Course: Engineering Artificial Intelligence

University College Cork (Ireland) Department of Civil and Environmental Engineering

Course:Engineering Artificial Intelligence

Dr. Radu Marinescu

Lecture 7


Today’s class

o Logic• Propositional logic• First order logic

R. Marinescu October-2009 page 2


What is logic?

o A formal language• Syntax – what expressions are legal• Semantics – what legal expressions mean• Proof system – a way of manipulating syntactic expressions to get

other syntactic expressions (which will tell us something new)



What is logic?

o A formal language• Syntax – what expressions are legal• Semantics – what legal expressions mean• Proof system – a way of manipulating syntactic expressions to get

other syntactic expressions (which will tell us something new)o Why proofs? Two kinds of inferences an agent

might want to make• Multiple percepts => conclusions about the world• Current state & operators => properties of next state



Propositional logic syntax

o Syntax: what you’re allowed to write•for (int t = 0; t == 100; t++) {…}• “Colorless green ideas sleep furiously” (Noam Chomsky)





o Sentences (WFF: well formed formulas)•true and false are sentences





o Sentences (WFF: well formed formulas)•true and false are sentences• Propositional variables are sentences: P, Q, R, Z





o Sentences (WFF: well formed formulas)•true and false are sentences• Propositional variables are sentences: P, Q, R, Z• If Φ and Ψ are sentences, then so are

(Φ), ¬ Φ, Φ /\ Ψ, Φ \/ Ψ, Φ → Ψ, Φ ↔ Ψ





o Sentences (WFF: well formed formulas)•true and false are sentences• Propositional variables are sentences: P, Q, R, Z• If Φ and Ψ are sentences, then so are

(Φ), ¬ Φ, Φ ˄ Ψ, Φ ˅ Ψ, Φ → Ψ, Φ ↔ Ψ• Nothing else is a sentence



Precedence

o Precedence rules enable “shorthand” form of sentences, but formally only the fully parenthesized form is legal

o Syntactically ambiguous forms allowed in shorthand only when semantically equivalent: A ˄ B ˄ C is equivalent to (A ˄ B) ˄ C and A ˄ (B ˄ C)


¬

˄

˅

→

↔

highest

lowest

A ˅ B ˄ C A ˅ (B ˄ C)

A ˄ B → C ˅ D (A ˄ B) → (C ˅ D)

A → B ˅ C ↔ D (A → (B ˅ C)) ↔ D


Semantics

o Meaning of a sentence is truth value {t, f}o Interpretation is an assignment of truth values to

propositional variables• holds(Φ, i) [Sentence Φ is t in interpretation i]• fails(Φ, i) [Sentence Φ is f in interpretation i]



Semantic rules

o holds(true, i) for all io fails(false, i) for all Io holds(¬Φ, i) if and only if fails(Φ, i)

o holds(Φ ˄ Ψ, i) iff holds(Φ, i) and holds(Ψ, i)

o holds(Φ ˅ Ψ, i) iff holds(Φ, i) or holds(Ψ, i)


(negation)

(conjunction)

(disjunction)


Semantic rules

o holds(true, i) for all io fails(false, i) for all Io holds(¬Φ, i) if and only if fails(Φ, i)

o holds(Φ ˄ Ψ, i) iff holds(Φ, i) and holds(Ψ, i)

o holds(Φ ˅ Ψ, i) iff holds(Φ, i) or holds(Ψ, i)

o holds(P, i) iff i(P) = to fails(P, i) iff i(P) = f


(negation)

(conjunction)

(disjunction)


Some important shorthand

o Φ → Ψ ≡ ¬Φ ˅ Ψ (conditional, implication)

o Φ ↔ Ψ ≡ (Φ → Ψ) ˄ (Ψ → Φ) (biconditional, equivalence)


antecedent → consequent







P Q ¬P P˄Q P˅Q P→Q Q→P P↔Q

f f t f f t t t

f t t f t t f f

t f f f t f t f

t t f t t t t t

Truth tables







P Q ¬P P˄Q P˅Q P→Q Q→P P↔Q

f f t f f t t t

f t t f t t f f

t f f f t f t f

t t f t t t t t

Truth tables

Note that implication is not “causality”, if P is f then P→Q is t


Terminology

o A sentence is valid iff its truth table value is t in all interpretations

• Valid sentences: true, ¬false, P ˅ ¬P

o A sentence is satisfiable iff its truth value is t in at least one interpretation

• Valid sentences: P, true, ¬P

o A sentence is unsatisfiable iff its truth value is f in all interpretations

• Unsatisfiable sentences: P ˄¬P, false, ¬true



Terminology

o A sentence is valid iff its truth table value is t in all interpretations

• Valid sentences: true, ¬false, P ˅ ¬P

o A sentence is satisfiable iff its truth value is t in at least one interpretation

• Valid sentences: P, true, ¬P

o A sentence is unsatisfiable iff its truth value is f in all interpretations

• Unsatisfiable sentences: P ˄¬P, false, ¬true


All are finitely decidable!


