Formale Methoden der InformatikIntroduction and a

Short Review of Propositional Logic

Uwe Egly

Vienna University of TechnologyInstitute of Information Systems

Knowledge-Based Systems Group

Outline

IntroductionMotivationThe Circuit ExampleFormalizing a Logical SystemInductive Definitions Recap

Syntax of PL0Tree Notation of Propositional Formulas

Semantics of PL0NotationsUseful EquivalencesEntailmentChecking Entailment

Satisfiability CheckingTranslation to Normal FormsDLL

Why Logic in CS Studies?

◮ ProgrammingPROLOG, boolean expressions in if-then, assertions, . . .

◮ Languages for and reasoning about specifications,dynamic behavior (temporal logics), consistency, . . .

◮ Knowledge representationWeb ontologies, description logics, reasoning underincomplete/uncertain information, . . .

◮ Hardwarecircuit specification, (automated) verification of circuits, . . .

◮ and much more!!!

Types of Logics Interesting for CS

There is not the logic for CS, but many of them

◮ Besides classical logic, there are non-classical logics◮ minimal and intuitionistic logic (“constructive” logics)◮ modal logics, temporal logics, . . .◮ many-valued logics, fuzzy logics

◮ Many of these logics come in different “levels”◮ Propositional logics (PL0)◮ First-order (PL1): (quantifiers over object variables)◮ Second-order (quantifiers over function/predicate symbols)◮ and even higher-order logics

Example: Correct Implementation of a Circuit

Does the implemented circuit compute specified function

Spec : ((p ∧ q) ∨ r) ∧ (¬(p ∧ q) ∨ s)

&

A1

p

q≥1

O1r

1

N1

≥1

O2s

&

A2

Example: Correct Implementation of a Circuit

Does the implemented circuit compute specified function

Spec : ((p ∧ q) ∨ r) ∧ (¬(p ∧ q) ∨ s)

&

A1

p

q

A1↔ p ∧ q≥1

O1

O1↔ A1 ∨ r

r

1

N1

N1↔ ¬A1≥1

O2s O2↔ N1 ∨ s

&

A2

A2↔ O1 ∧O2

How to Model and Solve the Problem?

◮ Modeling: Abstract a given problem P into logic PL0 ( P ′)◮ The implemented function Impl is:

Impl = (A1↔ p ∧ q) ∧ (O1↔ A1 ∨ r) ∧ (N1↔ ¬A1) ∧

(O2↔ N1 ∨ s) ∧ (A2↔ O1 ∧O2)→ A2

◮ Is the implementation correct (does Impl comply w. Spec)?◮ P ′ is now a reasoning problem, not a test problem!◮ Proof: Use automatic decision procedures to determine

the solution of P, i.e., compute satisfiability/validity of P ′

◮ SAT Solvers and BDDs◮ Model Checker◮ Theorem Provers

Formalizing a Logical System. . . or the ingredients for an (automated) inference system

◮ Old dream of mankind(at least of some “logicians” like G. W. Leibniz (1646–1716))

◮ What is needed?◮ Syntax (= a formal language for constructing formulas)◮ Semantics (= meaning of formulas)◮ Inference rules (syntacticly derive “correct” statements)◮ Correctness (no false conclusions are drawn)◮ Completeness (all correct conclusions are drawn)

◮ Is this all what we need?

Other Important Questions

◮ Expressive power of the representation language (RL)Are we able to represent our problem in the given RL?

◮ Decidability of the reasoning procedure (on the given RL)Is there a terminating algo providing the correct answer fora given reasoning task?

◮ ComplexityWhat amount of resources (time, space) are required (inthe worst case) to answer the question?

Basic Ideas Underlying Inductive Definitions

◮ Inductive definitions widely used in CS (recursion!!!)◮ Basic idea: given basic set A0 ⊆ B rule f : Bn 7→ B

◮ Incremental construction of sets

Ai+1 = Ai ∪ {f (a1, . . . ,an) |a1, . . . ,an ∈ Ai}

◮ Limes of Ai (when i goes to infinity)

A =⋃

i≥0

Ai

◮ A set A is closed under f , if the following holds:

a1, . . . ,an ∈ A ⇒ f (a1, . . . ,an) ∈ A

Basic Ideas Underlying Inductive Definitions (2)

