Can a logical formula that is a contradiction be represented as Disjunctive Normal Form?

Question

Hi I was trying to solve this math problem in mathematics.

Obtain the D.N.F, C.N.F, P.D.N.F, P.C.N.F of the following expression : $(Q \to P) \land (\lnot P \land Q)$

I understand that there is no Particular Disjunctive Normal Form for this statement, as the the statement is a contradiction.

However, I'm not sure if it's suffice to say that there is no D.N.F for this statement as well.

I understand that the definition of a Disjunctive Normal Form is "a disjunction of one or more conjunctions of one or more literals"(from wikipedia). And that "All logical formulas can be converted into an equivalent disjunctive normal form"(again from wikipedia).

But I was wondering if $(P \land \lnot P) \lor (Q \land \lnot Q)$ could be an answer. As each of the conjunctions hold more than one literals can I say that this is the D.N.F of the given contradiction?

P.S. Pls correct my errors in the argument above. It would be more than welcome. Thank you in advance!!

Answer 1

$P \land \neg P$ is in DNF

It may not look like it, but it really is.

Here's the thing: a statement is in DNF iff it is a generalized disjunction of generalized conjunctions of literals. And by generalized we mean that the disjunction/conjunction can have any number of disjuncts/conjuncts. So the above statement is in fact a disjunction ... of exactly one disjunct.

Indeed, even a statement like $P$ is in DNF: it is a disjunction of one disjunct $P$, which is itself a conjunction of one conjunct $P$, which is a literal. By similar reasoning, $P$ is in CNF as well.

In fact, $P \land \neg P$ is both a DNF and a CNF of your original formula. It is in CNF since it is a conjunction of two conjuncts, each of which is a disjunction of one literal.

Answer 2

$\begin{align}(Q \to P) \land (\lnot P \land Q) &\equiv (\lnot Q \lor P) \land(\lnot P \land Q)\tag{def. $\to$}\\ \\ &\equiv (\lnot Q \land (\lnot P \land Q)) \lor (P\land (\lnot P \land Q))\tag{distribution}\\ \\ &\equiv (\lnot Q \land \lnot P \land Q) \lor (P \land \lnot P \land Q)\tag {assoc.}\\ \\ &\equiv ((\lnot Q \land Q)\land \lnot P) \lor ((P\land \lnot P) \land Q))\tag{comm.+assoc.}\\ \\ &\equiv (\bot \land \lnot P) \lor (\bot\land Q)\\ \\ &\equiv \bot \lor \bot\\ \\ &\equiv \bot\end{align}$

The third line above is in Disjunctive Normal Form.