Examples


Sentence Valid?Interpretation that makesentence’s truth value = f

smoke → firesmoke ˅ ¬fire

valid


Examples



smoke → smoke¬ smoke ˅ smoke

valid

smoke → fire¬ smoke ˅ fire

satisfiable,not valid

smoke = t, fire = f


Examples



smoke → smoke¬smoke ˅ smoke

valid

smoke → fire¬smoke ˅ fire


smoke = t, fire = f

(s → f) → (¬s → ¬f) satisfiable,not valid

s = f, f = ts → f = t, ¬s → ¬f = f


Examples



smoke → smoke¬smoke ˅ smoke

valid

smoke → fire¬smoke ˅ fire


smoke = t, fire = f

(s → f) → (¬s → ¬f) satisfiable,not valid

s = f, f = ts → f = t, ¬s → ¬f = f

(s → f) → (¬f → ¬s) valid


Satisfiability

o Related to constraint satisfactiono Given a sentence S, try to find an interpretation i

such that holds(S, i)o Analogous to finding an assignment of values to

variables such that the constraints hold



Satisfiability

o Related to constraint satisfactiono Given a sentence S, try to find an interpretation i

such that holds(S, i)o Analogous to finding an assignment of values to

variables such that the constraints hold

o Brute force method: enumerate all interpretations and check

o Better methods• Heuristic search• Constraint propagation• Stochastic search



Satisfiability problems

o Scheduling nurses to work in a hospital• Propositional variables represent, for example, that Pat is working

on Tuesday at 2• Constraints on the schedule are represented using logical

expressions over the variables

o Finding bugs in software• Propositional variables represent states of the program• Use logic to describe how the program works and to assert there

is a bug• If the sentence is satisfiable, you’ve found a bug



A good lecture?

o Imagine we knew that:• If today is sunny, then Thomas will be happy (S → H)• If Thomas is happy, the lecture will be good (H → G)• Today is sunny (S)

Should we conclude that the lecture will be good?



Checking interpretations


S H G

t t t

t t f

t f t

t f f

f t t

f t f

f f t

f f f




S H G S→H H→G S

t t t t t t

t t f t f t

t f t f t t

t f f f t t

f t t t t f

f t f t f f

f f t t t f

f f f t t f




S H G S→H H→G S G

t t t t t t t

t t f t f t f

t f t f t t t

t f f f t t f

f t t t t f t

f t f t f f f

f f t t t f t

f f f t t f f

Good lecture!


Entailment

o A knowledge base (KB) entails a sentence S iff every interpretation that makes KB true also makes S true


KB Sholds

interpretations interpretationssubset

holds holds


Computing entailment

o Enumerate all interpretationso Select those in which all elements of KB are trueo Check to see if S is true in all those

interpretations


KB Sholds


holds holds


Computing entailment

o Enumerate all interpretationso Select those in which all elements of KB are trueo Check to see if S is true in all those

interpretations


KB Sholds


holds holds

Way too many interpretations in general !!


Entailment and Proof

o A proof is a way to test whether a KB entails a sentence, without enumerating all possible interpretations


KB Sholds


holds holds

proof


Proof

o Proof is a sequence of sentenceso First ones are premises (in KB)o Then, you can write down on the next line the result

of applying an inference rule to previous lineso When S is on a line, you have proved S from KB

o If inference rules are sound, then any S you can prove from KB is entailed by KB

o If inference rules are complete, then any S that is entailed by KB can be proved from KB



Natural deduction

o Some inference rules


α → β

α

β

Modusponens

α → β

¬β

¬α

Modustolens

α

β

α ˄ β

AND-introduction

α ˄ β

α

AND-elimination


Natural deduction example


Step Formula Derivation

1 P ˄ Q given

2 P → R given

3 (Q ˄ R) → S given

Prove S





1 P ˄ Q given

2 P → R given


4 P 1 And-Elim

Prove S





1 P ˄ Q given

2 P → R given


4 P 1 And-Elim

5 R 4,2 Modus Ponens

Prove S





1 P ˄ Q given

2 P → R given


4 P 1 And-Elim


6 Q 1 And-Elim

Prove S





1 P ˄ Q given

2 P → R given


4 P 1 And-Elim


6 Q 1 And-Elim

7 Q ˄ R 5,6 And-Intro

Prove S





1 P ˄ Q given

2 P → R given


4 P 1 And-Elim


6 Q 1 And-Elim

7 Q ˄ R 5,6 And-Intro

8 S 7,3 Modus Ponens

Prove S


Proof systems

o There are many natural deduction systems; they are typically “proof checkers”, sound but not complete

o Natural deduction uses lots of inference rules which introduces a large branching factor in the search for a proof

o In general, you need to do “proof by cases” which introduces even more branching


1 P ˅ R

2 Q ˅ R

3 P → R

Prove R


Propositional resolution

o Resolution rule:

o Single inference rule is a sound and complete proof system

o Requires all sentences to be converted to conjunctive normal form (CNF)


α ˅ β

¬β ˅ γ

α ˅ γ


Conjunctive normal form (CNF)

o Conjunctive normal form (CNF) formulas:


CBADBCBA


Conjunctive normal form (CNF)

o Conjunctive normal form (CNF) formulas:

• (A ˅ B ˅ ¬C) is a clause, which is a disjunction of literals

• A, B, and ¬C are literals, each of which is a variable or the negation of a variable

• Each clause is a requirement that must be satisfied and can be satisfied in multiple ways

• Every sentence in propositional logic can be written in CNF


CBADBCBA


Converting to CNF

1. Eliminate arrows using definitions

2. Drive in negations using De Morgan’s Laws

3. Distribute OR over AND

4. Every sentence can be converted to CNF, but it may grow exponentially in size


CABACBA


CNF conversion example


DCBA



1. Eliminate arrows


DCBA

DCBA



1. Eliminate arrows

2. Drive in negations


DCBA

DCBA

DCBA



1. Eliminate arrows

2. Drive in negations

3. Distribute


DCBA

DCBA

DCBA

DCBDCA



o Resolution rule:


α ˅ β

¬β ˅ γ

α ˅ γ



o Resolution rule:

o Resolution refutation:• Convert all sentences to CNF• Negate the desired conclusion (converted to CNF)• Apply resolution until either

Derive false (a contradiction) Can’t apply any more


α ˅ β

¬β ˅ γ

α ˅ γ



o Resolution refutation:• Convert all sentences to CNF• Negate the desired conclusion (converted to CNF)• Apply resolution until either

Derive false (a contradiction) Can’t apply any more

o Resolution refutation is sound and complete• If we derive a contradiction, then the conclusion follows from

axioms• If we can’t apply any more, then the conclusion cannot be proved

from the axioms



Propositional resolution example


1 P ˅ Q

2 P → R

3 Q → R

Prove R Step Formula Derivation




1 P ˅ Q

2 P → R

3 Q → R


1 P ˅ Q given

2 ¬P ˅ R given





1 P ˅ Q given

2 ¬P ˅ R given

1 P ˅ Q

2 P → R

3 Q → R





1 P ˅ Q given

2 ¬P ˅ R given

3 ¬Q ˅ R given

1 P ˅ Q

2 P → R

3 Q → R





1 P ˅ Q given

2 ¬P ˅ R given

3 ¬Q ˅ R given

4 ¬R negated conclusion

1 P ˅ Q

2 P → R

3 Q → R





1 P ˅ Q given

2 ¬P ˅ R given

3 ¬Q ˅ R given


5 Q ˅ R 1,2

1 P ˅ Q

2 P → R

3 Q → R





1 P ˅ Q given

2 ¬P ˅ R given

3 ¬Q ˅ R given


5 Q ˅ R 1,2

6 ¬P 2,4

1 P ˅ Q

2 P → R

3 Q → R





1 P ˅ Q given

2 ¬P ˅ R given

3 ¬Q ˅ R given


5 Q ˅ R 1,2

6 ¬P 2,4

7 ¬Q 3,4

1 P ˅ Q

2 P → R

3 Q → R





1 P ˅ Q given

2 ¬P ˅ R given

3 ¬Q ˅ R given


5 Q ˅ R 1,2

6 ¬P 2,4

7 ¬Q 3,4

8 R 5,7

1 P ˅ Q

2 P → R

3 Q → R





1 P ˅ Q given

2 ¬P ˅ R given

3 ¬Q ˅ R given


5 Q ˅ R 1,2

6 ¬P 2,4

7 ¬Q 3,4

8 R 5,7

9 [] 4,8

1 P ˅ Q

2 P → R

3 Q → R





1 P ˅ Q given

2 ¬P ˅ R given

3 ¬Q ˅ R given


5 Q ˅ R 1,2

6 ¬P 2,4

7 ¬Q 3,4

8 R 5,7

9 [] 4,8

1 P ˅ Q

2 P → R

3 Q → R

false ˅ R

¬R ˅ false

false ˅ false


Example problem


Prove R

1 (P → Q) → Q

2 (P → R) → R

3 (R →S) → ¬(S → Q)

Do it yourselves!


Proof strategies

o Unit preference: prefer a resolution step involving a unit clause (clause with one literal)

• Produces a shorter clause – which is good since we are trying to produce a zero-length clause, that is a contradiction



Proof strategies

o Unit preference: prefer a resolution step involving a unit clause (clause with one literal)

• Produces a shorter clause – which is good since we are trying to produce a zero-length clause, that is a contradiction

o Set of support: choose a resolution involving the negated goal or any clause derived from the negated goal

• We’re trying to produce a contradiction that follows from the negated goal, so these are “relevant” clauses

• If a contradiction exists, one can find one using the set-of-support strategy


Course: Engineering Artificial Intelligence

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