Chapter 4: Prolog (Substitution,

Chapter 4:  Prolog (Substitution, Chapter 4:  Prolog (Substitution, - Start

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Chapter 4: Prolog (Substitution, - Description

Unific. ation. . and . Resolution. ). Dr Youcef Djenouri. djenouri@imada.sdu.dk. D. M552: Part . 2 . Programing. . Logic. 2017-2018. Substitution: . Definition. (1/2). Substitution. . θ . is . an operation allowing . ID: 656674 Download Presentation

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Chapter 4: Prolog (Substitution,




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Presentations text content in Chapter 4: Prolog (Substitution,

Slide1

Chapter 4: Prolog (Substitution, Unification and Resolution)

Dr Youcef Djenouridjenouri@imada.sdu.dk

DM552: Part 2 Programing Logic

2017-2018

Slide2

Substitution: Definition (1/2)Substitution θ is an operation allowing to replace some variables occurring in a formula with terms.

The goal of applying a substitution is to make a certain formula more specific so that it matches another formula. Substitutions allow for unification of formulae (or

terms).

Slide3

Substitution: Definition (2/2)A substitution is any finite mapping of variables into terms of the form: θ

: V ⇒ TERAny (finite) substitution θ can be presented asθ={t

1/X1, t2/X2… tn/Xn}ti is the term to be substituted for a variable Xi, i= 1…n. No two elements in the set have the same variable after the / symbol,

Slide4

Substitution: Example{ y/X, f(y)/Z } is a substitution. { a/Y, b/X}

is a substitution.{a/X, b/X} is not a substitution.{X/f(Y)}

is not a substitution.

Slide5

Instance (1/2) Let θ ={t1/X1, t2/X2…

tn/Xn} be a substitution and let E be an expression. Then E

θ is an expression obtained by replacing all occurrences of every vi in E with the corresponding term ti, Eθ is an instance of E.

Slide6

Instance (2/2) θ ={ a/X, f(b)/Y, c/Z} , E=P(X, Y, Z)

Eθ= P(a, f(b), c) is an instance of E. θ ={f(f(a))/X, X/Y} , E

=(P(X ) ∨ Q(Y)) Eθ= (P(f(f(a))) ∨ Q(X)) is an instance of E. θ ={ Y/X, a/Y} , E=(P(X ) ∨ Q(Y)) Eθ= (P(Y) ∨ Q(a)) is an instance of E.

Slide7

Composition of substitutionsLet θ= {t1/X1, t2/X2…

tn/Xn} and Γ={

u1/Y1, u2/Y2… um/Ym} be two substitutions. The composition of θ and Γ is denoted by θ o Γ, and it is obtained by building the set {t1 Γ

/

X

1

,

t

2

Γ

/

X

2

t

n

Γ

/

X

n

,

u

1

/Y

1

,

u

2

/Y

2

u

m

/

Y

m

}

R

emove

the

following

elements

:

t

j

Γ

/

X

j

such that

t

j

Γ

=

X

j

u

i

/

Y

j

such that

Y

j

is in {X

1

,

X

2

X

n

}

Slide8

Exampleθ= {t1/X1, t2/X2}= {

f(Y)/X, Z/Y} Γ={u1

/Y1, u2/Y2,u3/Y3}= {a/X, b/Y, Y/Z}{t1 Γ /X1, t2 Γ /X2, u1/Y1, u2/Y2,u3/Y3

}={f(b

)/X,

Y

/Y

,

a/X

,

b/Y

,

Y

/Z}

θ

o

Γ

={f(b

)/X,

Y

/Z}

Slide9

ClassworkConsider θ = {f(Y)/X} and µ = {

f(Z)/Y} Compute θ o µ and µ o

θ ?

Slide10

UnificationA substitution θ is called a unifier for a set {E1…Ek} if and

only if: E1θ =E2θ = E

3θ =…= EkθThe set {E1…Ek} is unifiable if and only if there exists a unifier for it. A unifier θ is a most general unifier for a set {E1…Ek} if and only if for each

unifier

Γ

there

exisits

a substitution

μ

such

that

Γ

= θ o

μ

.

Slide11

ExampleE1=P(X), E2= P(f(Y)) are unifiable by

Γ={f(f(Z))/X, f(Z)/Y}E

1 Γ = E2 Γ = P(f(f(Z))). θ= {f(Y)/X} is a most general unifier, we can find μ ={f(Z)/Y} such that θ o μ= {f(f(Z))/X, f(Z)/

Y

}=

Γ

Slide12

ClassworkProve that the following expressions are all unifiers by θ = {f(b)/X, b/Y, u/Z}

E = f(X, b, g(Z)).

F = f(f(Y), Y, g(u)).

Slide13

Resolution: Used to prove a consequence from a set of logical formulaeStates: Sets of clauses L1

…  LmInitially: The consequence we want to prove is negated and added to the set of clauses representing the initial state

Goal: Derive the empty clause ()

Slide14

Example (1/6)¬P(X)  P(f(X))

¬Q(a, Y)  ¬R(Y, X

)  P(X)R(b, g(a, Z))Q(a, b)Goal: P(f(g(a, c)))

Slide15

Example (2/6)Negate goal, add to formulas1. ¬P(X) 

P(f(X))2. ¬Q(a, Y)  ¬R(Y, X)

 P(X)3. R(b, g(a, Z))4. Q(a, b)5. ¬P(f(g(a, c)))

Slide16

Example (3/6)Resolve (4) and (2) by applying {b/Y}

1. ¬P(X)  P(f(X))2. ¬Q(a, Y)

 ¬R(Y, X)  P(X)3. R(b, g(a, Z))4. Q(a, b)5. ¬P(f(g(a, c)))¬R(b, X)  P(X)

Slide17

Example (4/6)Resolve (6) and (3) by applying {g(a,Z

)/X}1. ¬P(X)

 P(f(X))2. ¬Q(a, Y)  ¬R(Y, X)  P(X)3. R(b, g(a, Z))4. Q(a, b)5. ¬P(f(g(a, c)))6. ¬R(b, X) 

P(X)

P(X)

Slide18

Example (5/6)Resolve (7) and (1) by applying {g(a,Z

)/X}1. ¬

P(X)  P(f(X))2. ¬Q(a, Y)  ¬R(Y, X)  P(X) 3. R(b, g(a, Z))4. Q(a, b) 5. ¬P(f(g(a, c)))6. ¬R(b, X)  P(X)

7.

P(g(a,

Z))

P(f(X))

Slide19

Example (6/6)Resolve (8) and (5) by applying {g(a,c)/

X}1. ¬P(X)  P(f(X))

2. ¬Q(a, Y)  ¬R(Y, X)  P(X)3. R(b, g(a, Z))4. Q(a, b)5. ¬P(f(g(a, c)))6. ¬R(b, X)  P(X)7. P(g(a, Z))

8.

P(f(X))

Slide20

SLDTreeFacts: grand_father(jacob, X)

father(john, merry).father(jacob, jones).father(jones, sylia

). father(jacob, Z), father(Z, X ) father(jacob, Z), mother(Z, X)mother(sylia, anes). mother(merry, relly).Rules: father(jacob, jones), father(jones, X)

Exit

grand_father

(X, Y)

:-

father

(X

, Z),

father

(Z

, Y).

grand_father

(X

, Y)

:-

father

(X

, Z

),

mother(Z

, Y

).

Question:

grand_father

(

jacob

,

X

)

father

(

jacob

,

jones

),

father

(

jones

, sylia)


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