1. Chng 1: i s mnh CHNG 1 : I S MNH 1.1. Tng quan Mc tiu ca chng 1 Hc xong chng ny, sinh vin phi nm bt c cc vn sau: - Th no l mnh , chn tr ca mnh , cc php ton mnh . - Thc hin c cc php ton mnh . - Hiu c cc ng dng ca php ton logic trong lp trnh v trong i sng hng ngy. Kin thc c bn cn thit Cc kin thc c bn trong chng ny bao gm: - Kin thc v php ton i s, php ton hnh hc c bn. - C kh nng suy lun. - Bit lp trnh bng ngn ng Pascal, C Ti liu tham kho Phm vn Thiu, ng Hu Thnh. Ton ri rc ng dng trong tin hc. Nh xut bn Khoa hc v K thut, H Ni - 1997 (chng 1, trang 6 - 28). Ni dung ct li - nh ngha mnh , biu thc mnh . - Cc php ton - V d ng dng - Gii thiu mt s thut ng chuyn dng - Tng ng logic v cch chng minh. 1.2. nh ngha mnh Mi cu pht biu l ng hay l sai c gi l mt mnh . (Definition proposition: Any statement that is either true or false is called a proposition.) Trang 5

2. Chng 1: i s mnh V d 1: Cc cu xc nh di y l mt mnh . 2+3=5 . 3*4 = 10 . . Tam gic u c 3 cnh bng nhau . Washington D.C. l th ca Hoa K . Toronto l th ca Canada Cu xc nh "2 + 3 = 5", "Tam gic u c 3 cnh bng nhau" v "Washington D.C. l th ca Hoa K" l cc mnh ng. Cn cc cu xc nh "3*4 = 10" v "Toronto l th ca Canada" l cc mnh sai. Nh vy, mt mnh c th l mnh ng hoc mnh sai. Hay ni cch khc, mt mnh ch c th la chn 1 trong 2 gi tr l ng hoc l sai. Mt mnh khng th va ng va sai. V d 2: Xt cc cu pht biu sau . Hm nay l th my ? . Mt s thc m khng phi l s chnh phng . Hy c k an ny . x+1=2 . x+y=z Cu "Hm nay l th my ? " khng l mnh v n ch l mt cu hi khng c gi tr ng, sai. Cu "Mt s m khng phi l s chnh phng" c chn tr l ng nu xt trn tp hp s thc R nhng li c chn tr sai khi xt trn tp hp s phc. Cu "x+1=2" v cu "x+y=z" khng phi l mnh v chng chng ng cng chng sai bi cc bin trong nhng cu cha c gn cho mt gi tr c th no. Gi tr ng, sai ca mt mnh c gi l chn tr ca mnh . Chn tr ca mnh ng k hiu l T (true), chn tr ca mnh sai k hiu l F (false). Bng chn tr ca mnh bao gm cc trng hp ng, sai c th xy ra ca mnh . Mc ch ca cc hat ng khoa hc l phn bit cc mnh xc nh chn tr ca n. S xc nh chn tr ny da vo thc nghim v l lun. L lun y l xc nh chn tr ca mnh bng cch kt hp cc mnh m ta bit Trang 6

3. Chng 1: i s mnh chn tr. Cc lut l ch ng cch kt hp mang tnh chnh xc ca php ton i s. V th, chng ta cn ni n "i s mnh ". 1.3. Cc php tnh mnh Trong php tnh mnh , ngi ta khng quan tm n ngha ca cu pht biu m ch ch n chn tr ca cc mnh . Do , khi thc hin cc php ton mnh thng thng ngi ta khng ghi r cc cu pht biu m ch ghi k hiu. Cc ch ci s c dng k hiu cc mnh . Nhng ch ci thng dng l P, Q, R,..... 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 c mnh phc hp l mnh lin kt (c cha php tnh mnh ). Cc php tnh mnh c s dng nhm mc ch kt ni cc mnh li vi nhau to ra mt mnh mi. Cc php ton mnh c trnh by trong chng ny bao gm : php ph nh, php hi, php tuyn, php XOR, php ko theo, php tng ng. 1.3.1. 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="20" Bng chn tr (truth table) p p TF F T Qui tc: Nu P c gi tr l T th ph nh P c gi tr l F. Trang 7

4. Chng 1: i s mnh 1.3.2. 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. V d : Cho 2 mnh P v Q nh sau P = " 2 > 0 " l mnh ng Q = " 2 = 0 " l mnh sai P Q = " 2> 0 v 2 = 0 " l mnh sai. Bng chn tr p q p q T T T T F F F T F F F F Qui tc : Hi ca 2 mnh ch ng khi c hai mnh l ng. Cc trng hp cn li l sai. 1.3.3. Php tuyn (DISJUNCTION) Cho hai mnh P, Q. Cu xc nh "P hay (hoc) Q" l mt mnh mi c gi l tuyn ca 2 mnh P v Q. K hiu P Q. V d : Cho 2 mnh P v Q nh sau P = " 2 > 0 " l mnh ng Q = " 2 = 0 " l mnh sai P Q = " 2 0 " l mnh ng. Bng chn tr p q pq T T T T F T F T T F F F Trang 8

