[PDF] - First-order Logic [RN2] Sec 7.1-7.6 Chap 8-9 [RN3] Sec 7.1-7.6 Chap PDF Document

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First-order Logic [RN2] Sec 7.1-7.6 Chap 8-9 [RN3] Sec 7.1-7.6 Chap 8-9

CS 486/686 University of Waterloo Lecture 20: November 15, 2012

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Outline

First-order logic

– Syntax and semantics

Inference

– Propositionalization with ground inference – Lifted resolution

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Logic

General language

– Can encode any kind of deterministic and discrete knowledge – Well defined syntax and semantics – Knowledge base: store domain knowledge

General algorithms

– Search by inference – Can infirm or confirm conclusions

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Syntax of a very simple logic

Sentence  AtomicSentence | ComplexSentence
AtomicSentence  True | False | Symbol
Symbol  P | Q | R | …
ComplexSentence  Sentence

| (Sentence  Sentence ) | (Sentence  Sentence) | (Sentence  Sentence ) | (Sentence  Sentence) Symbols are binary variables

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Logical connector semantics

P Q

P

P  Q P  Q P  Q P  Q false false true false false true true false true true false true true false true false false false true false false true true false true true true true

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Recall the Street Puzzle

1 2 3 4 5

Ni = {English, Spaniard, Japanese, Italian, Norwegian} Ci = {Red, Green, White, Yellow, Blue} Di = {Tea, Coffee, Milk, Fruit-juice, Water} Ji = {Painter, Sculptor, Diplomat, Violinist, Doctor} Ai = {Dog, Snails, Fox, Horse, Zebra} The Englishman lives in the Red house The Spaniard has a Dog The Japanese is a Painter The Italian drinks Tea The Norwegian lives in the first house on the left The owner of the Green house drinks Coffee The Green house is on the right of the White house The Sculptor breeds Snails The Diplomat lives in the Yellow house The owner of the middle house drinks Milk The Norwegian lives next door to the Blue house The Violinist drinks Fruit juice The Fox is in the house next to the Doctor’s The Horse is next to the Diplomat’s

Who owns the Zebra? Who drinks Water?

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Example – Street Puzzle

Symbols: Vin, Vic, Vid, Vij, Via

i  {1,2,3,4,5} n  {English, Spaniard, Japanese, Italian, Norwegian} c  {Red, Green, White, Yellow, Blue} d  {Tea, Coffee, Milk, Fruit-juice, Water} j  {Painter, Sculptor, Diplomat, Violinist, Doctor} a  {Dog, Snails, Fox, Horse, Zebra}

Example:

V2red = true: the 2nd house is red V2red = false: the 2nd house is not red

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Example – Street Puzzle

Constraint: the Spaniard has a dog
Constraint: the green house is on the

immediate left of the red house (V1spaniard  V1dog)  (V2spaniard  V2dog)  (V3spaniard  V3dog)  (V4spaniard  V4dog)  (V5spaniard  V5dog)

V1red  (V1green  V2red)  (V2green  V3red) 

(V3green  V4red)  (V4green  V5red)  V5green

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Example: 4-Queens problem

Symbols: Qij i,j{1,2,3,4}
Qij = true: queen at

location (i,j)

Qij = false: no queen at

location (i,j)

1 2 3 4 1 2 3 4

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Example: 4-Queens problem

At least one queen in row 1:
At most one queen in row 1:

1 2 3 4 1 2 3 4

Q11  Q12  Q13  Q14 Q11  Q12  Q13  Q14 Q12  Q11  Q13  Q14 Q13  Q11  Q12  Q14 Q14  Q11  Q12  Q13

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Knowledge Base (KB)

Knowledge base

– Encode problem description and any knowledge that could help solve the problem – Database of facts and constraints – Possible language: propositional logic