Schema of the inductive definition

A is the smallest set for which (a) and (b) holds:

(a) A0 ⊆ A

(b) a1, . . . ,an ∈ A ⇒ f (a1, . . . ,an) ∈ A(A is closed under f )

Often simple induction is used:

1. You want to prove ∀n :nat .P(n)(i.e., a property P holds for all natural numbers)

2. You show the base case P(0)

3. In the induction step, you prove ∀n :nat . (P(n)→ P(n + 1))(i.e., if you had a proof of P(n) (induction hypothesis), youcould prove P(n + 1))

The Syntax of PL0 (Alternative 1)

◮ Given a countable set, BV, of boolean variables◮ Boolean variables like p,q, r ,p1, . . . represent facts:

e.g., r represents “it is raining”

◮ Use p,q, . . . as metavariables for BVs and φ, ψ, . . . asmetavariables for formulas

◮ Inductive definition of the set of propositional formulas L0

B1: Every p ∈ BV is a formula (called atomic formula or atom)B2: ⊤ (verum) and ⊥ (falsum) are formulasS1: If φ is a formula, then so is (¬φ) (negation)S2: If φ1 and φ2 are formulas, then so are (φ1 ∧ φ2)

(conjunction), (φ1 ∨ φ2) (disjunction), (φ1 → φ2)(implication), (φ1 ↔ φ2) (equivalence) and (φ1 ⊕ φ2)(exclusive-or)

◮ What is A0, f , and A here?

The Syntax of PL0 (Alternative 2)

L ::= ⊤ |⊥ |BV | (¬L) | (L∧L) | (L∨L) | (L → L) | (L ↔ L) | (L⊕L)

◮ Read this as: L is the smallest set such that 1. and 2. hold:

1. If p ∈ BV, p is ⊤ or p is ⊥, then p ∈ L2. If φ, ψ ∈ L then

(¬φ), (φ ∧ ψ), (φ ∨ ψ), (φ→ ψ), (φ↔ ψ), (φ⊕ ψ) ∈ L

◮ To save parenthesis, use the following ranking of bindingstrength: ¬, ∧, ∨,→,↔, ⊕ (¬ binds stronger than ∧, etc.)

◮ Example: ¬p ∧ q → r ∨ s means (((¬p) ∧ q)→ (r ∨ s))

The Syntax of PL0 Cont’d

◮ A literal is ⊤, ⊥, an atom or the negation thereof◮ A clause is a disjunction of literals◮ Immediate subformula (relation)

◮ φ is an immediate subformula (isf) of ¬φ◮ φ1 and φ2 are isfs of (φ1 ◦ φ2) for ◦ ∈ {∧,∨,→,↔,⊕}

◮ Subformula relation: reflexive-transitive closure of theimmediate subformula relation

◮ Example: all subformulas of ¬p ∧ q → r ∨ s¬p ∧ q → r ∨ s, ¬p ∧ q, r ∨ s, ¬p, q, r , s, p

The Syntax of PL0 Cont’d

◮ A literal is ⊤, ⊥, an atom or the negation thereof◮ A clause is a disjunction of literals◮ Immediate subformula (relation)

◮ φ is an immediate subformula (isf) of ¬φ◮ φ1 and φ2 are isfs of (φ1 ◦ φ2) for ◦ ∈ {∧,∨,→,↔,⊕}

◮ Subformula relation: reflexive-transitive closure of theimmediate subformula relation

◮ Example: all subformulas of ¬p ∧ q → r ∨ s¬p ∧ q → r ∨ s, ¬p ∧ q, r ∨ s, ¬p, q, r , s, p

The Syntax of PL0 Cont’d

◮ A literal is ⊤, ⊥, an atom or the negation thereof◮ A clause is a disjunction of literals◮ Immediate subformula (relation)

◮ φ is an immediate subformula (isf) of ¬φ◮ φ1 and φ2 are isfs of (φ1 ◦ φ2) for ◦ ∈ {∧,∨,→,↔,⊕}

◮ Subformula relation: reflexive-transitive closure of theimmediate subformula relation

◮ Example: all subformulas of ¬p ∧ q → r ∨ s¬p ∧ q → r ∨ s, ¬p ∧ q, r ∨ s, ¬p, q, r , s, p

The Syntax of PL0 Cont’d

◮ A literal is ⊤, ⊥, an atom or the negation thereof◮ A clause is a disjunction of literals◮ Immediate subformula (relation)

