propositional logic resolution, dpll and tableau

Propositional LogicResolution, DPLL and tableau

Mario Alviano

University of Calabria, Italy

A.Y. 2014/2015

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Outline

1 Propositional resolutionResolutionRefutationsRefinements and examples

2 DPLL

3 Propositional Tableau

4 Exercises

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Outline

2 DPLL

4 Exercises

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Outline

2 DPLL

4 Exercises

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Propositional resolution (1)

Main observation

(a ∨ b) ∧ (¬a ∨ c) ≡ (a ∨ b) ∧ (¬a ∨ c) ∧ (b ∨ c)

(a ∨ b) ∧ (¬a ∨ c) |= (b ∨ c)

More general

C1 ∧ · · · ∧ Ck ∧ (a ∨ L11 ∨ . . . ∨ L1

n) ∧ (¬a ∨ L21 ∨ . . . ∨ L2

m)|= (L1

1 ∨ . . . ∨ L1n ∨ L2

1 ∨ . . . ∨ L2m)

This is known as resolution!

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Main observation

(a ∨ b) ∧ (¬a ∨ c) ≡ (a ∨ b) ∧ (¬a ∨ c) ∧ (b ∨ c)

(a ∨ b) ∧ (¬a ∨ c) |= (b ∨ c)

More general

C1 ∧ · · · ∧ Ck ∧ (a ∨ L11 ∨ . . . ∨ L1

n) ∧ (¬a ∨ L21 ∨ . . . ∨ L2

m)|= (L1

1 ∨ . . . ∨ L1n ∨ L2

1 ∨ . . . ∨ L2m)

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Main observation

(a ∨ b) ∧ (¬a ∨ c) ≡ (a ∨ b) ∧ (¬a ∨ c) ∧ (b ∨ c)

(a ∨ b) ∧ (¬a ∨ c) |= (b ∨ c)

More general

C1 ∧ · · · ∧ Ck ∧ (a ∨ L11 ∨ . . . ∨ L1

n) ∧ (¬a ∨ L21 ∨ . . . ∨ L2

m)|= (L1

1 ∨ . . . ∨ L1n ∨ L2

1 ∨ . . . ∨ L2m)

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Main observation

(a ∨ b) ∧ (¬a ∨ c) ≡ (a ∨ b) ∧ (¬a ∨ c) ∧ (b ∨ c)

(a ∨ b) ∧ (¬a ∨ c) |= (b ∨ c)

More general

C1 ∧ · · · ∧ Ck ∧ (a ∨ L11 ∨ . . . ∨ L1

n) ∧ (¬a ∨ L21 ∨ . . . ∨ L2

m)|= (L1

1 ∨ . . . ∨ L1n ∨ L2

1 ∨ . . . ∨ L2m)

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Additional observation

(a ∨ a ∨ B) ≡ (a ∨ B)

Remove duplicated literals in clausesIt is known as factorization

Resolution (set notation)

C1, . . . ,Ck , {a ∨ L11, . . . ,L

1n}, {¬a,L2

1, . . . ,L2m}

|= {L11, . . . ,L

1, . . . ,L2m}

Factorization comes “for free”!

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(a ∨ a ∨ B) ≡ (a ∨ B)

C1, . . . ,Ck , {a ∨ L11, . . . ,L

1n}, {¬a,L2

1, . . . ,L2m}

|= {L11, . . . ,L

1, . . . ,L2m}

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(a ∨ a ∨ B) ≡ (a ∨ B)

C1, . . . ,Ck , {a ∨ L11, . . . ,L

1n}, {¬a,L2

1, . . . ,L2m}

|= {L11, . . . ,L

1, . . . ,L2m}

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Derivation by resolution (1)

Resolvent

Given two clauses C1 and C2 such that a ∈ C1 and ¬a ∈ C2,(C1 \ {a}) ∪ (C2 \ {¬a}) is the resolvent of C1 and C2.

