.abc

.

Logic

David

Suenderm

ann

http://suendermann.com

Baden-W

uertte

mberg

Coopera

tiveSta

teU

nive

rsity

Stu

ttgart,

Germ

any

D.Suenderm

ann

Logic

January

28,2012

1

.G

enera

lre

mark

s.

The

most

up-to

-date

versio

nofth

isdocu

ment

as

well

as

auxiliary

mate

rialca

nbe

found

onlin

eat

http://suendermann.com

D.Suenderm

ann

Logic

January

28,2012

2

.O

utlin

e.

pro

positio

nallo

gic

first-o

rderlo

gic

Pro

log

D.Suenderm

ann

Logic

January

28,2012

3

.O

utlin

e.

pro

positio

nallo

gic

first-o

rderlo

gic

Pro

log

D.Suenderm

ann

Logic

January

28,2012

4

.Logica

lopera

tors

.

Pro

positio

nallo

gic

(aka

pro

positio

nalca

lculu

s)is

asyste

moffo

rmula

s

repre

sentin

gpro

positio

ns.

We

areusin

gth

efo

llow

ing

logica

lopera

tors

(aka

logica

lco

nnective

s):

¬(n

ot)

∧(a

nd)

∨(o

r)

→(if

...th

en)

↔(if

and

only

if(a

kaiff

))

D.Suenderm

ann

Logic

January

28,2012

5

.W

ell-fo

rmula

ted

form

ula

s.

In

ord

erto

build

valid

(well-fo

rmula

ted)

form

ula

s(W

FFs)

inpro

positio

nal

logic,

we

startw

itha

set

ofpro

positio

nalvaria

ble

sP

=p1,...,p

N.

Give

nP

,we

can

derive

the

set

ofW

WFs

Fin

ductive

lyas

follo

ws:

1.

1∈

F(tru

e)

2.

0∈

F(fa

lse)

3.

ifp∈

Pth

en

p∈

F(e

very

variable

isa

WFF)

4.

iff∈

Fth

en¬

f∈

F(th

enegatio

nofa

WFF

isa

WFF)

5.

iff,g∈

Fth

en

a)

(f∧

g)∈

F

b)

(f∨

g)∈

F

c)(f→

g)∈

F

d)

(f↔

g)∈

F

D.Suenderm

ann

Logic

January

28,2012

6

.W

ell-fo

rmula

ted

form

ula

s:exa

mple

s.

Assu

me

the

follo

win

gvaria

ble

s

P=p,q,r.

(1)

These

areW

WFs:

p(p∧

q)

((¬p→

q)∨

(q→¬

r))

these

arenot:

p¬

(p←

q)

→q

D.Suenderm

ann

Logic

January

28,2012

7

.Pre

cedence

and

asso

ciativity

oflo

gica

lco

nnective

s.

O

ute

rmost

pare

nth

esis

can

be

dro

pped:

(p∧

q)⇔

p∧

q(2)

Consid

erth

efo

llow

ing

pre

cedence

s:

.....

opera

tor

pre

cedence

¬1

(strongest)

∧2

∨3

→4

↔5

(weake

st)

E.g

.,we

have

:

(p∧

q)→

(q∨

r)⇔

p∧

q→

q∨

r(3)

Assu

me

opera

tors

ofth

esa

me

pre

cedence

tobe

left-a

ssocia

tive:

(p→

q)→

r⇔

p→

q→

r(4)

D.Suenderm

ann

Logic

January

28,2012

8

.Pre

cedence

and

asso

ciativity

oflo

gica

lco

nnective

s:exe

rcise.

Set

pare

nth

ese

sto

indica

teth

eord

erofeva

luatio

n:

p↔

q∨

r∧

s→

t∧¬

u∧

v↔

w(5)

Dro

pas

many

pare

nth

ese

sas

possib

le:

((¬¬

((((((p↔

q)∨

r)∨

s)∨

t)↔

u)∧

v)∧

w)∧

x)

(6)

D.Suenderm

ann

Logic

January

28,2012

9

.Som

eapplica

tions

ofpro

positio

nallo

gic

.

D

esig

nand

analysis

ofdig

italcircu

its

–Seve

ralm

illion

logica

lgate

s(p

hysica

lre

aliza

tions

oflo

gica

l

connective

s)are

imple

mente

din

nowadays’

micro

poce

ssors.

–A

circuit

com

pariso

nalg

orith

mch

eck

sw

heth

erth

epro

positio

nal

form

ula

sre

pre

sentin

gtw

ocircu

itsare

equiva

lent.

Pla

nnin

g

–In

many

pla

nnin

gta

sks

(such

as

inlo

gistics,

train

orairlin

e

schedulin

g),

multip

leco

mpulso

ryre

strictions

apply.

–These

restrictio

ns

can

ofte

nbe

exp

resse

din

term

sofpro

positio

nal

logic.

–O

ptim

also

lutio

ns

may

be

dete

rmin

ed

by

means

oflo

gica

lre

solu

tion

tech

niq

ues

(we

will

learn

about

this

late

rin

this

lectu

re).

D.Suenderm

ann

Logic

January

28,2012

10

.Som

eapplica

tions

ofpro

positio

nallo

gic

.

Com

pute

r-assiste

dpro

of

–O

ften,pro

ofs

ofco

nje

cture

sin

am

ath

em

atica

ldiscip

line

may

be

very

com

ple

x(1

00s

ofpages).

–If

the

discip

line’s

axio

ms

aregive

nin

form

oflo

gica

lfo

rmula

s(th

e

know

ledge

base

),th

e(in

)valid

ityofco

nje

cture

sm

aybe

pro

ven

by

reso

lutio

n.

–This

appro

ach

allo

ws

for

pro

ofs

ofa

com

ple

xityhum

ans

arenot

able

tohandle

(1,0

00,0

00s

ofpages).

Gam

eth

eory

–M

any

one-,

two-,

or

multi-p

layergam

es

can

be

exp

resse

din

term

sof

form

ula

sofpro

positio

nallo

gic.

–Exa

mple

sin

clude

chess,

the

8-p

uzzle

,or

the

8-q

ueens

puzzle

.

–Again

,re

solu

tion

may

be

applie

dto

solve

these

puzzle

sor

derive

“optim

al”

solu

tions.

D.Suenderm

ann

Logic

January

28,2012

11

.D

eep

Blu

e.

chess-p

laying

com

pute

rby

IBM

On

May

11,1997,D

eep

Blu

ewon

asix-

gam

em

atch

again

stGarry

Kasp

arov.

base

don

bru

te-fo

rceco

mputin

gpower

(30

nodes

with

480

VLSIch

ess

chip

s)

w

ritten

inC

underAIX

The

eva

luatio

nfu

nctio

nco

nta

ined

mul-

tiple

para

mete

rstu

ned

on

700,0

00

gra

ndm

aste

rgam

es.

pic:–

source

:http://flickr.com/photos/22453761@N00/592436598/

–auth

or:

Jam

es

the

photo

gra

pher

–lice

nse

:Cre

ativ

eCom

mons

Attrib

utio

n2.0

Generic

D.Suenderm

ann

Logic

January

28,2012

12

.8-p

uzzle

.

Da

s 8

-Pu

zzle

51

6

32

4

12

3

45

6

?

87

78

"NP

Vll

täd

i"

51

6

51

32

6

87

4

51

6

32

87

4...

...

8-P

uzzle

"NP

V

51

6

32

4

87

51

6

32

87

45

16

32

87

4

87

4

51

6

37

2

84

...

...

8P

uzzle

-8

75

16

32

4

87

...

84

56

31

2

87

4...

87

4

"NP

Vll

täd

i"

51

68

74

...8

-Pu

zzle

"NP

V

87

51

68

74

84

...8

Pu

zzle

-8

78

4

87

48

74

D.Suenderm

ann

Logic

January

28,2012

13

.8

queens

puzzle

.

Pla

ce8

chess

queens

on

ach

essb

oard

that

no

two

queens

atta

ckeach

oth

er.

There

are

648

=

4,426,165,368

possib

learra

ngm

ents.

But

only

92

solu

tions.

......

D.Suenderm

ann

Logic

January

28,2012

14

.Sem

antics

ofpro

positio

nalfo

rmula

s.

So

far,we

have

inve

stigate

dth

esyn

tax

(structu

re)

ofpro

positio

nal

form

ula

s.

N

ow

,we

want

tolo

ok

at

the

sem

antics,

that

isth

em

eanin

g,or

truth

,of

pro

positio

nalfo

rmula

s.

Ingenera

l,w

ithout

furth

erknow

ledge

about

the

variable

sofa

form

ula

,

we

cannot

tell

wheth

er

afo

rmula

istru

eor

false

.

E.g

.,w

ithout

know

ing

the

truth

valu

eof

pand

q,th

efo

rmula

p∨

q(7)

may

be

true

or

false

.

To

that

end,we

intro

duce

the

valu

atio

n,orin

terp

reta

tion

ofa

form

ula

as

the

functio

n

I:F→

Bw

ithB

=0,1.

(8)

D.Suenderm

ann

Logic

January

28,2012

15

.Valu

atio

n.

W

edefine

the

valu

atio

nI

inductive

lyas

follo

ws:

1.

I(1

)=

1

2.

I(0

)=

0

3.

I(p

)∈

Bfo

rall

p∈

P

(valu

es

forall

p∈

Pappearin

gin

the

form

ula

have

tobe

pro

vided)

4.

I(¬

f)

=I

¬(I

(f))

for

all

f∈

F

5.

I(f∧

g)

=I

∧(I

(f),

I(g

))fo

rall

f,g∈

F

6.

I(f∨

g)

=I

∨(I

(f),

I(g

))fo

rall

f,g∈

F

7.

I(f→

g)

=I

→(I

(f),

I(g

))fo

rall

f,g∈

F

8.

I(f↔

g)

=I

↔(I

(f),

I(g

))fo

rall

f,g∈

F

D.Suenderm

ann

Logic

January

28,2012

16

.Valu

atio

noflo

gica

lco

nnective

s.

I¬(p

):p

01

10

I∧(p

,q):

p

01

q0

00

10

1

I∨(p

,q):

p

01

q0

01

11

1

I→

(p,q):

p

01

q0

10

11

1

I↔

(p,q):

p

01

q0

10

10

1

D.Suenderm

ann

Logic

January

28,2012

17

.Valu

atio

n:exa

mple

.

W

eare

give

nth

efo

rmula

f:=

p→

q→

(¬p→

q→

q).

(9)

N

ow

,we

want

toeva

luate

ffo

rall

possib

lein

terp

reta

tions

of

pand

q.

A

handy

way

todo

sois

touse

atru

thta

ble

:

pq

g:=¬

ph

:=g→

qi:=

h→

qj

:=p→

qj→

i

00

10

11

1

01

11

11

1

10

01

00

1

11

01

11

1

with

p→

q︸

︷︷

︸

j

→(

h︷

︸︸

︷¬

p︸︷︷︸

g

→q→

q)

︸︷︷

︸

i

.(10)

D.Suenderm

ann

Logic

January

28,2012

18

.Valu

atio

n:exce

rcise.

Eva

luate

the

follo

win

gfo

rmula

sfo

rall

possib

lein

terp

reta

tions

(a)

r∨

r∧¬

q→

p∧

(¬q∧¬¬

r)

(b)

(r↔

p→

p→

r∧¬¬¬¬¬¬(¬

q↔

q))∧

r

Hin

t:You

do

not

need

touse

the

truth

table

exce

ssively

ifyo

uca

napply

simplifi

catio

ns

deriva

ble

from

the

valu

atio

nta

ble

oflo

gica

lco

nnective

s

such

as

¬¬

p⇔

p(11)

p∧

0⇔

0(12)

p∧¬

p⇔

0(13)

p→

p⇔

1(14)

p↔

0⇔¬

p(15)

p↔¬

p⇔

0(16)

D.Suenderm

ann

Logic

January

28,2012

19

.Turn

ing

natu

ralla

nguage

into

pro

positio

nalfo

rmula

s.

In

specto

rW

atso

nis

calle

dto

aje

welry

store

that

has

been

subje

ctto

a

robbery

where

thre

esu

bje

cts,Austin

,Bria

n,and

Colin

,were

arreste

d.

Afte

reva

luatio

nofall

facts,

this

isknow

n:

1.

At

least

one

ofth

esu

bje

ctsis

guilty:

f1

:=a∨

b∨

c.

(17)

2.

IfAustin

isguilty

he

had

exa

ctlyone

acco

mplice

:

f2

:=a→

b∧¬

c∨¬

b∧

c.

(18)

3.

IfBria

nis

innoce

nt,

sois

Colin

:

f3

:=¬

b→¬

c.

(19)

4.

Ifexa

ctlytw

osu

bje

ctsare

guilty,

Colin

isone

ofth

em

.H

ence

,out

of

thre

epossib

lepairs

ofsu

bje

cts,th

ere

isonly

one

impossib

le:

f4

:=¬

(a∧

b∧¬

c).

(20)

5.

IfColin

isin

noce

nt

then

Austin

isguilty:

f5

:=¬

c→

a.

(21)

Exe

rcise:W

ho

areth

ecu

lprits?

(Hin

t:co

nju

nctive

lyco

mbin

ef1 ,

...,f5)

D.Suenderm

ann

Logic

January

28,2012

20

.Tauto

logie

s.

W

ehave

seen

that

Form

ula

9is

true

for

eve

rypro

positio

nalva

luatio

n.

Such

afo

rmula

isca

lled

tauto

logy

(Ludw

igW

ittgenste

in,1921).

Iff

fis

ata

uto

logy

this

isalso

denote

das

|=f.

(22)

Wittg

enste

inon

logic

language

and

the

myste

ryofth

eworld

:

http://www.youtube.com/watch?v=Pv68v

reEQM

Exa

mple

s:

1.|=

p∨¬

p

2.|=

p→

p

3.|=

p∧

q→

p

4.|=

p→

p∨

q

5.|=

p→

0↔¬

p

6.|=

p∧

q↔

q∧

p

D.Suenderm

ann

Logic

January

28,2012

21

.Valid

ity,sa

tisfiability,

contin

gency

.

va

lidity:

fis

valid⇔

fis

ata

uto

logy,

i.e.,

all

inte

rpre

tatio

ns

make

ftru

e.

satisfi

ability:

At

least

one

inte

rpre

tatio

nm

ake

sf

true.

unsa

tisfiability,

contra

dictio

n:All

inte

rpre

tatio

ns

make

ffa

lse.

contin

gency:

Inte

rpre

tatio

ns

of

fare

contin

gent

upon

the

truth

valu

es

of

f’s

ato

mic

parts.

I.e.,

contin

gent

pro

positio

ns

areneith

ernece

ssarilytru

e

nor

nece

ssarilyfa

lse.

D.Suenderm

ann

Logic

January

28,2012

22

.Equiva

lence

.

Two

form

ula

sf

and

gare

equiva

lent

iff

|=f↔

g.

(23)

Exa

mple

s:

1)|=¬

0↔

1|=¬

1↔

0

2)|=

p∨¬

p↔

1|=

p∧¬

p↔

0te

rtium

non

datu

r

3)|=

p∨

0↔

p|=

p∧

1↔

pneutra

lele

ment

4)|=

p∨

1↔

1|=

p∧

0↔

0

5)|=

p∧

p↔

p|=

p∨

p↔

pid

em

pote

ncy

6)|=

p∧

q↔

q∧

p|=

p∨

q↔

q∨

pco

mm

uta

tivity

7)|=

p∧

(q∧

r)↔

|=p∨

(q∨

r)↔

asso

ciativity

|=↔

(p∧

q)∧

r|=↔

(p∨

q)∨

r

8)|=¬¬

p↔

pdouble

-negatio

n

9)|=

p∧

(p∨

q)↔

p|=

p∨

p∧

q↔

pabso

rptio

n

D.Suenderm

ann

Logic

January

28,2012

23

.Equiva

lence

(cont.)

.

10)|=

p∧

(q∨

r)↔

|=p∨

q∧

r↔

distrib

utivity

|=↔

p∧

q∨

p∧

r|=↔

(p∨

q)∧

(p∨

r)

11)|=¬

(p∧

q)↔¬

p∨¬

q|=¬

(p∨

q)↔¬

p∧¬

qde

Morg

an’s

rule

s

12)|=

(p→

q)↔¬

p∨

qelim

inatio

nof→

13)|=

(p↔

q)⇔

elim

inatio

nof↔

|=⇔

(¬p∨

q)∧

(¬q∨

p)

D.Suenderm

ann

Logic

January

28,2012

24

.Equiva

lence

:exe

rcise.