◮ φ is an immediate subformula (isf) of ¬φ◮ φ1 and φ2 are isfs of (φ1 ◦ φ2) for ◦ ∈ {∧,∨,→,↔,⊕}

◮ Subformula relation: reflexive-transitive closure of theimmediate subformula relation

◮ Example: all subformulas of ¬p ∧ q → r ∨ s¬p ∧ q → r ∨ s, ¬p ∧ q, r ∨ s, ¬p, q, r , s, p

The Syntax of PL0 Cont’d

◮ A literal is ⊤, ⊥, an atom or the negation thereof◮ A clause is a disjunction of literals◮ Immediate subformula (relation)

◮ φ is an immediate subformula (isf) of ¬φ◮ φ1 and φ2 are isfs of (φ1 ◦ φ2) for ◦ ∈ {∧,∨,→,↔,⊕}

◮ Subformula relation: reflexive-transitive closure of theimmediate subformula relation

◮ Example: all subformulas of ¬p ∧ q → r ∨ s¬p ∧ q → r ∨ s, ¬p ∧ q, r ∨ s, ¬p, q, r , s, p

Propositional Formulas as Trees

◮ Formulas can be depicted as formula trees◮ Example: ((p ∧ q)→ r)↔ (p → (q → r))

↔

→

∧

p q

r

→

p →

q r

The Semantics of PL0

◮ A boolean value (or truth value) is either 0 (false) or 1 (true)

◮ An interpretation function for a set P of boolean variables:

mapping I : P 7→ {0,1}

◮ Interpretation functions are often called truth assignmentse.g., I assigns 1 to p ∈ P, i.e., I(p) = 1

◮ Since we want to “evaluate” formulas under I, we have toextend I to formulas

The Extension of I to Formulas

◮ I(⊤) = 1 and I(⊥) = 0◮ I(¬φ) = 1 iff I(φ) = 0◮ I(φ ∧ ψ) = 1 iff I(φ) = I(ψ) = 1◮ I(φ ∨ ψ) = 1 iff I(φ) = 1 or I(ψ) = 1◮ I(φ→ ψ) = 1 iff I(φ) = 0 or I(ψ) = 1◮ I(φ↔ ψ) = 1 iff I(φ) = I(ψ)

◮ I(φ⊕ ψ) = 1 iff I(φ) 6= I(ψ)

Equivalent notations:φ is true under I iff I satisfies φ iff I(φ) = 1 iff I |= φ

φ is false under I iff I does not satisfy φ iff I(φ) = 0 iff I 6|= φ

Truth Tables

◮ Another representation of the extension of I from above◮ One table for each connective (What is the table for ⊕?)

¬

1 00 1

∧ 1 01 1 00 0 0

∨ 1 01 1 10 1 0

→ 1 01 1 00 1 1

↔ 1 01 1 00 0 1

Even More Notations

◮ If I satisfies φ, then we call I a model of φ◮ Mod(ψ) is the set of all models of ψ◮ φ is satisfiable (valid) if φ is true in some (all) interpretations◮ φ is unsatisfiable if φ is false in all interpretations◮ Formulas φ and ψ are equivalent, denoted by φ ≡ ψ, iff

they have exactly the same models, i.e., Mod(φ) = Mod(ψ)

◮ Example: (φ→ ψ) ≡ (¬ψ → ¬φ): Take I ∈ Mod(φ→ ψ)I(φ→ ψ) = 1 iff I(φ) = 0 or I(ψ) = 1

iff I(¬φ) = 1 or I(ψ) = 1iff I(¬φ) = 1 or I(¬ψ) = 0iff I(¬ψ → ¬φ) = 1

Some Useful Equivalences

Commutativity φ ◦ ψ ≡ ψ ◦ φ for ◦ ∈ {∨,∧,↔}Idempotence φ ◦ φ ≡ φ for ◦ ∈ {∨,∧}Tautology φ ∨ ⊤ ≡ ⊤Unsatisfiability φ ∧ ⊥ ≡ ⊥Neutrality φ ∧ ⊤ ≡ φ