5. Chng 1: i s mnh Qui tc : Tuyn ca 2 mnh ch sai khi c hai mnh l sai. Cc trng hp cn li l ng. 1.3.4. 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. Bng chn tr p q pq T T F T F T F T T F F F 1.3.5. Php ton trn bit Cc my tnh dng cc bit biu din thng tin. Mt bit c 2 gi tr kh d l 0 v 1. Bit cng c th c dng biu din chn tr. Thng ngi ta dng bit 1 biu din chn tr ng v bit 0 biu din chn tr sai. Cc php ton trn bit trong my tnh l cc php ton logic. Thng tin thng c bin din bng cch dng cc xu bit. Ta c nh ngha xu bit nh sau: nh ngha : Mt xu bit (hoc xu nh phn) l dy c mt hoc nhiu bit. Chiu di ca xu l s cc bit trong xu . V d : 101011000 l mt xu bit c chiu di l 9 C th m rng cc php ton trn bit ti cc xu bit. Ngi ta nh ngha cc OR bit, AND bit v XOR bit i vi 2 xu bit c cng chiu di l cc xu c cc bit ca chng l ca1c OR, AND, XOR ca cc bit tng ng trong 2 xu tng ng. Chng ta cng dng cc k hiu , , biu din cc php tnh OR bit, AND v XOR tng ng. Trang 9

6. Chng 1: i s mnh V d : Tm OR bit, AND bit v XOR bit i vi 2 xu sau y (mi xu c tch thnh 2 khi, mi khi c 5 bit cho d c) 01101 10110 11000 11101 11101 11111 OR bit 01000 10100 AND bit 10101 01011 XOR bit 1.3.6. 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 gi thit v Q c gi l kt lun. V d : Cho hai mnh P v Q nh sau P = " tam gic T l u " Q = " tam gic T c mt gc bng 60" xt chn tr ca mnh P Q, ta c nhn xt sau : - Nu P ng, ngha l tam gic T l u th r rng rng P Q l ng. - Nu P sai, ngha l tam gic T khng u v cng khng l cn th d Q l ng hay sai th mnh P Q vn ng. Sau y l bng chn tr ca v d v cng l bng chn tr ca mnh P Q. p q pq T T T T F F F T T F F T Qui tc : mnh ko theo ch sai khi gi thit ng v kt lun sai. Cc trng hp khc l ng. Trang 10

7. Chng 1: i s mnh T mnh P Q, chng ta c th to ra cc mnh ko theo khc nh l mnh Q P v Q P c gi l mnh o v mnh phn o ca mnh P Q. V d : Tm mnh o v phn o ca mnh sau " Nu ti c nhiu tin th ti mua xe hi" Mnh o l : " Nu ti mua xe hi th ti c nhiu tin" Mnh phn o l : " Nu ti khng mua xe hi th ti khng c nhiu tin" 1.3.7. Php tng ng (BICONDITIONAL) Cho P v Q l hai mnh . Cu "P nu v ch nu 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 Bng chn tr p q pq T T T T F F F T F F F T 1.4. 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 . Trang 11

8. Chng 1: i s 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 P P Q R QR P (Q R) chn tr ca biu thc T F T T T T mnh P (Q R ) T F T F F F T F F T F F T F F F F F F T T T T T F T T F F T F T F T F T F T F F F T 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 l mnh phc hp. C th nh ngha cc bin mnh nh sau: P: Michelle thng trong k thi Olympic Trang 12

9. Chng 1: i s mnh Q: mi ngi s khm phc c y R: c ta s tr nn giu c S: 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 : 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. Nu Michelle thng trong k thi Nu c ta khng thng th c ta s Olympic, mi ngi s khm phc AND mt tt c. c y, v c ta s tr nn giu c. Michelle Mi ngi s C ta khng C ta s thng trong khm phc c thng mt tt c. k thi y, v c ta s Olympic tr nn giu c. Mi ngi s khm C ta s tr C ta s mt AND NOT phc c y nn giu c. tt c. Trang 13

10. Chng 1: i s mnh 1.5. Cc ng dng ca Logic (EVERDAY LOGICAL) Ngy nay, logic mnh c ng dng nhiu trong cc lnh vc khc nhau nh: - Vit - Ni - Tm kim trn mng (search engines) - Ton hc - Cc chng trnh my tnh (logic in programming) Do , hiu bit cc qui tc s dng logic l rt hu ch. Sau y l mt vi v d ch ra cc ng dng . V d 1: Logic trong tm kim trn mng t vn : Bn mun tm ti liu trn mng c lin quan n hai t "disc golf". Nu bn g vo tm kim hai t "disc golf" ny, bn s tm thy cc ti liu v disc v cc ti liu v golf nhng khng tm thy cc cc ti liu v "disc golf". Cch gii quyt : Bn ch cn g vo tm kim l "disc AND golf" V d 2 : Logic trong lp trnh (Logic in programming) t vn : Bn mun t iu kin l nu 00 AND x < = 10 ) x++ ; V d 3 : Logic trong cch ni gia nh t vn : M ca b An ni rng : "Nu con ngoan th con c th c n kem hoc n bnh bng lan". B An hiu rng nu n ngoan th n s c n kem v n bnh bng lan. Tuy nhin, m ca b An tc gin v tht s b ta ch cho php n c n mt trong hai th m thi. Cch gii quyt l m ca b An phi ni nh th ny :"Nu con ngoan th con s c n hoc l kem hoc l bnh bng lan nhng khng c n c hai". Trang 14

