Logic - DHBW Stuttgarthladik/Logik/2015/logic - overlays.pdf · Logic, logic, logic... Why do...

Jan Hladik

DHBW Stuttgart

Wintersemester 2015/16

Jan Hladik (DHBW Stuttgart) Logic Wintersemester 2015/16 1 / 120

Introduction

Jan Hladik

Dipl.-Inform.: RWTH Aachen, 2001

Dr. rer. nat.: TU Dresden, 2007

Industry experience: SAP Research

Work in publicly funded research projectsCollaboration with SAP product groupsSupervision of Bachelor, Master, and PhD students

Professor: DHBW Stuttgart, 2014

Research

Semantic Web, Semantic Technologies, Automated Reasoning

Introduction

Jan Hladik

Dipl.-Inform.: RWTH Aachen, 2001

Dr. rer. nat.: TU Dresden, 2007

Industry experience: SAP Research

Work in publicly funded research projectsCollaboration with SAP product groupsSupervision of Bachelor, Master, and PhD students

Professor: DHBW Stuttgart, 2014

Research

Semantic Web, Semantic Technologies, Automated Reasoning

Outline

1 Why logic?

2 propositional and first-order logic

1 syntax and semantics of propositional logic2 reasoning

1 resolution2 tableaus

3 syntax and semantics of first-order logic4 reasoning

Logic, logic, logic. . .

Why do computer scientists need logic?

Mr Spock says:

Logic isthe beginning of wisdom,not the end.

Logic is not the ultimate purpose of a computer science course

but it lays the foundation for other topics

and is necessary to understand what computer science is about

Mr Spock says:

What we need logic for

You can be a hacker without knowing anything about logic

like you can drive a car without knowing about engines

You need to know about the theoretical foundations of computerscience if you want to

understand what’s going on in a computer (Boolean logic)ensure that your program computes the right functiondesign your communication protocol so that it does not allow fordeadlocksensure that your algorithm runs as fast as possible

Without logic, you will only be a

You can be a hacker without knowing anything about logiclike you can drive a car without knowing about engines

understand what’s going on in a computer (Boolean logic)

ensure that your program computes the right functiondesign your communication protocol so that it does not allow fordeadlocksensure that your algorithm runs as fast as possible

understand what’s going on in a computer (Boolean logic)ensure that your program computes the right function

design your communication protocol so that it does not allow fordeadlocksensure that your algorithm runs as fast as possible

understand what’s going on in a computer (Boolean logic)ensure that your program computes the right functiondesign your communication protocol so that it does not allow fordeadlocks

ensure that your algorithm runs as fast as possible

Propositional logic

most basic logic we consider in this course

limited expressivity

easy to understand and use

contains many concepts that are also used in more expressive logics

Propostional variables represent propositions

Example

A ; “it rains”

Operators represent relations between propositions

Example

A→ B ; “if it rains, the ground is wet”

Propositional logic

Example

A ; “it rains”

Example

Propositional logic

Example

A ; “it rains”

Example

Propositional variables

are referenced by capital letters A,B,X, Y, . . .

represent propositions that can be true or false

“It rains.”“Peter loves Mary.”“Socrates is mortal.”

Syntax of propositional logic

Definition (propositional formula)

every propositional variable is a propositional formula

> and ⊥ are propositional formulas

if ϕ and ψ are propositional formulas, then so are

¬ϕ (negation)ϕ ∧ ψ (conjunction)ϕ ∨ ψ (disjunction)ϕ→ ψ (material implication)ϕ↔ ψ (material equivalence)(ϕ) (brackets indicate precedence)

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

¬ϕ (negation)

ϕ ∧ ψ (conjunction)ϕ ∨ ψ (disjunction)ϕ→ ψ (material implication)ϕ↔ ψ (material equivalence)(ϕ) (brackets indicate precedence)

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

¬ϕ (negation)ϕ ∧ ψ (conjunction)

ϕ ∨ ψ (disjunction)ϕ→ ψ (material implication)ϕ↔ ψ (material equivalence)(ϕ) (brackets indicate precedence)

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

¬ϕ (negation)ϕ ∧ ψ (conjunction)ϕ ∨ ψ (disjunction)

ϕ→ ψ (material implication)ϕ↔ ψ (material equivalence)(ϕ) (brackets indicate precedence)

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

¬ϕ (negation)ϕ ∧ ψ (conjunction)ϕ ∨ ψ (disjunction)ϕ→ ψ (material implication)

