FOPL

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Topic 9.

The first-order predicate logic

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How can we represent the following

statements?

Fish live in water.

If he is sick he needs a doctor.

The SAT problem cannot be solved in polynomial time unlessP is equal to NP.

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Capabilities of an AI representation

language

Handle qualitative knowledge.

clear c

clear(a)

ontable(a)

ontable(b)

on(c,b)

a b

c

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cube(b)

cube(a)

pyramid(c)

Allow new knowledge to be inferred from a set of rules andfacts.

XYon(Y,X)clear(X)

Allow representation of general principles as well as specificsituations.

A useful AI representation needs variables.

Capture complex semantic meaning.

hassize(bluebird,small)hascolor(bluebird,blue)

hascovering(bluebird,feathers)

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sa ue r , r

isa(bluebird,vertebrate)

Allow for meta-level reasoning.

An intelligent system must be able to solve problems as well as explainhow it solved the problem and why it made certain decisions

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Examples of propositional() statements:

: arcoa s ac .Q: Sugar is hydrocarbon().R: Smith is married.

The symbols, such as P, Q, and R, used to

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symbols.

First order predicate logic (FOPL)(1)Can you represent the following statements by thepropositional logic?

Every man is mortal.

Confucius is a man. Therefore, Confucius is mortal.

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First order predicate logic (2)


Confucius is a man. man(confucius)

Confucius is mortal. mortal(confucius)

We must prove the following formula is valid in order to prove that Confucius ismortal.

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(X (man(X) mortal(X))) man(confucius) mortal(confucius)

The syntax of the predicate calculus (1)

Symbols begin with a letter and are followed by legalcharacters.

A..Z, a..z.,0..9,_.

Example

Legal symbolsGeorge, fire3, tom_and_jerry, XXX, friends_of

Illegal symbols

@#$, 3black, no blank allowed, hi!!

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The syntax of the predicate calculus (2)Predicate calculus symbols include

Truth symbols: true and false.

. Example: george, myMother.

Variable symbols: the first character is uppercase. Example: X, Man.

Function symbols: the first character is lowercase. A function is amapping that maps a list of constants to a constant. function_name(t1,t2,,tn)

If plus is a function of arity 2 with domain the integer, plus(2,3) is a function

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expression whose value is the integer 5.


Predicate calculus terms are defined as follows. A constant is a term.

A variable is a term.

Iffis an n-arity function symbol, and t1, , tn are terms, thenf(t1, , tn) is a term.

All terms are generated by applying the above rules.A predicate calculus term is either a constant, variable orfunction expression.

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xamp e.cat, times(2,3), X, blue, mother(jane), kate.

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A predicate names a relationship between zero or more objectsin the world. Predicate symbols begin with a lowercase letter.

Example:

like, equals, on, near, part_of

An atomic sentence is a predicate of arity n followed by n terms. Example:

likes(george, kate)

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es ,george r en s a er_o av , a er_o an rew The truth values, true and false, are atomic sentences.


First order predicate calculus includes two variable qualifiers and that constrain the meaning of a sentence containing a variable.

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The syntax of the predicate calculus (6) Universal qualifier

indicates that the sentence is true for all values of the variable.

Example: X like(X,ice_cream) is true for all values in the domain ofthe definition of X

Existential qualifier indicates that the sentence is true for at least one value of the

variable in the domain.

Example: Y friends(Y,peter) is true if there is at least one object,indicated by Y, that is a friend of peter.

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Predicate calculus sentences are defined recursively as follows.Ever atomic sentence is a sentence.

1. Ifs is a sentence, then s is a sentence.

2. Ifs1 and s2 are sentences, then s1 s2 is a sentence.

3. Ifs1 and s2 are sentences, then s1 s2 is a sentence.

4. Ifs1 and s2 are sentences, then s1 s2 is a sentence.5. Ifs1 and s2 are sentences, then s1 s2 is a sentence.

6. If X is a variable and s is a sentence, then X s is a sentence.

7. If X is a variable and s is a sentence, then X s is a sentence.

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. X foo(X,two,plus(two,three)) equal(plus(two,three),five) is a sentence.

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weather(rain, monday) go(tom,mountains)If it does not rain on Monday, Tom will go to the mountains.

X(basketball_player(X) tall(X))

All basketball player are tall.

X (person(X) likes(X,anchovies))

Some people like anchovies.

X likes(X,taxes)

Nobody likes taxes.

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,

For all X, X is clear if there does not exist a Y such that Y is on X.

