Cnf And Dnf In Discrete Mathematics Pdf

Conjunctive Normal Form

PROPOSITIONAL SATISFIABILITY AND CONSTRAINT SATISFACTION

Holger H. Hoos , Thomas Stützle , in Stochastic Local Search, 2005

Definition 6.1 Variable and Clause Dependencies

Given a CNF formula F and two variables x, x′ appearing in F, x′ is dependent on x (and vice versa) if, and only if, there is a clause in which both x and x′ appear. Furthermore, we define the set of variables dependent on x as

$V_{dep}(F,x):=\{x^{\prime }\in Var(F)|x{\}}^{\prime }\text{is}\text{dependent}\text{on}x$

A clause c of F is dependent on x, if, and only if, x appears in c, and the set of clauses dependent on x is defined as

$C_{dep}(F,x):=\left\{cisaclauseofF\right|c\}\text{is}\text{dependent}\text{on}x$

A clause c is critically satisfied by a variable x under assignment a if, and only if, x appears in c, c is satisfied under a, and flipping the value of x makes c unsatisfied. Finally, a variable x′ is critically dependent on a variable x under assignment a, if, and only if, there is a clause c that is dependent on x and x′, and flipping x results in the clause to change its satisfaction status from (i) satisfied to unsatisfied or vice versa, or (ii) satisfied to critically satisfied (by x′) or vice versa.

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Propositional Calculus

Martin D. Davis , ... Elaine J. Weyuker , in Computability, Complexity, and Languages (Second Edition), 1994

Exercises

1.

Find CNF and DNF formulas equal to each of the following.

(a)

$((p\wedge (q\vee r))\vee (q\wedge (p\vee r))).$ .

(b)

$((\neg p\vee (p\wedge \neg q))\wedge (r\vee (\neg p\wedge q))).$

(c)

$(p\supset (q\leftrightarrow r)).$

2.

Find a DNF formula that has the truth table

p	q	r
1	1	1	0
0	1	1	1
1	0	1	1
0	0	1	1
1	1	0	0
0	1	0	1
1	0	0	0
0	0	0	0

[Hint: The second row of the table corresponds to the ∧-clause (¬ p ∧ q ∧ r ). Each row for which the value is 1 similarly determines an ∧-clause.]

3.

Show how to generalize Exercise 2 to obtain a DNF formula corresponding to any given truth table.

4.

Describe a dual of the method of Exercise 3 which, for any formula α gives a DNF formula β such that α = ¬β. Then show how to turn ¬β into a CNF formula γ such that α = γ. Apply the method to the truth table in Exercise 2. [Hint: Each row in the truth table for which the value is 0 corresponds to an ∧-clause which should not be true.]

5.

Let $\mathcal{A}=\{p,q,r\}$ .

(a)

Give a DNF formula α over $\mathcal{A}$ such that α ^v = 1 for exactly three assignments v on $\mathcal{A}$ .

(b)

Give a CNF formula β over $\mathcal{A}$ such that β ^v = 1 for exactly three assignments v on $\mathcal{A}$ .

6.

(a)

Let α be

$\begin{array}{l}\left(p\wedge q\wedge r\right)\vee \left(p\wedge q\wedge \neg r\right)\vee \left(p\wedge \neg q\wedge r\right)\\ \vee \left(p\wedge \neg q\wedge \neg r\right)·\end{array}$

Give DNF formulas β, γ, δ with 3, 2, 1 ∧-clauses, respectively, such that α = β = γ =δ.

(b)

Let α be

$\begin{array}{l}\left(p\vee q\vee r\right)\wedge \left(p\vee q\vee \neg r\right)\wedge \left(p\vee \neg q\vee r\right)\\ \wedge \left(p\vee \neg q\vee \neg r\right)·\end{array}$

Give CNF formulas β, γ, δ with 3, 2, 1 ∨-clauses, respectively, such that α = β= γ=δ.

7.

Give a CNF formula α with two ∨-clauses such that α ≠β for all CNF formulas β with one ∨ -clause.

8.

Use a normal form to show the correctness of the inference

$\left(p\supset q\right),\left(r\vee \neg q\right),\neg \left(p\vee r\right)\models \neg p·$

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Multiagent Plan Recognition from Partially Observed Team Traces

Hankz Hankui Zhuo , in Plan, Activity, and Intent Recognition, 2014

9.2.1 MAX-SAT Problem

In computer science, satisfiability (often abbreviated SAT) is the problem of determining whether there exists an interpretation that satisfies the formula. In other words, it establishes whether the variables of a given Boolean formula can be assigned in such a way as to make the formula evaluate to true. Equally important is to determine whether no such assignments exist, which would imply that the function expressed by the formula is identically false for all possible variable assignments. In this latter case, we would say that the function is unsatisfiable; otherwise, it is satisfiable. To emphasize the binary nature of this problem, it is often referred to as Boolean or propositional satisfiability.

