and the clauses which follow, your "grouping" of terms is ambiguous, and it looks like you might have mixed up the distributive law, perhaps going "too far" with your work:
$$\begin{align} P \rightarrow \Big[Q \rightarrow (\neg P \wedge R)\Big] & \equiv \lnot P \lor \Big[\lnot Q \lor (\lnot P \land R)\Big]\tag{1}\\ \\
\equiv \lnot P \lor \lnot Q \lor (\lnot P \land R)\tag{2}\end{align}$$
In $(1)$, we simply applied, twice, the transformation of an implication to an equivalent disjunction: $$A \rightarrow B \equiv \lnot A \lor B$$
Next, $(2)$ is in disjunctive form: it is the disjunction of three clauses, two of which are the negation of atomic propositions, and one clause which is a simple conjunction. However, we can simplify further to obtain disjunctive normal form:
$$\lnot P \lor \lnot Q \lor (\lnot P \land R)\equiv \lnot P \lor \lnot Q$$
A: Hint: I would perhaps transform the implications one at a time and make use of extra brackets:
$$ \begin{align*}
P \rightarrow (Q \rightarrow (\neg P \wedge R))
&\equiv P \to (\neg Q \lor (\neg P \wedge R)) \\
&\equiv \neg P \lor (\neg Q \lor (\neg P \wedge R)) \\
&\equiv \neg P \lor \neg Q \lor (\neg P \wedge R) \\
\end{align*} $$
which is a disjunction of conjunctive clauses. I'm assuming you can convert to DNF from here.
