Converting compound statements to Conjunctive Normal Form

Question

$\newcommand{\tt}{\text{t}}\newcommand{\ff}{\text{f}}$Given a compound statement how do we convert it to conjunctive normal form? I have no idea where to begin.

a) $$(r\to \lnot p)\land(q\to \lnot r) :CNF=?$$

b) Find the value of this statement when $p$ is true, $q$ is false and $r$ is true:

What I have so far is I created a truth table:

$$\begin{array}{c|c|c} p&q&r&\neg p&\neg q&\neg r&r\to\neg p&q\to\neg r&(r\to\neg p)\land(q\to\neg r)\\ \hline \tt&\tt&\tt&\ff&\ff&\ff&\ff&\ff&\ff\\ \tt&\tt&\ff&\ff&\ff&\tt&\tt&\tt&\tt\\ \tt&\ff&\tt&\ff&\tt&\ff&\ff&\tt&\ff\\ \tt&\ff&\ff&\ff&\tt&\tt&\tt&\tt&\tt\\ \ff&\tt&\tt&\tt&\ff&\ff&\tt&\ff&\ff\\ \ff&\tt&\ff&\tt&\ff&\tt&\tt&\tt&\tt\\ \ff&\ff&\tt&\tt&\tt&\ff&\tt&\tt&\tt\\ \ff&\ff&\ff&\tt&\tt&\tt&\tt&\tt&\tt \end{array}$$

so when $p$ is true, $q$ is false and $r$ is true: the result is f. Have I done this correctly?

Answer 1

Use $a\to b \ \equiv\ \lnot a\lor b$ and perhaps some distributivity.

So that, e.g. $(r\to\lnot p)\equiv(\lnot r\lor\lnot p)$. This directly leads to $$(\lnot r\lor\lnot p)\ \land\ (\lnot q\lor\lnot r),$$ which is already in conjuctive normal form.