We

aregive

nth

efo

rmula

f:=

p→

q→

(¬p→

q→

q).

(24)

Pro

veth

at

fis

ata

uto

logy

usin

gequiva

lence

rule

s.

D.Suenderm

ann

Logic

January

28,2012

25

.Lite

rals

and

clause

s.

Tra

nsfo

rmatio

ns

such

as

the

ones

inth

ela

stexe

rciseca

nbe

applie

d

alg

orith

mica

llytu

rnin

gan

arbitrary

form

ula

into

anorm

alize

dfo

rm.

In

ord

erto

show

this,

we

need

aco

uple

ofdefinitio

ns.

A

pro

positio

nalfo

rmula

fis

calle

da

litera

liff

eith

erofth

efo

llow

ing

case

sapplie

s:

1.

f=

0or

f=

1.

2.

f=

pw

ithp∈

P( p

ositive

litera

l)

3.

f=¬

pw

ithp∈

P( n

egative

litera

l)

The

set

ofall

litera

lsis

denote

das

L.

Apro

positio

nalfo

rmula

cis

calle

da

clause

iffc

has

the

form

c=

l1∨···∨

lm

(25)

where

l1,...,lm∈

L.

That

is,a

clause

isa

disju

nctio

noflite

rals.

The

set

ofall

clause

sis

denote

das

C.

D.Suenderm

ann

Logic

January

28,2012

26

.Set

nota

tion

ofcla

use

s.

D

ue

toth

easso

ciativity,

com

muta

tivity,and

idem

pote

ncy

ofth

e∨

connective

,in

acla

use

c,th

eord

eroflite

rals

isarb

itrary,and

repeate

d

litera

lsca

nbe

dro

pped.

This

isw

hy

we

can

inte

rpre

teth

elite

rals

of

cas

the

ele

ments

ofa

set:

l1 ,

...,lm.

(26)

Due

toth

eequiva

lence

|=l1∨···∨

lm∨

0↔

l1∨···∨

lm

,(27)

we

have|=l1 ,

...,lm

,0↔l1

,...,lm

(28)

(dro

ppin

gof0

from

acla

use

).

Asp

ecia

lca

seis

|=0↔

(29)

(the

em

pty

clause

).

D.Suenderm

ann

Logic

January

28,2012

27

.Trivia

lcla

use

s.

A

clause

cis

calle

dtrivia

l,i.e

.

|=c

(30)

iffwe

have

one

ofth

efo

llow

ing

case

s:

1.

Due

toth

eequiva

lence

|=l1∨···∨

lm∨

1↔

1(31)

we

have|=l1 ,

...,lm

,1↔1

(32)

(clause

conta

ins

1).

2.

Due

toth

eequiva

lence

|=l1∨···∨

lm∨

p∨¬

p↔

l1∨···∨

lm∨

1(33)

we

have

(usin

gIte

m1)

|=l1 ,

...,lm

,p,¬

p↔1

(34)

(clause

conta

ins

com

ple

mentary

litera

ls).

D.Suenderm

ann

Logic

January

28,2012

28

.Conju

nctive

norm

alfo

rm.

A

form

ula

fis

inco

nju

nctive

norm

alfo

rm(C

NF)

ifff

has

the

form

f=

c1∧···∧

cn

with

c1,...,c

n∈

C.

(35)

Due

toth

easso

ciativity,

com

muta

tivity,and

idem

pote

ncy

ofth

e∧

connective

,in

afo

rmula

f,th

eord

er

ofcla

use

sis

arbitrary,

and

repeate

d

clause

sca

nbe

discard

ed.

This

isw

hy

we

can

inte

rpre

teth

ecla

use

sof

fas

the

ele

ments

ofa

set

which

,in

turn

,are

sets

oflite

rals:

c1,...,c

n.

(36)

D.Suenderm

ann

Logic

January

28,2012

29

.Conju

nctive

norm

alfo

rm:exa

mple

.

The

form

ula

(p∨

q∨¬

r)∧

(q∨¬

r∨

p∨

q)∧

(¬r∨

p∨¬

q)

(37)

isin

CN

F,and

itsse

tnota

tion

is

p,q,¬

r,p,¬

q,¬

r.

(38)

D.Suenderm

ann

Logic

January

28,2012

30

.Conju

nctive

norm

alfo

rm:ta

uto

logy

.

D

ue

toth

eequiva

lence

|=c1∧···∧

cm∧

1↔

c1∧···∧

lm

,(39)

we

have|=c1,...,c

m,1↔c1,...,c

m

(40)

(dro

ppin

gof1

from

afo

rmula

inCN

F).

Asp

ecia

lca

seis

|=1↔

(41)

(em

pty

CN

F).

That

is,if

f=c1,...,c

n

the

follo

win

gca

ses

areequiva

lent

a)|=

f

b)|=

ci

forall

i∈1,...,n

c)c

i↔1

forall

i∈1,...,n

d)

f↔

(tauto

logy

pro

pertie

s).

D.Suenderm

ann

Logic

January

28,2012

31

.Conju

nctive

norm

alfo

rm:alg

orith

m.

This

alg

orith

mtu

rns

an

arbitrary

pro

positio

nalfo

rmula

fin

toCN

F:

1.

Elim

inate↔

by

Equiva

lence

13:

|=(p↔

q)⇔

(¬p∨

q)∧

(¬q∨

p)

(42)

2.

Elim

inate→

by

Equiva

lence

12:

|=(p→

q)↔¬

p∨

q(43)

3.

Sim

plify

the

form

ula

usin

gth

eEquiva

lence

s1,8,11:

a)|=¬

0↔

1

b)|=¬

1↔

0

c)|=¬¬

p↔

p

d)|=¬

(p∧

q)↔¬

p∨¬

q

e)|=¬

(p∨

q)↔¬

p∧¬

q

Now

,th

eco

nnective

¬w

illonly

appear

infro

nt

ofpro

positio

nal

variable

s(th

efo

rmula

isin

negatio

nnorm

alfo

rm).

D.Suenderm

ann

Logic

January

28,2012

32

.Conju

nctive

norm

alfo

rm:alg

orith

m(co

nt.)

.

4.

Exp

and

the

form

ula

usin

gth

efo

llow

ing

variants

ofth

elaw

of

distrib

utivity

(Equiva

lence

10):

a)|=

p∨

q∧

r↔

(p∨

q)∧

(p∨

r)

b)|=

p∧

q∨

r↔

(p∨

r)∧

(q∨

r)

This

isto

transfo

rmth

efo

rmula

into

aco

nju

nctio

nofdisju

nctio

ns.

5.

Tra

nsfo

rmth

efo

rmula

into

set

nota

tion.

D.Suenderm

ann

Logic

January

28,2012

33

.Conju

nctive

norm

alfo

rm:alg

orith

m:exa

mple

.

Let

us

dem

onstra

teth

ealg

orith

musin

gth

efo

rmula

f:=

p→

q→

(¬p→¬

q).

(44)

1.

(skip

ped)

2.

.|=

f⇔¬

p∨

q→

(¬p→¬

q)

⇔¬

(¬p∨

q)∨

(¬p→¬

q)

⇔¬

(¬p∨

q)∨

(¬¬

p∨¬

q)

(45)

3.

.|=

f⇔¬

(¬p∨

q)∨

(p∨¬

q)

⇔¬¬

p∧¬

q∨

(p∨¬

q)

⇔p∧¬

q∨

(p∨¬

q)

(46)

4.

.|=

f⇔

(p∨

(p∨¬

q))∧

(¬q∨

(p∨¬

q))

(47)

5.

.|=

f⇔p,¬

q

(48)

D.Suenderm

ann

Logic

January

28,2012

34

.Conju

nctive

norm

alfo

rm:deriva

tion

from

truth

table

.

In

case

there

arenot

too

many

variable

sin

volve

d,th

eCN

Fca

nalso

be

derive

ddire

ctlyfro

mth

etru

thta

ble

.

To

dem

onstra

teth

is,le

tus

consid

erth

eabove

exa

mple

f:=

p→

q︸

︷︷

︸

g

→(¬

p→¬

q)

︸︷︷

︸

h

.(49)

pq

gh

f

00

11

1

01

10

0

10

01

1

11

11

1

Fro

mth

ose

row

sw

here

feva

luate

sto

0,we

can

derive

the

CN

Fas

|=f⇔¬

(¬p∧

q)⇔

p∨¬

q(50)

or,

alte

rnative

ly,fro

mro

ws

turn

ing

1th

edisju

nctive

norm

alfo

rm(D

NF)

|=f⇔¬

p∧¬

q∨

p∧¬

q∨

p∧

q.

(51)

D.Suenderm

ann

Logic

January

28,2012

35

.Conju

nctive

norm

alfo

rm:exe

rcise.

Tra

nsfo

rmin

toCN

F:

a)

f:=

p→

(q↔

r)

b)

g:=

p→

q→

r↔¬

s∨

t

D.Suenderm

ann

Logic

January

28,2012

36

.Form

alpro

ofs

.

Pro

ofs

areone

ofth

em

ain

notio

ns

oflo

gic.

Give

n

–a

set

offo

rmula

sf1,...,f

n

(the

know

ledge

base

)and

–a

form

ula

g(th

eco

nje

cture

),

we

ask

wheth

er

gca

nbe

pro

ven

base

don

(or

infe

rred

from

the

know

ledge

base

.

That

is,we

ask

wheth

er

f1∧

...∧

fn→

g(52)

isa

tauto

logy.

One

possib

ilityto

answ

erth

isquestio

nis

toco

nve

rtForm

ula

52

into

CN

F

and

toch

eck

wheth

er

all

itscla

use

sare

trivial.

D.Suenderm

ann

Logic

January

28,2012

37

.Form

alpro

ofs:

exa

mple

(hyp

oth

etica

lsyllo

gism

).

Give

nth

eknow

ledge

base

f1

:=p→

q

f2

:=q→

r(53)

pro

veth

eco

nje

cture

g:=

p→

r(54)

This

is,we

need

topro

vew

heth

er

|=(p→

q)∧

(q→

r)→

(p→

r)

(55)

This

pro

ofca

nbe

done

usin

ga

truth

table

or

by

conve

rsion

into

CN

F.

This

specifi

cru

leis

calle

dhyp

oth

etica

lsyllo

gism

.

D.Suenderm

ann

Logic

January

28,2012

38

.In

fere

nce

rule

s.

Anoth

erway

topro

vea

conje

cture

from

aknow

ledge

base

isto

use

infe

rence

rule

s.

An

infe

rence

rule

isa

pair

consistin

gofa

set

ofpre

mise

sf1,...,f

n

and

aco

nclu

sion

gw

ritten

as

f1 ,···

,f

n

∴g

read

as

Fro

mf1 ,

...,f

n,we

can

infer

that

g.

D.Suenderm

ann

Logic

January

28,2012

39

.In

fere

nce

rule

s:exa

mple

s.

sim

plifi

catio

n

p∧

q

∴p

modus

ponens

p→

q,p

∴q

bico

nditio

nalelim

inatio

n

p↔

q,p∨

q

∴p∧

q

modus

tolle

ns

¬q,p→

q

∴¬

p

hyp

oth

etica

lsyllo

gism

p→

q,q→

r

∴p→

r

re

solu

tion

p∨

q,¬

p∨

r

∴q∨

r

D.Suenderm

ann

Logic

January

28,2012

40

.Cut

rule

.

Retu

rnin

gto

the

set

repre

senta

tion

ofCN

F,a

handy

infe

rence

rule

for

clause

sis

the

cut

rule

.

Give

na

pro

positio

nalvaria

ble

pand

two

clause

sc1

and

c2,th

ecu

tru

leis:

p∪

c1,¬

p∪

c2

∴c1∪

c2

D.Suenderm

ann

Logic

January

28,2012

41

.Cut

rule

:pro

of

.

W

etra

nsfo

rmth

eco

nve

ntio

nalre

pre

senta

tion

ofth

ecu

tru

lein

toCN

F:

f:=

(p∨

c1)∧

(¬p∨

c2)→

c1∨

c2

⇔¬

((p∨

c1)∧

(¬p∨

c2))∨

c1∨

c2

⇔¬

(p∨

c1)∨¬

(¬p∨

c2)∨

c1∨

c2

⇔¬

p∧¬

c1∨

p∧¬

c2∨

c1∨

c2

︸︷︷

︸

a

⇔(¬

p∨

a)∧

(¬c1∨

a)

a⇔

(p∨

c1∨

c2)

︸︷︷

︸

b

∧(¬

c2∨

c1∨

c2)

︸︷︷

︸

c

f⇔

((¬p∨

b)∧

(¬p∨

c))∧

((¬c1∨

b)∧

(¬c1∨

c))

⇔¬

p,p,c1,c2,¬

p,¬

c2,c1,c2,

¬

c1,p,c1,c2,¬

c1,¬

c2,c1,c2

(56)

Sin

ceall

clause

sin

fare

trivial,

|=f.

(57)

D.Suenderm

ann

Logic

January

28,2012

42

.Cut

rule

:sp

ecia

lca

ses

.

U

sing

Equiva

lence

12

(elim

inatio

nof→

),th

efo

llow

ing

ofourin

fere

nce

rule

exa

mple

sca

nbe

regard

ed

as

specia

lca

ses

ofth

ecu

tru

le:

–M

odus

ponens.

With

c1

:=

and

c2

:=q,we

have

p∪,¬

p∪q

∴∪q

–M

odus

tolle

ns

(show

this).

–H

ypoth

etica

lsyllo

gism

(show

this).

–Reso

lutio

n.

D.Suenderm

ann

Logic

January

28,2012

43

.Pro

vability

.

Give

na

set

ofcla

use

sM

(the

assu

mptio

ns;

aknow

ledge

base

)and

a

single

clause

f,we

define

are

latio

n

M⊢

f(58)

read

as

fis

pro

vable

from

M.

This

rela

tion

isin

ductive

lydefined

as

a)

Mpro

ves

all

itsassu

mptio

ns:

Iff∈

Mth

en

M⊢

f.

(59)

b)

Iftw

ocla

use

sp∪

c1

and¬

p∪

c2

arepro

vable

from

Mth

en

the

clause

c1∪

c2

ispro

vable

(applyin

gth

ecu

tru

le):

IfM⊢p∪

c1

and

M⊢¬

p∪

c2

then

M⊢

c1∪

c2.

(60)

D.Suenderm

ann

Logic

January

28,2012

44

.Pro

vability:

exa

mple

.

W

ewant

tosh

ow

that

¬

p,q,¬

q,¬

p,¬

q,p,q,p⊢0.

(61)

This

isone

way

todo

so:

1.¬

p,q,¬

q,¬

p⊢¬

p

2.¬

q,p,q,p⊢p

3.¬

p,p⊢

D.Suenderm

ann

Logic

January

28,2012

45

.Pro

vability:

exe

rcise.

W

eare

give

nth

efo

llow

ing

know

ledge

base

f1

:=p→

q

f2

:=q→

r

f3

:=r→

s

f4

:=s→

t

f5

:=t→

u

f6

:=u→

p

f7

:=p∨

q∨

r∨

s∨

t∨

u

f8

:=¬

p∨¬

q∨¬

r∨¬

s∨¬

t∨¬

u(62)

Show

that

0ca

nbe

infe

rred

from

the

know

ledge

base

.

D.Suenderm

ann

Logic

January

28,2012

46

.Satisfi

ability

ofse

tsofcla

use

s.

In

multip

leapplica

tions,

itis

nece

ssaryto

dete

rmin

ea

variable

inte

rpre

tatio

nre

nderin

gall

the

clause

sin

ase

tC

0tru

e.

That

is,we

want

todete

rmin

ew

heth

er

C0

issa

tisfiable

and,if

so,a

solu

tion,i.e

.,a

satisfyin

gin

terp

reta

tion.