SAT is known to be NP-complete [9] , but the flip side is that it is very powerful in its representational ability: any propositional logic formula can be rewritten as a conjunctive normal form (CNF) formula. A CNF formula $f$ is a conjunction of clauses. A clause is a disjunction of literals. A literal $l_i$ is a variable $x_i$ or its negation $\neg x_i$ . A variable $x_i$ may take values 0 (for false) or 1 (for true). The length of a clause is the number of its literals. The size of $f$ , denoted by $\mid f\mid$ , is the sum of the length of all its clauses. An assignment of truth values to the variables satisfies a literal $x_i$ if $x_i$ takes the value 1, satisfies a literal $\neg x_i$ if $x_i$ takes the value 0, satisfies a clause if it satisfies at least one literal of the clause, and satisfies a CNF formula if it satisfies all the clauses of the formula. An empty clause contains no literals and cannot be satisfied. An assignment for a CNF formula $f$ is complete if all the variables occurring in $f$ have been assigned; otherwise, it is partial.

The MAX-SAT problem for a CNF formula $f$ is the problem of finding an assignment of values to variables that minimizes the number of unsatisfied clauses (or equivalently, that maximizes the number of satisfied clauses). Two CNF formulas, $f_1$ and $f_2$ , are equivalent if $f_1$ and $f_2$ have the same number of unsatisfied clauses for every complete assignment of $f_1$ and $f_2$ . There are many solvers for MAX-SAT problems (e.g., MaxSatz [17]); we used the weighted version of MaxSatz ¹ in our implementation.

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INTRODUCTION

Holger H. Hoos , Thomas Stützle , in Stochastic Local Search, 2005

Example 1.1 A Simple SAT Instance

Let us consider the following propositional formula in CNF:

$\begin{array}{l}F:=(\neg x_1\vee x_2)\\ \wedge (\neg x_2\vee x_1)\\ \wedge (\neg x_1\vee \neg x_2\vee \neg x_3)\\ \wedge (x_1\vee x_2)\\ \wedge (\neg x_4\vee x_3)\\ \wedge (\neg x_5\vee x_3)\end{array}$

For this formula, we obtain the variable set Var(F) = {x ₁, x ₂, x ₃, x ₄, x ₅}; consequently, there are 2⁵ = 32 different variable assignments. Exactly one of these, x ₁ = x ₂ = ⊤, x ₃ = x ₄ = x ₅ = ⊥, is a model, rendering F satisfiable.

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13th International Symposium on Process Systems Engineering (PSE 2018)

Qi Chen , ... Ignacio E. Grossmann , in Computer Aided Chemical Engineering, 2018

2.2 GDP Basic Steps

Basic steps in GDP bring a formulation incrementally from conjunctive normal form to disjunctive normal form by intersecting disjuncts with constraints or other disjuncts. Ruiz and Grossmann (2012) show that by doing so, the HR of the problem is tightened at the expense of a growth in the number of disjuncts. The reformulation that applies a basic step between disjunctions "d" and "e" is implemented with the following command:

TransformationFactory('gdp.basic_step').apply_to(m, targets=[m.d, m.e])

The result is a reformulation from a GDP model to another GDP model with a basic step applied. The tightened GDP could then be further reformulated to MILP/MINLP or used in the context of the hybrid BM/HR cutting plane algorithm.

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Handbook of Computability Theory

Piergiorgio Odifreddi , in Studies in Logic and the Foundations of Mathematics, 1999

2.1 Truth-table Reducibilities

Since every propositional formula can be written in disjunctive or conjunctive normal form, using only the complete set of connectives {¬, ∧, ∨}, a number of possible variations of tt-reducibility immediately comes to mind, e.g. by restricting that set to:

•

{∧, ∨}: positive reducibility (⩽ _p ),

•

{∧}: conjunctive reducibility (⩽ _c ),

•

{∨}: disjunctive reducibility (⩽ _d ),

•

{¬}: bounded truth-table reducibility with norm 1 (⩽ _btt(1)).

Prepositional connectives provide canonical ways of looking at Boolean functions, but they leave out a natural one: sum modulo 2, which gives rise to a further variation of tt-reducibility:

•

linear reducibility (⩽ _l ) reducibility.