Derivation

Given a set Γ of clauses, a derivation by resolution of a clauseC from Γ, denoted Γ `R C, is a sequence C1, . . . ,Cn such thatCn = C and for each Ci (1 ≤ i ≤ n) we have

1 Ci ∈ Γ, or2 Ci is a resolvent of Cj and Ck , where j < i and k < i .

If Γ `R C then Γ |= C.

Proof. By induction on the sequence C1, . . . ,Cn.

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Resolvent

Derivation

Proof. By induction on the sequence C1, . . . ,Cn.

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Resolvent

Derivation

Proof. By induction on the sequence C1, . . . ,Cn.7 / 36

Derivation by resolution (2)Consider

Γ = {rain→ streetwet , rain}

, or equivalentlyΓ = {¬rain ∨ streetwet , rain}, or equivalentlyΓ = {{¬rain, streetwet}, {rain}}

The following is a derivation by resolution:

C1 = {¬rain, streetwet}C2 = {rain}C3 = {streetwet}

{¬rain, streetwet} {rain}

{streetwet}

rain→ streetwet , rain `R streetwet

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Γ = {rain→ streetwet , rain}, or equivalentlyΓ = {¬rain ∨ streetwet , rain}

, or equivalentlyΓ = {{¬rain, streetwet}, {rain}}

{streetwet}

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Γ = {rain→ streetwet , rain}, or equivalentlyΓ = {¬rain ∨ streetwet , rain}, or equivalentlyΓ = {{¬rain, streetwet}, {rain}}

{streetwet}

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{streetwet}

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The following is a derivation by resolution:C1 = {¬rain, streetwet}

C2 = {rain}C3 = {streetwet}

{¬rain, streetwet}

{rain}

{streetwet}

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The following is a derivation by resolution:C1 = {¬rain, streetwet}C2 = {rain}

C3 = {streetwet}

{streetwet}

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The following is a derivation by resolution:C1 = {¬rain, streetwet}C2 = {rain}C3 = {streetwet}

{streetwet}

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The following is a derivation by resolution:C1 = {¬rain, streetwet}C2 = {rain}C3 = {streetwet}

{streetwet}

rain→ streetwet , rain `R streetwet8 / 36

Outline

2 DPLL

4 Exercises

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Refutation (1)

Our goal is to model Γ |= ⊥

Let � be the empty clause

� is like ⊥� is different from an empty set of formulas!

Refutation

A derivation by resolution of � from Γ is called a refutation of Γ.

Resolution Theorem

Γ `R � if and only if Γ is unsatisfiable.

Proof. Soundness: Γ `R � implies Γ |= � (by the previousLemma).Completeness: by induction over the number of variables in Γ.

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Refutation (1)

Our goal is to model Γ |= ⊥Let � be the empty clause

Refutation

Resolution Theorem

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Refutation (1)

� is like ⊥

� is different from an empty set of formulas!

Refutation

Resolution Theorem

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Refutation (1)

Refutation

Resolution Theorem

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Refutation (1)

Refutation

Resolution Theorem

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Refutation (1)

Refutation

Resolution Theorem

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Refutation (2)Goal: {rain→ streetwet , rain} |= streetwet

IFF: {rain→ streetwet , rain} ∪ {¬streetwet} |= ⊥

The following is a refutation of{rain→ streetwet , rain, ¬streetwet}:

C1 = {¬rain, streetwet}C2 = {rain}C3 = {¬streetwet}

C4 = {streetwet}C5 = {} = �

{¬rain, streetwet} {rain} {¬streetwet}

{streetwet}

rain→ streetwet , rain, ¬streetwet `R �

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Refutation (2)Goal: {rain→ streetwet , rain} |= streetwetIFF: {rain→ streetwet , rain} ∪ {¬streetwet} |= ⊥

{streetwet}

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{streetwet}

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C1 = {¬rain, streetwet}

C2 = {rain}C3 = {¬streetwet}

{¬rain, streetwet}

{rain} {¬streetwet}

{streetwet}

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C1 = {¬rain, streetwet}C2 = {rain}

