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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 $(P \lor \lnot Q) \land (P \lor \lnot R) \lor \lnot S$
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  • $\begingroup$ Step 4 : do not remove parentheses: $[(P ∨ ¬Q) ∧ R] ∨ [(P ∨ ¬Q) ∧ ¬S]$ $\endgroup$ – Mauro ALLEGRANZA Feb 25 at 9:24
  • $\begingroup$ @MauroALLEGRANZA I don't see how you progressed from 3. to 4. What law did you apply? $\endgroup$ – Bee Feb 25 at 12:10
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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!

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