The next result justifies the list just given.

Theorem 2.2 (Bulitko [1980], Selivanov [1982])

The only truth-table reducibilities on non-trivial sets (different from ∅ and ω) are the following seven: ⩽ _m , ⩽ _btt(1), ⩽ _c , ⩽ _d , ⩽ _p , ⩽ _l , ⩽ _tt .

On the non-trivial r.e. sets the truth-table reducibilities reduce to six, since ⩽ _btt(1) and ⩽ _m then coincide.

Theorem 2.3 (Post [1944], Shoenfield [1957], Jockusch [1968], Degtev [1973, 1981, 1982], Bulitko [1980])

On the positive side, the following implications hold on the r.e. sets, and no other does even on the complete sets:

On the negative side, there are non-recursive r.e. sets on which all the reducibilities collapse; more precisely, there is an r.e. tt-degree that consists of only one r.e. m-degree.

The degrees induced by the six reducibilities provide different pictures of the world of r.e. sets.

Theorem 2.4 (Degtev [1979, 1982, 1983a, 1983b])

The structures of the r.e. m-, c-, d-, p-, l- and tt-degrees are pairwise not elementarily equivalent.

Sketch of Proof

The sentence

$\left(\forall a\ne 0\right)\left(\exists b\ngeqq a\right)\left(\exists c\ngeqq a\right)\left(a⩽b\cup c\right)$

is true for the r.e. d-, p-, l- and tt-degrees, but false for the r.e. m- and c-degrees.

The sentence expressing distributivity for uppersemilattices, i.e.

$\left(\forall a\right)\left(\forall b\right)\left(\forall c\right)\left[a⩽b\cup c\to \left(\exists b_1⩽b\right)\left(\exists c_1⩽c\right)\left(a=b_1\cup c_1\right)\right]$

is true for the r.e. m-degrees but false for the r.e. c-degrees.

The sentence

$\left(\exists a\ne 0\right)\left(\forall b\right)\left(\forall c\right)\left(c⩽a\cup b\to c⩽b\vee a⩽b\cup c\right)$

is true for the r.e. l-degrees but false for the r.e. d-, p- and tt-degrees.

The sentence

$\left(\forall a\ne 0\right)\left(\exists b\ngeqq a\right)\left(\exists c\ngeqq a\right)\left(a⩽b\cup c\wedge b⩽a\cup c\right)$

is true for the r.e. tt-degrees but false for the r.e. d- and p-degrees.

Finally, the sentence

$\left(\forall a\ne 0\right)\left(\exists b\ne 0\right)\left[b⩽a\wedge \left(\forall c\right)\left(\forall d\right)\left(b⩽c\cup d\wedge c⩽b\cup d\right)\to \left(\exists e⩽b,c\right)\left(b=d\cup d\right)\right]$

is true for the r.e. d-degrees but false for the r.e. p-degrees.

On arbitrary sets, one has instead the following slightly more complicated picture.

Theorem 2.5 (Jockusch [1968], Degtev [1982])

On the positive side, the following implications hold on arbitrary sets, and no other does:

Moreover, p- and tt-reducibility never collapse in a non-trivial way, in the sense that each non-recursive tt-degree contains at least two p-degrees.

On the negative side, the following reducibilities can collapse in a non-trivial way: m- and p-; m- and l-; l- and tt-. More precisely, there are: non-recursive p-degrees consisting of only one m-degree; non-recursive l-degrees consisting of only one m-degree; and non-recursive tt-degrees consisting of only one l-degree.

Sketch of Proof

The collapse of p- and m-reducibility is achieved by considering a semirecursive set, i.e. a set A for which there is a recursive function f of two variables such that:

•

f(x, y) = x or f(x, y) = y,

•

if x ∈ A or y ∈ A then f(x, y) ∈ A.

If B ⩽ _p A then B ⩽ _m A, and B is semirecursive too; thus the p-degree of a semirecursive set consists of a single m-degree.

The separation of tt- and p-reducibility is achieved by associating to every set A its tt-cylinder A^tt , defined as

$A^{tt}=\left\{x:A\models {\sigma }_x\right\}$

Then A ≡ _ttA^tt , and it is enough to show that the tt-degree of A contains a semirecursive set B, and that if A^tt ⩽ _p B then A is recursive.

The collapse of l- and m-reducibility is achieved by considering a set A for which there is a recursive function f of two variables such that:

$f\left(x,y\right)\in A\iff \left(x\in A\wedge y\notin A\right)\vee \left(x\notin A\wedge y\in A\right)$

If B ⩽ _l A then B ⩽ _m A; thus if A has minimal m-degree, its l-degree consists of a single m-degree.