ϕ↔ ψ (material equivalence)(ϕ) (brackets indicate precedence)

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

¬ϕ (negation)ϕ ∧ ψ (conjunction)ϕ ∨ ψ (disjunction)ϕ→ ψ (material implication)ϕ↔ ψ (material equivalence)

(ϕ) (brackets indicate precedence)

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

Examples

A ∧B⊥ ∨A¬>

A ∧ (B ∨ C)

¬(A ∨B)

C → (B ∨D)

A→ (B∧(C ↔ D)∨(D → E))

A↔ (A↔ D)

((A↔ B)↔ (C ↔ D))→ E

Exercise: syntax of PL

Which of the following are propositional formulas?

a) A→ ⊥b) A ∧ (B ∨ C)

c) A¬Bd) (A→ C) ∧ (¬A→ C)→ C

e) ∨B ∨ C ∧Df) A→ (B ∨ ¬B)(C ∨ ¬C)

g) A→ A

h) A ∧ ¬Ai) A¬ ∧Bj) ¬A ∧Bk) (¬A) ∧Bl) ¬(A ∧B)

Precedence and associativity of logical operators

Precedence of operators:

operator precedence

¬ 1 (strongest)∧ 2∨ 3→ 4↔ 5 (weakest)

Example

A ∧B → B ∨ C means (A ∧B)→ (B ∨ C)

Operators of the same precedence are left-associative.

Example

A→ B → C means (A→ B)→ C

operator precedence

Example

A ∧B → B ∨ C means (A ∧B)→ (B ∨ C)

Example

operator precedence

Example

A ∧B → B ∨ C means (A ∧B)→ (B ∨ C)

Example

operator precedence

Example

A ∧B → B ∨ C means (A ∧B)→ (B ∨ C)

Example

Exercise: Precedence and brackets

Remove as many brackets as possible without changing the precedence.

a) ((A ∧B) ∨ ((C ∧D)→ (A ∨ C)))

b) ((((A ∧ (B ∨ C) ∧D))→ A) ∨ C)

c) (A ∧ (B ∨ (C ∧ (D → (A ∨ C)))))

Semantics of propositional logic

assigning meaning to a logical formula.

assigning 0 or 1 to the variables

making the formula true or false

An interpretation assigns a truth value to variables (and thus formulas)

Definition (interpretation)

Function I : V → B where

V is a set of propositional variables

B is the binary set {0, 1}

Example

A 7→ 0 ; “it does not rain”

