0
$\begingroup$

The formula I am trying to turn into conjunctive normal form and disjunctive normal form is:

$P \rightarrow (Q \land R)$

could anyone please help me give two answers, CNF and DNF? I have managed to get DNF which turned out to be:

$¬P \lor (Q \land R)$ also is my DNF correct?

$\endgroup$
1
  • $\begingroup$ still need help $\endgroup$ Commented Oct 15, 2015 at 21:07

1 Answer 1

0
$\begingroup$

As you have observed, $A \rightarrow B \equiv \neg A \rightarrow B$, so $$P \rightarrow (Q \land R) \equiv \neg P \lor (Q \land R),$$ which is in disjunctive normal form. Now use distributivity to get $$\neg P \lor (Q \land R) \equiv (\neg P \lor Q) \land (\neg P \lor R),$$ which is in conjunctive normal form.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .