# What is the conjunctive normal form for $(\neg Q\land P) \lor (\neg Q\land R) \lor (\neg P \land \neg R)$

$$(\neg Q\land P) \lor (\neg Q\land R) \lor (\neg P \land \neg R)$$
i have calculated this using wolframalpha and the output of CNF was
$$(\neg Q \lor\neg P) \land (\neg Q \lor\neg R)$$
but all i can reach out -after using the Absorption rule- is $$(\neg Q\land P) \lor (\neg Q\land R) \lor (\neg P \land \neg R)$$ and using Distribution the final result is $$\neg Q\land(R \lor P)\lor \neg P \land \neg R$$
so can someone help me getting what is wrong

• Your formula is equivalent, but it is not the conjunctive normal form. Jul 20, 2020 at 20:22
• what is wrong is you can't have $\color{red}{∨}$ here for CNF. $$¬Q∧(R∨P)\color{red}{∨}(¬P∧¬R)$$ Jul 20, 2020 at 22:18

The Redundacy Law or Reduction might be useful here $$\def\box#1#2{\boxed{\underline{\text{#1}}\\#2\\}}$$ \box{The Redundancy Law/Reduction} {(P\land \neg Q)\lor Q\equiv P\lor Q\\ \underline{\text{Proof.}}\\ \begin{align} &(P\land \neg Q)\lor Q\\ \equiv&(P\lor Q)\land (\neg Q\lor Q)\hspace{5ex}\text{Distributive law}\\ \equiv&(P\lor Q)\land \top\hspace{13ex}\text{Negation law}\\ \equiv&P\lor Q\hspace{19.3ex}\text{Identity law} \end{align} } Here is a possible approach for your question, which gives the Minimal CNF \begin{align} &(¬Q∧P)∨(¬Q∧R)∨(¬P∧¬R)\\ \equiv&(\neg Q\land(P\lor R))∨(¬P∧¬R)\tag*{Distributive law}\\ \equiv&(\neg Q\land\neg(\neg P\land\neg R))∨(¬P∧¬R)\tag*{De Morgan's law}\\ \equiv&\neg Q\lor (\neg P\land \neg R)\tag*{Redundancy law}\\ \equiv&(\neg Q\lor \neg P)\land(\neg Q\lor\neg R)\tag*{Distributive law} \end{align}