Example

Interpretation of complex formulas

⊥I >I0 1

ϕI 0 1

(¬ϕ)I 1 0

(ϕ ∧ ψ)I

ϕI 0 1ψI

0 0 01 0 1

(ϕ ∨ ψ)I

ϕI 0 1ψI

0 0 11 1 1

(ϕ→ ψ)I

ϕI 0 1ψI

0 1 01 1 1

(ϕ↔ ψ)I

ϕI 0 1ψI

0 1 01 0 1

⊥I >I0 1

ϕI 0 1

(¬ϕ)I 1 0

(ϕ ∧ ψ)I

ϕI 0 1ψI

0 0 01 0 1

(ϕ ∨ ψ)I

ϕI 0 1ψI

0 0 11 1 1

(ϕ→ ψ)I

ϕI 0 1ψI

0 1 01 1 1

(ϕ↔ ψ)I

ϕI 0 1ψI

0 1 01 0 1

⊥I >I0 1

ϕI 0 1

(¬ϕ)I 1 0

(ϕ ∧ ψ)I

ϕI 0 1ψI

0 0 01 0 1

(ϕ ∨ ψ)I

ϕI 0 1ψI

0 0 11 1 1

(ϕ→ ψ)I

ϕI 0 1ψI

0 1 01 1 1

(ϕ↔ ψ)I

ϕI 0 1ψI

0 1 01 0 1

⊥I >I0 1

ϕI 0 1

(¬ϕ)I 1 0

(ϕ ∧ ψ)I

ϕI 0 1ψI

0 0 01 0 1

(ϕ ∨ ψ)I

ϕI 0 1ψI

0 0 11 1 1

(ϕ→ ψ)I

ϕI 0 1ψI

0 1 01 1 1

(ϕ↔ ψ)I

ϕI 0 1ψI

0 1 01 0 1

⊥I >I0 1

ϕI 0 1

(¬ϕ)I 1 0

(ϕ ∧ ψ)I

ϕI 0 1ψI

0 0 01 0 1

(ϕ ∨ ψ)I

ϕI 0 1ψI

0 0 11 1 1

(ϕ→ ψ)I

ϕI 0 1ψI

0 1 01 1 1

(ϕ↔ ψ)I

ϕI 0 1ψI

0 1 01 0 1

⊥I >I0 1

ϕI 0 1

(¬ϕ)I 1 0

(ϕ ∧ ψ)I

ϕI 0 1ψI

0 0 01 0 1

(ϕ ∨ ψ)I

ϕI 0 1ψI

0 0 11 1 1

(ϕ→ ψ)I

ϕI 0 1ψI

0 1 01 1 1

(ϕ↔ ψ)I

ϕI 0 1ψI

0 1 01 0 1

Example: interpretations

The interpretation I = {A 7→ 1, B 7→ 1, C 7→ 0} makes the formula

B true;

A ∧B true;

A ∧ C false;

(A ∧B) ∨ (A ∧ C) true;

((A ∧B) ∨ (A ∧ C))→ C false.

Basic equivalences

Definition (equivalent)

Two propisitional formulas ϕ and ψ are equivalent if every interpretationassigns the same truth value to both formulas, i.e.

ϕI = ψI for every I

Due to the semantics, one can express some operators by using other ones

Theorem (equivalences)

ϕ↔ ψ is equivalent to (ϕ→ ψ) ∧ (ψ → ϕ) or (ϕ ∧ ψ) ∨ (¬ϕ ∧ ¬ψ)

ϕ→ ψ is equivalent to ¬ϕ ∨ ψϕ ∧ ψ is equivalent to ¬(¬ϕ ∨ ¬ψ)

ϕ ∨ ψ is equivalent to ¬(¬ϕ ∧ ¬ψ)

Basic equivalences

ϕ→ ψ is equivalent to ¬ϕ ∨ ψ

ϕ ∧ ψ is equivalent to ¬(¬ϕ ∨ ¬ψ)

Basic equivalences

Exercise: interpretations

Find two interpretations for each of the following formulas

one that makes them true

one that makes them false

a) A ∧B → C

b) (A ∨B) ∧ (A ∨ C)→ (B ∧ C)

c) A→ B ↔ ¬B → ¬A

Tautologies

Definition (tautology)

A tautology is a formula which is true in every interpretation.

Examples (tautologies)

A↔ A (identity)

¬(A ∧ ¬A) (noncontradiction)

A ∨ ¬A (excluded middle)

A→ B ↔ ¬B → ¬A (contraposition)

(A→ B) ∧ (A→ ¬B)→ ¬A (reductio ad absurdum)

(A→ B) ∧ (B → C)→ (A→ C) (syllogism)

(A ∨B) ∧ (A→ C) ∧ (B → C)→ C (proof by cases)

Tautologies

A↔ A (identity)

Tautologies

A↔ A (identity)

Tautologies

A↔ A (identity)

Tautologies

A↔ A (identity)

Tautologies

A↔ A (identity)

Tautologies

A↔ A (identity)

Tautologies

A↔ A (identity)

Further tautologies

(A ∧B) ∨ C ↔ (A ∨ C) ∧ (B ∨ C) (distributivity)

(A ∧B) ∧ C ↔ A ∧ (B ∧ C) (associativity)

A ∧B ↔ B ∧A (commutativity)

¬(A ∧B)↔ ¬A ∨ ¬B (de Morgan’s law)

A ∧A↔ A; A ∨A↔ A (idempotence)

A ∧ > ↔ A; A ∨ ⊥ ↔ A (neutral element)

A ∧ ⊥ ↔ ⊥; A ∨ > ↔ > (absorbing element)

¬¬A↔ A (double negation)

Distributivity, associativity, commutativity, and deMorgan’s law also hold if∧ is replaced with ∨ and vice versa.

Further tautologies

Models, validity, satisfiability

Definition (model)

A model for a formula ϕ is an interpretation I that makes ϕ true (I |= ϕ).

I = {A 7→ 1, B 7→ 0} is a model for A ∨B, but not for A ∧B.

Definition (satisfiable, valid)

A formula ϕ is called satisfiable if it has a model.A formula ϕ is called valid (|= ϕ) if every interpretation is a model (i.e. ifit is a tautology).

satisfiable A, A ∨B, A ∧ ¬B, A↔ B

valid A ∨ ¬A, A→ A, A↔ ¬(¬A)

Definition (model)

valid A ∨ ¬A, A→ A, A↔ ¬(¬A)

Definition (model)

A formula ϕ is called satisfiable if it has a model.