Consid

erth

efo

llow

ing

obvio

us

case

s

1)

C1

=p,¬

q,r,¬

s,¬

t

(satisfi

able

)

2)

C2

=,p,¬

q,r

(unsa

tisfiable

)

3)

C3

=p,¬

q,¬

p

(unsa

tisfiable

)

D.Suenderm

ann

Logic

January

28,2012

47

.U

nit

clause

and

trivialse

tofcla

use

s.

A

clause

isca

lled

unit

clause

iffit

conta

ins

one

litera

l:

c=p

or

c=¬

p

with

p∈

P.

(63)

Ase

tofcla

use

sC

0is

calle

dtrivia

lin

one

ofth

ese

case

s:

1)

C0

conta

ins

the

em

pty

clause

:

∈

C0 .

(64)

That

is,C

0is

unsa

tisfiable

.

2)

C0

conta

ins

only

unit

clause

sofdiff

ere

nt

pro

positio

nalvaria

ble

s:

∀c(c∈

C0→∃p(p∈

P∧

(c=p∨

c=¬

p)))∧

∀p(p∈

P→¬

(p∈

C0∧¬

p∈

C0 )).

(65)

That

is,C

0is

satisfi

able

as

dete

rmin

ed

by

C0 ’s

unit

clause

s.

D.Suenderm

ann

Logic

January

28,2012

48

.T

he

alg

orith

mofD

avis

and

Putn

am

:basic

prin

ciple

s.

The

Davis

and

Putn

am

(DP)

alg

orith

mis

tofind

asa

tisfying

clause

give

n

ase

tofcla

use

sby

applyin

g

1)

cut

rule

2)

subsu

mptio

n

3)

case

distin

ction

D.Suenderm

ann

Logic

January

28,2012

49

.T

he

alg

orith

mofD

avis

and

Putn

am

:cu

tru

le.

The

DP

alg

orith

muse

sa

specia

lca

seofth

ecu

tru

le:

l1 ,

...,ln,¬

l,l

∴l1 ,

...,ln

reducin

gth

eorig

inalcla

use

length

by

one.

Genera

lly,th

isopera

tion

isperfo

rmed

by

the

functio

n

cut:2

C×

L→

2C

(66)

defined

as

cut(C

0 ,l)

=c\¬

l|c∈

C0∧¬

l∈

c.

(67)

Exa

mple

:

cut(p,q,¬

p,r,¬

q,¬

p)

=q

(68)

D.Suenderm

ann

Logic

January

28,2012

50

.T

he

alg

orith

mofD

avis

and

Putn

am

:su

bsu

mptio

n.

Assu

me

the

exa

mple

C0

:=p,q,¬

r,p.

(69)

As

we

have

|=p→

p∨

q∨¬

r,(70)

we

can

dro

p(su

bsu

me)p,q,¬

r

from

C0 .

Form

ally,

we

intro

duce

the

functio

n

sub

:2

C×

L→

2C

(71)

that

subsu

mes

all

clause

sin

C0

conta

inin

gth

eunit

clausel

:

sub(C

0 ,l)

=c∈

C0 |l6∈

c∪l.

(72)

Exa

mple

:

sub(p,q,¬

p,r,¬

q,¬

p)

=p,q,¬

q,¬

p

(73)

D.Suenderm

ann

Logic

January

28,2012

51

.T

he

alg

orith

mofD

avis

and

Putn

am

:re

ductio

n.

Sin

cecu

tru

leand

subsu

mptio

nhave

the

sam

earg

um

ent

space

sth

ey

can

be

com

bin

ed

toa

genera

lre

ductio

nfu

nctio

n.

In

doin

gso

,co

nsid

erth

at

applyin

gth

ecu

tru

lere

turn

sonly

cut

versio

ns

ofth

ose

clause

sth

at

orig

inally

conta

ined¬

l.

Hence

,we

add

all

those

clause

snotco

nta

inin

g¬

lbefo

reapplyin

g

subsu

mptio

n.

This

yield

s

red(C

0,l)

=su

b(C

1∪

C2,l)

(74)

with

C1

=cu

t(C0,l)

=c\¬

l|c∈

C0∧¬

l∈

c

(75)

and

C2

=c∈

C0|¬

l6∈

c.

(76)

D.Suenderm

ann

Logic

January

28,2012

52

.T

he

alg

orith

mofD

avis

and

Putn

am

:re

ductio

n(co

nt.)

.

Assu

min

gC

0is

not

trivial,

we

know

that

applyin

gsu

bsu

mptio

nto

C1

does

not

affect

any

ofits

clause

ssin

ceth

ey

orig

inally

conta

ined¬

l,i.e

.,

they

do

not

conta

inl.

Hence

,su

bsu

mptio

nca

nonly

affect

C2

which

isw

hy

we

can

write

red(C

0,l)

=su

b(C

1∪

C2 ,

l)

=C

1∪

sub(C

2 ,l)

=c\¬

l|c∈

C0∧¬

l∈

c∪

c∈

C0 |l6∈

c∧¬

l6∈

c∪l.

(77)

Exa

mple

:

red(p,q,¬

p,r,¬

q,¬

p)

=q,¬

q,¬

p

(78)

D.Suenderm

ann

Logic

January

28,2012

53

.T

he

alg

orith

mofD

avis

and

Putn

am

:ca

sedistin

ction

.

This

prin

ciple

isbase

don

the

follo

win

gpro

positio

n:

C0

issa

tisfiable

iffC

0∪p

or

C0∪¬

p

aresa

tisfiable

.

Here

,C

0is

ase

tofcla

use

s;p

isa

pro

positio

nalvaria

ble

occu

rring

inC

0.

This

sounds

likea

trivialprin

ciple

since

ifth

ere

isa

variable

inte

rpre

tatio

n

satisfyin

gC

0it

will

inclu

de

the

variable

pw

hich

must

be

eith

er

0or

1in

this

specifi

cin

terp

reta

tion.

The

case

distin

ction

prin

ciple

is,howeve

r,esse

ntia

lto

the

DP

alg

orith

m

as

itin

jects

unit

clause

sre

quire

dfo

rcu

tru

leand

subsu

mptio

n.

D.Suenderm

ann

Logic

January

28,2012

54

.T

he

alg

orith

mofD

avis

and

Putn

am

.

W

eare

give

na

set

ofcla

use

sC

0and

search

for

apro

positio

nalva

luatio

n

satisfyin

gC

0 .

1.

Itera

tively

apply

Ci+

1:=

red(C

i ,l)

forall

.....l∈p,¬

p|p∈

P∩

Ci ,

i=

0,1,...

IfC

i+1

turn

sout

tobe

trivial,

stop.

2.

Oth

erw

ise(ca

sedistin

ction),

choose

avaria

ble

pfro

mC

i(sh

ould

not

be

aunit

clause

since

those

had

been

tried

in1.)

a)

Itera

tively

apply

Ci+

1:=

red(C

i∪p,l)

for

all

l∈p,¬

p|p∈

P∩

Ci .

IfC

i+1

isfo

und

tobe

satisfi

able

,sto

p.

b)

Itera

tively

apply

Ci+

1:=

red(C

i∪¬

p,l)

for

all

l∈p,¬

p|p∈

P∩

Ci .

IfC

i+1

isfo

und

tobe

satisfi

able

,sto

p.

c)O

therw

ise,

C0

isnot

satisfi

able

.

Ste

p2

mig

ht

need

tobe

recu

rsively

applie

dm

ultip

letim

es

until

a

solu

tion

(satisfi

ability

orco

ntra

dictio

n)

isfo

und.

D.Suenderm

ann

Logic

January

28,2012

55

.T

he

alg

orith

mofD

avis

and

Putn

am

:exa

mple

.

W

eare

give

n

C0

:=p,q,s,¬

p,r,¬

t,r,

s,¬

r,q,¬

p,¬

s,p

C0

:=¬

p,¬

q,s,¬

r,p,¬

q,s,¬

r,¬

s,¬

p,¬

s

(79)

1.

(skip

ped—

C0

conta

ins

no

unit

clause

s)

2.

pick

p

a)

.C

1:=

red(C

0∪p,p)

=r,¬

t,r,

s,¬

r,q,¬

q,s,¬

r,¬

r,¬

s,¬

s,p

C2

:=re

d(C

1 ,¬

s)

=r,¬

t,r,¬

r,q,¬

q,¬

r,¬

s,p

C3

:=re

d(C

2 ,r)

=r,q,¬

q,¬

s,p

C4

:=re

d(C

3 ,q)

=r,q,,¬

s,p

(unsa

tisfiable

)(80)

D.Suenderm

ann

Logic

January

28,2012

56

.T

he

alg

orith

mofD

avis

and

Putn

am

:exa

mple

.

b)

.C

5:=

red(C

0∪¬

p,¬

p)

=q,s,r,

s,¬

s,¬

q,s,¬

r,¬

s,¬

p

C6

:=re

d(C

5 ,¬

s)

=q,r,¬

s,¬

q,¬

p

C7

:=re

d(C

6 ,q)

=q,r,¬

s,,¬

p

(unsa

tisfiable

)

(81)

c)C

0is

unsa

tisfiable

(contra

dictio

n).

D.Suenderm

ann

Logic

January

28,2012

57

.T

he

alg

orith

mofD

avis

and

Putn

am

:exe

rcise.

Apply

the

DP

alg

orith

mto

the

pro

vability

exe

rcisew

ith

a)

C0

:=f1 ,

...,f8,

b)

C0

:=f1 ,

...,f7.

D.Suenderm

ann

Logic

January

28,2012

58

.Reso

lutio

nte

chniq

ue

.

In

ord

erto

pro

vea

conje

cture

base

don

aknow

ledge

base

,we

can

apply

the

reso

lutio

nte

chniq

ue:

1.

All

sente

nce

sin

the

know

ledge

base

and

the

negatio

nofth

ese

nte

nce

tobe

pro

ven

(the

conje

cture

)are

conju

nctive

lyco

nnecte

d.

2.

The

resu

lting

sente

nce

istra

nsfo

rmed

into

CN

F.

3.

Ifth

eem

pty

clause

can

be

derive

dafte

ran

applica

tion

ofth

eD

P

alg

orith

m(o

r,alte

rnative

ly,th

epro

vability

tech

niq

ue),

the

orig

inal

sente

nce

isunsa

tisfiable

,i.e

.,th

eco

nje

cture

follo

ws

from

the

know

ledge

base

.

D.Suenderm

ann

Logic

January

28,2012

59

.Reso

lutio

nte

chniq

ue:exe

rcise.

Let

us

revisit

the

crimin

alca

sefro

mearlie

r.

Rem

em

ber,

we

were

give

nth

efo

llow

ing

facts

(ourknow

ledge

base

):

f1

:=a∨

b∨

c

f2

:=a→

b∧¬

c∨¬

b∧

c

f3

:=¬

b→¬

c

f4

:=¬

(a∧

b∧¬

c)

f5

:=¬

c→

a(82)

Dr.

Watso

n’s

gut

feelin

gis

that

Bria

nand

Colin

areth

ecu

lprits.

Pro

vehis

conje

cture

usin

gth

ere

solu

tion

tech

niq

ue.

D.Suenderm

ann

Logic

January

28,2012

60

.n

queens

puzzle

:exe

rcise.

Usin

gth

eD

Palg

orith

m,sh

ow

that

–th

e2

queens

puzzle

has

no

solu

tion,

–th

e3

queens

puzzle

has

no

solu

tion,

–th

e4

queens

puzzle

has

aso

lutio

nand

which

one

itis.

D.Suenderm

ann

Logic

January

28,2012

61

.O

utlin

e.

pro

positio

nallo

gic

first-o

rderlo

gic

Pro

log

D.Suenderm

ann

Logic

January

28,2012

62

.N

otio

ns

offirst-o

rder

logic

.

Term

sdenote

obje

cts.

Term

sare

com

pose

dofvaria

ble

sor

functio

nsym

bols:

fath

er(x

),m

oth

er(isa

ac),

x+

7.

(83)

with

the

variable

xand

the

functio

nsym

bols

fath

er,

moth

er,

isaac,

+,7

Obje

ctsca

nbe

put

into

rela

tion

by

pre

dica

tesym

bols

pro

ducin

gato

mic

form

ula

s :

isBro

ther(a

dam

,fa

ther(b

rian)),

x+

7<

x·7,

n∈

I.(84)

with

the

pre

dica

tesym

bols

IsBro

ther,

<,∈

Form

ula

sca

nbe

com

bin

ed

usin

glo

gica

lco

nnective

s:

isHum

an(x

)→

isMorta

l(x),

x>

1→

x+

7<

x·7.

(85)

Form

ula

sca

nco

nta

inquantifi

ers

definin

gth

ese

mantics

(scope)

of

variable

s:

∀x(isH

um

an(x

)→∃y(y

=m

oth

er(x

))).(86)

D.Suenderm

ann

Logic

January

28,2012

63

.Sig

natu

res

and

term

s.

A

signatu

reis

aquatru

ple

definin

ga

specifi

csyste

moffirst-o

rderlo

gic:

Σ=〈V

,F

,P

,arity〉.

(87)

The

mem

bers

ofΣ

are

–th

ese

tofvaria

ble

sV

,

–th

ese

toffu

nctio

nsym

bols

F,

–th

ese

tofpre

dica

tesym

bols

P,and

–a

functio

nassig

nin

gan

arityto

all

functio

nand

pre

dica

tesym

bols:

arity:F∪

P→

I.(88)

Give

na

signatu

reΣ

,we

define

the

set

ofΣ

term

sT

Σin

ductive

lyas:

1.∀x(x∈

V→

x∈

TΣ).

2.

Iff∈

Fand

n=

arity(f)

and

t1 ,

...,tn∈

TΣ

then

f(t

1 ,...,tn)∈

TΣ

.(89)

Specia

lca

se:If

f∈

Fand

arity(f)=

0th

en

we

can

write

fin

stead

of

f()

(aka

consta

nt).

D.Suenderm

ann

Logic

January

28,2012

64

.Sig

natu

res

and

term

s:exa

mple

.

W

eare

give

nth

esig

natu

reΣ

a=〈V

,F

,P

,arity〉

with

V=x,y,z

F=0,1,+

,·

P==

,≤

arity=〈0

,0〉〈1

,0〉〈+

,2〉〈·,

2〉〈=

,2〉〈≤

,2〉

Now

,we

can

derive

Σa

term

sas

follo

ws:

1.

x,y,z∈

TΣ

a

2.0,1∈

TΣ

a(0

-aryfu

nctio

nsym

bols)

3.

+(1

,x)∈

TΣ

a

4.·(+

(1,x),

y)∈

TΣ

a

Inth

efo

llow

ing,we

will

use

an

infix

nota

tion

forbin

aryre

latio

nand

functio

nsym

bols.

E.g

.,Term

4would

read

(1+

x)·y

(90)

D.Suenderm

ann

Logic

January

28,2012

65

.Ato

mic

form

ula

s.

Give

nth

esig

natu

reΣ

=〈V

,F

,P

,arity〉,

ifp∈

Pand

n=

arity(p)

and

t1,...,tn∈

TΣ

then

p(t

1 ,...,tn)∈

AΣ

,(91)

the

set

ofall

ato

mic

form

ula

s.

Specia

lca

se:If

p∈

Pand

arity(p)

=0

then

we

can

write

pin

stead

of

p()

(aka

pro

positio

nalvaria

ble

).

Contin

uin

gw

ithourabove

exa

mple

:

=(·(+

(1,x),

y),0)∈

AΣ

a(92)

or

inin

fix

nota

tion

(1+

x)·y

=0∈

AΣ

a.

(93)

D.Suenderm

ann

Logic

January

28,2012

66

.Fre

eand

bound

variable

s.

Consid

eran

exa

mple

from

calcu

lus:

x∫0

yzdz

=12

yx

2.

(94)

Eq.94

featu

res

the

thre

evaria

ble

sx,

yand

t.

Tryin

gto

substitu

tevaria

ble

son

both

sides

ofth

eequatio

n,e.g

.,x

by

yy

∫0

yzdz

=12

yy

2(95)

or

zby

yx∫0

yydy

=13

x3

!6=12

xx

2(96)

show

sth

at

there

can

be

two

types

ofvaria

ble

sin

form

ula

s:fre

e(x

,y)

and

bound

(z)

ones.

The

variable

zin

this

exa

mple

isbound

by

the

diff

ere

ntia

lopera

tor

d.