Entailment

– What can we derive/conclude from KB? – KB |= :  follows from KB

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Semantics

Entailment: KB |= 

–  is true in all the models in which KB is true

models KB



models KB



models KB

 KB |=  KB |  KB | 

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Inference

Process of verifying the truth of a

formula

Simple inference algorithm:

– Build a truth table that enumerates all models – Verify that formula is true in all models where KB is true

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Truth table

P Q R KB true true true false true true false false true false true false true false false false false true true false false true false false false false true true false false false true KB: P  Q, Q  R,

Q

KB |= P? KB |= P? KB |= Q? KB |= Q? KB |= R? KB |= R? no yes no yes no no

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Efficiency

Building a truth table is inefficient:

exponential in the number of variables

Alternatives: backtracking, resolution

– Rewrite “KB |= ?” as KB’ = (KB and ) – Show that KB’ is not satisfiable (e.g., there doesn’t exist any model for which KB is true, but  is false)

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Convenience

KB may contain any formula that follows the

syntax of our simple logic.

– Inconvenient: too many possible formula – Can we simplify syntax?

Conjunctive normal form (CNF)

– Only use , ,  – Conjunction of clauses, where each clause is a disjunction of (possibly negated) literals – Example: (V1  V2  V3)  (V1  V2)  (V3  V2)

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Logical equivalence

How do we transform a KB in CNF?
Logical equivalence rules

– (A)  A – (A  B)  A  B – (A  B)  (A  B)  (B  A) – (A  B)  (A  B) – (A  B)  (A  B) – (A  (B  C))  ((A  B)  (A  C))

Can every KB be transformed in CNF in polynomial

time and space?

– No: last rule may yield exponentially many clauses

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Resolution

Idea: generate new sentences by implications

– Unit resolution: A  (A  B) |= B – General resolution: (A  B)  (B  C) |= (A  C)

How do we do this with KB in CNF?

– Unit resolution: A  (A  B) |= B – General resolution: (A  B)  (B  C) |= (A  C)

Algorithm: apply resolution to every possible pair of

clauses until

– Empty clause is generated: false KB – No more clauses can be generated: all generated clauses are entailed

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Concise Language

Propositional logic provides a general language

to encode deterministic domain knowledge.

While the encoding is precise and well defined

it is often tedious or complicated.

Simple English sentences often lead to long

logical formula.

Can we make propositional logic more concise

and more natural?

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Recall: 4-Queens problem

At least one queen in row 1:
At most one queen in row 1:

1 2 3 4 1 2 3 4

Q11  Q12  Q13  Q14 Q11  Q12  Q13  Q14 Q12  Q11  Q13  Q14 Q13  Q11  Q12  Q14 Q14  Q11  Q12  Q13

Repeat for every row, column and diagonal. This is too tedious! Can we be more concise?

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First-order Logic

World

– consist of objects and relations – Has many objects

First-order logic:

– Natural: use predicates to encode relations and constants to encode objects – Concise: use quantifiers (e.g., , ) to talk about many objects simultaneously

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First-order Logic Syntax

Sentence  Predicate(Term, …)

| Term = Term | (Sentence Connective Sentence) | Quantifier Variable, … Sentence | Sentence

Term  Function(Term, …) | Constant | Variable
Connective   |  |  | 
Quantifier   | 
Constant  A | X1 | John | …
Variable  a | x | s | …
Predicate  Before | HasColor | Raining | …
Function  Mother | LeftLeg | …

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Example: kinship domain

Elizabeth is the mother of Charles

Mother(Elizabeth,Charles)

Charles is the father of William

Father(Charles,William)

One’s grandmother is the mother of
ne’s parent

x,z y Grandmother(x,z)  Mother(x,y)  Parent(y,z)

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Symbols

Constant symbols: objects

– E.g. William, Elizabeth, Charles

Predicate symbols: relationships

– Binary: Mother( , ), Grandmother( , ) – Unary: Female( ) – Predicates have a truth value

Function symbols: functions

– Denote an object – E.g.: MotherOf(William) = Elizabeth

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Quantifiers

Universal quantifier: 