A formula ϕ is called valid (|= ϕ) if every interpretation is a model (i.e. ifit is a tautology).

valid A ∨ ¬A, A→ A, A↔ ¬(¬A)

Definition (model)

valid A ∨ ¬A, A→ A, A↔ ¬(¬A)

Definition (model)

valid A ∨ ¬A, A→ A, A↔ ¬(¬A)

Definition (model)

valid A ∨ ¬A, A→ A, A↔ ¬(¬A)

Outline

2 Prolog

3 modal and description logics

4 Semantic Web

Reasoning in propositional logic

Determine if a formula ϕ is

valid (a tautology, i.e. every interpretation is a model)satisfiable (there is a model)unsatisfiable (there are no models)

Determine if a formula ϕ implies another formula ψ (ϕ |= ψ),i.e every model for ϕ is a model for ψ

Reasoning in propositional logic

Determine if a formula ϕ is

valid (a tautology, i.e. every interpretation is a model)satisfiable (there is a model)unsatisfiable (there are no models)

Determine if a formula ϕ implies another formula ψ (ϕ |= ψ),i.e every model for ϕ is a model for ψ

Showing that a formula is valid

A formula can be shown to be a tautology by using known equivalences:

Example: reductio ad absurdum

(A→ B) ∧ (A→ ¬B)→ ¬A

definition of → (twice)(¬A ∨B) ∧ (¬A ∨ ¬B)→ ¬A definition of →¬((¬A ∨B) ∧ (¬A ∨ ¬B)) ∨ ¬A De Morgan¬(¬A ∨B) ∨ ¬(¬A ∨ ¬B) ∨ ¬A De Morgan (twice)

(A ∧ ¬B) ∨ (A ∧B) ∨ ¬A distributivity(A ∧ (¬B ∨B)) ∨ ¬A excluded middle

(A ∧ >) ∨ ¬A > is neutral element for ∧A ∨ ¬A excluded middle>

Drawbacks:

One has to guess which transformation to use.

Failure to prove ϕ does not imply that ϕ is not valid.

(A→ B) ∧ (A→ ¬B)→ ¬A definition of → (twice)(¬A ∨B) ∧ (¬A ∨ ¬B)→ ¬A

definition of →¬((¬A ∨B) ∧ (¬A ∨ ¬B)) ∨ ¬A De Morgan¬(¬A ∨B) ∨ ¬(¬A ∨ ¬B) ∨ ¬A De Morgan (twice)

Drawbacks:

(A→ B) ∧ (A→ ¬B)→ ¬A definition of → (twice)(¬A ∨B) ∧ (¬A ∨ ¬B)→ ¬A definition of →¬((¬A ∨B) ∧ (¬A ∨ ¬B)) ∨ ¬A

De Morgan¬(¬A ∨B) ∨ ¬(¬A ∨ ¬B) ∨ ¬A De Morgan (twice)

Drawbacks:

(A→ B) ∧ (A→ ¬B)→ ¬A definition of → (twice)(¬A ∨B) ∧ (¬A ∨ ¬B)→ ¬A definition of →¬((¬A ∨B) ∧ (¬A ∨ ¬B)) ∨ ¬A De Morgan¬(¬A ∨B) ∨ ¬(¬A ∨ ¬B) ∨ ¬A

De Morgan (twice)(A ∧ ¬B) ∨ (A ∧B) ∨ ¬A distributivity

(A ∧ (¬B ∨B)) ∨ ¬A excluded middle(A ∧ >) ∨ ¬A > is neutral element for ∧

A ∨ ¬A excluded middle>

Drawbacks:

(A→ B) ∧ (A→ ¬B)→ ¬A definition of → (twice)(¬A ∨B) ∧ (¬A ∨ ¬B)→ ¬A definition of →¬((¬A ∨B) ∧ (¬A ∨ ¬B)) ∨ ¬A De Morgan¬(¬A ∨B) ∨ ¬(¬A ∨ ¬B) ∨ ¬A De Morgan (twice)

(A ∧ ¬B) ∨ (A ∧B) ∨ ¬A

distributivity(A ∧ (¬B ∨B)) ∨ ¬A excluded middle

Drawbacks:

