logic menhde

Nguyn Quang Chu Khoa CNTT HCN Tp.HCM

Logic mnh

Nguyn Quang Chu Khoa CNTT HBK Tp.HCM.

Mnh l g?

Mi cu pht biu l ng hay l sai c gi lmt mnh .

(Definition proposition: Any statement that is either true or false is called a proposition)

K hiu: P, Q, v R.


Mnh phc hp. nh ngha :

Mnh ch c mt gi tr n (lun ng hoc sai) c gi l mnh nguyn t ( atomic proposition ). Cc mnh khng phi l mnh nguyn t c gi l mng phc hp (compound propositions). Thng thng, tt cmnh phc hp l mnh lin kt (c cha php tnh mnh ).


Cc php ton mnh Bao gm : Php ph nh () Php hi() Php tuyn ()

Php XOR () Php ko theo() Php tng ng()


Php ph nh (NEGATION)Cho P l mt mnh , cu "khng phi l P" l mt mnh khc c gi l ph nh ca mnh P. K hiu : P ( P ).

V d : P = " 2 > 0 " P = " 2 0 "

Bng chn tr (truth table)p p

T FF T

Qui tc: Nu P c gi tr l T th ph nh P c gi tr lF.


Php hi (CONJUNCTION)

Cho hai mnh P, Q. Cu xc nh "P v Q" l mt mnh mi c gi l hi ca 2 mnh P v Q.

- K hiu P Q.

Qui tc : Hi ca 2 mnh ch ng khi c hai mnh l ng. Cc trng hp cn li l sai.

P Q P Q

S S

S S

S S S

Bng chn tr


Php tuyn (DISJUNCTION)

Cho hai mnh P, Q. Cu xc nh "P hay (hoc) Q" lmt mnh mi c gi ltuyn ca 2 mnh P v Q. -- K hiu P Q.

Qui tc : Tuyn ca 2 mnh ch sai khi c hai mnh l sai. Cc trng hp cn li l ng.

P Q P Q

S

S

S S S

Bng chn tr


Php XOR

Cho hai mnh P v Q. Cu xc nh "loi tr P hoc lai tr Q", ngha l "hoc l P ng hoc Q ng nhng khng ng thi c hai l ng" l mt mnh mi c gi l P xor Q.

K hiu P Q.Qui tc : Tuyn ca 2 mnh ch

sai khi c hai mnh l sai. Cc trng hp cn li l ng.

P Q P Q S

S

S

S S S

Bng chn tr


Php ko theo (IMPLICATION)

Cho P v Q l hai mnh . Cu "Nu P th Q" l mt mnh mi c gi l mnh ko theo ca hai mnh P,Q.

K hiu P Q. P c gi l githit v Q c gi l kt lun.

Qui tc : mnh ko theo ch sai khi gi thit ng v kt lun sai. Cc trng hp khc l ng.

P Q P Q

S S

S

S S

Bng chn tr


Php tng ng (BICONDITIONAL)

Cho P v Q l hai mnh . Cu "P nu v chnu Q" l mt mnh mi c gi l P tng ng Q. K hiu P Q. Mnh tng ng l ng khi P v Q c cng chn tr.

P Q = (P Q) (Q P) c l : P nu v ch nu Q P l cn v i

vi Q Nu P th Q v ngc li.


Biu thc mnh (LOGICAL CONNECTIVES)

Cho P, Q, R,... l cc mnh . Nu cc mnh ny lin kt vi nhau bng cc php ton th ta c mt biu thc mnh .

Ch :. Mt mnh cng l mt biu thc mnh . Nu P l mt biu thc mnh th P cng l biu thc mnh

Chn tr ca biu thc mnh l kt qu nhn c t s kt hp gia cc php ton v chn tr ca cc bin mnh .



