Question

DNF of p->q and CNF of pnq. Please for the love of God give the correct answer and don't spam just because you want points, I'm spending the last of my remaining points on this.​

Answer 1

Answer:

It is sufficient to show that p Ú q can be written in terms of  and . Then using DNF, we can write every compound proposition in terms of  and . The DNF of p  q is (pq)  (p q)  (p q).

All conjunctions of literals and all disjunctions of literals are in CNF, as they can be seen as conjunctions of one-literal clauses and conjunctions of a single clause, respectively. As in the disjunctive normal form (DNF), the only propositional connectives a formula in CNF can contain are and, or, and not.