(A ∧ ¬B) ∨ (A ∧B) ∨ ¬A distributivity(A ∧ (¬B ∨B)) ∨ ¬A

excluded middle(A ∧ >) ∨ ¬A > is neutral element for ∧

A ∨ ¬A excluded middle>

Drawbacks:

(A ∧ >) ∨ ¬A

> is neutral element for ∧A ∨ ¬A excluded middle>

Drawbacks:

(A ∧ >) ∨ ¬A > is neutral element for ∧A ∨ ¬A

excluded middle>

Drawbacks:

The same result can be shown using a truth table:

A B C := A→ B D := A→ ¬B C ∧D ¬A C ∧D → ¬A0 0 1 1 1 1 1

0 1 1 1 1 1 11 0 0 1 0 0 11 1 1 0 0 0 1

Problems:

inefficient, especially with many propositional variables

redundant, since many valuation changes do not affect results

A B C := A→ B D := A→ ¬B C ∧D ¬A C ∧D → ¬A0 0 1 1 1 1 10 1 1 1 1 1 1

1 0 0 1 0 0 11 1 1 0 0 0 1

Problems:

A B C := A→ B D := A→ ¬B C ∧D ¬A C ∧D → ¬A0 0 1 1 1 1 10 1 1 1 1 1 11 0 0 1 0 0 1

1 1 1 0 0 0 1

Problems:

A B C := A→ B D := A→ ¬B C ∧D ¬A C ∧D → ¬A0 0 1 1 1 1 10 1 1 1 1 1 11 0 0 1 0 0 11 1 1 0 0 0 1

Problems:

A B C := A→ B D := A→ ¬B C ∧D ¬A C ∧D → ¬A0 0 1 1 1 1 10 1 1 1 1 1 11 0 0 1 0 0 11 1 1 0 0 0 1

Problems:

Exercise: reasoning

Prove that the following formula is a tautology

A ∧B → C ↔ A→ (B → C)

a) known equivalences,

b) a truth table.

Outline

Resolution

method to test satisfiability of a propositional formula ϕ

introduced in 1965 by John Robinson

can also be used to test

validity: |= ϕ holds iff ¬ϕ is usatisfiable.(logical) implication: ϕ |= ψ holds iff ϕ ∧ ¬ψ is unsatisfiable.

Resolution

method to test satisfiability of a propositional formula ϕ

introduced in 1965 by John Robinson

can also be used to test

validity: |= ϕ holds iff ¬ϕ is usatisfiable.(logical) implication: ϕ |= ψ holds iff ϕ ∧ ¬ψ is unsatisfiable.

Resolution principle

Find contradiction in ϕ

operate on set of disjunctions (clauses)

find pairs of contradictory literals L,L in clauses C1, C2

generate resolvent C3 by joining C1 and C2 (parent clauses) andeliminating L and L

C3 is not equivalent to C1 ∧ C2!C1 ∧ C2 satisfiable ; C3 satisfiableC3 unsatisfiable ; C1 ∧ C2 unsatisfiable

empty clause 2: contradiction

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬X

X ∨ ZX

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬X

X ∨ ZX

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬X

X ∨ ZX

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬XX ∨ Z

C3 is not equivalent to C1 ∧ C2!

C1 ∧ C2 satisfiable ; C3 satisfiableC3 unsatisfiable ; C1 ∧ C2 unsatisfiable

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬XX ∨ Z

C3 is not equivalent to C1 ∧ C2!C1 ∧ C2 satisfiable ; C3 satisfiable

C3 unsatisfiable ; C1 ∧ C2 unsatisfiable

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬XX ∨ Z

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬XX ∨ Z

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬XX ∨ Z

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬XX ∨ Z

Example (ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X)

X ∨ Y ¬Y ∨ Z ¬Z ¬XX ∨ Z

Literals and their negations

Definition (Literal)

A literal is a (possibly negated) propositional variable.

Example

For the set of variables {A,B}, the set of possible literals is{A,¬A,B,¬B}.

Definition (Negation of a literal)

The negation L of a literal L is defined as follows

for a variable A: A = ¬A;

for a negated variable ¬A: ¬A = A

Example

B = ¬B; ¬C = C, . . .

Example

B = ¬B; ¬C = C, . . .

Example

B = ¬B; ¬C = C, . . .

Example

B = ¬B; ¬C = C, . . .

Conjunctive Normal Form

A propositional formula ϕ is in conjunctive normal form (CNF) if ϕ is aconjunction of disjunctions of literals.