11. Chng 1: i s mnh V d 4 : Logic trong tnh ton t vn : Bn c 3 ln kim tra trong lp hc. Nu bn t c 2 ln im A, hoc ch mt ln im A nhng khng c c mt ln no rt trong 3 ln kim tra th bn s t im A cho ton kha hc. Bn l ngi khng c sing nng lm, vy th bn s chn cch no t im A cho ton kha hc ? Cch gii quyt : Bi v iu kin l OR nn cch gii quyt l bn c th t 2 im A v rt ln 3, hay l ch cn t mt im A v khng rt ln no. Bn s la chn t mt im A v khng rt ln no. V d 5 : Logic trong i sng t vn : Sau khi nng 1 chic bnh cho 2 a chu trai v 2 a chu gi n thm, D Nellie ly bnh ra khi l nng v ngui. Sau , c ri khi nh n ng ca hng gn . Lc tr v th c ai n 1/4 chic bnh v thm ch cn t li ci da d bn phn bnh cn li. V khng cn ai n nh D ngy hm tr 4 a chu nn D bit ngay l 1 trong 4 a n m cha c cho php. D Nellie bn hi 4 a th c cc cu tr li nh sau: - Charles : Kelly n phn bnh - Dawn : Con khng n bnh - Kelly : Tyler n bnh - Tyler : Con khng n, Kelly ni chi khi bo rng con n bnh. Nu ch 1 trong 4 cu tr li trn l ng v ch 1 trong 4 a chu l th phm, hy tm ra ngi m D Nellie phi pht ? Cch gii quyt : V ch 1 trong 4 cu tr li trn l ng nn chng ta c th dng php vt cn tm li gii. - Gi s Charles ni ng ngha l Kelly n bnh. Ba cu cn li l sai. Dawn ni "Con khng n bnh" l sai ngha l Dawn c n bnh. Vy c n 2 ngi n bnh, iu ny mu thun gi thit, gi s khng c chp thun. - Gi s Dawn ni ng ngha l Dawn khng n bnh v 3 cu cn li l sai. Nhn thy c mu thun gia Kelly v Tyler. Bi v Kelly ni "Tyler n bnh" l sai ngha l Tyler khng n. Trong khi , Tyler li ni rng "Con khng n..." l sai, vy thc t l n c n. Gi thuyt ny l khng chp nhn c. Trang 15

12. Chng 1: i s mnh - Gi s Kelly ni ng ngha l Tyler n bnh v 3 cu cn li l sai. Nh vy, cng c 2 th phm l Kelly v Dawn. Mu thun gi thit. - Gi s sau cng l Tyler ni ng ngha l n khng n bnh v 3 cu cn li l sai. Nhn thy ch c mt ngi n bnh chnh l Dawn. Vy gi thuyt ny l hp l v th phm chnh l Dawn. V d 6 : Logic trong ton hc t vn : Tm s t nhin a bit rng trong 3 mnh di y c 2 mnh l ng v 1 mnh l sai. 1/ a + 51 l s chnh phng 2/ Ch s tn cng ca a l 1 3/ a - 38 l s chnh phng Cch gii quyt : Trc ht, chng ta s phi xc nh xem 2 mnh ng v 1 mnh sai l mnh no ? Sau t 2 mnh ng tm ra s t nhin a. S chnh phng l s nguyn dng khi ly cn bc hai. Do , s chnh phng c cc ch s tn cng l 0, 1, 4, 5, 6, 9. - Nhn thy gia mnh 1 v 2 c mu thun. Bi v, gi s 2 mnh ny ng thi l ng th a+51 c ch s tn cng l 2 nn khng th l s chnh phng. Vy trong 2 mnh ny phi c 1 mnh l ng v 1 l sai. - Tng t, nhn thy gia mnh 2 v 3 cng c mu thun. Bi v, gi s mnh ny ng thi l ng th a-38 c ch s tn cng l 3 nn khng th l s chnh phng. Vy trong 3 mnh trn th mnh 1 v 3 l ng, cn mnh 2 l sai. Vi x > 0 v y > 0 . t : a + 51 = x2 - a - 38 = y2 ---------------- 89 = 1.89 = x2 - y2 = ( x + y )( x - y ) Suy ra : x+y=1 (loi v x, y l nguyn dng nn khng th c x + y = 1) x - y = 89 Trang 16

13. Chng 1: i s mnh Hay l : x + y = 89 x-y =1 Gii h phung trnh ny ta c x = 45 v y = 44. Vy a = 1974. Trn y l vi v d n gin. Hy vng rng cc v d ny cho chng ta thy c s quan trng ca logic khng ch trong ton hc, khoa hc my tnh m cn trong cuc sng hng ngy. 1.6. Cc thut ng chuyn ngnh (SOME TERMINOLOGY) 1.6.1. 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 P P PP T F T F T T Vy PP l mt hng ng. 1.6.2. 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 T F F F T F Trang 17

14. Chng 1: i s mnh Vy PP l mt hng sai. 1.6.3. nh ngha tip lin (Contingency): Mt tip lin l mt biu thc mnh khng phi l hng ng v khng phi l hng sai. V d : Tm chn tr ca biu thc mnh (P Q ) Q p q q p q (pq) q T T F T T T F T F T F T F F F F F T F T Vy (P Q ) Q l mt tip lin v n khng phi l hng ng v cng khng phi l hng sai. 1.7. 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 l hng ng. K hiu F | G V d : Cho F = ( P Q ) ( Q R ) G=PR Xt xem G c l mnh h qu ca F khng ? P Q R PQ QR F G FG T T T T T T T T T T F T F F F T T F T F T F T T T F F F T F F T F T T T T T T T Trang 18

15. Chng 1: i s mnh F T F T F F T T F F T T T T T T F F F T T T T T 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. 1.8. Tng ng Logic (LOGICALLY EQUIVALENT) nh ngha 1 : Mnh P v mnh Q c gi l tng ng logic nu php tng ng ca P v Q (PQ) l hng ng. nh ngha 2 : Hai mnh P v Q c gi l tng ng logic nu v ch nu chng c cng chn tr. Mnh P v Q tng ng logic c k hiu l P Q (hay P = Q) V d 1 : Cho F = P(QR) G = (PQ) (PR) Xt xem hai mnh trn l c tng ng logic khng ? Trang 19

16. Chng 1: i s mnh Vy F v G l tng ng logic hay F=G. V d 2: Cho F=PQ G = (PQ) Xt xem hai mnh trn l c tng ng logic khng ? p q pq p pq T T T F T T F F F F F T T T T F F T T T Vy F G hay P Q = (PQ) F G p q r qr pq pr FG T T T T T T T T T T T F F T T T T T T F T F T T T T T T F F F T T T T T F T T T T T T T T F T F F F T F F T F F T F F F T F T F F F F F F F F T Trang 20

