Let $\psi_{1} := (P \wedge Q \wedge R)\vee(\neg P\wedge Q\wedge\neg R)\vee(\neg P\wedge\neg Q\wedge R)\vee(\neg P \wedge\neg Q\wedge\neg R)$ and

Let $\psi_{2} := (\neg P \wedge \neg Q) \vee (\neg P \wedge \neg R) \vee (P \wedge Q \wedge R)$

I have proven so far, that the two formulas in DNF are equivalent to $P\Leftrightarrow(Q\wedge R)$

Now I need to simplify $\psi_{1}$ with classical equivalences to obtain $\psi_{2}$

My problem is, that I am struggling with the distribution of $\vee$ and $\wedge$ with three Propositional variables in the brackets.

Maybe you guys can give me some tips or help me out a bit on this one.


2 Answers 2





$P \Leftrightarrow (P \lor Q) \land (P \lor \neg Q)$

$P \Leftrightarrow (P \land Q) \lor (P \land \neg Q)$

So, for example, you can combine the terms $\neg P \land \neg Q \land R$ and $\neg P \land \neg Q \land \neg R$ to $\neg P \land \neg Q$

In fact, that gets you almost there. Indeed, you could likewise combine $\neg P \land Q \land R$ and $\neg P \land \neg Q \land \neg R$ to $\neg P \land \neg R$

The only issue is that doing both combinations requires two terms $\neg P \land \neg Q \land \neg R$ . But, that's easy to obtain using:


$P \Leftrightarrow P \land P$

$P \Leftrightarrow P \lor P$

  • $\begingroup$ I got the same used $(\neg P \land \neg Q \land \neg R)$ and $(\neg P \land \neg Q \land R)$ to obtain $(\neg P \land \neg Q)$. I just haven´t figuered out what to do with the other two terms yet. $\endgroup$ Jan 16, 2018 at 18:56

Hint: If you're familiar with distribution over two terms, break up $(X \vee Y \vee Z)$ into $((X \vee Y )\vee Z)$ and distribute twice.


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