Theorem 2.6 (Degtev [1979, 1985], Mal'cev [1985])

The structures of m- and btt(1)-degrees are isomorphic, and so are the structures of c- and d-degrees.

The structures of m-, c-, p-, l- and tt-degrees are pairwise not elementarily equivalent, with the only possible exception of p- and tt-degrees.

The isomorphism of c- and d-degrees is trivial, since

$A{⩽}_{c}B\iff \overline{A}{⩽}_d\overline{B}$

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Computing Small Clause Normal Forms

Andreas Nonnengart , Christoph Weidenbach , in Handbook of Automated Reasoning, 2001

3.6 Clause Normal Form

Finally, we move all universal quantifiers outwards and compute a conjunctive normal form.

$\begin{array}{c}\begin{array}{llll}\begin{array}{r}\hfill \forall x\varphi \wedge \psi \\ \hfill \forall x\varphi \vee \psi \\ \hfill \varphi \vee ({\psi }_1\wedge {\psi }_2)\end{array}\hfill & \begin{array}{c}\to \\ \to \\ \to \end{array}\hfill & \begin{array}{l}\forall x(\varphi \wedge \psi )\hfill \\ \forall x(\varphi \vee \psi )\hfill \\ (\varphi \vee {\psi }_1)\wedge (\varphi \vee {\psi }_2)\hfill \end{array}\hfill & \begin{array}{c}\text{if}x\notin free(\psi )\\ \text{if}x\notin free(\psi )\\ \end{array}\hfill \end{array}\end{array}$

All three rules are equivalence preserving and terminating since they either decrease the length of subformulae positions starting with a quantifier or the number of conjunctions that occur below the position of a disjunction.

For our running example this means

$\begin{array}{c}\forall y_{1}R(y_1,y_1)\wedge (P(a)\vee \forall y_{3}R(y_3,f(y_3))\vee \forall y_{5}Q(y_5))\\ \downarrow \\ \forall y_1,y_3,y_5(R(y_1,y_1)\wedge (P(a)\vee R(y_3,f(y_3))\vee Q(y_5)))\\ \downarrow \\ \forall y_1,y_3,y_5((R(y_1,y_1)\vee P(a))\wedge (R(y_1,y_1)\vee R(y_3,f(y_3)))\wedge (R(y_1,y_1)\vee Q(y_5)))\end{array}$

where we left out some intermediate steps.

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Resolution in the Propositional Calculus

Nils J. Nilsson , in Artificial Intelligence: A New Synthesis, 1998

Exercises

14.1

Prove that the matrix procedure for converting from DNF to CNF preserves equivalence.

14.2

Heads, I win; tails, you lose. Express these statements (plus other statements you might need) in the propositional calculus, and then use resolution to prove that I win. (Alternatively, you can change the problem if you'd like to heads, you win; tails, I lose.)

14.3

The following wffs are instances of axioms that are sometimes used in the propositional calculus:

1.

Implication introduction: P ⊃ (Q ⊃ P)

2.

Implication distribution: (P ⊃ (Q ⊃ R)) ⊃ ((P ⊃ Q) ⊃ (P ⊃ R))

3.

Contradiction realization: (Q ⊃ ¬ρ) ⊃ ((Q ⊃ P) ⊃ ¬Q)

Use resolution refutation to prove each of these formulas.

14.4

Consider the following unsatisfiable set of clauses:

$\begin{array}{l}\text{P}\vee \text{Q}\hfill \\ \text{P}\vee \neg \text{Q}\hfill \\ \neg \text{P}\vee \text{Q}\hfill \\ \neg \text{P}\vee \neg \text{Q}\hfill \end{array}$

1.

Produce resolution refutations for each of the following strategies:

(a)

Set-of-support resolution (in which the set of support is the last clause in the preceding list of clauses)

(b)

ancestry-filtered form resolution

(c)

A strategy that violates both set of support and ancestry filtering

2.

Prove that there is no linear-input resolution refutation of this unsatisfiable set of clauses.