Example

(X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X is in CNF;(X ∧ Y ) ∨ ¬(¬Y ∨ Z) is not in CNF.

Example

(X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X is in CNF;

(X ∧ Y ) ∨ ¬(¬Y ∨ Z) is not in CNF.

Example

(X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X is in CNF;(X ∧ Y ) ∨ ¬(¬Y ∨ Z) is not in CNF.

Transformation into CNF

Theorem

Every propositional formula can be transformed into an equivalent formulain CNF.

Goal: (L1 ∨ L2 ∨ L3) ∧ (L4 ∨ L5) ∧ L6 . . .

1. eliminate material implication ϕ→ ψ ; ¬ϕ ∨ ψ1. and material equivalence ϕ↔ ψ ; ϕ ∧ ψ ∨ ¬ϕ ∧ ¬ψ2. De Morgan’s laws ¬(ϕ ∨ ψ) ; ¬ϕ ∧ ¬ψ3. eliminate double negation ¬¬ϕ ; ϕ4. distributivity ϕ ∨ (ψ ∧ χ) ; (ϕ ∨ ψ) ∧ (ϕ ∨ χ)5. associativity ϕ ∨ (ψ ∨ χ) ; ϕ ∨ ψ ∨ χ

Example

(X ∧ Y )∨¬(¬Y ∨Z) ; (X ∨ Y )∧ (Y ∨ Y )∧ (X ∨¬Z)∧ (Y ∨¬Z)

Theorem

Goal: (L1 ∨ L2 ∨ L3) ∧ (L4 ∨ L5) ∧ L6 . . .

Example

(X ∧ Y )∨¬(¬Y ∨Z) ; (X ∨ Y )∧ (Y ∨ Y )∧ (X ∨¬Z)∧ (Y ∨¬Z)

Theorem

Goal: (L1 ∨ L2 ∨ L3) ∧ (L4 ∨ L5) ∧ L6 . . .

1. eliminate material implication ϕ→ ψ ; ¬ϕ ∨ ψ1. and material equivalence ϕ↔ ψ ; ϕ ∧ ψ ∨ ¬ϕ ∧ ¬ψ

2. De Morgan’s laws ¬(ϕ ∨ ψ) ; ¬ϕ ∧ ¬ψ3. eliminate double negation ¬¬ϕ ; ϕ4. distributivity ϕ ∨ (ψ ∧ χ) ; (ϕ ∨ ψ) ∧ (ϕ ∨ χ)5. associativity ϕ ∨ (ψ ∨ χ) ; ϕ ∨ ψ ∨ χ

Example

(X ∧ Y )∨¬(¬Y ∨Z) ; (X ∨ Y )∧ (Y ∨ Y )∧ (X ∨¬Z)∧ (Y ∨¬Z)

Theorem

Goal: (L1 ∨ L2 ∨ L3) ∧ (L4 ∨ L5) ∧ L6 . . .

1. eliminate material implication ϕ→ ψ ; ¬ϕ ∨ ψ1. and material equivalence ϕ↔ ψ ; ϕ ∧ ψ ∨ ¬ϕ ∧ ¬ψ2. De Morgan’s laws ¬(ϕ ∨ ψ) ; ¬ϕ ∧ ¬ψ

3. eliminate double negation ¬¬ϕ ; ϕ4. distributivity ϕ ∨ (ψ ∧ χ) ; (ϕ ∨ ψ) ∧ (ϕ ∨ χ)5. associativity ϕ ∨ (ψ ∨ χ) ; ϕ ∨ ψ ∨ χ

Example

(X ∧ Y )∨¬(¬Y ∨Z) ; (X ∨ Y )∧ (Y ∨ Y )∧ (X ∨¬Z)∧ (Y ∨¬Z)

Theorem

Goal: (L1 ∨ L2 ∨ L3) ∧ (L4 ∨ L5) ∧ L6 . . .

1. eliminate material implication ϕ→ ψ ; ¬ϕ ∨ ψ1. and material equivalence ϕ↔ ψ ; ϕ ∧ ψ ∨ ¬ϕ ∧ ¬ψ2. De Morgan’s laws ¬(ϕ ∨ ψ) ; ¬ϕ ∧ ¬ψ3. eliminate double negation ¬¬ϕ ; ϕ

4. distributivity ϕ ∨ (ψ ∧ χ) ; (ϕ ∨ ψ) ∧ (ϕ ∨ χ)5. associativity ϕ ∨ (ψ ∨ χ) ; ϕ ∨ ψ ∨ χ

Example

(X ∧ Y )∨¬(¬Y ∨Z) ; (X ∨ Y )∧ (Y ∨ Y )∧ (X ∨¬Z)∧ (Y ∨¬Z)

Theorem

Goal: (L1 ∨ L2 ∨ L3) ∧ (L4 ∨ L5) ∧ L6 . . .