17. Chng 1: i s mnh Bng cc tng ng logic thng dng t T= hng ng, F = hng sai Equivalence Name pT T Domination laws pF F pT p Identity laws pF p pp p Idempotent laws pp p (p) p Double negation law pp T Cancellation laws pp F (Not an offical name) pq qp Commutative laws pq qp (pq)r p(qr) Associative laws (pq)r p(qr) p(qr) (pq)(pr) Distributive laws p(qr) (pq)(pr) (pq) pq De Morgans laws (pq) pq (pq) (pq) Implication law Lu : Domination laws : lut nut Identity laws : lut ng nht Idempotent laws : lut ly ng Trang 21

18. Chng 1: i s mnh Double negation law : lut ph nh kp Cancellation laws : lut xa b Commutative laws : lut giao hon Associative laws : lut kt hp Distributive laws : lut phn b De Morgans laws : lut De Morgan Ngoi cc tng ng thng dng trong bng trn, c mt tng ng logic khc m chng ta cng s hay gp trong cc chng minh. l : P(PQ)=P P(PQ)=P ( sinh vin t chng minh xem nh bi tp ) 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 (q q) Associative law Cancellation Law p T T Domination Law V d 2 : Chng minh rng (q p ) (p q ) = q Trang 22

19. Chng 1: i s mnh ((q p)) ( p q) ((q p)) ( p q) Implication law (q p) ( p q) Commutative law (q p) (q p) Distributive law q (p p) Cancellation law qT Identity law q V d 3 : p dng trong lp trnh Gi s trong chng trnh c cu lnh sau : while(NOT(A[i]!=0 AND NOT(A[i]>= 10))) Ta c th vit li cu lnh ny mt cch n gin hn bng cch s dng cng thc De Morgan. while( A[i]==0 OR A[i]>= 10) V d 4: Gi s trong chng trnh c cu lnh sau : while( (i10) OR (i= 10))) Trc ht chng ta s p dng cng thc De Morgan bin i biu thc sau cng nh sau : while( (i10) OR (i= 10) ) Sau , chng ta li s dng cng thc v tnh phn b ca php hi i vi php tuyn rt gn biu thc pha trc. Ta c cu lnh sau cng l : while( (i10 OR A[i]= 10) ) 1.9. Tng kt chng 1 Trong chng ny sinh vin cn nm vng nh ngha mnh cng cc php ton logic. Ngoi ra, cc thut ng chuyn ngnh cng rt quan trng. Sinh vin Trang 23

20. Chng 1: i s mnh phi bit cch p dng cc php ton logic trong lp trnh. Tuy nhin, c vn cn lu khi p dng tnh giao hon. Trong mt vi ngn ng lp trnh, v d nh C, Java, C++ th vic s dng tnh cht giao hon c th khng l mt tng hay. V d : Nu A l mt mng c n phn t th cu lnh : if(i5 then n:=n+2 ; b/ if ((n+2 = 8) or (n-3=6)) then n:= 2*n + 1 ; c/ if ((n-3=16) and (n div 5=1)) then n:= n + 3 ; d/ if ((n21) and (n-7=15)) then n:= n - 4 ; e/ if ((n div 5 = 2) or (n+1=20)) then n:=n+1 ; Ban u bin nguyn n c gn tr l 7. Hy xc nh gi tr n trong cc trng hp sau : Trang 24

21. Chng 1: i s mnh - Sau mi cu lnh ( ngha l khi qua cu lnh mi th gn li n = 7) - Sau tt c cc lnh ( s dng kt qu ca cu lnh trc tnh ton cho cu sau) 4/ Cho on chng trnh sau : a/ if n-m = 5 then n:= n-2 ; b/ if ((2*m=n) and (n div 4 =1) then n:= 4*m - 3 ; c/ if ((n0) and (t=3)) ; Vi mi cch gn gi tr bin nh sau, hy xc nh trong trng hp no th vng lp kt thc. a/ x= 7, y= 2, w= 5, t= 3 b/ x= 0, y= 2, w= -3, t= 3 c/ x= 0, y= -1, w= 1, t= 3 d/ x= 1, y= -1, w= 1, t= 3 6/ Trong mt phin ta x n 3 b can c lin quan n vn ti chnh, trc ta c 3 b co u tuyn th khai ng s tht v li khai nh sau : Anh A: Ch B c ti v anh C v ti Ch B : Nu anh A c ti th anh C cng c ti Anh C: Ti v ti nhng mt trong hai ngi kia l c ti Trang 25

22. Chng 1: i s mnh Hy xt xem ai l ngi c ti ? 7/ Cho cc mnh c pht biu nh sau, hy tm s ln nht cc mnh ng thi l ng. a/ Quang l ngi khn kho b/ Quang khng gp may mn c/ Quang gp may mn nhng khng khn kho d/ Nu Quang l ngi khn kho th n khng gp may mn e/ Quang l ngi khn kho khi v ch khi n gp may mn f/ Hoc Quang l ngi khn kho, hoc n gp may mn nhng khng ng thi c hai. 8/ Cho a v b l hai s nguyn dng. Bit rng, trong 4 mnh sau y c 3 mnh ng v 1 mnh sai. Hy tm mi cp s (a, b) c th c. 1/ a+1 chia ht cho b 2/ a = 2b + 5 3/ a+b chia ht cho 3 4/ a+7b l s nguyn t 9/ Khng lp bng chn tr, s dng cc cng thc tng ng logic, chng minh rng cc biu thc mnh sau l hng ng a/ (PQ)P b/ P( P P) c/ P((Q (PQ)) d/ (P Q) P e/ ((PQ) (QR)) (PR) 10/ Khng lp bng chn tr, s dng cc cng thc tng ng logic, xt xem biu thc mnh G c l h qu ca F khng ? a/ F = P(QR) G = (PQ)R b/ F = (PQ)(QR) G = P (Q R) c/ F = PQ G = (PQ) (P Q) 11/ Tng t bi tp 9 v 10, chng minh cc tng ng logic sau y: a/ (PQ) (PQ) P Trang 26

