I have the following conditional statement: ((p → q) → p) → p
Knowing that CNF is a conjunction of disjunctions (and that DNF is a disjunction of conjunctions), we will obviously have to introduce conjunctions and disjunctions.
Knowing that p → q can be alternatively expressed as ¬p ∨ q, it seems (to me) fairly easy how we could convert all of these conditionals into disjunctions.
However, I'm completely clueless as to how to introduce conjunctions. How could we do this?
Thanks in advance!
Almost. You are thinking of Implication Equivalence. There is also Implication's Negation Equivalence.
$$\begin{align}\text{I.E.}~&&\phi\to\psi~~\iff \lnot \phi\vee\psi\tag{note: disjunction}\\\text{I.N.E.}~&&\lnot(\phi\to\psi)\iff\phi\land\lnot\psi\tag{note: conjunction}\end{align}$$
So using these, step by step::$$\begin{align}1.&&((p\to q)\to p)\to p\tag{Premise}\\2.&&\lnot((p\to q)\to p)\lor p\tag{by I.E. on 1}\\3.&&((p\to q)\land\lnot p)\lor p\tag{by I.N.E on 2}\\4.&&((\lnot p\lor q)\land\lnot p)\lor p\tag{by I.E. on 3}\end{align}$$
Then we may distribute one way, aiming for a DNF::
$$\begin{align}5.&&(\lnot p\land\lnot p)\lor (q\land\lnot p)\lor p\tag{Distribute $\land/\lor$ on 4}\end{align}$$
Or we may distribute the other way, aiming for a CNF::
$$\begin{align}6.&&(\lnot p\lor q\lor p)\land(\lnot p\lor p)\tag{Distribute $\lor/\land$ on 4}\end{align}$$
Of course, neither of these are their final form. You will still need to reduce them by using a few other Equivalence Rules.