C3 = {¬streetwet}

{¬streetwet}

{streetwet}

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{streetwet}

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C4 = {streetwet}

C5 = {} = �

{streetwet}

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{streetwet}

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{streetwet}

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Validity, equivalence, entailment, ...Validity

|= φ iff ¬φ |= ⊥Test whether ¬φ `R �

Equivalence

φ ≡ ψ iff ¬(φ↔ ψ) |= ⊥Test whether ¬(φ↔ ψ) `R �

Entailment

Γ |= φ iff Γ, ¬φ |= ⊥Test whether Γ, ¬φ `R �

Satisfiability

φ is satisfiable iff ¬φ is not validTest whether φ 6`R �

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Equivalence

Entailment

Satisfiability

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Equivalence

Entailment

Satisfiability

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Equivalence

Entailment

Satisfiability

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Implementing resolution (1)Algorithm: SAT by resolutionInput : a set Γ of wffsOutput: true if Γ is SAT; false otherwise

1 begin2 ΓCNF := trasformToCNF (Γ);3 repeat4 if � ∈ ΓCNF then5 return false;

6 Γold := ΓCNF ;7 ΓCNF := ΓCNF ∪ resolveAll(ΓCNF );8 until Γold = ΓCNF ;9 return true;

Function resolveAll(ΓCNF : set of clauses)1 begin2 Γres := ∅;3 foreach C1 ∈ ΓCNF and foreach atom a ∈ C1 do4 foreach C2 ∈ ΓCNF such that ¬a ∈ C2 do5 Γres := Γres ∪ {C1 \ {a} ∪ C2 \ {¬a}};

6 return Γres

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Implementing resolution (2)

Complexity

Deciding Γ `R � requires up to an exponential number of steps(with respect to the size of the formula)

Since unsatisfiability of a formula is coNP-complete,this is “reasonable”

Example

Is the following formula satisfiable?(A ∨ B) ∧ (A↔ B) ∧ (¬A ∨ ¬B)

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Complexity

Example

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Complexity

Example

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Outline

2 DPLL

4 Exercises

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Refinements (1)

Drop tautological clauses

A clause C is a tautology if there is a ∈ V such thata ∈ C and ¬a ∈ C

Drop subsumed clauses

A clause C1 subsumes a clause C2 if C1 ⊆ C2

Linear Resolution

Any intermediate derivation uses the clause obtained in theprevious step.

Theorem

Linear resolution is refutation complete: If a set of wffs isunsatisfiable then a refutation by linear resolution exists.

{{A,B}, {A,¬B}, {¬A,B}, {¬A,¬B}}

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Refinements (1)

Linear Resolution

Theorem

{{A,B}, {A,¬B}, {¬A,B}, {¬A,¬B}}

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Refinements (1)

Linear Resolution

Theorem

{{A,B}, {A,¬B}, {¬A,B}, {¬A,¬B}}

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Refinements (1)

Linear Resolution

Theorem

{{A,B}, {A,¬B}, {¬A,B}, {¬A,¬B}}

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Refinements (1)

Linear Resolution

Theorem

{{A,B}, {A,¬B}, {¬A,B}, {¬A,¬B}}16 / 36

Refinements (2)

Linear Input Resolution

Any intermediate derivation uses the clause obtained in theprevious step and a clause of the original formula.

Theorem

Linear input resolution is refutation complete for (sets of) Hornclauses, where a Horn clause is a clause containing at mostone positive atom.