No unique mapping of sentences into predicate calculus expressions.

meaning definition

either p or q (PQ) (PQ)

p unless q QP

p because q (PQ) (QP)

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no p is q PQ

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A semantics for the predicate calculus (1)

The database of predicate calculus expressions, each havingtruth value T, describes the state of the world.

Example.mother(eve,abel), mother(eve,cain), father(adam,abel),father(adam, cain).

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A semantics for the predicate calculus (2)The truth value of an expression E with respect to aninterpretation I over a nonempty domain D is determined by:

e va ue o a cons an s e e emen o s ass gne o y .

The value of a variable is the set of elements of D it is assigned to by I.

The value of a function expression is the element of D obtained by evaluating the function forthe parameter values assigned by I.

The value of truth symbol true is T and false is F. The value of an atomic sentence is either T or F, as determined by I.

The truth value of expressions using , , , , and is determined from the values oftheir operands as defined previously.

The value of X s is T ifs is T for all assi nments to X under I otherwise it is F.

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The value of X s is T if there is an assignment to X in I under which s is T; otherwise, it is F.

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Example.Consider the formula XY p(X,Y).

.

D={1,2} (the domain of X and Y)

p(1,1) p(1,2) p(2,1) p(2,2)

T F F T

Assignment for predicate p.

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In this interpretation, for every X, there always exists aY such that p(X,Y) is true. Therefore, XY p(X,Y) is Tin this interpretation.

Example.Consider the formula X(p(X) q(f(X),a)).Let us define an interpretation as follows. D={1,2}

Assignment for function f Assignment for constant a

p(1) p(2) q(1,1) q(1,2) q(2,1) q(2,2)

Assignment for predicates p and q

f(1) f(2)

2 1

a

1

Therefore, X(p(X) q(f(X),a)) is T in this interpretation.

If X=1, p(1) q(f(1),1))= p(1) q(2,1)=F F=T.If X=2, p(2) q(f(2),1))= p(2) q(1,1)=T T=T.

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For a predicate calculus expression s and an interpretation I: Ifs has a value ofT under I and a particular variable assignment, then

I is said to satisfy s .

s is satisfiable if and only if there exist an interpretation and variableassignment that satisfy it; otherwise, it is unsatisfiable.

Ifs has a value T for all possible interpretations, s is said to be valid.

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A semantics for the predicate calculus (3)

Some relationships between negation and the universal andexistential quantifiers.

X p(X) X p(X)

X p(X) X p(X)

X q(X) Y q(Y) X q(X) Y q(Y)

X (p(X) q(X)) X p(X) Y q(Y)

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p q p q

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First-order predicate calculusallowsquantified variables to refer to objects in the

oma n o scourse an no o pre ca es orfunctions.

Example. (L) L(george,kate) is not a well-formed expression in the first-order predicatecalculus.

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Inference rulesAn inference rule is a way to produce additional wffs fromother ones.

n erence ru es prov e a compu a ona y eas e way o e erm newhether an expression, a component of an interpretation, logicallyfollows for that interpretation.

When every sentence X produced by a set of inference rules,R, operating on a set S of expressions logically follows fromS, R is said to be sound.

Example: modus ponens is a sound inference rule.

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from S, then R is said to be complete.

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Useful inference rules Modus ponens: If P and P Q are true, then we can infer that Q is true.

Modus tollens: If P Q is true and Q is false, we can infer P.

And elimination: If P Q is true, we can infer that P and Q are true.

And introduction: If P and Q are true, then P Q is true.

Universal instantiation: IfX p(X) is true and a is from the domain of X, wecan infer p(a).

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If it is raining then the ground will be wet.

P QIt is raining.

P.


X (man(X) mortal(X))

Confucius is a man.

.

Applying modus ponens, we can infer that Q is true.

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Confucius is mortal.

Applying universal instantiation, we can infer that man(confucius) mortal(confucius).

Applying modus ponens, we can infer that mortal(confucius).

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All dogs are animals.All animals will die.Fido is a dog.We want to prove that Fido will die rom above statements.

All dogs are animals.

X (dog(X) animal(X))Fido is a dog.

dog(fido).Modus ponens and universal instantiation ({fido/X}) give animal(fido).

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an ma s w e.Y (animal(Y) die(Y)).Modus ponens and universal instantiation ({fido/Y}) give die(fido).

X likes(george,X)

X likes eor e,X

If the domain of X is infinite, for example, X

ranges over the set of all humans, the testof the sentences may never halt.

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Decidability A logical system is decidable if it is possible to produce an

algorithm that will determine whether any wff is a theorem.

The propositional logic is decidable.

FOPL is not decidable.

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Monotonicity

A lo ical s stem is monotonic if a valid roof in the s stemcannot be made invalid by adding additional premises orassumptions. If we can prove

then we can also proveC-|B}A,{

C-|BABA ''

A and B can be anything, including A and B.

The propositional logic and FOPL are monotonic.

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Deductive, Abduction and Inductive

reasoning

Deductive reasoninA A B

Based on the use of modus pones and the other deductive rules of reasoning.

Dealing with certainties.

Abductive reasoning Not logically sound

Reasoning from observed facts to the best explanation.

Inductive reasoning

B A B

A

. Induction is the process of inferring the general case from the specific.

Useful for dealing with uncertainties.

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re ca e a cu us eso u on

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Exam lewffs

1. p(f(Y),a) q(b,c)

2. p(X,a) r(X,c) s(a,b)

3. p(f(Y),a) r(f(Y),c) s(a,b) (substituting f(Y) for X in 2)

4. q(b,c) r (f(Y),c) s(a,b) (from resolving 1 and 3)

The appropriate substitution is computed by a process called unification.

ange

symbols

(Unification)

Resolution

Convert arbitrary wffs to clause

forms

Example. Clause form

1. p(f(Y),a) q(b,c)

2. p(X,a) r(X,c) s(a,b)

wffswff to be

proved

Convert to clauseform

Unification add

Resolution

Produce ? New resolvent?

yes

no

yes

proved

no

stop

Flowchart of resolution refutation in the predicate calculus

You maybe neverreach this step

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Example.

All dogs are animals.All animals will die.Fido is a dog.We want to prove that Fido will die from above statements.

All dogs are animals. CLAUSE FORM

X (dog(X) animal(X)) dog(X)animal(X)All animals will die.

Y (animal(Y) die(Y)) animal(Y)die(Y)Fido is a dog.

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dog(fido). dog(fido)Negate the conclusion that Fido will die.

die(fido)

substitution

Resolution proof for the dead dog problem

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Convert to clause form (1/2)1. Eliminate implication signs.

P Q P Q

2. Reduce sco e of ne ation si ns.

(P Q) P Q

3. Standardize variables.

Each quantifier is renamed so that each has its own variable symbol.

(X P(X) ( XQ(X))) (X P(X) (YQ(Y)))

4. Move all qualifiers to the left without changing their order.

This form is called theprenex normalform that all the qualifiers are in frontas a prefix and the expression or matrix after.

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(X)(Y) (a(X) b(Y))

prefix matrix

Convert to clause form (2/2)

5. Eliminate existential quantifiers.

XY mother X,Y skolen function X mother X,f X Y depends on X.

Y g(Y) skolen constant g(sk) The existential quantifier being eliminated is not within the scope of

any universal quantifiers. Skolen functions must be new. Put the matrix in conjunctive form.

6. Eliminate universal quantifiers.

7. Convert the ex ression to the con unct of dis uncts forms.

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8. Call each conjunct a separate clause.

9. Rename variables.

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Example(X)((a(X) b(X)) (c(X,l) (Y)((Z)(c(Y,Z)) d(X,Y)))) ( X)(e(X))

1. Eliminate implication signs. (X)( (a(X) b(X)) (c(X,l) (Y)((Z)( c(Y,Z)) d(X,Y)))) ( X)(e(X))

2. Reduce the scopes of negation signs. (X)( ( a(X) b(X)) (c(X,l) (Y)((Z)(c(Y,Z)) d(X,Y)))) ( X)(e(X))

3. Standardize the variables; that is, rename variables so that variables bound byeren quan ers ave un que names.

( X)( ( a(X) b(X)) (c(X,l) ( Y)(( Z)(c(Y,Z)) d(X,Y)))) ( W)(e(W))

4. Convert to prenex normal form. (X)( Y)( Z)( W)( ( a(X) b(X)) (c(X,l) ( c(Y,Z)) d(X,Y)))) (e(W))

5. Eliminate existential qualifiers. ( X) ( W)( ( a(X) b(X)) (c(X,l) (c(f(X),g(X))) d(X, f(X))))) (e(W))

6. Remove all universal qualifiers. ( ( a(X) b(X)) (c(X,l) (c(f(X),g(X))) d(X, f(X))))) (e(W))

7. Transform to conjunctive normal form.

( a(X) b(X) c(X,l) e(W)) ( a(X) b(X) c(f(X),g(X)) d(X, f(X)) e(W))8. Call each conjunct a separate clause.

a) a(X) b(X) c(X,l) e(W)b) a(X) b(X) c(f(X),g(X)) d(X, f(X)) e(W)

9. Standardize the variables apart again.a) a(X) b(X) c(X,l) e(W)b) a(U) b(U) c(f(U),g(U)) d(U, f(U)) e(V)

Unification

Unification is an al orithm for determinin the substitutionsneeded to make two predicate calculus expressions match.