1. eliminate material implication ϕ→ ψ ; ¬ϕ ∨ ψ1. and material equivalence ϕ↔ ψ ; ϕ ∧ ψ ∨ ¬ϕ ∧ ¬ψ2. De Morgan’s laws ¬(ϕ ∨ ψ) ; ¬ϕ ∧ ¬ψ3. eliminate double negation ¬¬ϕ ; ϕ4. distributivity ϕ ∨ (ψ ∧ χ) ; (ϕ ∨ ψ) ∧ (ϕ ∨ χ)

5. associativity ϕ ∨ (ψ ∨ χ) ; ϕ ∨ ψ ∨ χ

Example

(X ∧ Y )∨¬(¬Y ∨Z) ; (X ∨ Y )∧ (Y ∨ Y )∧ (X ∨¬Z)∧ (Y ∨¬Z)

Theorem

Goal: (L1 ∨ L2 ∨ L3) ∧ (L4 ∨ L5) ∧ L6 . . .

Example

(X ∧ Y )∨¬(¬Y ∨Z) ; (X ∨ Y )∧ (Y ∨ Y )∧ (X ∨¬Z)∧ (Y ∨¬Z)

Theorem

Goal: (L1 ∨ L2 ∨ L3) ∧ (L4 ∨ L5) ∧ L6 . . .

Example

(X ∧ Y )∨¬(¬Y ∨Z) ; (X ∨ Y )∧ (Y ∨ Y )∧ (X ∨¬Z)∧ (Y ∨¬Z)

Exercise: Conjunctive Normal Form

Transform the following formulas into CNF:

(X ∨ Y ) ∧ (A ∧B)

(X ∨ Y )→ (A ∨B)

X ∨ (¬A ∧ ¬(B ∧ ¬C))

Clauses

Definition (Clause)

A clause is a disjunction of literals, written as a set.

Example

(X ∨ ¬Y ∨ Z) ; {X,¬Y,Z}

Every propositional formula can be written as a set of clauses.

ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬XK(ϕ) = {{X,Y }, {¬Y, Z}, {¬Z}, {¬X}}

Intuitively:

∧ between the sets

∨ between the elements of the sets

Clauses

Definition (Clause)

Example

(X ∨ ¬Y ∨ Z) ; {X,¬Y,Z}

Intuitively:

Clauses

Definition (Clause)

Example

(X ∨ ¬Y ∨ Z) ; {X,¬Y,Z}

ϕ = (X ∨ Y ) ∧ (¬Y ∨ Z) ∧ ¬Z ∧ ¬X

K(ϕ) = {{X,Y }, {¬Y, Z}, {¬Z}, {¬X}}

Intuitively:

Clauses

Definition (Clause)

Example

(X ∨ ¬Y ∨ Z) ; {X,¬Y,Z}

Intuitively:

Clauses

Definition (Clause)

Example

(X ∨ ¬Y ∨ Z) ; {X,¬Y,Z}

Intuitively:

Resolution of clauses

We have two clauses C1 = {X,Y } and C2 = {¬Y , Z}.We are looking for a model for C1 ∧ C2.

In an interpretation I, Y I is either 0 or 1.

if Y I = 1 holds, then CI1 = 1 and CI2 = 1 iff ZI = 1;

if Y I = 0 holds, then CI2 = 1 and CI1 = 1 iff XI = 1.

Every model for C1 and C2 is also a model for C3 = {X,Z}.If {C1, C2} is satisfiable, then {C1, C2, C3} is also satisfiable.

Definition (Resolvent)

Let C1 = {L1, L2, L3, . . .} and C2 = {L1, L6, L7, . . .} be propositionalclauses. Then C3 = {L2, L3, . . . , L6, L7, . . .} is called the resolvent of C1

and C2 ({C1, C2} ` C3).

If a clause C can be obtained from a clause set S using several resolutionsteps, we write S `∗ C.

and C2 ({C1, C2} ` C3).