23. Chng 1: i s mnh b/ (((PQ)R) Q) QR c/ ((PQ) (P Q)) Q PQ d/ (PQ) ((P Q) Q) (QP) e/ (PQ) (Q (R Q)) (QP) f/ P (P (PQ) P g/ P Q (P Q R) PQR h/ ((P Q) (PQR ) PQ i/ P ((Q (RR)) (Q (RS) (R S))) P j/ (PQR) (P S Q) (P S R) P (R (S Q) Trang 27

24. Chng 1: i s mnh CHNG 1 : I S MNH .................................................................................5 1.1. Tng quan .........................................................................................................5 1.2. nh ngha mnh ..........................................................................................5 1.3. Cc php tnh mnh .....................................................................................7 1.3.1. Php ph nh (NEGATION) ...................................................................7 1.3.2. Php hi (CONJUNCTION) .....................................................................8 1.3.3. Php tuyn (DISJUNCTION) ...................................................................8 1.3.4. Php XOR..................................................................................................9 1.3.5. Php ton trn bit.......................................................................................9 1.3.6. Php ko theo (IMPLICATION).............................................................10 1.3.7. Php tng ng (BICONDITIONAL) ................................................11 1.4. Biu thc mnh (LOGICAL CONNECTIVES)........................................11 1.5. Cc ng dng ca Logic (EVERDAY LOGICAL)........................................14 1.6. Cc thut ng chuyn ngnh (SOME TERMINOLOGY) .............................17 1.6.1. nh ngha Hng ng (Tautologie): ......................................................17 1.6.2. nh ngha Hng sai (Contradiction): .....................................................17 1.6.3. nh ngha tip lin (Contingency):........................................................18 1.7. Mnh h qu...............................................................................................18 1.8. Tng ng Logic (LOGICALLY EQUIVALENT)...................................19 1.9. Tng kt chng 1 ..........................................................................................23 1.10. Bi tp chng 1 .........................................................................................24 Trang 28

25. Chng 2: Suy lun ton hc & Cc phng php chng minh CHNG 2 : SUY LUN TON HC & CC PHNG PHP CHNG MINH 2.1. Tng quan Mc tiu ca chng 1 Hc xong chng ny, sinh vin phi nm bt c cc vn sau: - Khi nim v suy lun ton hc - Cc phng php chng minh v bit vn dng cc phng php ny chng minh mt bi ton c th. Kin thc c bn cn thit Cc kin thc c bn trong chng ny bao gm: - Cc php ton i s, hnh hc c bn c th a ra v d minh ha trong tng phng php. - Hiu r qui tc ca php ko theo chng 1. Ti liu tham kho Phm vn Thiu, ng Hu Thnh. Ton ri rc ng dng trong tin hc. Nh xut bn Khoa hc v K thut, H Ni - 1997 (chng 3, trang 208 - 228). Ni dung ct li - Khi nim v suy lun ton hc - Trnh by cc phng php chng minh bao gm: . Chng minh rng . Chng minh tm thng . Chng minh trc tip . Chng minh gin tip . Chng minh phn chng . Chng minh qui np Trang 28

26. Chng 2: Suy lun ton hc & Cc phng php chng minh 2.2. Suy lun ton hc 2.2.1. Khi nim Suy lun c xem l mt trong nhng nn tng xy dng nn cc ngnh khoa hc t nhin. T xa n nay, nh suy lun m ngi ta c th nhn thc c ci cha bit t nhng ci bit. Suy lun cn l c s ca s sng to. T cc phn on, a n cc chng minh chp nhn hay bc b mt vn no . Suy lun ton hc da trn nn tng ca cc php ton mnh , ch yu l php ko theo. chng minh mt vn no , thng thng ngi ta phi xc nh im ban u (c th gi l gi thit) v im kt thc (gi l kt lun). Qu trnh i t gi thit n kt lun gi l qu trnh chng minh v qu trnh ny c thc thi bng cch no th gi l phng php chng minh. Cc phng php chng minh l rt quan trng v khng nhng chng thng c s dng trong ton hc m cn c p dng nhiu trong tin hc. V d, s kim tra tnh ng n ca mt chng trnh, ca mt h iu hnh, xy dng cc lut suy din trong lnh vc tr tu nhn to... Do , chng ta cn phi nm vng cc phng php chng minh. Tuy nhn, c nhng phng php chng minh ng v n c da trn c s ca mt mnh ng (hng ng) v c nhng phng php chng minh sai. Cc phng php chng minh sai ny l c hoc v . Khi phng php chng minh da trn mt hng sai th s mang li kt qu sai nhng ngi ta vn cho l ng th c gi l c . i khi c nhng phng php chng minh da trn mt tip lin (c khi mnh l ng nhng cng c lc sai) m ngi ta tng lm l hng ng nn cho l kt qu bao gi cng ng th trng hp ny gi l v (hay ng nhn). Sau y, chng ta s i tm hiu cc qui tc suy lun. 2.2.2. Cc qui tc suy lun Nh gii thiu trn, nhng suy lun c dng cc qui tc suy din gi l suy lun c c s. Khi tt c cc suy lun c c s l ng th s dn n mt kt lun ng. Mt suy lun c c s c th dn n mt kt lun sai nu mt trong cc mnh dng trong suy din l sai. Sau y l bng cc qui tc suy lun ng. Trang 29

