Is there a way i can go from dnf to cnf quickly? Can I simply negate dnf to get cnf? and get negate cnf to get dnf?
e.g.
$P \to (Q\wedge \neg (P \to R))$
= $\neg P \vee (Q \wedge \neg ( \neg P \vee R))$
= $\neg P \vee (Q \wedge (P \wedge \neg R))$
= $\neg P \vee (Q \wedge P \wedge \neg R)$ — is this correctly in DNF?
How could I go from CNF from here? I'm slightly stuck.
Best Answer
Yes. You've got it in DNF. To convert to CNF use the distributive law: $A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C)$
$$\neg P \vee (Q \wedge P \wedge \neg R) \leftrightarrow (\neg P \vee Q) \wedge (\neg P \vee (P \wedge \neg R)) \leftrightarrow (\neg P \vee Q) \wedge (\neg P \vee P) \wedge (\neg P \wedge \neg R) $$
(of course the $\neg P \vee P$ can be canceled).
For more information...
https://en.wikipedia.org/wiki/Conjunctive_normal_form