φ ∨ ⊥ ≡ φ

Negation φ ∨ ¬φ ≡ ⊤φ ∧ ¬φ ≡ ⊥

Double Negation ¬¬φ ≡ φ

Implication φ→ ψ ≡ ¬φ ∨ ψ

Some Useful Equivalences Cont’d

De Morgan ¬(φ ∨ ψ) ≡ ¬φ ∧ ¬ψ¬(φ ∧ ψ) ≡ ¬φ ∨ ¬ψ

Absorption φ ∨ (φ ∧ ψ) ≡ φ

φ ∧ (φ ∨ ψ) ≡ φ

Distributivity φ ∧ (ψ ∨ χ) ≡ (φ ∧ ψ) ∨ (φ ∧ χ)φ ∨ (ψ ∧ χ) ≡ (φ ∨ ψ) ∧ (φ ∨ χ)

Associativity φ ∨ (ψ ∨ χ) ≡ (φ ∨ ψ) ∨ χφ ∧ (ψ ∧ χ) ≡ (φ ∧ ψ) ∧ χ

Connections Between the Different Notations

◮ Distinguish between◮ tautologies: all interpretations are models◮ satisfiable formulas: some interpretations are models◮ contradictions: no interpretation is a model

◮ A formula φ is valid iff ¬φ is unsatisfiableRecall that this is the basis for a “proof by contradiction”

◮ A formula φ is satisfiable iff ¬φ is not valid◮ Two formulas φ and ψ are equivalent iff φ↔ ψ is valid◮ A formula φ is valid iff φ is equivalent to ⊤◮ A formula φ is unsatisfiable iff φ is equivalent to ⊥

Entailment (or Logical Implication)

◮ So far, |= relates an interpretation and a formula◮ Allow also a set of formulas on the left side◮ Important: a set of formulas coincides with the conjunction

of its elements, i.e., {φ1, . . . , φn} is∧n

i=1 φi

◮ Important: an empty conjunction is 1 in all interpretationsi.e., it is equivalent to ⊤

◮ W entails φ, W |= φ, iff Mod(W ) ⊆ Mod(φ)

◮ W |= φ iff I |= φ for all models I of W (for all I ∈ Mod(W ))

Properties of Entailment

◮ W |= ψ implies W ∪ {φ} |= ψ Monotonicity for PL0◮ W ∪ {φ} |= ψ iff W |= φ→ ψ Deduction Thm◮ W ∪ {φ} |= ¬ψ iff W ∪ {ψ} |= ¬φ Contraposition Thm◮ W ∪ {φ} is unsatisfiable iff W |= ¬φ Contradiction Thm

Important Questions to Answer

Q1 How can we algorithmicly check whether I |= φ holds?➥ Evaluation of a formula φ under a given interpretation I

Q2 When does W |= φ hold?Q2’ When does |= φ hold?

◮ This is a special case of Q2 with empty W !◮ Q2’ asks: when is φ a tautology (1 in all interpretations)?

Q3 Can we define deduction (based on syntactic rules), s.t.the results of deduction and entailment coincide?

◮ Write W ⊢i φ if φ is derived from W by deduction proc. i◮ i is sound: if W ⊢i φ holds, then W |= φ also holds◮ i is complete: if W |= φ holds, then W ⊢i φ also holds

Evaluation of a Formula Under an Interpretation

Evaluate φ : (p → q) ∧ (p ∧ q → r)→ (p → r) underI = {p 7→ 1,q 7→ 0, r 7→ 1} starting with r

formula value1 (p → q) ∧ (p ∧ q → r)→ (p → r) 12 p → r 13 (p → q) ∧ (p ∧ q → r) 04 p ∧ q → r 15 p → q 06 p ∧ q 07 p p p 18 q q 09 r r 1

Try other evaluation orders!

Evaluation of a Formula Under an Interpretation

Evaluate φ : (p → q) ∧ (p ∧ q → r)→ (p → r) underI = {p 7→ 1,q 7→ 0, r 7→ 1} starting with r

formula value1 (p → q) ∧ (p ∧ q → r)→ (p → r) 12 p → r 13 (p → q) ∧ (p ∧ q → r) 04 p ∧ q → r 15 p → q 06 p ∧ q 07 p p p 18 q q 09 r r 1

Try other evaluation orders!

