How to convert (F⇒(H⇒G))⇒(F⇒(G⇒H)) to Conjunctive Normal Form?

Question

I've expanded the formula to:

$$(F \wedge H \wedge\neg G) \vee (\neg F \vee\neg G \vee H)$$

but don't know how to switch the first set to or and the middle or to an and.

Answer 1

Use the distributive law: $(A \wedge B) \vee C = (A \vee C) \wedge (A \vee C)$.

Answer 2

\begin{align} &\phantom{\equiv}(F\implies(H\implies G))\implies(F\implies(G\implies H))\\ &\equiv\neg(\neg F\lor(\neg H\lor G))\lor(\neg F\lor(\neg G\lor H))\\ &\equiv\neg(\neg F\lor \neg H\lor G)\lor(\neg F\lor \neg G\lor H)\\ &\equiv(F\land H\land \neg G)\lor(\neg F\lor \neg G\lor H)\\ &\equiv(F\lor (\neg F\lor \neg G\lor H))\land (H\lor (\neg F\lor \neg G\lor H))\land (\neg G\lor (\neg F\lor \neg G\lor H))\\ &\equiv(F\lor \neg F\lor \neg G\lor H)\land (H\lor \neg F\lor \neg G\lor H)\land (\neg G\lor \neg F\lor \neg G\lor H)\\ &\equiv 1\land (\neg F\lor \neg G\lor H)\land (\neg F\lor \neg G\lor H)\\ &\equiv\neg F\lor \neg G\lor H \end{align}

Answer 3

Indeed distributive law is one approach, another method is use Absorption law:

\begin{align} &(F∧H∧¬G)∨(¬F∨¬G∨H)\\ \equiv&(((F∧¬G)∧H)∨H)∨(¬F∨¬G)\tag*{Reordering}\\ \equiv&H∨(¬F∨¬G)\tag*{Absorption law}\\ \end{align}