D.Suenderm

ann

Logic

January

28,2012

67

.Fre

eand

bound

variable

s(co

nt.)

.

Infirst-o

rderlo

gic,

variable

sare

bound

by

the

quantifi

ers∀

and∃.

Acco

rdin

gly,

inth

efo

llow

ing

exa

mple

∀x,y(P

(x)→

Q(x

,f(x

),z))

(97)

xand

yare

bound

variable

s,and

zis

free.

Afo

rmula

with

no

free

variable

sis

calle

da

sente

nce

.

D.Suenderm

ann

Logic

January

28,2012

68

.Fre

eand

bound

variable

s:exe

rcise.

Consid

erth

efo

llow

ing

exa

mple

s:

a)∀x(F

(x)→∃y(G

(y,z)))

b)∀x(F

(x)→∃y(G

(x,y)))

c)∃x(y

+x≤

y)

Which

variable

sare

free,and

which

ones

arebound?

Which

ofth

ese

form

ula

sare

sente

nce

s?

D.Suenderm

ann

Logic

January

28,2012

69

.Form

ula

s.

The

set

ofvaria

ble

svar(t)

occu

rring

ina

term

tis

inductive

lydefined

as

1.

var(x)

:=x

for

all

x∈

V.

2.

var(f(t

1 ,...,tn))

:=var(t

1 )∪···∪

var(tn).

Give

na

signatu

reΣ

,in

the

follo

win

g,we

denote

–th

ese

toffo

rmula

sas

FΣ,

–th

ese

tofall

bound

variable

sofa

form

ula

f∈

FΣ

as

bound(f

),and

–th

ese

tofall

free

variable

sof

fas

as

free(f

).

These

sets

arein

ductive

lydefined

as

1.

0,1∈

FΣ

(truth

valu

es)

and

free(0

)=

free(1

)=

bound(0

)=

bound(1

)=

(98)

2.∀f(f∈

AΣ→

f∈

FΣ)

(ato

mic

form

ula

s)and

free(f

)=

var(t1 )∪···∪

var(tn);

bound(f

)=

with

f=

p(t

1,...,tn)(99)

D.Suenderm

ann

Logic

January

28,2012

70

.Form

ula

s(co

nt.)

.

3.∀f(f∈

FΣ→¬

f∈

FΣ)

(negatio

n)

and

free(¬

f)

=fre

e(f

);bound(¬

f)

=bound(f

)(100)

4.∀f,g(f

,g∈

FΣ∧

free(f

)∩

bound(g

)=∧

bound(f

)∩

free(g

)=

→

(f∨

g)∈

FΣ)

(logica

lco

nnective

s)and

free(f∨

g)

=fre

e(f

)∪fre

e(g

);bound(f∨

g)

=bound(f

)∪bound(g

)(101)

applyin

gto

all

logica

lco

nnective

s(∨

,∧

,→

,↔

)

5.∀f,x(f∈

FΣ∧

x∈

V\bound(f

)→

(∀x(f

)),(∃

x(f

))∈

FΣ)

(quantifi

ers)

and

free(∀

x(f

))=

free(∃

x(f

))=

free(f

)\x;

(102)

bound(∀

x(f

))=

bound(∃

x(f

))=

bound(f

)∪x

(103)

D.Suenderm

ann

Logic

January

28,2012

71

.A

dditio

nalco

nve

ntio

ns

.

In

litera

ture

you

enco

unte

rexp

ressio

ns

ofth

efo

rm

∀x∈

R:∃n∈

N:x

<n

.(104)

These

abbre

viatio

ns

can

be

transfo

rmed

into

the

form

erly

intro

duce

d

term

inolo

gy

by

∀x∈

M:f

def

⇐⇒∀x(x∈

M→

f)

∃x∈

M:f

def

⇐⇒∃x(x∈

M∧

f)

(105)

Acco

rdin

gly,

Form

ula

104

can

be

writte

nas

∀x(x∈

R→∃n(n∈

N∧

x<

n)).

(106)

Sequence

sofid

entica

lquantifi

ers

can

be

abbre

viate

d:

∀x,y(f

)d

ef

⇐⇒

∀x(∀

y(f

)).(107)

Quantifi

ers

have

ahig

her

pre

cedence

than

logica

lco

nnective

s:

∀x(f

)∧

gd

ef

⇐⇒

(∀x(f

))∧

g.

(108)

D.Suenderm

ann

Logic

January

28,2012

72

.Stru

cture

s.

So

far,we

have

learn

ed

how

tocre

ate

(well-fo

rmula

ted)

form

ula

sin

first-o

rderlo

gic,

i.e.,

we

have

deale

dw

ithsyn

tax.

The

next

step

isth

ese

mantics

offirst-o

rderlo

gic.

To

that

end,give

na

signatu

reΣ

,we

intro

duce

the

notio

nofa

structu

re

S=〈U

,J〉

(109)

with

1.

the

unive

rseU

,a

non-e

mpty

set

conta

inin

gall

the

valu

es

that

can

occu

rw

hen

eva

luatin

gte

rms,

2.

the

inte

rpre

tatio

nJ

ofall

functio

nand

pre

dica

tesym

bols

ofΣ

.

D.Suenderm

ann

Logic

January

28,2012

73

.In

terp

reta

tion

.

Form

ally,

Jis

defined

as

1.

Eve

ryfu

nctio

nsym

bolf∈

Fw

ithth

earity

n=

arity(f)

ism

apped

toan

n-ary

functio

n

fJ

:U×···×

U︸

︷︷

︸

ntim

es

→U

.(110)

2.

Eve

rypre

dica

tesym

bolp∈

Pw

ithth

earity

n=

arity(f)is

mapped

toan

n-ary

rela

tion

pJ⊆

Un.

(111)

3.

If=∈

Pth

en

itsin

terp

reta

tion

should

be

natu

ral,

i.e.

=J

=〈u

,v〉|u

,v∈

U∧

u=

v.

(112)

D.Suenderm

ann

Logic

January

28,2012

74

.Stru

cture

s:exa

mple

.

U

sing

Σa

as

inth

eabove

exa

mple

,we

define

astru

cture

Sa

=〈U

,J〉

(113)

as

follo

ws:

1.

U=a,b

2.0

J=

a

3.1

J=

b

4.

+J

=〈〈a

,a〉,

a〉,〈〈a

,b〉,

b〉,〈〈b

,a〉,

b〉,〈〈b

,b〉,

a〉

5.·J

=〈〈a

,a〉,

a〉,〈〈a

,b〉,

a〉,〈〈b

,a〉,

a〉,〈〈b

,b〉,

b〉

6.

=J

=〈a

,a〉,〈b

,b〉

7.≤

J=〈a

,a〉,〈a

,b〉,〈b

,b〉

D.Suenderm

ann

Logic

January

28,2012

75

.Varia

ble

assig

nm

ent

.

Give

na

signatu

reΣ

and

astru

cture

S=〈U

,J〉,

we

define

avaria

ble

assig

nm

ent

I:V→

U(114)

assig

nin

ga

valu

efro

mth

eunive

rseU

toeve

ryvaria

ble

inV

.

E.g

.,usin

gth

esig

natu

reΣ

aand

the

structu

reS

a,a

possib

levaria

ble

assig

nm

ent

is

Ia

=〈x

,a〉,〈y

,b〉,〈z

,a〉

.(115)

D.Suenderm

ann

Logic

January

28,2012

76

.Varia

ble

assig

nm

ent

.

Furth

erm

ore

,we

intro

duce

the

variable

repla

cem

ent

I[x

1/c1]···[x

n/c

n](y

)=

c1

:y

=x

1;

...

cn

:y

=x

n;

I(y

):

oth

erw

ise

(116)

with

x1,...,x

n,y∈

Vand

c1,...,c

n∈

Ure

pla

cing

the

valu

eofa

variable

yby

ci

ifth

evaria

ble

happens

tobe

identica

lto

xi .

Usin

gth

eabove

exa

mple

,we

have

Ia[y

/a][z

/b]=〈x

,a〉,〈y

,a〉,〈z

,b〉

.(117)

D.Suenderm

ann

Logic

January

28,2012

77

.Eva

luatio

nofte

rms

.

Give

na

signatu

reΣ

,a

structu

reS

=〈U

,J〉,

and

avaria

ble

assig

nm

ent

I,fo

reve

ryte

rmt,

itsva

lue

(writte

nas

S(I

,t))

can

be

derive

d

inductive

lyas

1.∀x(x∈

V→

S(I

,x)

=I(x

))

2.

Iff∈

Fand

n=

arity(f)

and

t1 ,

...,tn∈

TΣ

then

S(I

,f(t

1 ,...,tn))

=f

J(S

(I,t1 ),

...,S(I

,tn)).

(118)

Exe

rcise:U

sing

the

above

signatu

reΣ

a,stru

cture

Sa,and

variable

assig

nm

ent

Ia,w

hat

isth

eeva

luatio

nofTerm

4:

S(I

a,(1

+x)·y)?

(119)

D.Suenderm

ann

Logic

January

28,2012

78

.Eva

luatio

nofato

mic

form

ula

s.

Give

na

signatu

reΣ

,a

structu

reS

=〈U

,J〉,

and

avaria

ble

assig

nm

ent

I,fo

reve

ryato

mic

form

ula

p(t

1 ,...,tn),

itsva

lue

can

be

derive

das

S(I

,p(t

1 ,...,tn))

=

1:〈S

(I,t1 ),

...,S(I

,tn)〉∈

pJ,

0:

oth

erw

ise.

(120)

Exe

rcise:U

sing

the

above

signatu

reΣ

a,stru

cture

Sa,and

variable

assig

nm

ent

Ia,w

hat

isth

eeva

luatio

nofForm

ula

93:

S(I

,(1

+x)·y

=0)?

(121)

D.Suenderm

ann

Logic

January

28,2012

79

.Eva

luatio

noffo

rmula

s.

Give

na

signatu

reΣ

,a

structu

reS

=〈U

,J〉,

and

avaria

ble

assig

nm

ent

I,fo

reve

ryfo

rmula

f∈

FΣ,its

valu

eca

nbe

derive

din

ductive

lyas

1.

S(I

,0)

=0;S(I

,1)

=1

2.

S(I

,¬

f)

=I

¬(S

(I,f))

3.

S(I

,f∨

g)

=I

∨(S

(I,f),

S(I

,g))

4.

S(I

,f∧

g)

=I

∧(S

(I,f),

S(I

,g))

5.

S(I

,f→

g)

=I

→(S

(I,f),

S(I

,g))

6.

S(I

,f↔

g)

=I

↔(S

(I,f),

S(I

,g))

7.

S(I

,∀x(f

)=

1:∀c(c∈

U→

S(I

[x/c],

f)

=1)

0:

oth

erw

ise

8.

S(I

,∃x(f

)=

1:∃c(c∈

U∧

S(I

[x/c],

f)

=1)

0:

oth

erw

ise

D.Suenderm

ann

Logic

January

28,2012

80

.Eva

luatio

noffo

rmula

s:exe

rcise.

U

sing

the

above

signatu

reΣ

a,stru

cture

Sa,and

variable

assig

nm

ent

Ia,

dete

rmin

eth

eeva

luatio

nofth

efo

llow

ing

form

ula

s:

1.∃x((1

+x)·y

=0),

2.∀x((1

+x)·y

=0),

3.∀x,y((1

+x)·y

=0),

4.∀x∃y((1

+x)·y

=0),

5.∀y∃x((1

+x)·y

=0→∀z(x·z

=0)).

D.Suenderm

ann

Logic

January

28,2012

81

.U

nive

rsalva

lidity

.

If

fis

afo

rmula

resu

lting

in

S(I

,f)

=1

(122)

for

eve

rypossib

levaria

ble

assig

nm

ent

I,th

en

fis

unive

rsally

valid

.

Bein

gth

eequiva

lent

toa

tauto

logy

inpro

positio

nallo

gic,

unive

rsal

valid

ityofa

form

ula

fis

writte

nas

|=f.

(123)

Iffre

e(f

)=,th

en

S(I

,f)

does

not

depend

on

I.In

this

case

,f

is

calle

da

close

dfo

rmula

.

For

close

dfo

rmula

s,we

use

an

abbre

viate

dte

rmin

olo

gy:

S(f

):=

S(I

,f)

iffre

e(f

)=.

(124)

Also

,in

the

case

that

S(f

)=

1,

(125)

we

sayth

at

the

structu

reS

isa

modelfo

rf

writte

nas

S|=

f.

(126)

D.Suenderm

ann

Logic

January

28,2012

82

.Equiva

lence

,(u

n)sa

tisfiability

.

Also

the

notio

ns

ofequiva

lence

and

(un)sa

tisfiability

arein

herite

dfro

m

pro

positio

nallo

gic.

Two

form

ula

sf

and

gare

equiva

lent

iff

|=f↔

g.

(127)

Ase

toffo

rmula

sM⊆

FΣ

issa

tisfiable

,if

there

isa

variable

assig

nm

ent

Isu

chth

at

∀m

(m∈

M→

S(I

,m

)=

1).

(128)

Oth

erw

ise,

Mis

calle

dunsa

tisfiable

writte

nas

M|=

0.

(129)

D.Suenderm

ann

Logic

January

28,2012

83

.Is

afo

rmula

unsa

tisfiable

?.

O

ur

next

goalis

todefine

an

alg

orith

mth

at

check

sw

heth

er

M|=

0.

(130)

In

genera

l,th

isquestio

nis

undecid

able

(as

show

nla

ter).

We

will,

howeve

r,be

able

todefine

aca

lculu

s⊢

for

which

we

have

M⊢

0↔

M|=

0.

(131)

This

calcu

lus

will

be

base

don

ase

mi-d

ecisio

nalg

orith

m.

I.e.,

ifin

deed

M|=

0,th

ealg

orith

mw

illeve

ntu

ally

disco

verth

isfa

ct.

If,howeve

r,M

issa

tisfiable

,th

ealg

orith

mm

ayru

nete

rnally.

Motiva

tion

ofse

mi-d

ecid

ability:

Gold

bach

conje

cture

.

D.Suenderm

ann

Logic

January

28,2012

84

.Im

porta

nt

equiva

lence

s.

In

ord

erto

define

the

calcu

lus⊢,we

will

transfo

rmfo

rmula

sin

to

first-o

rdercla

use

s.

This

transfo

rmatio

nw

illm

ake

use

ofth

efo

llow

ing

equiva

lence

s(in

additio

nto

the

ones

we

know

from

pro

positio

nallo

gic):

21.|=¬∀x(f

)↔∃x(¬

f)

22.|=¬∃x(f

)↔∀x(¬

f)

23.|=∀x(f

)∧∀x(g

)↔∀x(f∧

g)

24.|=∃x(f

)∨∃x(g

)↔∃x(f∨

g)

25.|=∀x,y(f

)↔∀y,x(f

)

26.|=∃x,y(f

)↔∃y,x(f

)

27.

Ifx∈

V∧

x6∈

free(f

)th

en

a)|=∀x(f

)↔

f

b)|=∃x(f

)↔

f

D.Suenderm

ann

Logic

January

28,2012

85

.Im

porta

nt

equiva

lence

s(co

nt.)

.

28.

Ifx∈

V∧

x6∈

free(g

)∪

bound(g

)th

en

a)|=∀x(f

)∨

g↔∀x(f∨

g)

b)|=∃x(f

)∧

g↔∃x(f∧

g)

c)|=

g∨∀x(f

)↔∀x(g∨

f)

d)|=

g∧∃x(f

)↔∃x(g∧

f)

Ince

rtain

situatio

ns,

itis

nece

ssaryto

renam

ebound

variable

s.

Iff∈

FΣ

and

x,y∈

Vth

en

f[x

/y]is

the

form

ula

which

we

obta

inby

repla

cing

eve

ryoccu

rrence

of

xby

z.

For

exa

mple

(∀u∃v(P

(u,v)))[u

/z]=∀z∃v(P

(z,v)).

(132)

This

leads

us

toourla

stequiva

lence

:

29.

Ifx∈

bound(f

)∧

y6∈

free(f

)∪

bound(f

)th

en

|=f↔

f[x

/y]

D.Suenderm

ann

Logic

January

28,2012

86

.Pre

nex

norm

alfo

rm.