– For all – x P(x)  P(const1)  P(const2)  …

Existential quantifier: 

– There exists – x P(x)  P(const1)  P(const2)  …

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Nested Quantifiers

Order of identical quantifiers doesn’t

matter

Brothers are siblings

– x y Brother(x,y)  siblings(x,y) – y x Brother(x,y)  siblings(x,y)

Similarly for existential quantifiers

– x y P(x,y)  y x P(x,y)

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Nested Quantifiers

Order of different quantifiers matters
x y Loves(x,y)

– Everyone loves someone – Conjunction of disjunctions (CNF)

x y Loves(x,y)

– There is a person that loves everyone – Disjunction of conjunction (DNF)

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Connections between  and 

We only need one of the quantifiers
De Morgan’s rule

– x P  x P

P  Q  (P  Q)

– x P  x P

(P  Q)  P  Q

– x P  x P P  Q  (P  Q) – x P  x P P  Q  (P  Q)

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Equality

Special relation
We could define an “Equality” predicate

– x = y  Equality(x,y)

True: when x and y are the same
False: otherwise

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Practice: 4-Queens problem

4-queens problem in first-order logic
Predicates:

– Queen( , )

Constants:

– 1, 2, 3, 4 (column and row numbers)

Column constraints:

– i,j1,j2 Queen(i,j1)  j1  j2  Queen(i,j2) – i j Queen(i,j)

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Practice: street puzzle

Predicates:

– House(), Person( ), Color( ), Drink( ), Job( ), Animal( ) – Attr( , ) (attribute of)

Constants:

– 1, 2, 3, 4, 5 (house number) – English, Spaniard, Japanese, Italian, Norwegian – Red, Green, White, Yellow, Blue – Tea, Coffee, Milk, Juice, Water – Painter, Sculptor, Diplomat, Violinist, Doctor – Dog, Snails, Fox, Horse, Zebra

Function:

– Left(x): number of the house to the immediate left of x

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Practice: street puzzle

The Spaniard has a dog

– x House(x)  (Attr(x,Spaniard)  Attr(x,Dog))

the green house is on the immediate left of

the red house

– Attr(1,Red)  – Attr(5,Green)  – x House(x)  x1  (Attr(x,Red)  Attr(Left(x),Green))

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Propositional vs first-order logic

Propositional logic:

– variables

First-order logic:

– Quantifiers, predicates, constants, functions

Syntactically different!
Are they equally expressive?

– Prop logic  1st-order logic: – 1st-order logic  prop logic: yes yes (finite domains) no (infinite domains)

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Propositional  first-order logic

Variables  predicates
Indices  constants
4-queens problem:

– Q11  Q12  Q13  Q14 – Q(1,1)  Q(1,2)  Q(1,3)  Q(1,4)

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First-order  propositional logic

Quantifiers

–   conjunction –   disjunction

Predicates  variables
Constants  indices
Functions of constants  functions of

indices

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First-order  propositional logic

x House(x)  x1 

(Attr(x,Red)  Attr(Left(x),Green))

x Housex  x1  (Attrx,RedAttrLeft(x),Green)
(House1  11  (Attr1,Red  AttrLeft(1),Green))

(House2  21  (Attr2,Red  AttrLeft(2),Green)) (House3  31  (Attr3,Red  AttrLeft(3),Green)) (House4  41  (Attr4,Red  AttrLeft(4),Green)) (House5  51  (Attr5,Red  AttrLeft(5),Green))

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Inference

Propositionalize KB

– Run favorite ground inference algorithm (e.g., backtracking (DPLL), resolution) – Efficient?

