# Propositional Logic CNF

gefragt 2017-08-06 13:25:56 +0200

How can I compute this in a faster way?

d<->a<->b


I want to know if I can directly apply any formulas here to reduce it to CNF. Right now it takes me much time to compute it.

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## 3 Antworten

geantwortet 2017-08-07 09:55:39 +0200

The first answer of contradictioned is great. Thanks for answering that! However, there was a small typo. The first two connectives in the the right most formula should be switched. So it should be $(\neg x \wedge \neg y) \vee (x \wedge y)$. (which has been fixed now)

For the given example $d \leftrightarrow a \leftrightarrow b$ , I want to point out that connectives are left associative (slide 6 in the VRS-Chapter on propositional logic).

Your formula is hence equivalent to $(d \leftrightarrow a) \leftrightarrow b$. If you look at the inner subformula, it can be written as $(\neg d \vee a) \wedge(d \vee \neg a)$ or $(d \wedge a) \vee (\neg d \wedge \neg a)$. Thus, you could transform the formula like this:

• $d \leftrightarrow a \leftrightarrow b =$
• $(d \leftrightarrow a) \leftrightarrow b =$
• $(\neg (d \leftrightarrow a) \vee b) \wedge ((d \leftrightarrow a) \vee \neg b) =$
• $(\neg (d \leftrightarrow a) \vee b) \wedge (((\neg d \vee a) \wedge (d \vee \neg a)) \vee \neg b) =$
• $(\neg (d \leftrightarrow a) \vee b) \wedge (\neg d \vee a \vee \neg b) \wedge (d \vee \neg a \vee \neg b) =$
• $(\neg ((d \wedge a) \vee (\neg d \wedge \neg a)) \vee b) \wedge (\neg d \vee a \vee \neg b) \wedge (d \vee \neg a \vee \neg b) =$
• $(((\neg d \vee \neg a) \wedge (d \vee a)) \vee b) \wedge (\neg d \vee a \vee \neg b) \wedge (d \vee \neg a \vee \neg b) =$
• $(\neg d \vee \neg a \vee b) \wedge (d \vee a \vee b) \wedge (\neg d \vee a \vee \neg b) \wedge (d \vee \neg a \vee \neg b)$
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## Kommentare

1

Thanks for spotting :)

( 2017-08-07 10:42:52 +0200 )bearbeiten
1

Thanks for the explanation!

( 2017-08-07 11:44:28 +0200 )bearbeiten

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( 2017-08-07 11:46:21 +0200 )bearbeiten

geantwortet 2017-08-09 10:49:10 +0200

The answers so far are great and explain to really compute the CNF. As you were asking actually for a faster way to do that, here are two:

Fast Way #1:

If you just need the result, then you may remember the parity function: Given n variables x1,...,xn the parity function is the boolean function associated with the formula x1⊕...⊕xn and it is true if and only if an odd number of the variables is true. Next, observe that (a⊕b) is equivalent to ¬(a↔︎b), so that (d↔︎a)↔︎b is equivalent to (d⊕a)⊕b. Hence, the formula is true if and only if an odd number of variables is true, i.e., either exactly one or all three.

Hence, its DNF is obtained by listing these assignments:

a∧b∧d ∨ a∧¬b∧¬d ∨ ¬a∧b∧¬d ∨ ¬a∧¬b∧d,


and the DNF of its negation is obtained by listing the other assignments (where none or exactly two variables are true):

a∧b∧¬d ∨ a∧¬b∧d ∨ ¬a∧b∧d ∨ ¬a∧¬b∧¬d


The CNF of (d↔︎a)↔︎b can now be obtained by negating the latter, which clearly yields

(¬a∨¬b∨d)  ∧ (¬a∨b∨¬d) ∧ (a∨¬b∨¬d) ∧ (a∨b∨d)


Fast Way #2:

BDDs can also compute CNFs and DNFs by simply enumerating all paths from the root node to the 1 or 0 leaves, respectively. Hence, you may first transform the formula into a BDD, i.e., if-then-else form which is then a=>(d↔︎b)|¬(d↔︎b) by making case distinctions on a, and then a=>(b=>d|¬d)|(b=>¬d|d) by a further case distinction on b. From here you can read the DNF a∧b∧d ∨ a∧¬b∧¬d ∨ ¬a∧b∧¬d ∨ ¬a∧¬b∧d and proceed as above.

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geantwortet 2017-08-07 09:44:16 +0200

If you are free to choose a procedure to do that, I would probably (up to four variables) just write down a truth-table of the formula and read the CNF from that. That is in my opinion the least error prone way.

If you are asked to transform it by applying formulas, then why not use iteratively

$$x \leftrightarrow y \equiv (\neg x \lor y) \land (x \lor \neg y) \equiv (\neg x \land \neg y) \lor (x \land y)$$

until you have all other operators eliminated and then bring it into CNF? Again, using many simple steps--in my opinion--reduces the amount of mistakes. However, you have to practice these small steps ;)

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Gefragt: 2017-08-06 13:25:56 +0200

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Letztes Update: Aug 09 '17