V d : Tm chn tr ca biu thc mnh P (Q R )



Do biu thc mnh l s lin kt ca nhiu mnh bng cc php ton nn chng ta c th phn tch biu din cc biu thc mnh ny bng mt cy mnh .V d : Xt cu pht biu sau :" Nu Michelle thng trong k thi Olympic, mi ngi s khm phc c y, v c ta s tr nn giu c. Nhng, nu c ta khng thng th c ta s mt tt c."y l mt biu thc mnh v php ton chnh l php hi. C th vit li nh sau :"Nu Michelle thng trong k thi Olympic, mi ngi s khm phc c y, v c ta s tr nn giu c.Nhng, nu c ta khng thng th c ta s mt tt c. "



C hai mnh chnh trong biu thc mnh ny lmnh phc hp. C th nh ngha cc bin mnh nh sau:

P: Michelle thng trong k thi OlympicQ: mi ngi s khm phc c yR: c ta s tr nn giu cS: c ta s mt tt c

Biu din cu pht biu trn bng cc mnh v cc php ton, ta c biu thc mnh sau :

( P (Q R)) (P S)


Biu din cu pht biu trn thnh mt cy ng ngha nh sau


Cc thut ng chuyn ngnh (SOME TERMINOLOGY)

nh ngha Hng ng (Tautologie):Mt hng ng l mt mnh lun c chn tr l ng.Mt hng ng cng l mt biu thc mnh lun c chn tr l ng

bt chp s la chn chn tr ca bin mnh . V d : xt chn tr ca biu thc mnh P P

Vy PP l mt hng ng.

P P P P

F T T

T F T


Cc thut ng chuyn ngnh (SOME TERMINOLOGY)

nh ngha Hng sai (Contradiction):Mt hng sai l mt mnh lun c chn tr l sai.Mt hng sai cng l mt biu thc mnh lun c chn tr

l sai bt chp s la chn chn tr ca bin mnh . V d : xt chn tr ca biu thc mnh P P

P P PP

F T F

T F F


Quines Method

P Q PP=true P=false

True Q true false Q falseTrue Q true

QQ=true Q=false

True false


Hm s tht (Truth function)

L 1 hm m cc i s ca n ch c thnhn gi tr hoc true hoc false

Bt k 1 wff no cng u l 1 hm truth V d: g(P,Q) = P Q Mi hm truth c phi l 1 wff?


Hm s tht (Truth function)

V d: cho 1 hm truth f(P,Q), hm c gi tr true khi P v Q c gi tr ngc nhau. Hy tm xem c 1 wff no c cng bng chn tr vi hm f?

P Q f(P,Q) wffT T FT F T To P QF T T To P QF F F


ng vi mi hng trong bng m hm f cgi tr true, ta to ra 1wff.

Mi wff l php hi ca cc k t i shoc php ph nh ca cc k t ny theo 2 quy tc sau:Nu P l true th t P vo php hi

(conjunction)Nu P l false th t P vo php hi


P Q f(P,Q) P Q P QT T F F

T

FF

FT F T FF T T TF F F F

Bng chn tr ca P P Q vQ v P P Q cQ c ng 1 hng 1 hng cng cgigi trtr true, vtrue, v ccc gic gi trtr true ntrue ny xy xy ra cy ra cng hng hng ng vvi ci cc gic gi trtr true ctrue ca ha hm fm f

f(P,Q) f(P,Q) ((P P Q) v (Q) v (P P Q)Q)


Mnh h qu nh ngha : Cho F v G l 2 biu thc mnh . Ngi ta ni rng

G l mnh h qu ca F hay G c suy ra t F nu F G lhng ng.

K hiu F | G V d : Cho F = ( P Q ) ( Q R )

G = P RXt xem G c l mnh h qu ca F khng ?


Mnh h qu

Vy G l mnh h qu ca F Nhn xt : Nu G l h qu ca F th khi F l ng th bt

bt buc G phi ng.Ngc li, nu G l ng th cha c kt lun g v chn tr

ca F.


Tng ng Logic (LOGICALLY EQUIVALENT)

Vy F v G l tng ng logic hay F=G. V d 2: Cho F = P Q

G = PQ Xt xem hai mnh trn l c tng ng logic

khng ?