Example

1 {{A,B}, {A,¬B}, {¬A,B}, {¬A,¬B}}2 {{A}, {B}, {A,¬B}, {¬A,B}, {¬A,¬B}}

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Refinements (2)

Theorem

Example

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Refinements (2)

Theorem

Example

1 {{A,B}, {A,¬B}, {¬A,B}, {¬A,¬B}}

2 {{A}, {B}, {A,¬B}, {¬A,B}, {¬A,¬B}}

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Refinements (2)

Theorem

Example

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Examples

1 Is (A1 ∨ A2) ∧ (¬A2 ∨ ¬A3) ∧ (A3 ∨ A4) ∧ (¬A4 ∨ ¬A1)satisfiable?

2 Does A follow from (A ∨ B ∨ C) ∧ (¬C ∨ B) ∧ (A ∨ ¬B)?3 Does ¬A follow from (A ∨ B ∨ C) ∧ (¬C ∨ B) ∧ (A ∨ ¬B)?4 Does A∧B follow from (¬A→ B)∧ (A→ B)∧ (¬A→ ¬B)?

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Outline

2 DPLL

4 Exercises

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DPLL algorithm

By ¬` we denote the opposite literal of `if ` = ¬a then ¬` = a

Simplify is often called Unit PropagationCall-by-name parameters (references)

Algorithm: DPLLInput : a set Γ of clausesOutput: true if Γ is SAT; false otherwise

1 begin2 Simplify(Γ);3 if Γ = ∅ then return true ;4 if � ∈ Γ then return false ;5 ` := ChooseLiteral(Γ);6 return DPLL(Γ ∪ {`}) or DPLL(Γ ∪ {¬`});

Procedure Simplify(Γ)1 begin2 while {`} ∈ Γ do3 foreach c ∈ Γ do4 if ` ∈ c then Γ := Γ \ {c} ;5 else if ¬` ∈ c then Γ := (Γ \ {c}) ∪ (c \ {¬`}) ;

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Properties

DPLL(Γ) returns true if Γ is satisfiable, and false otherwise

DPLL(Γ) can be (easily) modified in order to compute one(or all) models of Γ

DPLL is sound and complete

DPLL(Γ) works in polynomial-space

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Properties

DPLL(Γ) returns true if Γ is satisfiable, and false otherwiseDPLL(Γ) can be (easily) modified in order to compute one(or all) models of Γ

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Properties

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Properties

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Features

Simplification

The input set of clauses is simplified at each branch using (atleast) unit clause propagation

Branching

When no further simplification is possible, a literal is selectedusing some heuristic criterion (ChooseLiteral) and assumed asa unit clause in the current set of clauses

Backtracking

When a contradiction (empty clause) arises, the searchresumes from some previous assumption ` by assuming ¬`instead

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Features

Simplification

Branching

Backtracking

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Features

Simplification

Branching

Backtracking

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}:

Γ = {{x1,¬x3}, {x4,¬x3}}ChooseLiteral returns ¬x3: Γ = ∅, i.e., the formula is SAT!

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}: Γ = {{x1,¬x3,¬x4}, {x4,¬x3}}If ChooseLiteral returns ¬x3, the process is the same asbefore; otherwise, if x1 is returned, another choice has tobe made

Branching order (ChooseLiteral) can make big differences!

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}: Γ = {{x1,¬x3}, {x4,¬x3}}

ChooseLiteral returns ¬x3: Γ = ∅, i.e., the formula is SAT!

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}: Γ = {{x1,¬x3}, {x4,¬x3}}ChooseLiteral returns ¬x3:

Γ = ∅, i.e., the formula is SAT!

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}: Γ = {{x1,¬x3}, {x4,¬x3}}ChooseLiteral returns ¬x3: Γ = ∅, i.e., the formula is SAT!