The

above

equiva

lence

sca

nbe

use

dto

rew

ritean

arbitrary

form

ula

such

that

all

quantifi

ers

appear

at

the

begin

nin

gofth

efo

rmula

.

This

repre

senta

tion

isca

lled

pre

nex

norm

alfo

rm.

Exa

mple

1:

∀x(p

(x))→∃x(p

(x))

(close

d)

12↔¬∀x(p

(x))∨∃x(p

(x))

21↔∃x(¬

p(x

))∨∃x(p

(x))

24↔∃x(¬

p(x

)∨

p(x

))

2↔∃x(1

)

27b↔

1(u

nive

rsally

valid

)(133)

D.Suenderm

ann

Logic

January

28,2012

87

.Pre

nex

norm

alfo

rm:exa

mple

.

Exa

mple

2:

∃x(p

(x))→∀x(p

(x))

(close

d)

12↔¬∃x(p

(x))∨∀x(p

(x))

21↔∀x(¬

p(x

))∨∀x(p

(x))

29↔∀x(¬

p(x

))∨∀y(p

(y))

28a↔∀x(¬

p(x

)∨∀y(p

(y)))

28c↔∀x,y(¬

p(x

)∨

p(y

))(134)

This

form

ula

says,if

there

isat

least

one

xtu

rnin

gp(x

)tru

eth

en

p(x

)

isalw

aystru

e.

Intu

rn,if

there

isno

xtu

rnin

gp(x

)tru

eth

en

p(x

)is,

logica

lly,alw

ays

false

.

Conse

quently,

p(x

)is

independently

of

xeith

ertru

eorfa

lse(a

pro

positio

nalvaria

ble

).

D.Suenderm

ann

Logic

January

28,2012

88

.Equisa

tisfiability

.

The

next

norm

aliza

tion

step

require

sa

wid

er

notio

nofequiva

lence

.

W

ew

illre

fer

toit

as

equisa

tisfiability.

Consid

erth

eexa

mple

form

ula

s

f1

=∀x∃y(p

(x,y))

(135)

and

f2

=∀x(p

(x,s(x

))).(136)

f1

and

f2

arenot

equiva

lent.

E.g

.,p(x

,y)

:=x

>y.

They

do

not

eve

nuse

the

sam

esig

natu

re(f

2use

sth

efu

nctio

nsym

bols

not

existin

gin

f1 ).

D.Suenderm

ann

Logic

January

28,2012

89

.Equisa

tisfiability

(cont.)

.

H

oweve

r,we

can

rela

tef1

and

f2

as

follo

ws:

If

astru

cture

S1

isa

modelfo

rf1,i.e

.,

S1(f

1 )=

1(137)

then

itca

nbe

exte

nded

toa

structu

reS

2bein

ga

modelfo

rf2 ,

S2(f

2 )=

1.

(138)

This

can

be

done

by

definin

gan

inte

rpre

tatio

ns

Jsu

chth

at

p(x

,s(x

))(139)

istru

efo

rall

possib

lex

ofth

eunive

rse.

Inour

above

exa

mple

,th

isco

uld

for

insta

nce

be

s(x

):=

x−

1.

Form

ally,

two

form

ula

sf1

and

f2

areequisa

tisfiable

if

|=S

1(f

1 )↔

S2(f

2 )(140)

also

writte

nas

f1≈

f2.

(141)

D.Suenderm

ann

Logic

January

28,2012

90

.Sko

lem

izatio

n.

W

eare

give

na

signatu

reΣ

=〈V

,F

,P

,arity〉

and

aclo

sed

form

ula

fof

the

formf

=∀x

1,...,x

n∃y(g

(x1,...,x

n,y)).

(142)

W

ech

oose

anew

n-ary

functio

nsym

bols6∈

Fand

exte

nd

the

signatu

re:

Σ′=〈V

,F∪s,P

,arity∪〈s

,n〉〉.

(143)

Now

,we

define

the

form

ula

f′as

follo

ws:

f′:=

skole

m(f

):=∀x

1,...,x

n(g

(x1,...,x

n,s(x

1,...,x

n))).

(144)

Here

,we

have

dro

pped

the

existe

ntia

lquantifi

er.

Eve

ryoccu

rrence

ofth

evaria

ble

yhas

been

repla

ced

by

the

term

s(x

1,...,x

n).

For

the

resu

lting

form

ula

f′,

we

have

f′≈

f.

(145)

D.Suenderm

ann

Logic

January

28,2012

91

.Sko

lem

izatio

n:exa

mple

s.

f1

=∀x,y∃z(p

(x,y,z)),

skole

m(f

1 )=∀x,y(p

(x,y,s1z(x

,y)));

f2

=∀x∃y∀z(p

(x,y,z)),

skole

m(f

2 )=∀x,z(p

(x,s2y(x

),z));

f3

=∃x∀y∃z(p

(x,y,z)),

skole

m(f

3 )=∀y∃z(p

(s3x,y,z)),

skole

m(sko

lem

(f3 ))

=∀y(p

(s3x,y,s3z(y

)))(146)

D.Suenderm

ann

Logic

January

28,2012

92

.Cla

usa

lnorm

alfo

rm.

This

alg

orith

mtu

rns

an

first-o

rderfo

rmula

fin

tocla

usa

lnorm

alfo

rm:

1.

Conve

rtf

into

pre

nex

norm

alfo

rm.

2.

Elim

inate

all

existe

ntia

lquantifi

ers

by

skole

miza

tion.The

resu

ltis

a

form

ula

ofth

efo

rm

f′=∀x

1,...,x

n(g

)(147)

where

gco

nta

ins

no

quantifi

ers.

3.

Sin

ceg

conta

ins

only

ato

mic

form

ula

sco

nnecte

dby

logica

l

connective

s,it

can

be

turn

ed

into

conju

nctive

norm

alfo

rmw

hich

pro

duce

s

f′′=∀x

1,...,x

n(d

1∧···∧

dm

)(148)

where

di

aredisju

nctio

ns

oflite

rals

(infirst-o

rderlo

gic,

litera

lsare

ato

mic

form

ula

sor

their

negatio

ns).

D.Suenderm

ann

Logic

January

28,2012

93

.Cla

usa

lnorm

alfo

rm(co

nt.)

.

4.

Applyin

gEquiva

lence

23,th

eunive

rsalquantifi

ers

can

be

distrib

ute

d

onto

all

di s

resu

lting

inth

ecla

usa

lnorm

alfo

rm

f′′′=

c1∧···∧

cm

with

ci=∀x

1,...,x

n(d

i ),(149)

aco

nju

nctio

nofth

efirst-o

rdercla

use

sc1,...,c

m.

5.

This

nota

tion

can

furth

erbe

simplifi

ed

by

agre

ein

gth

at

all

free

variable

sare

implicite

lyunive

rsally

quantifi

ed:

f(4)

=d

1∧···∧

dm

⇔d

1,...,d

m.

(150)

D.Suenderm

ann

Logic

January

28,2012

94

.Cla

usa

lnorm

alfo

rm:exe

rcise.

Turn

the

follo

win

gfo

rmula

sin

tocla

usa

lnorm

alfo

rm:

1.

f1

=∀y∃x((1

+x)·y

=0→∀z(x·z

=0)),

2.

f2

=∃x,y(p

(x)∧

p(y

)↔∃z(p

(z))),

3.¬

f2.

For

Task

1,give

nth

estru

cture

Sa,find

afu

nctio

nin

gin

terp

reta

tion

of

the

skole

mfu

nctio

nre

pla

cing

the

variable

x.

D.Suenderm

ann

Logic

January

28,2012

95

.Pro

vability

.

O

ur

goalis

tofind

out

wheth

er

afo

rmula

fis

unive

rsally

valid

:

|=f.

(151)

This

isth

esa

me

as

tote

llw

heth

er

f’s

negatio

nis

unsa

tisfiable

:

¬

f|=

0.

(152)

This

can

be

done

by

1.

transfo

rmin

g¬

fin

tocla

usa

lnorm

alfo

rm

c1∧···∧

cm≈¬

f,

(153)

2.

trying

topro

vein

consiste

ncy

from

the

set

ofcla

use

s:

c1,...,c

m⊢.

(154)

D.Suenderm

ann

Logic

January

28,2012

96

.Pro

vability:

exa

mple

.

W

ewant

tofind

out

wheth

erth

efo

rmula

f:=∃x∀y(p

(x,y))→∀y∃x(p

(x,y))

(155)

isunive

rsally

valid

.

This

can

be

done

by

usin

gth

epro

vability

alg

orith

m:

1.

Tra

nsfo

rmin

g¬

fin

tocla

usa

lnorm

alfo

rm:

1.1

Conve

rsion

into

pre

nex

norm

alfo

rm:

¬(∃

x∀y(p

(x,y))→∀y∃x(p

(x,y)))

12↔¬

(¬∃x∀y(p

(x,y))∨∀y∃x(p

(x,y)))

11↔∃x∀y(p

(x,y))∧¬∀y∃x(p

(x,y))

21↔∃x∀y(p

(x,y))∧∃y¬∃x(p

(x,y))

22↔∃x∀y(p

(x,y))∧∃y∀x(¬

p(x

,y))

29↔∃x∀y(p

(x,y))∧∃v∀u(¬

p(u

,v))

28b↔∃x(∀

y(p

(x,y))∧∃v∀u(¬

p(u

,v)))

D.Suenderm

ann

Logic

January

28,2012

97

.Pro

vability:

exa

mple

(cont.)

.

28d↔∃x,v(∀

y(p

(x,y))∧∀u(¬

p(u

,v)))

27a↔∃x,v(∀

y(p

(x,y))∧∀y,u(¬

p(u

,v)))

23↔∃x,v∀y(p

(x,y)∧∀u(¬

p(u

,v)))

27a↔∃x,v∀y(∀

u(p

(x,y))∧∀u(¬

p(u

,v)))

23↔∃x,v∀y,u(p

(x,y)∧¬

p(u

,v))

=:

f′

(156)

1.2

Sko

lem

izatio

n:

–f

′is

ofth

efo

rm

f′=∃x(g

(x))

with

g=∃v∀y,u(p

(x,y)∧¬

p(u

,v)).

(157)

–H

ere

,g

is1-ary

since

there

areno

unive

rsalquantifi

ers

infro

nt

of

the

existe

ntia

lquantifi

er.

D.Suenderm

ann

Logic

January

28,2012

98

.Pro

vability:

exa

mple

(cont.)

.

–Conse

quently,

skole

miza

tion

intro

duce

sth

e0-ary

functio

ns

x

repla

cing

x:

f′′

=sko

lem

(f′)

=g(s

x())

=g(s

x)

=∃v∀y,u(p

(sx,y)∧¬

p(u

,v)).

(158)

–In

ase

cond

step,th

eexiste

ntia

lquantifi

erin

front

of

vis

repla

ced

by

skole

miza

tion:

f′′′

=sko

lem

(f′′)

=∀y,u(p

(sx,y)∧¬

p(u

,s

v)

︸︷︷

︸

h

).(159)

1.3

-5Cla

usa

lnorm

alfo

rm:

Sin

ceh

isalre

ady

inco

nju

nctive

norm

alfo

rm,we

have

f(4)=

p(s

x,y)∧¬

p(u

,s

v).

(160)

D.Suenderm

ann

Logic

January

28,2012

99

.Pro

vability:

exa

mple

(cont.)

.

2.

Pro

ving

inco

nsiste

ncy:

–W

eare

give

ncla

usa

lnorm

alfo

rm

M:=p(s

x,y)

,¬

p(u

,s

v).

(161)

–Sin

ceall

variable

sin

first-o

rdercla

use

sare

implicite

lyunive

rsally

bound,we

can

substitu

tey

by

sv

and

uby

sx

resu

lting

in

M⊢p(s

x,s

v)

,¬

p(s

x,s

v).

(162)

–Applica

tion

ofth

ecu

tru

legive

s

M⊢.

(163)

–Conse

quently,

we

have

show

nth

at

|=f.

(164)

D.Suenderm

ann

Logic

January

28,2012

100

.Substitu

tion

.

In

the

pro

vability

exa

mple

,we

have

“guesse

d”

the

term

ss

xand

sv

tobe

able

toapply

the

cut

rule

.

W

enow

want

tostu

dy

am

eth

od

todo

system

atica

llyca

lcula

tete

rms

which

the

cut

rule

can

be

applie

dto

.

Inord

erto

do

so,fo

ra

give

nsig

natu

reΣ

,we

define

asu

bstitu

tion

σ=〈x

1,t1 〉,

...,〈x

n,tn〉

=:[x

17→

t1 ,

...,x

n7→

tn]

(165)

with

xi∈

V,ti∈

TΣ

for

i∈1,...,n

and

xi6=

xj

for

i6=

j.(166)

The

dom

ain

ofth

esu

bstitu

tion

isdefined

as

dom

(σ)

=x

1,...,x

n.

(167)

D.Suenderm

ann

Logic

January

28,2012

101

.A

pplica

tion

ofa

substitu

tion

.

Give

na

term

tand

asu

bstitu

tion

σ,we

define

the

applica

tion

of

σto

t

(writte

ntσ

)in

ductive

lyas

1.

xi σ

:=ti

for

xi∈

dom

(σ),

2.

yσ

:=y

for

y∈

V∧

y6∈

dom

(σ),

3.

f(s

1,...,s

m)σ

:=f(s

1σ,...,s

mσ)

oth

erw

ise.

Exa

mple

:W

eare

give

nth

esu

bstitu

tion

σ=

[x7→

c,y7→

f(d

)].(168)

Then,we

have

the

follo

win

gapplica

tions:

1.

zσ

=z,

2.

f(y

)σ=

f(f

(d)),

3.

h(x

,g(y

))σ=

h(c

,g(f

(d))),

4.p(y

),q(d

,h(z

,x))

σ=p(f

(d)),

q(d

,h(z

,c))

.

D.Suenderm

ann

Logic

January

28,2012

102

.Syn

tactica

lequatio

ns

and

unifi

ers

.

A

synta

cticalequatio

n(S

E)

isa

constru

ctofth

efo

rms

.=t

where

sand

tare

eith

er

both

term

sor

both

ato

mic

form

ula

s.

A

system

ofsyn

tactica

lequatio

ns

(SSE)

isa

set

ofSEs:

E:=s1

.=t1 ,

...,s

n.=

tn.

(169)

Give

nan

SSE

Eand

asu

bstitu

tion

σ,we

define

the

applica

tion

of

σto

Eas

Eσ

:=s1σ

.=t1 σ

,...,s

nσ

.=tnσ.

(170)

Asu

bstitu

tion

σso

lves

an

SE

s.=

tiff

we

have

sσ

=tσ

.

IfE

isan

SSE

then

the

substitu

tion

σis

calle

da

unifi

eriff

itso

lves

eve

ry

SE

inE

.

Exa

mple

:Show

that

σ=

[x17→

x2,x

37→

f(x

4)]

(171)

isa

unifi

erofth

eSSE

E=p(f

(x4))

.=p(x

3),

q(x

1,x

2)

.=q(x

2,x

1)

.(172)

D.Suenderm

ann

Logic

January

28,2012

103

.U

nifi

catio

n.

W

enow

want

tostu

dy

an

alg

orith

mca

lcula

ting

aunifi

er

σfo

ra

give

n

SSE

E.

A

num

berofre

ductio

nru

les

will

transfo

rma

tuple

consistin

gofan

SSE

and

asu

bstitu

tion

into

anoth

ersu

chtu

ple

:

〈E1,σ

1〉 〈E

2,σ

2〉.

(173)

Here

com

eth

ere

ductio

nru

les:

1.

Ify∈

V∧

t∈

TΣ∧

y6∈

var(t)th

en

〈E∪y

.=t

,σ〉 〈E

[y7→

t],σ∪〈y

,t〉〉.

(174)

–If

an

SSE

conta

ins

an

SE

ofth

efo

rmy

.=t

where

yis

avaria

ble

not

conta

ined

inte

rmt

then

the

SE

isso

lved

by

the

substitu

tion

[y7→

t].

–Conse

quently,

the

SE

can

be

dro

pped

infa

vor

ofapplyin

gth

e

substitu

tion

toall

oth

erSEs

inth

eSSE

and

addin

git

toσ.

–E.g

.,x

.=s(y

).

