Transform into DNF 
Transform into a DNF form
$$[(p \lor q) \to (q \lor r)] \to [(p \to q) \land \neg r]$$

I transformed it into following formula, but don't know what to do next:
$$[(p\lor q) ∧ (\neg q∧\neg r)]\lor[(\neg p\lor q) ∧ \neg r]$$
 A: I will write everything in detail just to make sure all the readers understand my notes and maybe suggest something. I prefer $\implies$ instead of $\rightarrow\;\;\;$   (:
$$((p \lor q) \implies (q \lor r)) \implies((p \implies q) \land \neg r)$$
contrapositive:
$$\equiv\neg((p \implies q) \land \neg r)\implies\neg((p \lor q) \implies (q \lor r))$$
$$\equiv((p\land\neg q)\lor r)\implies((p\lor q)\land(\neg q\land \neg r))$$
$$\equiv((p\land\neg q)\lor r)\implies((p\land \neg q\land \neg r)\lor(q\land \neg q\land \neg r))$$
$$\equiv((p\land\neg q)\lor r)\implies(p\land \neg q\land \neg r)$$
$$\equiv((p\land\neg q)\lor r)\implies((p\land \neg q)\land \neg r)$$
$$$$
$$\equiv\neg(((p\land\neg q)\lor r)\land\neg((p\land \neg q)\land \neg r))$$
$$\equiv\neg(((p\land\neg q)\lor r)\land(\neg (p\land \neg q)\lor r))$$
$$\equiv\neg(((p\land \neg q)\land\neg (p\land\neg q))\lor r)\equiv\neg r$$
$$
\begin{array} {|r|r|}\hline p & q & r & \neg r \\ \hline 0 & 0 & 0 & 1 \\ \hline 0 & 0 & 1 & 0 \\ \hline 0 & 1 & 0 & 1 \\ \hline 0 & 1 & 1 & 0 \\ \hline 1 & 0 & 0 & 1 \\ \hline 1 & 0 & 1 & 0 \\ \hline 1 & 1 & 0 & 1 \\ \hline 1 & 1 & 1 & 0 \\ \hline  \end{array}$$
DNF:
$$\bar p\cdot\bar q \cdot\bar r+\bar p\cdot q \cdot\bar r+p\cdot\bar q\cdot\bar r+p\cdot q\cdot\bar r$$
Hope it helps, and please, if you notice a mistake, correct me.
