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I tried to solve the following question, but I don't know if my solution is correct!

Q: Transform the following proposition into conjunction normal form (CNF): 

( ∧ ) ∨ ( ∧ ) ∨ ( → )

A: Let P ≡ The Whole Proposition, 1 ≡ ( ∧ ), 2 ≡ ( ∧ ), and 3 ≡ ( → )

≡ ¬¬( ( ∧ ) ∨ ( ∧ ) ∨ ( → ) )                [Negation Law (P)]
≡ ¬¬( ( ∧ ) ∨ ( ∧ ) ∨ (¬ ∨ ) )               [Material Implication Law (3)]
≡ ¬ ( (¬ ∨ ¬) ∧ (¬ ∨ ¬) ∧ ( ∧ ¬) )           [De Morgan Law (P)]
≡ ¬ ( (¬ ∨ ¬) ∧ (¬ ∨ ¬) ∧ ¬¬( ∧ ¬) )         [Negation Law (3)]
≡ ¬ ( (¬ ∨ ¬) ∧ (¬ ∨ ¬) ∧ ¬(¬ ∨ ) )          [De Morgan Law (3)]
≡ CNF

UPDATE

The correct answer is as follow:

   ( ∧ ) ∨ ( ∧ ) ∨ ( → )

≡  ( ∧ ) ∨ ( ∧ ) ∨ (¬ ∨ )                          [Material Implication Law]

≡  [( ∧ ) ∨ ] ∧ [( ∧ ) ∨ ] ∨ (¬ ∨ )              [Distribution Law]

≡  ( ∨ ) ∧ ( ∨ ) ∧ (( ∨ ) ∧ ( ∨ ) ∨ (¬ ∨ )    [Distribution Law]

≡  ( ∨ ) ∧ ( ∨ ) ∧ ( ∨ ) ∧ ( ∨  ∨ ¬ ∨ )       [Associative Law]

≡  CNF

Thanks to (@Bram28).

  • 2
    That's not in CNF -- a CNF must have all the conjunction as the _topmost_ logical connective; you have them all beneath a negation. – 2017-02-05
  • 0
    If not, would you please show me how to solve it then? @HenningMakholm – 2017-02-05
  • 1
    Don't negate; distribute – 2017-02-05
  • 0
    Would you be more specific? @GrahamKemp – 2017-02-05
  • 0
    @Badger Bram28 has it covered. Give it a try. – 2017-02-05

1 Answers 1

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HINT

The equivalence rule you want to use is

Distribution

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

Using this you can turn disjunctions of conjunctions into conjunctions of disjunctions, which is what you want to do to get things into CNF.

  • 1
    Also $\rm ~(S\wedge T)\vee(U\wedge V) ~\iff~ (S\vee U)\wedge(S\vee V)\wedge(T\vee U)\wedge(T\vee V)~$ and so on. – 2017-02-05
  • 0
    yes, the good old FOIL works here as well! – 2017-02-05
  • 0
    I tried that before, but the last proposition with an arrow couldn't figure it out! – 2017-02-06
  • 0
    @Bader Rewrite the conditional as a disjunction, which is what you did already. And then know that if you have something like $(A \land B) \lor C \lor D$, you should treat the $C \lor D$ as one claim when doing distribution, so you get: $(A \land B) \lor C \lor D \Leftrightarrow (A \land B) \lor (C \lor D) \Leftrightarrow (A \lor (C \lor D)) \land (B \lor (C \lor D)) \Leftrightarrow (A \lor C \lor D) \land (B \lor C \lor D)$. – 2017-02-06
  • 0
    Ohhh, I see!! The answer was in front of my eyes all the time but I guess I made it look rather complicated. Thanks @Bram28 – 2017-02-10