D.Suenderm

ann

Logic

January

28,2012

104

.U

nifi

catio

n(co

nt.)

.

2.

Ify∈

V∧

t∈

TΣ∧

y∈

var(t)∧

y6=

tth

en

〈E∪y

.=t

,σ〉

Ω.

(175)

–If

an

SSE

conta

ins

an

SE

ofth

efo

rmy

.=t

where

yis

avaria

ble

conta

ined

inte

rmt

but

not

titse

lfth

en

the

SE

isnot

solva

ble

.

–That

isbeca

use

eve

rypossib

lete

rmwe

can

substitu

tefo

ry

will

also

affect

tand

render

itdiff

ere

nt.

–E.g

.,x

.=s(x

).

3.

Ify∈

V∧

t∈

TΣ∧

t6∈

Vth

en

〈E∪t

.=y,σ〉 〈E∪y

.=t

,σ〉.

(176)

–M

ove

sth

evaria

ble

toth

efro

nt.

–E.g

.,s(y

).=

x.

4.

〈E∪t

.=t

,σ〉 〈E

,σ〉.

(177)

–D

rops

trivialSEs.

–E.g

.,s(y

).=

s(y

).

D.Suenderm

ann

Logic

January

28,2012

105

.U

nifi

catio

n(co

nt.)

.

5.

Iff∈

F∪

P∧

n=

arity(f)∧

s1,...,s

n,t1 ,

...,tn∈

TΣ

then

〈E∪f(s

1,...,s

n)

.=f(t

1 ,...,tn)

,σ〉

〈E∪s1

.=t1 ,

...,s

n.=

tn,σ〉.

(178)

–Fro

mth

earg

um

ents

ofan

n-ary

functio

n(p

redica

te)

pre

sent

on

both

sides

ofan

SE,

nin

divid

ualSEs

arederive

d.

–E.g

.,s(s

(x),

y)

.=s(y

,z).

6.

Iff,g∈

F∪

P∧

f6=

g∧

m=

arity(f)∧

n=

arity(g)

∧s1,...,s

m,t1 ,

...,tn∈

TΣ

then

〈E∪f(s

1,...,s

m)

.=g(t

1 ,...,tn)

,σ〉

Ω.

(179)

–D

iffere

nt

functio

ns

(pre

dica

tes)

cannot

be

unifi

ed.

–E.g

.,f(x

).=

g(x

).

D.Suenderm

ann

Logic

January

28,2012

106

.U

nifi

catio

nalg

orith

m.

Give

nan

SSE

E,to

dete

rmin

ea

unifi

er,

we

startoff

with

an

em

pty

substitu

tion:

〈E,[]〉.

(180)

Now

,we

apply

the

reductio

nru

les

until

we

have

eith

er

ofth

efo

llow

ing

case

s:

a)

We

can

derive

Ωm

eanin

gth

at

Eis

not

solva

ble

.

b)

We

can

show

that

〈E,[]〉 〈,σ〉.

(181)

Here

,σ

isth

eunifi

erof

Ealso

writte

nas

σ=

mgu(E

)(182)

(the

most

genera

lunifi

er)

with

the

specia

lca

se

mgu(s

,t)

:=m

gu(

s.=

t).

(183)

D.Suenderm

ann

Logic

January

28,2012

107

.U

nifi

catio

n:exa

mple

.

W

eare

give

nan

SSE

with

asin

gle

SE:

E=p(x

1,f(x

4))

.=p(x

2,x

3)

with

(184)

V=x

1,...

,F

=f,P

=.

N

ow

,we

atte

mpt

todete

rmin

ea

unifi

er:

〈p(x

1,f(x

4))

.=p(x

2,x

3)

,[]〉

5 〈

x1

.=x

2,f(x

4 ).=

x3,[]〉

1 〈

f(x

4)

.=x

3,[x

17→

x2]〉

3 〈

x3

.=f(x

4)

,[x

17→

x2]〉

1 〈,[x

17→

x2,x

37→

f(x

4)]〉

(185)

That

is,we

have

mgu(E

)=

mgu(p

(x1,f(x

4)),

p(x

2,x

3))

=[x

17→

x2,x

37→

f(x

4)].

(186)

D.Suenderm

ann

Logic

January

28,2012

108

.U

nifi

catio

n:exe

rcise.

D

ete

rmin

e,if

possib

le,a

unifi

er

ofth

efo

llow

ing

SSEs:

a)

E1

=p(d

,x

4)

.=p(x

2,h(c

,d)),

q(h

(d,x

1))

.=q(x

4)

,

b)

E2

=p(h

(x1,c))

.=p(x

2),

q(x

2,d)

.=q(h

(d,c),

x4)

,

c)E

3=p(h

(x2,d))

.=p(h

(x1,d)),

q(x

1,c)

.=q(h

(x2,d),

c)

.

Inth

ese

exe

rcises,

assu

me

V=x

1,...

,F

=c,d,h,P

=p,q.

D.Suenderm

ann

Logic

January

28,2012

109

.First-o

rderre

solu

tion

.

The

first-o

rderca

lculu

sm

ake

suse

oftw

oin

fere

nce

rule

swe

will

define

next.

The

first-o

rder

reso

lutio

nru

leexp

ects

1.

two

first-o

rdercla

use

sc1

and

c2

and

2.

two

ato

mic

form

ula

sp(s

1,...,s

n)

and

p(t

1 ,...,tn)

whose

synta

cticalequatio

nis

solva

ble

,i.e

.,we

can

dete

rmin

eth

eunifi

er

µ=

mgu(p

(s1,...,s

n),

p(t

1,...,tn)).

(187)

Then,th

isis

the

definitio

nofth

ere

solu

tion

rule

:

c1∪p(s

1,...,s

n)

,c2∪¬

p(t

1 ,...,tn)

∴c1µ∪

c2µ

The

reso

lutio

nru

lem

ake

suse

ofth

ecu

tru

leand

the

substitu

tion

rule

c

∴cσ

where

cis

afirst-o

rdercla

use

and

σis

asu

bstitu

tion.

D.Suenderm

ann

Logic

January

28,2012

110

.First-o

rder

reso

lutio

n:varia

ble

renam

ing

.

W

hen

applyin

gth

ere

sultio

nru

le,it

isso

metim

es

nece

ssaryto

renam

e

variable

s.

In

the

follo

win

gfirst-o

rdercla

use

set

M=p(x

),¬

p(f

(x)),

(188)

unifi

catio

nw

illre

sult

inΩ

since

both

invo

lved

clause

suse

the

sam

e

variable

x.

Substitu

ting

xby

yin

one

ofth

ecla

use

sdoes

not

change

sem

antics

M′=p(x

),¬

p(f

(y))

(189)

simply

beca

use

∀x(¬

p(f

(x)))

29↔∀y(¬

p(f

(y))).

(190)

Now

,we

areable

todete

rmin

eth

eunifi

er

µ=

mgu(p

(x),

p(f

(y)))

=[x7→

f(y

)](191)

Inco

nclu

sion,we

can

pro

veco

ntra

dictio

nby

applyin

gth

ere

sultio

nru

le:

p(x

),¬

p(f

(y))

res.

[x7→

f(y)]

⊢.

(192)

D.Suenderm

ann

Logic

January

28,2012

111

.Facto

rizatio

n:m

otiva

tion

.

Consid

erth

efo

llow

ing

exa

mple

M=p(x

),p(y

),¬

p(x

),¬

p(y

)

(193)

Applica

tion

ofth

ere

solu

tion

rule

usin

gc1

=p(x

)and

c2

=¬

p(x

)and

the

unifi

er

µ=

mgu(p

(y),

p(y

))=

[]pro

duce

s

p(x

),p(y

),¬

p(x

),¬

p(y

)

res.

[]

⊢p(y

),¬

p(y

)

⊢1.

(194)

D.Suenderm

ann

Logic

January

28,2012

112

.Facto

rizatio

n:m

otiva

tion

(cont.)

.

This

resu

ltdoes

not

help

since

we

areatte

mptin

gto

pro

veco

ntra

dictio

n.

In

stead,one

could

facto

rizeM

as

follo

ws:

p(x

),p(y

),¬

p(x

),¬

p(y

)

↔p(x

),p(y

),¬

p(v

),¬

p(w

)

↔∀x,y,v,w

((p(x

)∨

p(y

))∧

(¬p(v

)∨¬

p(w

)))

↔∀x,y,v,w

(p(x

)∧¬

p(v

)∨

p(x

)∧¬

p(w

)∨

p(y

)∧¬

p(v

)∨

p(y

)∧¬

p(w

))

↔∀x,v(p

(x)∧¬

p(v

))∨∀x,w

(p(x

)∧¬

p(w

))∨

∀y,v(p

(y)∧¬

p(v

))∨∀y,w

(p(y

)∧¬

p(w

)))

↔∀x(p

(x))∧∀x(¬

p(x

))∨∀x(p

(x))∧∀x(¬

p(x

))∨

∀y(p

(y))∧∀y(¬

p(y

))∨∀y(p

(y))∧∀y(¬

p(y

)))

↔∀x(p

(x)∧¬

p(x

))∨∀y(p

(y)∧¬

p(x

))

↔0.

(195)

D.Suenderm

ann

Logic

January

28,2012

113

.Facto

rizatio

n:m

otiva

tion

(cont.)

.

Anoth

erpossib

ilityis,

as

inpro

positio

nallo

gic,

toapply

case

distin

ction

toacco

unt

for

multip

leoccu

rrence

sofnegate

dlite

rals:

Pickp(x

):

p(x

),p(x

),p(y

),¬

p(x

),¬

p(y

)

res.

[]

⊢p(x

),p(x

),p(y

),¬

p(y

)

res.

[x7→

y]

⊢,p(x

),p(y

)

⊢0.

(196)

D.Suenderm

ann

Logic

January

28,2012

114

.Facto

rizatio

n:m

otiva

tion

(cont.)

.

Pick¬p(x

)↔¬∀x(p

(x))↔∃x(¬

p(x

))≈¬

p(s

)↔¬

p(s

):

¬

p(s

),p(x

),p(y

),¬

p(x

),¬

p(y

)

res.

[x7→

s]

⊢¬

p(s

),p(y

),¬

p(x

),¬

p(y

)

res.

[y7→

s]

⊢,¬

p(x

),¬

p(y

)

⊢0.

(197)

So,in

conclu

sion,we

have

p(x

),p(y

),¬

p(x

),¬

p(y

)⊢

0.

(198)

D.Suenderm

ann

Logic

January

28,2012

115

.Facto

rizatio

n.

In

genera

l,we

acco

unt

for

multip

leoccu

rrence

sofunifi

able

pre

dica

tes

in

acla

use

by

means

offa

ctoriza

tion

rule

s.

Give

n

1.

the

first-o

rdercla

use

cand

2.

two

ato

mic

form

ula

sp(s

1,...,s

n)

and

p(t

1 ,...,tn)

whose

synta

cticalequatio

nis

solva

ble

,i.e

.,we

can

dete

rmin

eth

eunifi

er

µ=

mgu(p

(s1,...,s

n),

p(t

1,...,tn)).

(199)

Then,th

ese

areth

edefinitio

ns

ofth

efa

ctoriza

tion

rule

s:

c∪p(s

1,...,s

n),

p(t

1 ,...,tn)

∴cµ∪p(s

1,...,s

n)µ

and

c∪¬

p(s

1,...,s

n),¬

p(t

1 ,...,tn)

∴cµ∪¬

p(s

1,...,s

n)µ

D.Suenderm

ann

Logic

January

28,2012

116

.Facto

rizatio

n:exa

mple

.

Retu

rnin

gto

the

above

exa

mple

,applica

tion

ofth

efa

ctoriza

tion

rule

s

pro

duce

s

p(x

),p(y

),¬

p(x

),¬

p(y

)

fact.

⊢p(y

),¬

p(x

),¬

p(y

)

fact.

⊢p(y

),¬

p(y

)

res.

[]

⊢0.

(200)

D.Suenderm

ann

Logic

January

28,2012

117

.U

nifi

catio

n/re

solu

tion/fa

ctoriza

tion:exe

rcise.

Can

you

derive

contra

dictio

nfro

mth

efo

llow

ing

clause

sets?

a)

M1

=¬

p(y

,y)

,p(f

(x),

y),

p(y

,g(v

))

b)

M2

=p(x

,y),

p(g

(u),

v),

p(z

,v)

,¬

p(x

,y),¬

p(g

(u),

v)

c)M

3=p(x

,f(z

)),p(g

(u),

v),

p(g

(u),

f(z

))

d)

M4

=p(f

(g(u

)),v),¬

p(f

(z),

v)

,¬

p(x

,y),¬

p(g

(u),

v)

D.Suenderm

ann

Logic

January

28,2012

118

.First-o

rderve

rsion

ofD

avis

and

Putn

am

:re

solu

tion

functio

n.

Sim

ilarlyto

the

functio

ncu

tin

troduce

dfo

rth

edefinitio

nofth

eD

P

alg

orith

min

pro

positio

nallo

gic,

we

intro

duce

the

functio

nre

sfo

r

first-o

rderlo

gic.

The

functio

nre

sapplie

sth

ere

solu

tion

rule

toa

set

ofcla

use

sC

0w

ith

resp

ect

toa

litera

l(u

nit

clause

).

As

oppose

dto

cut,

itdoes

not

dro

pany

clause

sfro

mC

0but

simply

adds

conclu

sions

from

the

reso

lutio

nru

leto

the

clause

set.

The

reaso

nth

at

clause

sare

not

dro

pped

isth

at

diff

ere

nt

ato

mic

form

ula

s

may

be

base

don

the

sam

epre

dica

tebut

diff

ere

nt

argum

ents

which

may

pro

duce

adiff

ere

nt

unifi

er

and,hence

,a

diff

ere

nt

resu

ltcla

use

.

Ifth

eorig

inalcla

use

would

have

been

dro

pped,th

isdiff

ere

nt

clause

would

have

been

misse

d.

D.Suenderm

ann

Logic

January

28,2012

119

.First-o

rder

versio

nofD

avis

and

Putn

am

:re

solu

tion

functio

n(e

xam

ple

s).

Exa

mple

1:

res(p(x

),¬

q(f

(t)),q(x

),q(x

))=

p(x

),¬

q(f

(t)),p(f

(t)),q(x

)

(201)

Exa

mple

2:

res(p(y

),¬

q(x

),p(f

(y)),¬

q(x

),¬

p(x

),¬

p(x

))=

(202)

p(y

),¬

q(x

),¬

q(x

),p(f

(y)),¬

q(x

),¬

q(f

(y))

,¬

p(x

)

Exa

mple

3(in

finite

loop):

C0

:=¬

p(x

),p(f

(x))

,p(x

)

C1

:=re

s(C0 ,

p(x

))

=¬

p(x

),p(f

(x))

,p(f

(x))

,p(x

)

=¬

p(x

),p(f

(x))

,p(f

(y))

,p(x

)

C2

:=re

s(C1 ,

p(f

(y)))

=¬

p(x

),p(f

(x))

,p(f

(f(y

))),p(f

(y))

,p(x

)

C3

:=re

s(C2 ,

p(f

(f(y

))))...

(203)

D.Suenderm

ann

Logic

January

28,2012

120

.First-o

rder

calcu

lus:

exa

mple

.

W

ehave

now

learn

ed

all

the

pie

ces

nece

ssaryto

apply

first-o

rderca

lculu

s

ina

system

atic

way.

Let

us

reite

rate

on

the

invo

lved

steps

by

means

ofan

exa

mple

.

This

isourknow

ledge

base

:

a)

Every

dra

gon

ishappy

ifall

his

child

renknow

how

tofly.

b)

Every

reddra

gon

know

show

tofly.

c)Child

renofred

dra

gons

arered

.

Now

,le

tus

tryto

pro

vew

heth

er

ornot

d)

All

reddra

gons

arehappy.

D.Suenderm

ann

Logic

January

28,2012

121

.First-o

rderca

lculu

s:exa

mple

(cont.)

.

To

get

started,we

define

asig

natu

reΣ

a=〈V

,F

,P

,arity〉

with

V=x,y

F=

P=r,

f,h

,c

arity=〈r,

1〉〈f

,1〉〈h

,1〉〈c

,2〉

We

assu

me

that

the

unive

rseco

nta

ins

all

dra

gons.

