could someone please be so kind to answer this question and explain the answer? Which of the following formulas are in CNF?

  1. ¬(A∨B∨C)∧(A∨B)
  2. (A∧B∧C)∨(A∧B)
  3. (A∨B∨¬C)∧(A∨B)
  4. (A∧B∧¬C)∨(A∧B)
  • $\begingroup$ $1$ and $3$ are in CNF, Since they connected with $\wedge$ $\endgroup$ – Manx Sep 24 '19 at 20:22
  • $\begingroup$ @Manx Option 1 is not in CNF as there is a negation of the disjunction $A \vee B \vee C$. $\endgroup$ – RyRy the Fly Guy Sep 24 '19 at 21:16
  • $\begingroup$ I see, thank you sir :) $\endgroup$ – Manx Sep 24 '19 at 22:12

CNF is when a Boolean expression is written as a conjunction of clauses, where a clause is defined as a disjunction of literals. And... a literal is either a simple proposition or its negation. In other words, the expression must be a conjunction of disjunctions of some combination of the variables $p,q,r,...$ or their negations. Think of CNF as a chain of conjunctions (ANDs) that join chains of disjunctions (ORs) which join propositions or their negations. As a helpful mnemonic, CNF is conjunctions of disjunctions.

For example, the following is in CNF:

$(p \vee q) \wedge s \wedge (\neg a \vee b \vee \neg c) \wedge t \wedge \neg w$

Options 2 and 4 are disjunctions of conjunctions, so they are not in CNF. Option 1 seems to be a good candidate, but if you inspect it carefully you will notice there is a negation of a set of parentheses and not a simple proposition. If a negation is present, then it must be the negation of a simple proposition. If you were to apply DeMorgan's rule (by distributing the negation and changing the disjunctions in the parantheses to conjunctions), then 1 would be in CNF.

Option 3 is the only option composed of pure literals written in terms of conjunctions of disjunctions, so...

3 is in CNF.


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