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}:

Γ = {{x1,¬x3,¬x4}, {x4,¬x3}}If ChooseLiteral returns ¬x3, the process is the same asbefore; otherwise, if x1 is returned, another choice has tobe made

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}: Γ = {{x1,¬x3,¬x4}, {x4,¬x3}}

If ChooseLiteral returns ¬x3, the process is the same asbefore; otherwise, if x1 is returned, another choice has tobe made

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}: Γ = {{x1,¬x3,¬x4}, {x4,¬x3}}If ChooseLiteral returns ¬x3,

the process is the same asbefore; otherwise, if x1 is returned, another choice has tobe made

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}: Γ = {{x1,¬x3,¬x4}, {x4,¬x3}}If ChooseLiteral returns ¬x3, the process is the same asbefore;

otherwise, if x1 is returned, another choice has tobe made

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

Simplify using {¬x2}: Γ = {{x1,¬x3,¬x4}, {x4,¬x3}}If ChooseLiteral returns ¬x3, the process is the same asbefore; otherwise, if x1 is returned,

another choice has tobe made

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

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Examples (1)

Example

Γ = {{x1, x2,¬x3}, {¬x2}, {x4,¬x3}}

Example

Γ = {{x1, x2,¬x3,¬x4}, {¬x2}, {x4,¬x3}}

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Examples (2)

1 {{x1, x2, x3}, {x1, x2,¬x3}, {x1,¬x2, x3},{x1,¬x2,¬x3}, {¬x1, x4}, {x1,¬x4,¬x5, x6}, {¬x1, x7}}

2 {{x1, x2}, {x1,¬x2}, {¬x1, x2}, {¬x1,¬x2}}

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Examples (2)

1 {{x1, x2, x3}, {x1, x2,¬x3}, {x1,¬x2, x3},{x1,¬x2,¬x3}, {¬x1, x4}, {x1,¬x4,¬x5, x6}, {¬x1, x7}}

2 {{x1, x2}, {x1,¬x2}, {¬x1, x2}, {¬x1,¬x2}}

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Refinements

How to efficiently detect unit clauses?

2-watched literalsHow to implement ChooseLiteral?

Look-ahead heuristicsLook-back heuristics

How to take advantage from conflicts?

LearningBackjumping

Can we reuse something from a previous computation?

Progressive SAT

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Refinements

How to efficiently detect unit clauses?2-watched literals

How to implement ChooseLiteral?

LearningBackjumping

Progressive SAT

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Refinements

How to implement ChooseLiteral?

LearningBackjumping

Progressive SAT

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Refinements

How to implement ChooseLiteral?Look-ahead heuristics

Look-back heuristicsHow to take advantage from conflicts?

LearningBackjumping

Progressive SAT

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Refinements

How to implement ChooseLiteral?Look-ahead heuristicsLook-back heuristics

LearningBackjumping

Progressive SAT

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Refinements

LearningBackjumping

Progressive SAT

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Refinements

How to take advantage from conflicts?Learning

BackjumpingCan we reuse something from a previous computation?

Progressive SAT

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Refinements

How to take advantage from conflicts?LearningBackjumping

Progressive SAT

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Refinements

Progressive SAT

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Refinements

Can we reuse something from a previous computation?Progressive SAT

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Outline

2 DPLL

4 Exercises

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Propositional Tableau (1)

Yet another decision procedure for unsatisfiability

The idea is to break down logical connectives into theirconstituent partsWe consider the tableau for formulas in Negation NormalForm (NNF)

Negation Normal Form

1 Eliminate >, ⊥,→,↔2 Apply double negation equivalences:

¬¬φ ≡ φ3 Apply De Morgan equivalences:

¬(φ ∨ ψ) ≡ ¬ψ ∧ ¬φ¬(φ ∧ ψ) ≡ ¬ψ ∨ ¬φ

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Yet another decision procedure for unsatisfiabilityThe idea is to break down logical connectives into theirconstituent parts

We consider the tableau for formulas in Negation NormalForm (NNF)

¬(φ ∨ ψ) ≡ ¬ψ ∧ ¬φ¬(φ ∧ ψ) ≡ ¬ψ ∨ ¬φ

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Yet another decision procedure for unsatisfiabilityThe idea is to break down logical connectives into theirconstituent partsWe consider the tableau for formulas in Negation NormalForm (NNF)

¬(φ ∨ ψ) ≡ ¬ψ ∧ ¬φ¬(φ ∧ ψ) ≡ ¬ψ ∨ ¬φ

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Yet another decision procedure for unsatisfiabilityThe idea is to break down logical connectives into theirconstituent partsWe consider the tableau for formulas in Negation NormalForm (NNF)

¬(φ ∨ ψ) ≡ ¬ψ ∧ ¬φ¬(φ ∧ ψ) ≡ ¬ψ ∨ ¬φ

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Propositional Tableau (2)Let Γ be a set of wffs in NNF.

1 Start with a chain made of the formulas in Γ

2 If a branch contains a formula φ ∧ ψ then add to its leaf thechain φ — ψ

3 If a branch contains a formula φ ∨ ψ then add to its leaf twochildren φ and ψ

4 If a branch contains a formula and its negation then closethe branch

5 If all branches are closed then Γ is unsatisfiable

Notation

φ ∧ ψφψ

(∧)φ ∨ ψφ | ψ

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Notation

φ ∧ ψφψ

(∧)φ ∨ ψφ | ψ

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Notation

φ ∧ ψφψ

(∧)φ ∨ ψφ | ψ

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Notation

φ ∧ ψφψ

(∧)φ ∨ ψφ | ψ

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Notation

φ ∧ ψφψ

(∧)φ ∨ ψφ | ψ

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Notation

φ ∧ ψφψ

(∧)φ ∨ ψφ | ψ

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Example: Γ := {(A ∨ ¬B) ∧ B,¬A}

(A ∨ ¬B) ∧ B

A ∨ ¬B

Γ is unsatisfiable!

We can simplifythe notation usedfor a closure byputting a markeron the leaf of theclosed path!

On the blackboard

A ∧ C,¬A ∨ B

Theorem

Propositional tableau is sound and complete, i.e.,if the tableau for Γ is closed then Γ |= ⊥, andif Γ |= ⊥ then the tableau for Γ is closed.

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Example: Γ := {(A ∨ ¬B) ∧ B,¬A}

(A ∨ ¬B) ∧ B

A ∨ ¬B

On the blackboard

A ∧ C,¬A ∨ B

Theorem

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Example: Γ := {(A ∨ ¬B) ∧ B,¬A}

(A ∨ ¬B) ∧ B

A ∨ ¬B

On the blackboard

A ∧ C,¬A ∨ B

Theorem

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Example: Γ := {(A ∨ ¬B) ∧ B,¬A}

(A ∨ ¬B) ∧ B

A ∨ ¬B

On the blackboard

A ∧ C,¬A ∨ B

Theorem

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Example: Γ := {(A ∨ ¬B) ∧ B,¬A}

(A ∨ ¬B) ∧ B

A ∨ ¬B

On the blackboard

A ∧ C,¬A ∨ B

Theorem

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Example: Γ := {(A ∨ ¬B) ∧ B,¬A}

(A ∨ ¬B) ∧ B

A ∨ ¬B

On the blackboard

A ∧ C,¬A ∨ B

Theorem

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Example: Γ := {(A ∨ ¬B) ∧ B,¬A}

(A ∨ ¬B) ∧ B

A ∨ ¬B

On the blackboard

A ∧ C,¬A ∨ B

Theorem

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Example: Γ := {(A ∨ ¬B) ∧ B,¬A}

(A ∨ ¬B) ∧ B

A ∨ ¬B

On the blackboard

A ∧ C,¬A ∨ B

Theorem

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Example: Γ := {(A ∨ ¬B) ∧ B,¬A}

(A ∨ ¬B) ∧ B

A ∨ ¬B

On the blackboard

A ∧ C,¬A ∨ B

Theorem

Propositional tableau is sound and complete, i.e.,if the tableau for Γ is closed then Γ |= ⊥, andif Γ |= ⊥ then the tableau for Γ is closed. 29 / 36

Outline

2 DPLL

4 Exercises

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Exercises (1)(From Logic for Computer Science: Foundations of AutomaticTheorem Proving)

1 Show that the following set of clauses are unsatisfiableusing the resolution method:

1 {{A,B,¬C}, {A,B,C}, {A,¬B}, {¬A}}2 {{A,¬B,C}, {B,C}, {¬A,C}, {B,¬C}, {¬B}}3 {{A,¬B}, {A,C}, {¬B,C}, {¬A,B}, {B,¬C}, {¬A,¬C}}4 {{A,B}, {¬A,B}, {A,¬B}, {¬A,¬B}, }

2 Find all resolvents of the following pairs of clauses:1 {A,B}, {¬A,¬B}2 {A,¬B}, {B,C,D}3 {¬A,B,¬C}, {B,C}4 {A,¬A}, {A,¬A}

3 Find all resolvents of the following sets of clauses:1 {{A,¬B}, {A,B}, {¬A}}2 {{A,B,C}, {¬B,¬C}, {¬A,¬C}}3 {{¬A,¬B}, {B,C}, {¬C,A}}4 {{A,B,C}, {A}, {B}}

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Exercises (2)

1 Show using resolution whether the following statementshold:

1 x ∨ y ∨ ¬z |= (x ∨ z)↔ (¬y → x)2 ((¬X ∨ ¬Y )→ ¬(¬Y ∨ X )) is satisfiable

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Exercises (3)

(From Logic for Computer Science: Foundations of AutomaticTheorem Proving)

1 Show that the following set of clauses are unsatisfiableusing the DPLL algorithm:

1 {{A,B,¬C}, {A,B,C}, {A,¬B}, {¬A}}2 {{A,¬B,C}, {B,C}, {¬A,C}, {B,¬C}, {¬B}}3 {{A,¬B}, {A,C}, {¬B,C}, {¬A,B}, {B,¬C}, {¬A,¬C}}4 {{A,B}, {¬A,B}, {A,¬B}, {¬A,¬B}, }

2 Show using DPLL whether the following statements hold:1 x ∨ y ∨ ¬z |= (x ∨ z)↔ (¬y → x)2 ((¬X ∨ ¬Y )→ ¬(¬Y ∨ X )) is satisfiable

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Exercises (4)

(From Logic for Computer Science: Foundations of AutomaticTheorem Proving)

Give proof trees (by means of tableau) for the followingtautologies:

1 A→ (B → A)

2 (A→ B)→ ((A→ (B → C))→ (A→ C))

3 A→ (B → A ∧ B)

4 A→ A ∨ B5 B → A ∨ B6 (A→ B)→ ((A→ ¬B)→ ¬A)

7 A ∧ B → A8 A ∧ B → B9 (A→ C)→ ((B → C)→ (A ∨ B → C))

10 ¬¬A→ A

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Exercises (5)(From Logic for Computer Science: Foundations of AutomaticTheorem Proving)

Give proof trees (by means of tableau) for the followingequivalences:

1 (A ∨ B) ∨ C ≡ A ∨ (B ∨ C) (associativity)2 (A ∧ B) ∧ C ≡ A ∧ (B ∧ C) (associativity)3 A ∨ B ≡ B ∨ A (commutativity)4 A ∧ B ≡ B ∧ A (commutativity)5 A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C) (distributivity)6 A ∧ (B ∨ C) ≡ (A ∧ B) ∨ (A ∧ C) (distributivity)7 ¬(A ∨ B) ≡ ¬A ∧ ¬B (De Morgan)8 ¬(A ∧ B) ≡ ¬A ∨ ¬B (De Morgan)9 A ∨ A ≡ A (idempotency)

10 A ∧ A ≡ A (idempotency)11 ¬¬A ≡ A (double negation)12 (A ∨ B) ∧ (¬A ∨ C) ≡ (A ∨ B) ∧ (¬A ∨ C) ∧ (B ∨ C)

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END OF THELECTURE

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propositional logic resolution, dpll and tableau

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