The

pre

dica

tes

have

the

follo

win

gin

terp

reta

tions:

–r(x

)is

1iff

xis

red.

–f(x

)is

1iff

xknow

show

tofly.

–h(x

)is

1iff

xis

happy.

–c(x

,y)

is1

iffx

isy’s

child

.

D.Suenderm

ann

Logic

January

28,2012

122

.First-o

rderca

lculu

s:exa

mple

(cont.)

.

Form

alizin

gknow

ledge

base

and

claim

:

a)

f1

:=∀x(∀

y(c

(y,x)→

f(y

))→

h(x

))

b)

f2

:=∀x(r

(x)→

f(x

))

c)f3

:=∀x(∃

y(c

(x,y)∧

r(y

))→

r(x

))

d)

f4

:=∀x(r

(x)→

h(x

))

Now

,to

see

wheth

erth

ecla

imd)

can

be

derive

dfro

mth

eknow

ledge

base

a)

toc)

isto

pro

veth

at

f:=

f1∧

f2∧

f3→

f4

(204)

isunive

rsally

valid

.

D.Suenderm

ann

Logic

January

28,2012

123

.First-o

rderca

lculu

s:exa

mple

(cont.)

.

This

isth

esa

me

as

topro

veth

at¬

fis

aco

ntra

dictio

n:

¬f↔¬

(f1∧

f2∧

f3→

f4)

↔¬

(¬(f

1∧

f2∧

f3 )∨

f4)

↔¬¬

(f1∧

f2∧

f3 )∨¬

f4)

↔¬¬

(f1∧

f2∧

f3 )∧¬

f4

↔f1∧

f2∧

f3∧¬

f4

(205)

Next,

we

need

totu

rn¬

fin

toa

set

ofcla

use

s.

This

can

be

done

by

turn

ing

f1,

f2 ,

f3 ,

and¬

f4

individ

ually

into

clausa

l

norm

alfo

rm.

D.Suenderm

ann

Logic

January

28,2012

124

.First-o

rderca

lculu

s:exa

mple

(cont.)

.

f1↔∀x(∀

y(c

(y,x)→

f(y

))→

h(x

))

↔∀x(∀

y(¬

c(y

,x)∨

f(y

))→

h(x

))

↔∀x(¬∀y(¬

c(y

,x)∨

f(y

))∨

h(x

))

↔∀x(∃

y(¬

(¬c(y

,x)∨

f(y

)))∨

h(x

))

↔∀x(∃

y(¬¬

c(y

,x)∧¬

f(y

))∨

h(x

))

↔∀x(∃

y(c

(y,x)∧¬

f(y

))∨

h(x

))

↔∀x∃y(c

(y,x)∧¬

f(y

)∨

h(x

))

≈∀x(c

(s(x

),x)∧¬

f(s

(x))∨

h(x

))

↔∀x((c

(s(x

),x)∨

h(x

))∧

(¬f(s

(x))∨

h(x

)))

↔c(s

(x),

x),

h(x

),¬

f(s

(x)),

h(x

)

(206)

f2↔∀x(r

(x)→

f(x

))

↔∀x(¬

r(x

)∨

f(x

))

↔¬

r(x

),f(x

)

(207)

D.Suenderm

ann

Logic

January

28,2012

125

.First-o

rderca

lculu

s:exa

mple

(cont.)

.

f3↔∀x(∃

y(c

(x,y)∧

r(y

))→

r(x

))

↔∀x(¬∃y(c

(x,y)∧

r(y

))∨

r(x

))

↔∀x(∀

y(¬

(c(x

,y)∧

r(y

)))∨

r(x

))

↔∀x(∀

y(¬

c(x

,y)∨¬

r(y

))∨

r(x

))

↔∀x,y(¬

c(x

,y)∨¬

r(y

)∨

r(x

))

↔¬

c(x

,y),¬

r(y

),r(x

)

(208)

¬f4↔¬∀x(r

(x)→

h(x

))

↔¬∀x(¬

r(x

)∨

h(x

))

↔∃x(¬

(¬r(x

)∨

h(x

)))

↔∃x(r

(x)∧¬

h(x

))

≈r(t)∧¬

h(t)

↔r(t)

,¬

h(t)

(209)

D.Suenderm

ann

Logic

January

28,2012

126

.First-o

rderca

lculu

s:exa

mple

(cont.)

.

Applyin

gth

eD

Palg

orith

m:

C0

:=c(s

(x),

x),

h(x

),¬

f(s

(x)),

h(x

),¬

r(x

),f(x

),

¬

c(x

,y),¬

r(y

),r(x

),r(t)

,¬

h(t)

C1

:=re

s(C0 ,

r(t))

=c(s

(x),

x),

h(x

),¬

f(s

(x)),

h(x

),¬

r(x

),f(x

),f(t)

,

¬

c(x

,y),¬

r(y

),r(x

),¬

c(x

,t),

r(x

),r(t)

,¬

h(t)

C2

:=re

s(C1 ,¬

h(t))

=c(s

(x),

x),

h(x

),c(s

(t),t)

,¬

f(s

(x)),

h(x

),¬

f(s

(t)),

¬

r(x

),f(x

),f(t)

,¬

c(x

,y),¬

r(y

),r(x

),¬

c(x

,t),

r(x

),

r(t)

,¬

h(t)

(210)

D.Suenderm

ann

Logic

January

28,2012

127

.First-o

rderca

lculu

s:exa

mple

(cont.)

.

C3

:=re

s(C2 ,

c(s

(t),t))

=c(s

(x),

x),

h(x

),c(s

(t),t)

,¬

f(s

(x)),

h(x

),¬

f(s

(t)),

¬

r(x

),f(x

),f(t)

,¬

c(x

,y),¬

r(y

),r(x

),

¬

r(t),

r(s

(t)),¬

c(x

,t),

r(x

),r(s

(t)),r(t)

,¬

h(t)

C4

:=re

s(C3 ,

r(s

(t)))

=c(s

(x),

x),

h(x

),c(s

(t),t)

,¬

f(s

(x)),

h(x

),¬

f(s

(t)),

¬

r(x

),f(x

),f(s

(t)),f(t)

,¬

c(x

,y),¬

r(y

),r(x

),

¬

c(x

,s(t)),

r(x

),¬

r(t),

r(s

(t)),¬

c(x

,t),

r(x

),r(s

(t)),

r(t)

,¬

h(t)

(unsa

tisfiable

)(211)

D.Suenderm

ann

Logic

January

28,2012

128

.First-o

rder

calcu

lus:

exe

rcises

.

W

eknow

that

Sets

can

only

be

conta

ined

insets

not

conta

ined

inth

eSuen

derm

ann

set.

Now

,pro

veth

at

The

Suen

derm

ann

setdoes

not

conta

initself.

Give

nth

eth

ree

pro

pertie

sofa

tota

lord

er

1)

a<

band

b<

cim

plie

sa

<c

(transitivity),

2)

a<

b,

b<

a,or

a=

bis

true

(trichoto

my),

and

3)

a<

ais

not

true

(anti-re

flexivity),

pro

veth

ere

flexivity

ofth

eequiva

lence

(i.e.

a=

a).

D.Suenderm

ann

Logic

January

28,2012

129

.Pro

log

.

Pro

log

(pro

gra

mm

ing

inlo

gic)

ispro

gra

mm

ing

language

asso

ciate

dw

ith

artificia

lin

tellig

ence

as

well

as

com

pute

rlin

guistics.

In

acco

rdance

with

the

archite

cture

ofXPSs,

the

main

com

ponents

of

logica

lpro

gra

mm

ing

are

1.

aknow

ledge

base

(facts

and

rule

s),

2.

an

infe

rence

engin

e.

Adva

nta

ge

oflo

gica

lpro

gra

mm

ing

isth

at

one

does

not

have

todeve

lop

an

alg

orith

mto

solve

the

pro

ble

msin

ceth

isjo

bis

done

by

the

infe

rence

engin

e.

Inste

ad,we

describ

eth

epro

ble

mby

means

oflo

gica

lfo

rmula

s.

The

open-so

urce

SW

I-Pro

log

isava

ilable

as

part

ofth

em

ajo

rLin

ux

distrib

utio

ns

as

well

as

Cyg

win

(http://cygwin.com)

orca

nbe

obta

ined

fromhttp://www.swi-prolog.org

D.Suenderm

ann

Logic

January

28,2012

130

.Facts

and

rule

s.

Facts

areato

mic

form

ula

sw

ithth

ePro

log

synta

x

p(t

1 ,...,tn)

(212)

featu

ring

the

pre

dica

tep

and

the

term

st1 ,

...,tn.

All

the

variable

sin

facts

areunive

rsally

bound,i.e

.,Eq.212

repre

sents

the

logica

lfo

rmula

∀x

1,...,x

m(p

(t1,...,tn)).

(213)

Rule

sare

conditio

nalpro

positio

ns

with

the

Pro

log

synta

x

A:−

B1,...,B

n.

(214)

featu

ring

the

ato

mic

form

ula

sA

,B

1,...,B

n.

Again

,all

the

variable

sin

rule

sare

unive

rsally

bound,so

,Eq.214

repre

sents

the

form

ula

∀x

1,...,x

m(B

1∧

...∧

Bn→

A).

(215)

This

genera

llyre

quire

sfo

rmula

sto

be

give

nas

Horn

clause

s.

D.Suenderm

ann

Logic

January

28,2012

131

.Som

eco

nve

ntio

ns

.

The

first

chara

cter

ofvaria

ble

sis

aca

pita

lle

tteror

an

undersco

re.

The

first

chara

cter

ofpre

dica

tes

or

functio

ns

isa

lower-ca

sele

tter.

The

pre

dica

tetrue

repre

sents

valid

ity.

The

symbols+,-,*,/,.

arefu

nctio

nsym

bols

you

can

use

inin

fix

nota

tion.

The

symbols<,>,=,=<,>=,\=,==,\==

arepre

dica

tesym

bols

you

can

use

inin

fix

nota

tion.N

ote

that

==

tests

forequality,

\==

tests

for

inequality,

and

=is

the

unifi

catio

nopera

tor.

The

symbol\+

(or,

alte

rnative

ly,not())

isth

enegatio

nopera

tor.

The

symbol%

isuse

dfo

rco

mm

ents.

The

symbols,

and;

isuse

dfo

rco

nju

nctio

nand

disju

nctio

n,re

spective

ly.

D.Suenderm

ann

Logic

January

28,2012

132

.O

nPro

log’s

disju

nctio

n.

The

follo

win

gderiva

tion

show

sth

at

disju

nctio

ns

inPro

log

rule

sare

effective

lyno

additio

nalfe

atu

re:

A:−

B1;...;B

n.⇔

B1∨

...∨

Bn→

A.

⇔¬

(B1∨

...∨

Bn)∨

A

⇔¬

B1∧

...∧¬

Bn∨

A

⇔(¬

B1∨

A)∧

...∧

(¬B

n∨

A)

⇔A

:−

B1.

···

A:−

Bn.

(216)

D.Suenderm

ann

Logic

January

28,2012

133

.A

nexa

mple

.

Let

us

now

consid

era

realistic

exa

mple

:

–All

students

aresm

art.

–W

hoeve

ris

smart

ispowerfu

l.

–W

hoeve

ris

com

pute

rscie

ntist

and

pro

fesso

ris

powerfu

l.

–Com

pute

rscie

ntists

arecra

zy.

–Ala

nis

astu

dent.

–Bra

dis

astu

dent.

–Colin

isa

com

pute

rscie

ntist.

–Colin

isa

pro

fesso

r.

D.Suenderm

ann

Logic

January

28,2012

134

.A

nexa

mple

(cont.)

.

This

isth

ere

spective

Pro

log

code

(seestudent.pl

inth

eauxiliary

packa

gekbs

*.zip):

1smart(X):-student(X).

2powerful(X):-smart(X).

3powerful(X):-cs(X),prof(X).

4crazy(X):-cs(X).

5student(alan).

6student(brad).

7cs(colin).

8prof(colin).

D.Suenderm

ann

Logic

January

28,2012

135

.A

nexa

mple

(cont.)

.

W

ewant

tofind

out

wheth

er

there

isa

powerfu

land

crazy

individ

ual.

The

resp

ective

logica

lfo

rmula

is

∃x(powerful(x

)∧crazy(x

)).(217)

Inord

erto

find

out,

we

first

launch

Pro

log

with

the

com

mand

pl

and

get

the

com

mand

pro

mpt

?-

To

load

ourknow

ledge

base

,we

type

consult(student).

Now

,we

can

use

the

Pro

log

synta

xofEq.217

toch

eck

the

valid

ityof

our

conje

cture

:

powerful(X),crazy(X).

D.Suenderm

ann

Logic

January

28,2012

136

.A

nexa

mple

(cont.)

.

W

eobta

inth

ere

sponse

X=

colin

tellin

gus

that

Colin

isa

powerfu

land

crazy

individ

ual.

Inord

erto

identify

oth

erpote

ntia

lca

ndid

ate

s,we

type

;

resu

lting

inth

ere

sponse

No

which

indica

tes

that

there

areno

more

solu

tions

toth

epro

ble

m.

D.Suenderm

ann

Logic

January

28,2012

137

.Pro

log’s

infe

rence

alg

orith

m.

W

eare

give

nth

ePro

log

pro

gra

mP

consistin

gofa

num

berofru

les

of

the

formR

:=A

:−

B1,...,B

m(218)

and

aquery

ofth

efo

rm

G=

Q1,...,Q

n.

(219)

Here

,fa

ctsare

exp

anded

toru

les

by

A↔

A:−true.

(220)

The

infe

rence

alg

orith

mwork

sas

follo

ws:

1.

Search

(inord

er

ofappeara

nce

)all

the

rule

sA

inP

,fo

rw

hich

there

exists

aunifi

er

µ=

[]if

Q1

=true

mgu(Q

1,A

)oth

erw

ise(221)

D.Suenderm

ann

Logic

January

28,2012

138

.Pro

log’s

infe

rence

alg

orith

m(co

nt.)

.

2.

Inca

seth

ere

arem

ultip

lesu

chru

les,

a)

sele

ctth

efirst

rule

(inord

er

ofappeara

nce

),

b)

set

ach

oice

poin

t(C

P)

toperfo

rma

diff

ere

nt

sele

ction

at

this

poin

tin

case

itbeco

mes

nece

ssaryat

ala

ter

mom

ent.

3.

Here

,tw

oca

ses

aredistin

guish

ed:

a)

m+

n=

1:This

means

succe

ss,and

Pro

log

retu

rns

the

last

non-e

mpty

µ.

b)

Oth

erw

ise,we

recu

rsively

contin

ue

with

the

query

G:=

B1µ

,...,B

mµ

,Q

2µ

,...,Q

nµ

.(222)

Ifwe

do

not

find

aso

lutio

n,we

retu

rnto

the

last

choice

poin

t

reve

rsing

the

repla

cem

ents

G:=

Gµ

acco

rdin

gly.

Negatio

nis

imple

mente

din

Pro

log

as

negatio

nas

failu

re.

I.e.,

ifQ

1in

1is

ofth

esyn

taxnot(Q

′1 )th

ealg

orith

mtrie

sto

pro

veQ

′1 .

Ifit

succe

eds,

we

know

that

Q1

isfa

lse,oth

erw

ise,we

assu

me

itis

true.

D.Suenderm

ann

Logic

January

28,2012

139

.Pro

log’s

infe

rence

alg

orith

m:exa

mple

.

IDCP

GR

µ

11

powerful(X

),crazy(X

)powerful(X

):−smart(X

)[]

21

smart(X

),crazy(X

)smart(X

):−student(X

)[]

33

student(X

),crazy(X

)student(alan):−true

[X7→

alan]

43

true,crazy(alan)

[]

53

crazy(alan)

crazy(X

):−cs(X

)[X7→

alan]

63

cs(alan)

cs(colin):−true

Ω

71

student(X

),crazy(X

)student(brad):−true

[X7→

brad]

83

true,crazy(brad)

[]

91

crazy(brad)

crazy(X

):−cs(X

)[X7→

brad]

10

1cs(brad)

cs(colin):−true

Ω

D.Suenderm

ann

Logic

January

28,2012

140

.Pro

log’s

infe

rence

alg

orith

m:exa

mple

(cont.)