14.5

Convert the following propositional calculus wff into clauses:

$\neg \left[\left(\left(\text{P}\vee \neg \text{Q}\right)\supset \text{R}\right)\supset \left(\text{P}\wedge \text{R}\right)\right]$

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Tableaux and Related Methods

Reiner Hähnle , in Handbook of Automated Reasoning, 2001

Index

A

ancestor cancellation159

ancestry family159

atom104

B

branch111

closed112, 126

open112

subsumption145

C

cancellation pruning rule159

chain153

admissible153

clause105

empty105

Horn105

instance106

ground106

new106

positive105

relevant131

unit105

CNFsee conjunctive normal form complete

calculus117

strongly117

completion mode119

computation rule117

fair117

conjunctive normal form105

connected extension step130

connection122

connection method122

consolution165

constraint model generation143

cut147, 148, 163

D

Davis-Putnam procedure163

destructive118

dilemma rule164

disjunctive logic programming157

domain106

E

extension rule with local lemmas149

F

fact105

folding up rule150

formula

complement105

first-order104

prepositional104

formula instantiation problem165

G

H

Herbrand complexity107

Hintikka set120

hyper tableau extension140

hypergraph124

I

inheritance near-Horn Prolog159

integrity constraint143

interpretation106

L

level cut124

literal104

ancestor153

body157

head157

negative104

ordering136

positive104

type A153

type B153

local lemma149

logical consequence107

M

magic set124

mated122

mating122

matrix122

spanned123

MGUsee most general unifier minimal proof length165

minimally unsatisfiable131

model107

model elimination153

restart161

strict161

model generation143

most general unifier105

N

negation normal form104

NNFsee negation normal form

P

p-equivalent165

p-simulation165

path122

principle of bivalence163

proof confluent119

pure clause rule163

Q

R

range-restricted141

reduction124

reduction step130

relevancy testing124

resolution

linear155

Prolog-SLD156

SLWV137

unit163

rule105

S

satisfiability107

saturation134

scope104

selection function135

complete140, 143

consistent137, 143

stable137

sentence105

signature104

simplified problem reduction format size160

of formula104

of tableau112

Skolem term110

Stålmarck's procedure164

structure

canonical113

first-order106

Herbrand107

term domain107

subformula106

immediate106

negative occurrence106

positive occurrence106

proper106

substitution105

grounding105

idempotent105

renaming105

T

tableau111

clause126

closed126

connection129

EP141

hyper138

positive140

KE162

proof112, 126

proof procedure117

regular131

restart162

strict162

satisfaction113

semantic semantic140

subsumption145

weak connection131

with merging151

with regressive merging151

with unification111

tableau calculus111

tautology105

term104

ground104

truth106

U

unifier105, 115

unit extension step163

unit near-Horn Prolog159

V

validity107

variable104

bound105

free105

rigid114

universal 115

variable assignment106

variant106

W

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Mathematical background

Barry Dwyer , in Systems Analysis and Synthesis, 2016

2.1.3 Conjunctive Normal Form

To prove that two expressions are equal, a frequently used technique is to transform both expressions to a standard form. One such standard form is called conjunctive normal form or CNF. An expression in CNF is a 'product of sums'. The 'sums' are literals (simple propositions or negated propositions, e.g., $P$ , or $\neg Q$ ) linked by $\vee$ , which are then formed into a 'product' using $\wedge$ . ⁴

Consider the expression

(2.1.23) $(P\iff Q)$

Its conjunctive normal form is

(2.1.24) $(\neg P\vee Q)\wedge (P\vee \neg Q)$

To get this result, (using Axiom (2.1.9)) we reduce all the operators to $\wedge$ , $\vee$ , and $\neg$

(2.1.25) $(P\wedge Q)\vee (\neg P\wedge \neg Q)$

We then use the first distribution law, three times:

(2.1.26) $(P\wedge Q)\vee (\neg P\wedge \neg Q)=(P\vee \neg P\wedge \neg Q)\wedge (Q\vee \neg P\wedge \neg Q)=(P\vee \neg P)\wedge (P\vee \neg Q)\wedge (Q\vee \neg P\wedge \neg Q)=(P\vee \neg P)\wedge (P\vee \neg Q)\wedge (Q\vee \neg P)\wedge (Q\vee \neg Q)$

Finally, we eliminate the sums $(P\vee \neg P)$ and $(Q\vee \neg Q)$ , which are always $\mathrm{True}$ , leaving

(2.1.27) $(P\vee \neg Q)\wedge (Q\vee \neg P)$

Normalisation is a purely mechanical process that a computer can do (although it is NP-hard).

We can prove the theorem

(2.1.28) $(P\iff Q)=(P\Rightarrow Q)\wedge (P\Leftarrow Q)$

by converting both sides to CNF. We have already dealt with the left-hand side above. Normalising its right-hand side is left as a simple exercise for the reader. ⁵

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Cnf And Dnf In Discrete Mathematics Pdf

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