Conjuctive normal form of $(p\wedge(q\implies r))\implies s$

Question

I am asked to write this in CNF without using truth tables.

This is what I worked out so far:

$$(p\wedge(q\implies r))\implies s \\ \neg(p\wedge (q\implies r)) \vee s\\ (\neg p \vee \neg(\neg q \vee r))\vee s \\ \neg p \vee (q \wedge \neg r) \vee s $$

How do I move on from here?

Answer 1

Use the distributive law to move the conjunctions above the disjunctions.

(But first fix the error in your rewriting so far -- it looks like $r$ should be $\neg r$).