Every model for C1 and C2 is also a model for C3 = {X,Z}.

If {C1, C2} is satisfiable, then {C1, C2, C3} is also satisfiable.

and C2 ({C1, C2} ` C3).

Exercise: Resolution

Find as many resolvents as possible for the clause set S.

S = {{A,B,¬C}, {B,C,D}, {¬D}, {¬A,D}, {¬B,¬E}, {D,E}}

Using resolution to test satisfiability

Let S ` C.

Every model for S is also a model for C.

If S is satisfiable, then so is S ∪ {C}.If S ∪ {C} is unsatisfiable, then so is S.

The empty clause 2 is unsatisfiable.(It is obtained by resolving A and ¬A for some A.)

Theorem (soundness of resolution)

Let S be a set of clauses.If 2 can be obtained from S by resolution, then S is unsatisfiable.

Let S ` C.

If S is satisfiable, then so is S ∪ {C}.

If S ∪ {C} is unsatisfiable, then so is S.

Let S ` C.

Resolution algorithm for propositional logic

input: propositional formula ϕoutput: “satisfiable” or “unsatisfiable”

1: transform ϕ into CNF2: initialise C with disjunctions of ϕ3: while there is C3 = res(C1, C2) with C3 /∈ C do4: if C3 = 2 then5: output “unsatisfiable”6: else7: C := C ∪ {C3}8: output “satisfiable”

1: transform ϕ into CNF

2: initialise C with disjunctions of ϕ3: while there is C3 = res(C1, C2) with C3 /∈ C do4: if C3 = 2 then5: output “unsatisfiable”6: else7: C := C ∪ {C3}8: output “satisfiable”

1: transform ϕ into CNF2: initialise C with disjunctions of ϕ

3: while there is C3 = res(C1, C2) with C3 /∈ C do4: if C3 = 2 then5: output “unsatisfiable”6: else7: C := C ∪ {C3}8: output “satisfiable”

1: transform ϕ into CNF2: initialise C with disjunctions of ϕ3: while there is C3 = res(C1, C2) with C3 /∈ C do

4: if C3 = 2 then5: output “unsatisfiable”6: else7: C := C ∪ {C3}8: output “satisfiable”

1: transform ϕ into CNF2: initialise C with disjunctions of ϕ3: while there is C3 = res(C1, C2) with C3 /∈ C do4: if C3 = 2 then5: output “unsatisfiable”

6: else7: C := C ∪ {C3}8: output “satisfiable”

1: transform ϕ into CNF2: initialise C with disjunctions of ϕ3: while there is C3 = res(C1, C2) with C3 /∈ C do4: if C3 = 2 then5: output “unsatisfiable”6: else7: C := C ∪ {C3}

8: output “satisfiable”

1: transform ϕ into CNF2: initialise C with disjunctions of ϕ3: while there is C3 = res(C1, C2) with C3 /∈ C do4: if C3 = 2 then5: output “unsatisfiable”6: else7: C := C ∪ {C3}8: output “satisfiable”

Properties of the resolution algorithm

Theorem (soundness)

K(ϕ) `∗ 2 implies unsatisfiability of ϕ.

Theorem (refutation completeness)

If ϕ is unsatisfiable, then K(ϕ) `∗ 2 holds.

Theorem (termination)

The resolution algorithm terminates for every input.

Proof.

Only new clauses are added.

New clauses contain only literals from the input.

Clauses are never removed.

Theorem (soundness)

Proof.

Theorem (soundness)

Proof.

Theorem (soundness)

Proof.

Theorem (soundness)

Proof.

Theorem (soundness)

Proof.

Resolution and non-determinism

If several possibilities for new resolvents exist:

the algorithm will terminate for every choice

. . . but some choices will be faster than others

Definition (don’t-care-nondeterminism)

An algorithm is called don’t-care-nondeterministic if at any point withseveral choices every choice will lead to a solution.

no backtracking required

Turning natural language into propositional formulas

1 If John is not sick and he is invited to the meeting, he attends it.¬S ∧ I → A ; S ∨ ¬I ∨A

2 If the boss wants John at the meeting, he invites him.B → I ; ¬B ∨ I

3 If the boss does not want John at the meeting, John will be fired.¬B → F ; B ∨ F

4 John did not attend the meeting. ¬A5 John is not sick. ¬S6 Conjecture: John will be fired. F

To test if the conjecture is valid, we check if the premises together withthe negation of the conjecture are satisfiable.