The statement is $$\lnot((\lnot P \land Q) \lor \lnot(R \land \lnot S))$$ and the answer is $$(p \land r) \lor (p \land \lnot s) \lor (\lnot q \land r) \lor (\lnot q \land \lnot s)$$ according to the dCode Boolean Expressions Calculator.

I'm trying to get to the solution myself but I got stuck. I don't know what else to do past what I've done below:

  1. Start with $\lnot((\lnot P \land Q) \lor \lnot(R \lor \lnot S))$
  2. De morgan's law $\lnot(\lnot P \land Q) \land \lnot(\lnot(R \lor \lnot S))$
  3. De morgan's law $\lnot(\lnot P) ∨ \lnot Q \land \lnot(\lnot(R \lor \lnot S))$
  4. Double negation law $P \lor \lnot Q \land R \lor \lnot S$
  5. Distributive laws $(P \lor \lnot Q) \land (P \lor \lnot R) \lor \lnot S$
  • $\begingroup$ Step 4 : do not remove parentheses: $[(P ∨ ¬Q) ∧ R] ∨ [(P ∨ ¬Q) ∧ ¬S]$ $\endgroup$ Commented Feb 25, 2021 at 9:24
  • $\begingroup$ @MauroALLEGRANZA I don't see how you progressed from 3. to 4. What law did you apply? $\endgroup$
    – Bee
    Commented Feb 25, 2021 at 12:10

1 Answer 1


The result of the second DeMorgan should be $$\color{red}(\neg (\neg P) \lor \neg Q\color{red}) \land \neg (\neg (R \lor \neg S))$$ which by two Double Negations gives you $$(P \lor \neg Q) \land (R \lor \neg S)$$

Now do Distributive Laws (which in this case works just like the FOIL principle, if you're familiar with that), and you're right at the correct answer.

The important point is: you are dropping parentheses, but those parentheses are really important! For example, it is not clear if a statement like $$P \lor Q \land R$$ means $$P \lor (Q \land R)$$ or $$(P \lor Q) \land R$$ Those are two different statements, and so you really need to use parentheses to disambiguate!


You must log in to answer this question.

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