P Q P Q P P QT T T F

FTT

TT F F FF T T TF F T T

Vy F G hay P Q = (PQ)


Bng cc tng ng logic thng dngEquivalence NamepT TpF F

Domination laws : lut nut

pT ppF p

Identity laws : lut ng nht

pp ppp p

Idempotent laws : lut ly ng

(p) p Double negation law : lut ph nh kp

pp Tpp F

Cancellation laws : lut xa b


Bng cc tng ng logic thng dngEquivalence Name

pq qppq qp

Commutative laws : lut giao hon

(pq)r p(qr)(pq)r p(qr)

Associative laws : lut kt hp

p(qr) (pq)(pr)p(qr) (pq)(pr)

Distributive laws : lut phn b

(pq) pq(pq) pq

De Morgans laws : lut De Morgan

(pq) (pq) Implication : lut ko theo


V d

V d 1 : Khng lp bng chn tr, s dng cc tng ng logic chng minh rng

(P Q) Q l hng ng. ((p q) q) ( p q) q Implication law

(p q)q De Morgans law p (qq) Associative law p T Cancellation Law T Domination Law


V d

V d 2 : Chng minh rng (q p ) (p q ) = q

((q p))(pq) ((q p))(pq) Implication law(qp)(pq) Commutative law(qp)(q p) Distributive law q (p p) Cancellation law q T Identity law q


nh l: Mi hm truth u tng ng vi 1 wff mnh

Mt vi nh ngha: Literal: l 1 k t mnh hay ph nh ca n, nh

P,Q, Q,.. Hi s cp (fundamental conjunction) l 1 literal hoc

hi ca 2 hay nhiu literal V d: P Q

Tuyn s cp (fundamental disjunction) l 1 literal hoc tuyn ca 2 hay nhu literal V d: P v Q


Dng chun tc tuyn(Disjunctive normal form -DNF)

DNF hoc l 1 hi s cp hay tuyn ca 2 hay nhiu hi s cp

V d: P (P Q), (P Q R) (P Q R)

Nhng wff dng xy dng hm truth u dng DNF



Dang chuan tuyn co dang :(P1 Pn)1 (Q1 Qm)p, vi n, m, p 1.

Rt gn DNF: tm 1 DNF n gin hn tng ng vi DNF ang xt.

V d: (PQ R) (P Q R) (PR) (P Q R) (QP R) (PR) (P Q R) (QP R) (PR) (P Q R) (PR)



Mi wff u tng ng vi 1 DNF Cch xy dng 1 DNF tng ng vi 1 wff: Xo tt c cc php ko theo bng php tung ng

AB A B Chuyn php ph nh vo trong ngoc bng nh lut De

Morgan(A B) A B(A B) A B

p dng php phn phi tng ng c DNF



V d: Xy dng DNF cho wff sau:((P Q) R) S

((P Q) R) S ((P Q) R) S (P Q R) S (P S)(Q S)(R S)



Gi s 1 wff W c n k t mnh phn bit, DNF cho W c gi l DNF y (Full DNF) nu mi hi s cp u c n literal, mi literal tng ng vi 1 k t ca W.

V d: (PQ R) (P Q R) DNF y P (P Q R) khng phi l DNF y

nh l: mi wff m khng phi l mnh mu thun ( contradiction) u tng ng vi 1 DNF y


Dng chun tc hi(Conjunctive Normal Form CNF) CNF l 1 tuyn s cp hay hi ca 2 hay nhiu tuyn s

cp V d: P (P Q),

(P Q R) (P Q R) Ga s wff W c n k t mnh phn bit. CNF cho W c gi l y nu mi tuyn s cp c n literal, mi literal tng ng vi 1 k t ca W

V d: (P Q R) (P Q R) CNF y P (P Q R) khng phi l CNF y


Dng chun tc hi(Conjunctive Normal Form CNF)

Dang chuan CNF co dang :(P1 Pn)1 (Q1 Qm)p, vi n, m, p 1. nh l: Mi wff u tng ng vi 1 CNF Mi wff m khng phi l 1 mnh hng ng u tng ng vi 1 CNF y


V d xy dng DNF y

Cho 1 wff P Q. Tm DNF y .


V d xy dng CNF y

Cho 1 wff P Q. Tm CNF y .