Evaluation of a Formula Under an Interpretation

Evaluate φ : (p → q) ∧ (p ∧ q → r)→ (p → r) underI = {p 7→ 1,q 7→ 0, r 7→ 1} starting with r

formula value1 (p → q) ∧ (p ∧ q → r)→ (p → r) 12 p → r 13 (p → q) ∧ (p ∧ q → r) 04 p ∧ q → r 15 p → q 06 p ∧ q 07 p p p 18 q q 09 r r 1

Try other evaluation orders!

Evaluation of a Formula Under an Interpretation

Evaluate φ : (p → q) ∧ (p ∧ q → r)→ (p → r) underI = {p 7→ 1,q 7→ 0, r 7→ 1} starting with r

formula value1 (p → q) ∧ (p ∧ q → r)→ (p → r) 12 p → r 13 (p → q) ∧ (p ∧ q → r) 04 p ∧ q → r 15 p → q 06 p ∧ q 07 p p p 18 q q 09 r r 1

Try other evaluation orders!

How to Check Entailments W |= φ and |= φ

How can we use the evaluation procedure to answer Q2’?

◮ Consider formula φ with {p1, . . . ,pn} as prop. variables◮ Generate all 2n I over these n variables one after the other◮ Test each of the generated interpretation I

◮ Evaluate with the above procedure φ under the current I◮ If I(φ) = 0 then STOP and answer “φ is not a tautology”

◮ If I(φ) = 1 for all I then STOP and answer “φ is a tautology”

How is Q2 handled (i.e., W |= φ)?

◮ Assume W is {ψ1, . . . , ψm} and apply the Deduction Thm◮ Obtain, e.g., |= ψ1 → (ψ2 → (· · · (ψm → φ) · · · )

Decision Procedure for PL0

◮ Check proc. is based on enumeration of interpretations◮ For n atoms in the input formula: 2n interpretations possible◮ Worst case: all 2n interpretations have to be generated

◮ Check procedure for validity is a Decision Procedure◮ Possible answer is either yes or no◮ Decides entailment (and validity)◮ Decides satisfiability by reduction approach:φ is satisfiable iff ¬φ is not valid (i.e., ¬φ is not a tautology)

◮ Can we do better/differently?

Reduction to Satisfiability

Reduce validity, entailment, equivalence to satisfiability

1 Validity◮ ¬φ is unsatisfiable iff φ is valid

2 Entailment◮ φ entails ψ (φ |= ψ) iff φ→ ψ is valid (apply Deduction Thm)◮ Hence, φ |= ψ iff φ ∧ ¬ψ (i.e., ¬(φ→ ψ)) is unsatisfiable

3 Equivalence◮ φ is equivalent to ψ (φ ≡ ψ) iff φ↔ ψ is valid◮ Hence, φ ≡ ψ iff φ |= ψ and ψ |= φ hold◮ Consequently, φ ≡ ψ iff φ ∧ ¬ψ and ψ ∧ ¬φ are unsatisfiable

Sound and complete procedure for satisfiability is sufficient!

Table of Synonym Notions

All four statements in each line amount the same

entailment(s) validity satisfiability equivalenceφ |= ψ φ→ ψ valid φ ∧ ¬ψ unsat (φ→ ψ) ≡ ⊤

⊤ |= ψ ψ valid ¬ψ unsat ψ ≡ ⊤

⊤ 6|= ¬ψ ¬ψ not valid ψ sat ¬ψ 6≡ ⊤

φ |= ψ and ψ |= φ φ↔ ψ valid φ↔ ¬ψ unsat φ ≡ ψ

Normal Forms: NNF and CNF

◮ Restricted syntax, but retain the important properties(application-dependent, e.g., provability)

◮ Negation Normal Form (NNF): A formula in NNF has◮ negation signs only in front of atoms and◮ the only connectives are ∧ and ∨

◮ Conjunctive Normal Form (CNF): A formula in CNF◮ consists of a conjuction of clauses (=disjunction of literals)

(often represented as a set of clauses)◮ Clauses sometimes represented as sets of literals

(can cause problems)

◮ Every PL0-formula has an equivalent formula in NNF/CNF!◮ But how do we construct them?

Preparatory Concept: Tree Replacements

◮ Let φ[ψ] denote that ψ occurs in φ◮ This means that ψ occurs as zero, one or more subtree(s)

in the formula tree of φ◮ Possibility to perform subtree replacements◮ Example: φ : ¬(p ∧ q) ∨ r and ψ1 : ¬(p ∧ q)

◮ Then φ[ψ1] indicates occurrence(s) of ψ1 in φ◮ Construct φ[ψ2] with ψ2 : ¬p ∨ ¬q by a tree replacement

∨

¬

∧

p q

r=⇒

∨

∨

¬

p

¬

q

r

Equivalent Replacement

Lemma (Equivalent Replacement Lemma)Let I be an interpretation and I |= ψ1 ↔ ψ2. ThenI |= φ[ψ1]↔ φ[ψ2].

Theorem (Equivalent Replacement Theorem)Let ψ1 ≡ ψ2. Then φ[ψ1] ≡ φ[ψ2].

Basics of the NNF Translation

◮ Replace↔ by→ using (φ↔ ψ) ≡ ((φ→ ψ) ∧ (ψ → φ))

◮ Replace ⊕ using (φ⊕ ψ) ≡ ((φ ∨ ψ) ∧ (¬φ ∨ ¬ψ)) (why?)◮ Replace→ using (φ→ ψ) ≡ (¬φ ∨ ψ)

◮ Replace the left side of the following equivalences by theright side (order does not matter!)

¬(φ ∨ ψ) ≡ ¬φ ∧ ¬ψ ¬(φ ∧ ψ) ≡ ¬φ ∨ ¬ψφ ∨ ⊤ ≡ ⊤ ⊤ ∨ φ ≡ ⊤φ ∧ ⊥ ≡ ⊥ ⊥ ∧ φ ≡ ⊥φ ∧ ⊤ ≡ φ ⊤ ∧ φ ≡ φ

φ ∨ ⊥ ≡ φ ⊥ ∨ φ ≡ φ

¬¬φ ≡ φ

◮ Translation process is terminating with the NNF◮ For any PL0-formulas λ, nnf(λ)↔ λ holds

Basics of the CNF Translation

◮ Based on the application of distributivity laws◮ Start with the formula φ and translate it to NNF◮ Take nnf(φ) and replace the left side of the following

equivalences by the right side (order does not matter!)

1 φ ∨ (ψ ∧ χ) ≡ (φ ∨ ψ) ∧ (φ ∨ χ)2 (ψ ∧ χ) ∨ φ ≡ (ψ ∨ φ) ∧ (χ ∨ φ)

◮ Observe that nnf(φ) ≡ cnf(φ) ≡ φ holds

Exa: Transform φ : (p ∧ q → r)→ (q → r) to CNF

nnf(φ) : (p ∧ q ∧ ¬r) ∨ (¬q ∨ r) (why?)

Formula Rule(p ∧ q ∧ ¬r) ∨ (¬q ∨ r) 2(p ∨ ¬q ∨ r) ∧ (q ∧ ¬r ∨ (¬q ∨ r)) 2(p ∨ ¬q ∨ r) ∧ (q ∨ ¬q ∨ r) ∧ (¬r ∨ ¬q ∨ r)

Two disadvantages:

1. Disruption of the formula’s structure

2. cnf(φ) can be exponentially longer than φTake a formula in disjunctive NF and translate it to CNF!

Can be avoided: use definitional translation ( circuit exa)

The Method of Davis, Logeman and Loveland

◮ M. Davis, G. Logeman, D. Loveland A machine program fortheorem proving, CACM, Vol. 5, No. 7, pp. 394-397, 1962

◮ Basic framework for all modern SAT solvers(require additional features like learning andnon-chronological backtracking to be competitive)

◮ DLL is search-based (rather than resolution-based as DP)◮ Basic Idea: Given a clause set C

◮ Try all (partial) assignments and test whether C is sat◮ Stop enlarging the current partial assignment I when there

is a conflict (i.e., C cannot be satisfied with extensions of I)◮ Stop if I satisfying C is found or all possible I have been tried

The Basic DLL Procedure: An Example

a′ ∨ b ∨ ca ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ da ∨ c′ ∨ d ′

b′ ∨ c′ ∨ da′ ∨ b ∨ c′

a′ ∨ b′ ∨ c

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ ca ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ da ∨ c′ ∨ d ′

b′ ∨ c′ ∨ da′ ∨ b ∨ c′

a′ ∨ b′ ∨ c

a

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ c ←a ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ da ∨ c′ ∨ d ′

b′ ∨ c′ ∨ da′ ∨ b ∨ c′ ←a′ ∨ b′ ∨ c ←

a0

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ c ←a ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ da ∨ c′ ∨ d ′

b′ ∨ c′ ∨ da′ ∨ b ∨ c′ ←a′ ∨ b′ ∨ c ←

a

b0

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ c ←a ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ da ∨ c′ ∨ d ′

b′ ∨ c′ ∨ d ←a′ ∨ b ∨ c′ ←a′ ∨ b′ ∨ c ←

a

b0

0

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ c ←a ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ da ∨ c′ ∨ d ′

b′ ∨ c′ ∨ d ←a′ ∨ b ∨ c′ ←a′ ∨ b′ ∨ c ←

a

b0

c0

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ c ←a ∨ c ∨ d ←a ∨ c ∨ d ′ ←a ∨ c′ ∨ d ←a ∨ c′ ∨ d ′ ←b′ ∨ c′ ∨ d ←a′ ∨ b ∨ c′ ←a′ ∨ b′ ∨ c ←

a

b0

c0

0

Conflict between← clauses wrt d and d ′

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ c ←a ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ d ←a ∨ c′ ∨ d ′ ←b′ ∨ c′ ∨ d ←a′ ∨ b ∨ c′ ←a′ ∨ b′ ∨ c ←

a

b0

c0

�

0

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ c ←a ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ d ←a ∨ c′ ∨ d ′ ←b′ ∨ c′ ∨ d ←a′ ∨ b ∨ c′ ←a′ ∨ b′ ∨ c ←

a

b0

c0

�

0 1

Conflict between← clauses wrt d and d ′

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ c ←a ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ da ∨ c′ ∨ d ′

b′ ∨ c′ ∨ d ←a′ ∨ b ∨ c′ ←a′ ∨ b′ ∨ c ←

a

b0

c0

�

0�

1

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ ca ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ da ∨ c′ ∨ d ′

b′ ∨ c′ ∨ da′ ∨ b ∨ c′

a′ ∨ b′ ∨ c

a

b0

c0

�

0�

1c1

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0�

1

b1

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0�

1

I(c) = I(d) = 1 forced by I(a) = I(b) = 1 by UNIT

p′ means ¬p

The Basic DLL Procedure: An Example

a′ ∨ b ∨ ca ∨ c ∨ da ∨ c ∨ d ′

a ∨ c′ ∨ da ∨ c′ ∨ d ′

b′ ∨ c′ ∨ da′ ∨ b ∨ c′

a′ ∨ b′ ∨ c

a

b0

c0

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0�

1c1

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0�

1

b1

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0�

1

Easy to check: I(a) = I(b) = I(c) = I(d) = 1 is a model

p′ means ¬p

The Basic DLL Procedure

◮ Inference rules in DLL:◮ UNIT: In an unsatisfied clause C ∈ C (by the current partial

assignment), if there is only one literal unassigned, then Cis unit (Value of unit clause is implied: Make the literal true!)

◮ Boolean Constraint Propagation (BCP)Apply UNIT until no more unit clause is available in C

◮ PURE: (Seldom used in practice)If a variable occurs in one polarity only: Make its literal true

The Basic DLL Procedure: The Destructive Version

◮ Sometimes, “destructive” version of DLL is considered(seen in former lecture on theoretical CS)

◮ Apply assignment of variable to formula and simplify◮ Simplification: Use rules below to eliminate truth constants

φ ∨ ⊤ ≡ ⊤ ⊤ ∨ φ ≡ ⊤φ ∧ ⊥ ≡ ⊥ ⊥ ∧ φ ≡ ⊥φ ∧ ⊤ ≡ φ ⊤ ∧ φ ≡ φ

φ ∨ ⊥ ≡ φ ⊥ ∨ φ ≡ φ

◮ Two possible results on clause set C◮ Delete clauses with true literal from C◮ Delete false literal from clauses in C

The Features of the DLL Procedure

◮ Polynomial memory requirement (exponential in DP!)◮ Exponential time requirement remains

(except for restricted inputs like 2-CNF)

◮ Chronological backtrackingOk for random instances, bad for practical problems w. structure

◮ Need extensions to be of practical value◮ Heuristics to select variables to branch next

(including the assignment to consider first)◮ “Intelligent” (non-chronological) backtracking◮ Learning from successful and unsuccessful decisions