Converting formula to disjunctive normal form

Question

Convert $$\neg(P\to Q)\lor\neg(P\lor\neg(R\lor S))$$ to DNF.

This is what I've already done: $$\neg(\neg P\lor\neg Q)\lor(\neg P\lor\neg(\neg R\lor\neg S))$$

And from this point, I'm not sure how to proceed. Help would be appreciated.

Answer 1

You've made mistakes in failing to use DeMorgans.

First, I use, like you did, the equivalence of $P\rightarrow Q \equiv \lnot P \lor Q$, and second, I use DeMorgan's Law in step $(1);$

DeMorgan's is used also in step $(2)$ below;

and in step $(3)$ I use the distributive property:

$$\begin{align}\neg(P\to Q)\lor\neg(P\lor\neg(R\lor S)) &\equiv \lnot(\lnot P \lor Q) \lor (\lnot P \land \lnot \lnot (R\lor S))\tag{1}\\ \\ &\equiv (P \land \lnot Q) \lor (\lnot P \land (R \lor S))\tag{2}\\ \\ &\equiv (P\land \lnot Q) \lor (\lnot P \land R) \lor (\lnot P \land S)\tag{3}\\ \\ \end{align}$$

And we now have Disjunction Normal Form, in $(3)$.