H thng suy din hnh thc(Formal Reasoning System)

tm gi tr cho 1 mnh , c th dng bng chn tr

Nu mnh c nhiu hn 3 bin, vic xy dng bng chn tr kh phc tp

Nu dng php chng minh tng ng thay cho bng chn tr ?

Cn h thng suy din hnh thc


H thng suy din hnh thc(Formal Reasoning System)

Tin (Axiom): l 1 wff c dng lm c s suy din

H thng suy din hnh thc cn: Mt tp hp cc wff (well-formed formulas- wff) biu din cc

pht biu, nh ngha c lin quan. Mt tp hp cc tin Mt s lut (rule) gip kt lun cc s vic.

Lut suy din ( inference rule) s nh x 1 hay nhiu wff, c gi gi thit (premise, hypothese hay antecedent) thnh 1 wff duy nht gi l kt ln ( conclusion, consequent)


M hnh suy din (Modus ponens MP)

MP l tn gi hnh thc cho dng suy din khng nh rt thng dng v thng c dng nh sau: If P, then Q.P. Therefore, Q.

V d: Xt 1 MP nh sau: If today is Tuesday, then I will go to work. Today is Tuesday. Therefore, I will go to work.


M hnh suy din(Modus ponens MP)

Nu MP nh x 2 wff A v A B thnh 1 wff B, k hiu:

MP(A, AB) = B Nu R l 1 lut suy din, v C c suy din t

P1,..,Pk bi R, c k hiu: R(P1,Pk)=C Hoc P1,,Pk

C C ngha l therefore, thus, hence, so,..


Lut suy din R(P1,..,Pk)= C c lun duy tr gi tr nu wff sau l hng ng (tautology):

P1 Pk C


Modus tollens - MT MT l tn gi hnh thc cho phung php chng minh

gin tip hay chng minh bng phn o(contrapositive proof), v thng c dng nh sau:

If P, then Q. Q is false. Therefore, P is false.

V d v 1 MT:

Con ngi th phi cht. Lizzy khng cht. V vy, Lizzy khng phi l con ngi.


Modus tollens - MT

K hiu ca MTAB, B

A

Edited by Foxit ReaderCopyright(C) by Foxit Software Company,2005-2007For Evaluation Only.


Mt s lut suy din thng dng

Conjunction (Conj): A,B A B

Simplification(Simp) : A BA

Addition (Add): AA v B

Disjunctive Syllogism (DS) : A v B, AB

Hypothetical Syllogism (HS): AB, BCAC


Mt s lut suy din thng dng

Add

Simp

MP

MT

HS

DS

WelcomeTypewriter,~P

WelcomeTypewriter-----------------

WelcomeTypewriter(PvQ),~P ->Q


WelcomeTypewriter(a,b)->(a^b) Conj


Mi lut trn u c th kim chng bng cch ch ra:

C D l tautology (hng ng) vi C l gi thit v D l kt lun


Chng minh l g?(What is a proof?)

Chng minh l 1 chui hu hn cc wff m mi wff hoc l 1 tin hoc c suy din t 1 wff trc . Wff cui cng trong chng minh c gi l nh l (theorem)

V d: gi s chui cc wff sau l 1 proof:W1,..,Wn

Sao cho cui cng Wn = W


C 2 k thut chng minh:Chng minh c iu kin ( conditional proof)Chng minh gin tip (indirect proof)


Chng minh theo iu kin Hu ht cc pht biu cn chng minh u c dng iu kin nh sau:

A B C D Quy lut chng minh c iu kin

(conditional proof rule CP): gi s cn chng minh A1 A2 .. An BBt u chng minh bng cch vit mi gi

thit A1, A2, ,An trn cc dng ring vi k t P trong ct suy din

Dng cc gi thit nh cc tin , v cc lut chng minh dn n kt lun B


Chng minh theo iu kin Cu trc ca php chng minh c iu kin:

Proof 1. A P2. B P3. C P. . .. . .k. D .