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Substitution (1)Definition.

A substitution is a finite set of the form {t1/v1,,tn/vn},where every vi is a variable, every ti is a term different from vi,and no two elements in the set have the same variable after thestroke symbol. When t1,,tn are ground terms, the substitution is called a ground

substitution.

X/Y indicates that X is substituted for Y in the original

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expression.

Examples. Some instance of the expression foo(X,a,goo(Y)) generated by

legal substitutes are as follows.1. foo(fred,a,good(Z)) foo(X,a,goo(Y)) {fred/X,Z/Y}

2. foo(W,a,goo(jack)) foo(X,a,goo(Y)) {W/X,jack/Y}

3. foo(Z,a,goo(moo(Z))) foo(X,a,goo(Y)) {Z/X,moo(Z)/Y}

X/Y indicates that X is substituted for Y in the originalexpression.

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Substitution (2)Definition.

Let ={t1/v1,,tn/vn} be a substitution and Ebe an expression.Then E is an expression obtained from Eby replacingsimultaneously each occurrence of the variable vi,

i=1,2,n, in E by the term ti.

E is called an instance ofE.

Example

Let ={a/X,f(b)/Y,c/Z} and E=p(X,Y,Z). Then E = p(a,f(b),c).

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Definition.

Composition of unification

substitutions

Let ={t1/x1,,tn/xn} and ={u1/y1,,um/ym} be twosubstitutions.Then the composition of and is the substitution,

denoted by , that is obtained from the set{t1/x1,, tn/xn, u1/y1,,um/ym}

by deleting any element tj/xj for which tj=xj, and any elementui/yi such that yi is among {x1,,xn}.

The composition of substitutions is associative. ()= ()

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Exam le. {f(Y)/X,Z/Y}{a/X,b/Y,Y/Z}

{f(b)/X,Y/Y,a/X,b/Y,Y/Z}={f(b)/X ,b/Y,Y/Z}

{g(X,Y)/Z}{a/X,b/Y,c/W,d/Z}={g(a,b)/Z, a/X,b/Y,c/W,d/Z}

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UnifierDefinition.

A substitution is called a unifier for a set {E1, E2, , Ek} ifand only if E1 =E2 ==Ek . The set {E1, E2, , Ek} is saidto be unifiable if there is a unifier for it.

Example.The set {p(a,Y),p(X,f(b)} is unifiable because the substitution={a/X,f(b)/Y} is a unifier for the set.

{p(a,Y)}={p(a,f(b))}

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p , = p a,

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Most General Unifier (MGU)Definition.

A unifier for a set {E1, E2, , Ek} of expressions is a mostgeneral unifier (mgu) if and only if for each unifier for theset there is a substitution such that =.

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Unification algorithm

The unification algorithm can

find the most general unifier for a finite unifiable set ofnonempty expressions.

determine whether or not a set of expressions is unifiable.

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function unify(E1,E2)

begin // E1 and E2 are in the list representation

caseE1 and E2 are constants or the empty list:

if E1=E2 then return {} else return FAIL;

either E1 or E2 is empty then return FAIL.

E1 is a variable:

Predicate calculus

syntax List syntax

p(a,b) (p a b)

p(f(a),g(X,Y)) (p (f a) (g X Y))

.

E2 is a variable:

if E2 occurs in E1 then return FAIL else return {E1/E2}.

otherwise:

begin

HE1 := first element of E1;

HE2 := first element of E2;

SUBS1 := unify(HE1,HE2);

if SUBS1== FAIL then return FAIL.

TE1 := apply(SUBS1, rest of E1); // substitution: (rest of E1)SUBS1TE2 := apply(SUBS1, rest of E2); // substitution:(rest of E2)SUBS1

SUBS2 := unify(TE1,TE2);

if SUBS2==FAIL then return FAIL.

else return composition(SUBS1,SUBS2);// composition

end

end

end

Example

Find an mgu for W={p(a,X,f(g(Y))),p(Z,f(Z),f(U))}.