.

IDCP

GR

µ

11

powerful(X

),crazy(X

)powerful(X

):−cs(X

),prof(X

)[]

12

cs(X

),prof(X

),crazy(X

)cs(colin):−true

[X7→

colin]

13

true,prof(colin),crazy(colin)

[]

14

prof(colin),crazy(colin)

prof(colin):−true

[]

15

true,crazy(colin)

[]

16

crazy(colin)

crazy(X

):−cs(X

)[X7→

colin]

17

cs(colin)

cs(colin):−true

[]

18

true

[]

Pro

log’s

resp

onse

ishence

:[X7→

colin].

D.Suenderm

ann

Logic

January

28,2012

141

.D

rawback

sofPro

log’s

infe

rence

alg

orith

m.

Consid

erth

ePro

log

pro

gra

m:

1a:-not(true).

Let

us

query

wheth

era:

IDCP

GR

µ

1a

a:−not(true)

[]

2∗

true

[]

The

query

infe

rsfa

lseby

negatio

nas

failu

re(in

dica

ted

by∗

which

means

that

are

sult

derive

dfro

mth

isste

pneeds

tobe

inve

rted).

This,

howeve

r,does

not

coin

cide

with

ourundersta

ndin

gofth

ese

mantics

ofth

eim

plica

tion:⊥→

ais

true

independent

ofw

heth

era

ornot.

The

reaso

nis

Pro

log’s

close

d-w

orld

assu

mptio

n:It

assu

med

the

data

base

isco

mple

te;I.e

.,if

the

answ

er

cannot

be

deduce

d,it

isfa

lse.

Eve

nworse

,th

ere

sponse

toth

equery

not(a

)is

true

due

totw

o

applica

tions

ofin

versio

n.

D.Suenderm

ann

Logic

January

28,2012

142

.D

rawback

sofPro

log’s

infe

rence

alg

orith

m(co

nt.)

.

Consid

erth

ePro

log

pro

gra

m:

1a:-b.

2b:-a.

Let

us

query

wheth

era,b:

IDCP

GR

µ

1a,b

a:−b

[]

2b,b

b:−a

[]

3a,b

a:−b

[]

4b,b

b:−a

[]

···

The

pro

gra

mente

rsan

infinite

loop

eve

nth

ough

the

query

could

be

pro

ven

true

ina

few

steps:

(b→

a)∧

(a→

b)→

a∧b⇔⊤

.(223)

The

natu

reofPro

log

bein

gbase

don

Horn

logic

and

itsnegatio

nand

loop

handlin

gsh

ow

aco

nsid

era

ble

weakness

ofits

infe

rence

alg

orith

m.

D.Suenderm

ann

Logic

January

28,2012

143

.Lists

.

Apart

from

Pro

log’s

infe

rence

engin

e,a

pre

dom

inant

featu

reis

itslist

handlin

g.

Lists

can

be

writte

nin

thre

eways:

1.

.(s,t)

defines

alist

with

the

ele

ment

sand

the

tail

t;

2.

[s|t]

does

the

sam

e;

3.

[s1,...,s

n]defines

alist

with

the

ele

ments

s1,...,s

n

Acco

rdin

gly,

these

areequiva

lent

lists:

.(1,.(2

,.(3

,[])))

(224)

[1|[2|[3|[]]]]

(225)

[1,2,3]

(226)

D.Suenderm

ann

Logic

January

28,2012

144

.Lists:

exa

mple

functio

n.

W

ewant

todesig

na

functio

ncat

that

conca

tenate

stw

olists

L1

andL2

resu

lting

inth

elist

L2.

In

the

world

oflo

gica

lpro

gra

mm

ing,th

isco

uld

be

conce

ived

as

the

3-ary

functio

ncat(L1,L2,L3)

which

beco

mes

true

iffL3

isth

eco

nca

tenatio

nof

L2

andL2.

Are

spective

Pro

log

pro

gra

mis:

1cat([X|L1],L2,[X|L3]):-cat(L1,L2,L3).

2cat([],L,L).

This

pro

gra

mre

ads

An

em

pty

listco

nca

tenate

dw

itha

listL

resu

ltsin

the

sam

elist

L

(Fact

2).

Furth

erm

ore

,if

the

conca

tenatio

nofth

elists

L1

and

L2

resu

ltsin

L3 ,

then

L1

with

an

pre

cedin

gele

ment

Xco

nca

tenate

d

with

L2

must

resu

ltin

L3

with

the

sam

epre

cedin

gele

ment

X

(Rule

1).

Inth

efo

llow

ing,we

run

an

exa

mple

toundersta

nd

the

pro

gra

m’s

functio

nality.

D.Suenderm

ann

Logic

January

28,2012

145

.Lists:

exa

mple

functio

n(co

nt.)

.

IDCP

GR

µ

1cat([1

,2],

[3,4],

Y)cat([X|L

1],

L2,[X|L

3])

:−

[X7→

[1],

L17→

[2],

cat(L

1,L

2,L

3)

L27→

[3,4],

Y7→

[1|L

3]]

2cat([2

],[3

,4],

L3)

cat([X

′|L′1],

L′2,[X

′|L′3])

:−

[X′7→

[2],

L′17→

[],

cat(L

′1,L

′2,L

′3)

L′27→

[3,4],

L37→

[2|L

′3]]

3cat([],

[3,4],

L′3)

cat([X

′′|L

′′

1],

L′′

2,[X

′′|L

′′

3])

:−

Ω

cat(L

′′

1,L

′′

2,L

′′

3)

4cat([],

[3,4],

L′3)

cat([],

L,L)

:−

[L7→

[3,4],

L′37→

[3,4]]

Pro

log’s

resp

onse

ishence

:

Y7→

[1|L

3]

7→[1|[2|L

′3]]

7→[1|[2|[3

,4]]]

=[1

,2,3,4]

(227)

D.Suenderm

ann

Logic

January

28,2012

146

.Lists:

exa

mple

functio

n(co

nt.)

.

You

may

perce

iveso

me

flavo

rofPro

log’s

ele

gance

ifyo

uco

nsid

erw

hich

use

case

sth

eabove

exa

mple

functio

nfe

atu

res:

–Conca

tenate

two

lists:

cat([1

,2],

[3,4],

Y).

(228)

–Check

wheth

era

listre

sulte

dfro

manoth

erlist

by

way

of

conca

tenatio

n:

cat([1

,2],

Y,[1

,2,3,4]).

(229)

–Fin

dall

possib

lesp

litsofa

listin

totw

olists:

cat(X

,Y

,[1

,2,3,4]).

(230)

D.Suenderm

ann

Logic

January

28,2012

147

.H

ow

ord

er

matte

rs.

Consid

erth

efo

llow

ing

pro

gra

m:

1a.

2a:-b.

3b:-a.

W

eget

the

query

resu

lt

a.→

Yes.

(231)

Now

,we

reord

erth

eru

les:

1a:-b.

2a.

3b:-a.

This,

time,th

equery

resu

ltis

a.→

ERROR

:Outoflocalstack.

(232)

The

infe

rence

alg

orith

mke

eps

acce

ssing

Rule

1ove

rand

ove

ragain

.

Oth

erth

an

the

exa

mple

on

Page

145,th

istim

e,we

do

not

get

an

infinite

loop

but

asta

ckove

rflow

.

This

isbeca

use

Pro

log

has

tocre

ate

ach

oice

poin

tfo

reve

ryre

cursio

n

due

toth

epre

sence

ofth

ealte

rnative

Rule

3.

D.Suenderm

ann

Logic

January

28,2012

148

.H

ow

ord

er

matte

rs(co

nt.)

.

Consid

erth

efo

llow

ing

pro

gra

m:

1s([X],[X]).

2s([A,B],[A,D]):-s([B],[D]),A<D.

W

eget

the

query

resu

lt

s([1

,2],X).→

X=

[1,2].

(233)

Now

,we

switch

the

ele

ments

inRule

2’s

body:

1s([X],[X]).

2s([A,B],[A,D]):-A<D,s([B],[D]).

This

time,we

get

s([1

,2],X).→

ERROR

:Argumentsarenotsufficientlyinstantiated.

The

infe

rence

alg

orith

mtrie

sto

eva

luate

A<

Dfirst,

befo

reD

had

been

dete

rmin

ed

by

way

ofeva

luatin

gs([B

],[D

]).

D.Suenderm

ann

Logic

January

28,2012

149

.Sym

bolic

vs.num

erica

lco

mputa

tion

.

Consid

erth

efo

llow

ing

pro

gra

m:

1p(X,Y):-Y==X+1.

2q(X,Y):-Y>X+0.9999999,Y<X+1.0000001.

3r(X,Y):-Y

is

X+1.

4s(X,Y):-Y=X+1.

We

get

the

follo

win

gquery

resu

lts:

p(1

,2).→

No.

q(1

,2).→

Yes.

r(1

,2).→

Yes.

s(1

,2).→

No.

p(1

,Y).→

No.

q(1

,Y).→

ERROR

:Argumentsarenotsufficientlyinstantiated.

r(1

,Y).→

Y=

2.

s(1

,Y).→

Y=

1+

1.

D.Suenderm

ann

Logic

January

28,2012

150

.Sym

bolic

vs.num

erica

lco

mputa

tion

(cont.)

.

The

check

forequality

(==

)fa

ilsdue

toissu

es

with

Pro

log’s

num

erica

l

pre

cision.

Rath

er

than

forequality,

qch

eck

sfo

ra

small

range

around

the

exp

ecte

d

valu

eand

there

by

succe

eds.

When

querie

dw

ithth

efre

epara

mete

rY,

howeve

r,Pro

log

isnot

able

tolim

itth

e(re

al-va

lued)

search

space

and

com

pla

ins

about

insu

fficie

nt

insta

ntia

tion.

Pro

log’s

keyw

ord

is

assig

ns

the

exa

ctva

lue

ofX

+1

toY

and

there

fore

succe

eds.

Acco

rdin

gly,

the

free

para

mete

rY

gets

assig

ned

the

sum

of1

and

1.

The

unifi

catio

nopera

tor=

tries

toso

lveth

esyn

tactica

lequatio

n

2=

1+

1w

hich

isnot

possib

lesin

cediff

ere

nt

functio

nsym

bols

cannot

be

unifi

ed.H

ence

,it

fails.

When

querie

dw

itha

free

para

mete

r,howeve

r,

the

synta

cticalequatio

nisY

=1

+1

whose

solu

tion

isth

ere

sult

set.

D.Suenderm

ann

Logic

January

28,2012

151

.Pro

log:exe

rcises

.

W

ritepro

gra

ms

to

1)

dete

rmin

eth

em

axim

um

oftw

onum

bers

(2lin

es)

2)

calcu

late

the

facto

rial(2

lines)

3)

uniq

alist

(3lin

es)

4)

find

identica

lele

ments

intw

olists

(3lin

es)

5)

sort

alist

(4lin

es)

D.Suenderm

ann

Logic

January

28,2012

152

.A

ppendix

.

Solu

tions

tose

lecte

dexe

rcises

D.Suenderm

ann

Logic

January

28,2012

153

.Conju

nctive

norm

alfo

rm:so

lutio

nto

exe

rcise.

Tra

nsfo

rmin

toCN

F:

g:=

p→

q→

r︸

︷︷

︸

a

↔¬

s∨

t︸

︷︷

︸

b

(234)

1.

g⇔

(¬a∨

b)

︸︷︷

︸

c

∧(¬

b∨

a)

︸︷︷

︸

d

(235)

2.

.a⇔¬

p∨

q→

r

⇔¬

(¬p∨

q)∨

r(236)

3.

.a⇔

p∧¬

q∨

r(237)

¬a⇔¬

(p∧¬

q)∧¬

r

⇔(¬

p∨

q)∧¬

r(238)

¬b⇔

s∧¬

t(239)

D.Suenderm

ann

Logic

January

28,2012

154

.Conju

nctive

norm

alfo

rm:so

lutio

nto

exe

rcise(co

nt.)

.

4.

.c⇔

(¬p∨

q)∧¬

r∨

b

⇔(¬

p∨

q∨

b)∧

(¬r∨

b)

(240)

⇔(¬

p∨

q∨¬

s∨

t)∧

(¬r∨¬

s∨

t)

d⇔

s∧¬

t∨

a

⇔(s∨

a)

︸︷︷

︸

e

∧(¬

t∨

a)

︸︷︷

︸

f

(241)

e⇔

p∧¬

q∨

r∨

s

⇔(p∨

r∨

s)∧

(¬q∨

r∨

s)

(242)

f⇔

(p∨

r∨¬

t)∧

(¬q∨

r∨¬

t)(243)

5.

.g⇔¬

p,q,¬

s,t

,¬

r,¬

s,t

,p,r,

s,¬

q,r,

s,

p,r,¬

t,¬

q,r,¬

t

(244)

D.Suenderm

ann

Logic

January

28,2012

155

.First-o

rder

calcu

lus:

solu

tion

toexe

rcise.

W

eknow

that

Everyb

ody

loves

only

those

peo

ple

who

do

not

love

agard

ener.

Let

us

use

the

pre

dica

tes

–l(x

,y)

which

is1

iffx

love

sy

and

–g(x

)w

hich

is1

iffx

isa

gard

ener.

Let

us

furth

eruse

an

auxiliary

pre

dica

te

–n(x

)w

hich

is1

iffx

does

not

love

agard

ener.

Usin

gn(x

)’sdefinitio

n,th

eabove

axio

mco

uld

be

writte

nas

Every

xlo

vesonly

those

yfo

rw

hich

n(y

)

which

can

be

form

ally

exp

resse

das

f:=∀x,y(l(x

,y)→

n(y

)).(245)

D.Suenderm

ann

Logic

January

28,2012

156

.First-o

rder

calcu

lus:

solu

tion

toexe

rcise(co

nt.)

.

Exp

ressin

gn(y

)in

term

sof

l(y,z)

and

g(z

):

n(y

)↔¬∃z(l(y

,z)∧

g(z

)).(246)

H

ence

,Form

ula

248

beco

mes

f↔∀x,y(l(x

,y)→¬∃z(l(y

,z)∧

g(z

)))

↔∀x,y(¬

l(x,y)∨¬∃z(l(y

,z)∧

g(z

)))

↔∀x,y(¬

l(x,y)∨∀z(¬

(l(y,z)∧

g(z

))))

↔∀x,y,z(¬

l(x,y)∨¬

l(y,z)∨¬

g(z

))

↔¬

l(x,y),¬

l(y,z),¬

g(z

)

(247)

Our

conje

cture

isth

at

Gard

eners

do

not

love

them

selves

which

can

be

form

ally

exp

resse

das

g:=∀x(g

(x)→¬

l(x,x)).

(248)

D.Suenderm

ann

Logic

January

28,2012

157

.First-o

rder

calcu

lus:

solu

tion

toexe

rcise(co

nt.)

.

To

pro

veg,we

need

toco

nju

nctive

lyco

mbin

ef

with

g’s

negatio

n,so

,

let

us

atta

ckth

ela

tternow

:

¬g↔¬

(∀x(g

(x)→¬

l(x,x)))

↔¬

(∀x(¬

g(x

)∨¬

l(x,x)))

↔∃x(¬

(¬g(x

)∨¬

l(x,x)))

↔∃x(g

(x)∧

l(x,x))

≈g(s

)∧

l(s,s)

↔g(s

),l(s

,s)

(249)

D.Suenderm

ann

Logic

January

28,2012

158

.First-o

rder

calcu

lus:

solu

tion

toexe

rcise(co

nt.)

.

Applyin

gth

eD

Palg

orith

m:

C0

:=¬

l(x,y),¬

l(x,z),¬

g(z

),g(s

),l(s

,s)

C1

:=re

s(C0 ,

g(s

))

=¬

l(x,y),¬

l(x,z),¬

g(z

),¬

l(x,y),¬

l(x,s)

,

g(s

),l(s

,s)

C2

:=re

s(C1 ,

l(s,s))

=¬

l(x,y),¬

l(x,z),¬

g(z

),¬

l(s,z),¬

g(z

),

¬

l(x,y),¬

l(x,s)

,¬

l(s,s)

,g(s

),l(s

,s)

(250)

This

pro

ves

that

the

conje

cture

istru

e.

D.Suenderm

ann

Logic

January

28,2012

159