27. Chng 2: Suy lun ton hc & Cc phng php chng minh Quy Tc Hng ng Tn Lut P P(PQ) Cng P Q PQ (PQ)P Rt gn P P (P(PQ))Q Modus Ponens PQ Q Q (Q(PQ)) P Modus Tollens PQ P PQ ((PQ)(QR)) Tam on lun gi QR nh (PR) P R PQ (PQ) Q Tam on lun tuyn Q Trong cc phn s ca qui tc th cc gi thit c vit trn t s, kt lun c vit di mu s. K hiu c ngha l "vy th", "do ",... V d : Qui tc suy lun no l c s ca suy din sau : " Nu hm nay tri ma th c ta khng n, Nu c ta khng n th ngy mai c ta n, Vy th, nu hm nay tri ma th ngy mai c ta n." y l suy din da trn qui tc tam on lun gi nh. "Nu hm nay tuyt ri th trng i hc ng ca. Hm nay trng i hc khng ng ca. Do , hm nay khng c tuyt ri " y l suy din da trn qui tc Modus Tollens " Alice gii ton. Do , Alice gii ton hoc tin" y l suy din da trn qui tc cng. Ngy bin Trang 30

28. Chng 2: Suy lun ton hc & Cc phng php chng minh Cc phng php chng minh sai cn c gi l ngy bin. Ngy bin ging nh qui tc suy lun nhng khng da trn mt hng ng m ch l mt tip lin. y chnh l s khc nhau c bn gia suy lun ng v suy lun sai. Loi suy lun sai ny c gi l ng nhn kt lun. V d : Xt xem suy din sau l c c s ng khng ? " Nu bn gii ht bi tp trong sch ton ri rc 2 ny th bn nm vng logic. Bn nm vng logic vy th bn gii ht bi tp trong sch ton ri rc 2 ny". Nhn thy suy din ny l da trn mnh sau : ((PQ) Q) P Trong : P = "Bn gii ht bi tp trong sch ton ri rc 2" Q = "Bn nm vng logic" Mnh ((PQ) Q) P khng phi l hng ng v n s sai khi P l F v Q l T. Do , suy din ny khng hon ton c c s ng. Bi v, khi Q l T ngha l bn nm vng logic nhng khng chc l bn gii ht bi tp trong sch ton ri rc 2 ny m c th gii sch khc (P l F). 2.3. Cc phng php chng minh Nh gii thiu trong phn trn, mi bi ton cn chng minh thng thng u c hai phn chnh l gi thit v kt lun. Vic ch ra c ci no l gi thit, ci no l kt lun s gip cho vic chng minh d dng hn thng qua vic s dng phng php chng minh thch hp. Do , cc phng php chng minh trong dng bi ton ny l c lin quan n mnh ko theo. Vy, trc khi tm hiu cc phng php chng minh, chng ta hy xem li bng chn tr ca mnh P ko theo Q ( vi P l gi thit v Q l kt lun). Cc trng hp cho mnh P ko theo Q l ng cng chnh l cc phng php chng minh bi ton ng. p q pq Trang 31

29. Chng 2: Suy lun ton hc & Cc phng php chng minh T T T T F F F T T F F T Nhn thy rng, PQ l ng c 3 trng hp. Cc trng hp ny chnh l cc phng php chng minh s c trnh by di y. Trc khi i vo cc phng php chng minh, c mt khi nim m chng ta cn tm hiu, l khi nim v "hm mnh ". Hm mnh : Cho A l mt tp hp khng rng sao cho ng vi mi xA ta c mt mnh , k hiu l P(x). By gi ta ni P (hay P(x)) l mt hm mnh theo bin xA. Nh vy, khi ni ng vi mi xA, ta c mt mnh P(x), ngha l khi tnh ng sai ca P(x) c hon ton xc nh ph thuc vo tng gi tr ca xA. V d : Cho hm mnh P(x) = { x l s l } ; xN Ta c : P(1) l mnh ng P(2) l mnh sai. Tng qut, vi cc tp hp khng rng A1, A2, ..., An, sao cho ng vi mi x1A1, x2A2, ..., xnAn, ta c mt mnh , k hiu P(x1, x2, ...,xn ). Ta ni P(x1, x2, ...,xn ) l mt hm mnh theo n bin x. V d : Cho hm mnh P(x,y,z) = { 2x + y - z = 0 } x,y,zZ Ta c : P(x,y,z) l mnh ng khi x = 1, y = -1, z = 1. P(x,y,z) l mnh sai khi x = 1, y = 1, z = 1. 2.3.1. Chng minh rng ( P l sai) Da vo 2 dng cui ca bng chn tr, nhn thy rng khi P sai, bt chp kt lun Q th no th mnh PQ l lun ng. Vy, chng minh mnh Trang 32

30. Chng 2: Suy lun ton hc & Cc phng php chng minh PQ l ng, ngi ta ch cn chng minh rng P l sai. Phng php chng minh ny c gi l chng minh rng. Phng php chng minh rng thng c s dng chng minh cc trng hp c bit ca nh l. Trng hp tng qut th nh l ny lun ng vi mi s n nguyn dng. V d : Cho hm mnh P(n) = " Nu n>1 th n2 >n " Chng minh rng P(1) l ng. Gii : Ta c P(1) = { Nu 1 >1 th 12 >1 } Nhn thy rng gi thit 1>1 l sai, bt chp kt lun 12 >1 l ng hay sai th P(1) l ng. 2.3.2. Chng minh tm thng (Q l ng) Da vo dng 1 v dng 3 ca bng chn tr, nhn thy rng khi Q ng, bt chp gi thit P l ng hay sai th mnh PQ l lun ng. Vy, chng minh mnh PQ l ng, ngi ta ch cn chng minh rng Q l ng. Phng php chng minh ny c gi l chng minh tm thng. Phng php chng minh tm thng cng c s dng chng minh cc trng hp c bit ca nh l. Trng hp tng qut th nh l ny lun ng vi mi s n nguyn dng. V d : Cho hm mnh P(n) = { Nu a v b l 2 s nguyn dng v a b th an bn } Chng minh rng P(0) l ng. Gii : Ta c a0 = b0 =1. Do a0 b0 l ng. Vy P(0) l ng bt chp gi thit ab l ng hay sai. 2.3.3. Chng minh trc tip Trong dng 1 ca bng chn tr, mnh P ko theo Q c th c chng minh bng cch ch ra rng nu P ng th Q cng phi ng. Ngha l t hp P ng Q sai khng bao gi xy ra. Phng php ny c gi l chng minh trc tip. Vy thc hin phng php chng minh trc tip, ngi ta gi s rng P l ng, sau s dng cc qui tc suy lun hay cc nh l ch ra rng Q l ng v kt lun PQ l ng. Trang 33