Answer: {a/Z,f(a)/X, g(Y)/U}

W={(p a X (f (g Y))), (p Z (f Z) (f U)) }{}{a/Z}{f(a)/X}{}{g(Y)/U}=

{a/Z}{f(a)/X}{g(Y)/U}=

{a/Z,f(a)/X}{g(Y)/U}=

{a/Z,f(a)/X,g(Y)/U}{}{a/Z}{f(a)/X}{}{g(Y)/U}

,

Unify(p, p)

{}

unify{(a X (f (g Y))), (Z (f Z) (f U)) }

Unify(a, Z){a/Z}

unify{(X (f (g Y))), ( (f a) (f U)) }

{f(a)/X}{}{g(Y)/U}

{a/Z}{f(a)/X}{}{g(Y)/U}

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Unify(X,(f a)){f(a)/X}

unify{ (f (g Y)), (f U) }

g(Y)/U

Unify(f,f)

{}

unify{((g Y)), (U) }

{}{g(Y)/U}

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ExampleConsider the following set of formulas:F1: X (c(X)(w(X) r(X)))

F2: X (c(X) o(X)).

Show that G is a logical consequence of F1 and F2.

1. c(X) w(X) from F12. c(Y) r(Y)3. c(a) from F24. o(a)

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. 6. r(a) (5) and (4)7. r(a) (3) and (2)8. (6) and (7)

Therefore, G is a logical consequence of F1 and F2.

Example:Premises:1. The custom officials searched everyone who entered this country who

was not a VIP.2. Some of the drug pushers entered this country and they were only

searched by drug pushers.3. No drug pusher was a VIP.Conclusions: Some of the officials were drug pushers.

Let

e(X) mean X entered this country,v(X) mean X was a VIP,s(X,Y) mean Y searched X,c(X) mean X was a custom official, and

X mean X was a dru usher.

Inference and Resolution for Problem Solving52

Premises

X(e(X) v(X) Y (s(X,Y) c(Y)))

X(e(X) p(X) Y (s(X,Y) p(Y)))

X(p(X)v(X))Conclusions

X(c(X) p(X)).

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The resolutionproof is as follows.

Premises

X(e(X) v(X) Y (s(X,Y) c(Y))) X(e(X) p(X) Y (s(X,Y) p(Y)))

X(p(X) v(X))

(1) e(X) v(X) s(X,f(X))

(2) e(Y) v(Y) c(f(Y))

(3) p(g(Y1))(4) e(g(Y2))(5) s(g(Y3),Y3) p(Z)(6) p(W) v(W)(7) p(T) c(T)

(8) v(g(Y1)) from (3) and (6) {g(Y1)/W}

X(c(X) p(X)).

(9) v(g(Y2)) c(f(g(Y2))) from (2) and (4) {g(Y2)/Y}(10) c(f(g(Y2))) from (8) and (9) {Y2/Y1}(11) v(g(Y2)) s(g(Y2),f(g(Y2))) from (1) and (4) {g(Y2)/X}(12) s(g(Y2),f(g(Y2))) from (8) and (11) {Y2/Y1}(13) e(g(Y2) v(g(Y2)) from (1) and (12) {g(Y2)/X}(14) e(g(Y2) from (8) and (13) {Y2/Y1 }(15) from (4) and (14)

Predicate-calculus resolution

Predicate-calculus resolution is sound.

Predicate-calculus resolution is refutation complete.

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Example.Anyone passing his history exams and winning the lottery is happy. Butanyone who studies or is lucky can pass all his exams. John did not studybut he is lucky. Anyone who is lucky wins the lottery. Is John happy?

CLAUSE FORM

Inference and Resolution for Problem Solving55

THE CLAUSE FORM OF THE NEGATION OF CONCLUSION

One resolution refutation for the happy student problem

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Example.All people who are not poor and are smart are happy. Those people whoread are not stupid. John can read and is wealthy. Happy people haveexciting lives. Can anyone be found with an exciting life

CLAUSE FORM

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Answerextraction

One resolution refutation for the exciting life problem

{Z/W}{X/Z}{Y/X}{john/Y}{}={john/W,john/Z,john/X,john/Y}

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Example.Fido the dog goes whenever John, his master, goes. John is at thelibrary. Where is Fido?

Negation of the conclusion

substitution

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substitution

{X/Z}{library/X}={library/Z,library/X}

Example.Everyone has a parent. The parent of a parent is a grandparent.Given the person John, prove that John has a grandparent.

Conclusion

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CLAUSE FORM

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{john/X,pa(john)/Y}

{pa(john)/X,pa(pa(john))/V)}

{john/W,V/Z}{john/X,pa(john)/Y}{pa(john)/X,pa(pa(john))/V}={john/W,pa(X)/Z,john/X,pa(john)/Y}{pa(john)/X,pa(pa(john))/V)}={john/W, pa(pa(john)) /Z,john/X,pa(pa(john))/Y, pa(pa(john))/V}

FOPL

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