QED 1,2,3,k,CP

QED vit-tt ca ting La tinh quod erat demonstrandum) iu c chng minh


V d chng minh theo iu kin

Chng minh pht biu sau:(A vB) (AvC) A B C

Proof 1. AvB P2. AvC P3. A P4. B 1,3,DS5. C 2,3,DS6. B C 4,5,ConjQED 1,2,3,6,CP


Subproof

Chng minh theo iu kin thng l 1 phn ca 1 php chng minh khc, v c gi l subproof.

ch ra l subproof, cc dng ca nnn tht vo


V d v subproof

Chng minh pht biu sau:((A vB) (B C)) (BC) v D

Nhn xt??Kt lun ca wff trn li cha iu kin th

2 BC


V d v subproofProof: 1.(A vB) (B C) P

2. B P Bt u subproof ca BC3. A vB 2,Add4. B C 1,3,MP5. C 4,Simp6. BC 2,5,Simp k/thc subproof7. (B C) v D 6, Add

QED 1,7,CP

WelcomeTypewriterP for CP

WelcomeTypewriter2,5,CP



Subproof

Quy tc quan trng dnh cho Subproof: Nhng dng tht vo dnh cho chng minh

Subproof khng c dng suy din cho 1 s dng nm sau subproof.

Ngoi l l khi nhng dng tht vo ny khng ph thuc, hoc trc tip hoc gin tip, vo cc gi thit ca subproof.


n gin ha trong chng minh theo iu kin

Nu W l 1 nh l (theorem), ta c thdng n chng minh cc nh l khct W vo 1 dng ca php chng minh v

x l n nh 1 tin (axiom) HocKhng W tham gia vo chui chng minh

nhng vn s dng n trong ct suy din (reason) cho 1 s dng ca php chng minh (xem v d)


V d

Chng minh pht biu sau:(A B) (B C) (C D) (A D)


V dProof:1. (A B) P2. (B C) P3. (C D) P4. A B 1, T5. A P6. B 4,5,DS7. C 2,6,DS8. D 3,7,MP9. AD 5,8,CP

QED 1,2,3,9,CPDng 4 l OK v ta a nh l vo ct reason


n gin ha trong chng minh theo iu kin

Thay v phi vit ra 1 hng ng (tautology) hay nh l (theorem) trong ct reason, ta c th vit n gin biu tng T ngm ch hng ng

hoc nh l c dng Mt s chng minh d dng, nhng khng t chng minh rt kh, v c th sai ngay lc khi

u trc khi chng minh c n


V d Chng minh wff sau:

A ((AB) (CD)) (B->C)Proof:1. A P2. (AB) (CD) P3. B P4. A B 1,3,Conj5. A B 4,T6. (A B) 5,T7. (AB) 6,T8. CD 2,7,DS9. C 8, Simp10. B->C 3,9,CP

QED 1,2,10,CP


V d Xt nhm cu sau: The team wins or I am sad.

If the team wins, then I go to a movie. If I am sad, then my dog barks. My dog is quiet. Therefore I go to a movie

t k hiu cho cc mnh sau:W: The team wins S: I am sadM: I go to a movie B: My dog barks

To biu tng cho nhm cu trn(W S) (W M) (SB) B M

Hy chng minh wff trn l 1 nh l


V dProof1. W S P2. W M P3. SB P4. B P5. S 3,4,MT6. W 1,5,DS7. M 2,6,MP

QED 1,2,3,4,7,CP


Chng minh gin tip(Indirect proof)

Gi s cn chng minh A B, nhng khng tm ra cch no chng minh nu dng phng php chng minh theo iu kin

Cch 1 (contrapositive method): Hy tm cch chng minh s tng phn ca A B

A B B ANgha l bt u vi B nh 1 gi thit (premise) ,

tm cch chng minh A


Chng minh gin tip(Indirect proof)

Cch 2 (proof by contradiction): chng minh bng s phn o

A B A B falseNgha l c thm 2 gi thit (premise) A vB

Nn p dng cch ny khi:Khng c thng tin t cc gi thit c

cho Khng bit cch no chng minh


Quy tc chng minh gin tip(Indirect proof rule IP)

Gi s mun chng minh gin tip cho biu thc iu kin

A1 A2 An B Vit mi gi thit (premise) A1, An trn mi dng

ring bit vi P trong ct suy lun (reason ) t wff B vo dng k tip v vit P for IP vo ct

reason ch B l gi thit ca chng minh gin tip S dng cc gi thit ny dn n kt qu

false


Cu trc ca chng minh IP

Gi s cn chng minh A B C DProof: 1. A P2. B P3. C P4. D P for IP5. k. False .