No: propositionalizing may yield an exponentially large

formula

x y z P(x,y,z)  conjunction of disjunctions of

conjunctions

O(nm) where n is # of constants and m is # of quantifiers
Alternative: lifted inference

– Work directly with first-order formula

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Lifted Inference

First-order logic:

– Quantifiers allow us to characterize several

bjects simultaneously
Lifted inference:

– Do inference by reasoning about several terms simultaneously – Ground terms only when necessary – Hope: determine truth value of goal formula without grounding all terms

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Lifted Resolution

Two phases 1. Reduce to lifted conjunctive normal form 2. Resolution with unification

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Reduction to Lifted CNF

Six steps 1. Eliminate implications 2. Move negations inwards 3. Standardize variables 4. Skolemize 5. Drop universal quantifiers 6. Distribute  over 

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Reduction to Lifted CNF

Example: x [y Animal(y)  Loves(x,y)] [y Loves(y,x)] 1. Eliminate implications x [y Animal(y)  Loves(x,y)]  [y Loves(y,x)] 2. Move negations inwards (apply DeMorgan’s rules) x [y (Animal(y)  Loves(x,y))]  [y Loves(y,x)] x [y Animal(y)  Loves(x,y)]  [y Loves(y,x)] x [y Animal(y)  Loves(x,y)]  [y Loves(y,x)]

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Reduction to Lifted CNF

x [y Animal(y)  Loves(x,y)] [y Loves(y,x)] 3. Standardize variables (use different names) x [y Animal(y)  Loves(x,y)]  [z Loves(z,x)] 4. Skolemize (replace existential variables by new constants or functions) x [Animal(F(x))  Loves(x,F(x))]  [Loves(G(x),x)]

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Skolemization

x P(x)

– There is at least one term x that satisfies P(x) – We don’t care what that term is

Idea: create a new constant

– Let A be a new constant which could correspond to any object, but since we don’t care what that

bject is, just denote it by this new constant.

– x P(x)  P(A)

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Skolemization

x y P(x,y)

– For each x, there is at least one term y that satisfies P(x,y) – We don’t care what that term is but it may be different for different x

Idea: create a new function

– Let f be a new function that denotes the satisfying term for each x – x y P(x,y)  P(x,f(x))

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Reduction to Lifted CNF

x [Animal(F(x))  Loves(x,F(x))]  [Loves(G(x),x)] 5. Drop universal quantifiers (since all remaining variables are universally quantified) [Animal(F(x))  Loves(x,F(x))]  [Loves(G(x),x)] 6. Distribute  over  (to get a CNF) [Animal(F(x))  Loves(G(x),x)]  [Loves(x,F(x))  Loves(G(x),x)]

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Resolution with Unification

Recall

– Unit resolution:

A  (A  B) |= B
A  (A  B) |= B

– General resolution:

(A  B)  (B  C) |= (A  C)
(A  B)  (B  C) |= (A  C)
Lifted resolution: terms may not match

exactly because of variables

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Unification

Variables are like “wild cards” that can be

anything

Find unifier:

– Unify Knows(John,x) and Knows(John,Jane)  Knows(John,Jane) {x/Jane} – Unify Knows(John,x) and Knows(y,Bill)  Knows(John,Bill) {x/Bill, y/John} – Unify Knows(John,x) and Knows(y,Mother(y))  Knows(John,Mother(John)) {x/Mother(John), y/John}

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Most General Unifier

Find unifier that is as general as possible (i.e.,

preserves as many variables as possible)

Find most general unifier:

– Unify Knows(John,x) and Knows(y,Mother(z))  Knows(John,Mother(z)) {y/John, x/Mother(z)}

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Resolution with Unification

Find clauses with positive and negative terms that can

be unified

Eliminate unifying terms and perform substitutions in

remaining terms according to most general unifier

Unit resolution:

– A(f(x),x))  [A(y,g(z))  B(y,z)] |= B(f(g(z)),z) – {x/g(z),y/f(g(z))}

General resolution:

– [A(x,y)  B(y,f(z))]  [B(u,v)  C(v,w)] |= [A(x,y)  C(f(z),w)] – {u/y,v/f(z)}

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Next class

Markov Logic Networks