Find conjunctive normal form of propositional formula

Question

I need to find conjunctive normal form (CNF) of this propositional formula:

$$\lnot((p \implies (q \implies r)) \implies ((p \implies \lnot r) \implies (p \implies \lnot q)))$$

How could i do that? I got stuck at this point: \begin{align} &\quad \lnot((p \implies (q \implies r)) \implies ((p \implies \lnot r) \implies (p \implies \lnot q))) \\ &= (p \implies (q \implies r)) \land \lnot((p \implies \lnot r) \implies (p \implies \lnot q)) \\ &= (p \implies (q \implies r)) \land ((p \implies \lnot r) \land \lnot (p \implies \lnot q)) \\ &= ((p \land q) \implies r)) \land ((p \implies \lnot r) \land (p \land q)) \\ &= ((\lnot p \land \lnot q) \lor r)) \land ((\lnot p \lor \lnot r) \land (p \land q)) \end{align}

Answer 1

\begin{align} &\quad \lnot((p \implies (q \implies r)) \implies ((p \implies \lnot r) \implies (p \implies \lnot q))) \\ &\equiv \lnot(\lnot(\lnot p \lor (\lnot q \lor r)) \lor (\lnot(\lnot p \lor \lnot r) \lor (\lnot p \lor \lnot q))) \\ &\equiv (\lnot p \lor (\lnot q \lor r)) \land \lnot (\lnot(\lnot p \lor \lnot r) \lor (\lnot p \lor \lnot q)) \\ &\equiv (\lnot p \lor \lnot q \lor r) \land ((\lnot p \lor \lnot r) \land \lnot (\lnot p \lor \lnot q)) \\ &\equiv (\lnot p \lor \lnot q \lor r) \land (\lnot p \lor \lnot r) \land p \land q \\ \end{align} Looks like you just have an error in the last line. This is already in CNF, but by distributing $\land$ over $\lor$ you can further reduce to identically false.