QED 1,2,3,4,k,IP


V d Hy chng minh bi ton xem phim bng IP

(W S) (W M) (SB) B M1. W S P2. W M P3. SB P4. B P5. M P for IP6. W 2,5,MT7. S 3,4,MT8. W S 6,7,Conj9. (W S) 8,T10. (W S) (W S) 1,9,Conj11. False 10,T

QED 1-4,5,11,IP


Mt s ch khi chng minh

Khng s dng cc gi thit d tha khng cn thit

V d: chng minh AC. Bt u vi gi thit A, gi s thm gi thit ph B, ri s dng A v B suy din C bng chng minh trc tip hay gin tip

Kt qu l chng minh A B C thay v chng minh A C


Mt s ch khi chng minh Khng p dng quy lut suy din vo biu thc

conQuy lut suy din ch p dng cho 1 wff, khng p

dng cho cc biu thc con ca wff V d: gi s c 1 wff nh sau:

A ((A B) C) BR rng wff khng phi l tautology.

(khi A= true, B=false, C=true, wff false)Hy th chng minh wff trn l 1 nh l (theorem)


Mt s ch khi chng minh1. A P2. (A B) C P3. ~A (B C ) 2,T 4. (B C ) 1,3,DS5. B 1,2, MP (sai khi dng

MP cho biu thc con)QED 1,2,3,CP

Logic mnh Mnh l g?Mnh phc hp.Cc php ton mnh Php ph nh (NEGATION)Php hi (CONJUNCTION) Php tuyn (DISJUNCTION)Php XORPhp ko theo (IMPLICATION)Php tng ng (BICONDITIONAL)Biu thc mnh (LOGICAL CONNECTIVES)Biu thc mnh (LOGICAL CONNECTIVES)Biu thc mnh (LOGICAL CONNECTIVES)Biu thc mnh (LOGICAL CONNECTIVES)Biu din cu pht biu trn thnh mt cy ng ngha nh sauCc thut ng chuyn ngnh (SOME TERMINOLOGY)Cc thut ng chuyn ngnh (SOME TERMINOLOGY)Quines MethodHm s tht (Truth function)Hm s tht (Truth function)Mnh h quMnh h quTng ng Logic (LOGICALLY EQUIVALENT)Bng cc tng ng logic thng dngBng cc tng ng logic thng dngV dV dDng chun tc tuyn(Disjunctive normal form -DNF)Dng chun tc tuyn(Disjunctive normal form -DNF)Dng chun tc tuyn(Disjunctive normal form -DNF)Dng chun tc tuyn(Disjunctive normal form -DNF)Dng chun tc tuyn(Disjunctive normal form -DNF)Dng chun tc hi(Conjunctive Normal Form CNF)Dng chun tc hi(Conjunctive Normal Form CNF)V d xy dng DNF y V d xy dng CNF y H thng suy din hnh thc(Formal Reasoning System)H thng suy din hnh thc(Formal Reasoning System)M hnh suy din (Modus ponens MP)M hnh suy din (Modus ponens MP)Modus tollens - MTModus tollens - MTMt s lut suy din thng dngMt s lut suy din thng dngChng minh l g?(What is a proof?)Chng minh theo iu kinChng minh theo iu kinV d chng minh theo iu kinSubproofV d v subproofV d v subproofSubproofn gin ha trong chng minh theo iu kinV dV dn gin ha trong chng minh theo iu kinV dV dV dChng minh gin tip(Indirect proof)Chng minh gin tip(Indirect proof)Quy tc chng minh gin tip(Indirect proof rule IP)Cu trc ca chng minh IPV dMt s ch khi chng minhMt s ch khi chng minhMt s ch khi chng minh

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