31. Chng 2: Suy lun ton hc & Cc phng php chng minh V d 1: Chng minh rng { Nu n l s l th n2 l s l } Gii : Gi s rng gi thit ca nh l ny l ng, tc l n l s l. Ta c n = 2k + 1 ( k=0,1,2,...) n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k + 2k) + 1 l l. Vy nu n l s l th n2 l s l. V d 2 : Cho hm mnh P(n) = " Nu n>1 th n2 >n " Chng minh rng P(n) l ng vi n l s nguyn dng. Gii : Gi s n > 1 l ng, ta c : n=1+k ( k 1) n2 = ( 1 + k )2 = 1 + 2k + k2 = (1 + k) + k + k2 > n Vy Nu n>1 th n2 >n . 2.3.4. Chng minh gin tip V mnh PQ Q P. Do , chng minh mnh PQ l ng, ngi ta c th ch ra rng mnh Q P l ng. V d : Chng minh nh l { Nu 3n + 2 l s l th n l s l } Gii : Gi s ngc li kt lun ca php ko theo l sai, tc n l chn. Ta c n = 2k ( kN ) 3n + 2 = 3.2k + 2 = 2( 3k + 1 ) l s chn Vy Nu 3n + 2 l s l th n l s l Nhn xt C nhng bi ton c th s dng phng php chng minh trc tip hay gin tip u c c. Tuy nhin, c nhng bi ton khng th s dng phng php chng minh trc tip c hoc s dng trc tip th bi gii s di dng phc tp hn l s dng chng minh gin tip ( hoc ngc li). y chnh l s khc bit ca chng minh trc tip v chng minh gin tip. V d 1 : S dng chng minh gin tip chng minh rng " Nu n>1 th n2 >n " Gii : Gi s ngc li kt lun ca php ko theo l sai, tc l n2 < n. Trang 34

32. Chng 2: Suy lun ton hc & Cc phng php chng minh V n l nguyn dng nn ta c th chia 2 v cho n m bt ng thc khng i chiu. Ta c : n < 1. Vy t Q dn n P. Do , Nu n>1 th n2 >n. V d 2 : S dng chng minh trc tip chng minh rng " Nu 3n + 2 l s l th n l s l ". Gii : Gi s 3n + 2 l s l l ng. Nhn thy rng v 2 l s chn nn suy ra c 3n l s l. V 3 l s l do n l s l. Vy Nu 3n + 2 l s l th n l s l. y chng ta phi chng minh thm nh l l tch ca 2 s l l mt s l th bi gii cht ch hn. Do , trong bi ton ny vic s dng chng minh gin tip l hay hn dng trc tip. chng minh mnh c dng : (P1P2...Pn) Q Chng ta c th s dng hng ng sau : ((P1P2...Pn) Q) ((P1Q)(P2Q)....(PnQ)) Cch chng minh ny gi l chng minh tng trng hp. V d 3: Chng minh rng: " Nu n khng chia ht cho 3 th n2 khng chia ht cho 3". Gii : Gi P l mnh "n khng chia ht cho 3" v Q l mnh "n2 khng chia ht cho 3". Khi , P tng ng vi P1 P2. Trong : P1 = " n mod 3 =1" P2 = " n mod 3 =2" Vy, chng minh P Q l ng, c th chng minh rng: (P1 P2) Q hay l (P1 Q ) ( P2 Q) Gi s P1 l ng. Ta c, n mod 3 = 1. t n = 3k + 1 ( k l s nguyn no ). Suy ra n2 = ( 3k+1)2 = 9k2 + 6k + 1 = 3(3k2 + 2k) + 1 khng chia chn cho 3. Do , P1 Q l ng. Trang 35

33. Chng 2: Suy lun ton hc & Cc phng php chng minh Tng t, gi s P2 l ng. Ta c, n mod 3 = 2. t n = 3k + 2 ( k l s nguyn no ). Suy ra n2 = ( 3k+2)2 = 9k2 + 12k + 4 = 3(3k2 + 4k + 1) + 1 khng chia chn cho 3. Do , P2 Q l ng. Do P1 Q l ng v P2 Q l ng, hay l (P1 Q ) ( P2 Q). Vy (P1 P2) Q. 2.3.5. Chng minh phn chng Chng minh phn chng thng c s dng chng minh mnh P l ng. Trc ht, ngi ta gi s ngc li rng P l sai hay P l ng. T mnh P l ng dn n kt lun Q sao cho PQ phi ng. Khi , ngi ta ch ra rng Q l mt mu thun, ngha l : Q = R R. (S d c mu thun ny l do ta gi s P l sai) V PQ phi ng v Q l F, suy ra rng P = F P = T. Phng php chng minh phn chng thng c s dng chng minh nhng vn c bn v iu quan trng trong k thut ny l tm ra c mu thun RR. V d 1: Chng minh rng " 2 l s v t ". Gii : Gi P l mnh " 2 l s v t ". Gi s ngc li P l ng. Vy, 2 l s hu t ( v tp s thc gm 2 tp con l tp s v t v tp s hu t. Hai tp con ny khng c 3 giao nhau). Khi a,b (a,bN) sao cho: a 2 = ( vi a, b khng c c chung hay phn s ny l ti gin (mnh b R)) a2 Bnh phng hai v : 2 = 2b2 = a2 a2 l s chn a l s b2 chn. t a = 2c, c N. Ta c 2b2 = 4c2 b2 = 2c2 b2 l s chn b l s chn. Vy a, b u c c chung l 2 (mnh R). Trang 36

34. Chng 2: Suy lun ton hc & Cc phng php chng minh iu ny mu thun v a/b l ti gin. T P RR. S d c mu thun ny l do ta gi s 2 l s hu t. Vy 2 phi l s v t. V d 2 : Mt trong nhng cch gii bi ton tn ti l dng lp lun phn chng. Cho 7 on thng c di ln hn 10 v nh hn 100. Chng minh rng lun tm c 3 on c th ghp thnh mt tam gic. Gii : Trc ht sp xp cc on cho theo th t tng dn ca di a1, a2, ..., a7, v chng minh rng trong dy xp lun tm c 3 on lin tip sao cho tng ca 2 on u ln hn on cui (v iu kin 3 on c th ghp thnh mt tam gic l tng ca 2 on nh hn on th ba). Gi s iu cn chng minh l khng xy ra, ngha l ng thi xy ra cc bt ng thc sau: a1 + a2 a3 a2 + a3 a4 a3 + a4 a5 a4 + a5 a6 a5 + a6 a7 T gi thit a1 , a2 c gi tr ln hn 10, ta nhn c a3 > 20 . T a2 >10 v a3 > 20 ta nhn c a4 > 30 , a5 > 50, a6 > 80 v a7 > 130. iu a7 > 130 l mu thun vi gi thit cc di nh hn 100. C mu thun ny l do gi s iu cn chng minh khng xy ra. Vy, lun tn ti 3 on lin tip sao cho tng ca 2 on u ln hn on cui. Hay ni cch khc l 3 on ny c th ghp thnh mt tam gic. 2.3.6. Chng minh qui np Gi s cn tnh tng n s nguyn l u tin. Vi n = 1,2,3,4,5 ta c : n = 1: 1 = 1 = 12 n = 2: 1 + 3 = 4 = 22 n = 3: 1 + 3 + 5 = 9 = 32 n = 4: 1 + 3 + 5 + 7 = 16 = 42 Trang 37

35. Chng 2: Suy lun ton hc & Cc phng php chng minh n = 5: 1 + 3 + 5 + 7 + 9 = 25 = 52 T cc kt qu ny ta d on tng n s nguyn l u tin l n2. Tuy nhin, chng ta cn c phng php chng minh d on trn l ng. Qui np ton hc l mt k thut chng minh rt quan trng. Ngi ta dng n chng minh nhng kt qu c da trn s suy lun no nh v d trn. Tuy nhin, qui np ton hc ch dng chng minh cc kt qu nhn c bng mt cch no ch khng l cng c pht hin ra cng thc. Nguyn l chng minh qui np yu Nhiu nh l pht biu rng P(n) l ng n nguyn dng, trong P(n) l hm mnh , k hiu nP(n). Qui np ton hc l mt k thut chng minh cc nh l thuc dng trn. Ni cch khc qui np ton hc thng s dng chng minh cc mnh dng nP(n). Nguyn l chng minh qui np yu bao gm 2 bc : - Kim tra P(x0) l ng vi x0 l gi tr u tin ca dy s n - Gi s rng P(k) l ng khi n=k. T suy ra rng P(k+1) l ng. Ta c cch vit ca suy lun trn nh sau: [P(x0) (P(k)P(k+1))] nP(n) V d 1: Chng minh rng n n(n + 1) i = 1 + 2 + 3 + ... + n = i =1 2 n n(n + 1) Gii : t P(n) = i = i =1 2 1(1 + 1) - Vi n= 1 : 1 = P(1) l ng 2 k k (k + 1) - Gi s P(k) l ng khi n=k. Ta c : i = i =1 2 Cn chng minh rng P(k+1) l ng. Ngha l k +1 (k + 1)(k + 2) i = i =1 2 (iu phi chng minh) Trang 38

36. Chng 2: Suy lun ton hc & Cc phng php chng minh K +1 K k (k + 1) (k + 1)(k + 2) Ta c : i = i + (k + 1) = i =1 i =1 2 + (k + 1) = 2 (pcm) Vy nP(n). V d 2: Chng minh rng n i 1 P(n) = = 1 i =1 (i + 1)! (n + 1)! 1 1 - Vi n=1 : = 1 P(1) l ng 2 2 - Gi s P(k) l ng khi n= k. Ta c : K i 1 (i + 1)! = 1 (k + 1)! i =1 Cn chng minh rng : K +1 i 1 (i + 1)! = 1 (k + 2)! i =1 Ta c : K +1 i K i k +1 1 k +1 (i + 1)! i =1 (i + 1)! + (k + 2)! = 1 (k + 1)! + (k + 2)! i =1 = (k + 2) (k + 1) 1 = 1 = 1 (pcm) (k + 2)! (k + 2)! Vy nP(n) V d 3 : Chng minh bt ng thc sau : n < 2n vi n nguyn dng. - Khi n=1 : 1 < 2 mnh ng - Gi s mnh ng khi n=k, ta c k < 2k . Cn chng minh rng k + 1< 2k+1 . Tht vy, v k < 2k k +1 < 2k +1 < 2k + 2k = 2k+1. Do , n < 2n vi n nguyn dng. Ch 1: Trang 39

37. Chng 2: Suy lun ton hc & Cc phng php chng minh Khi s dng nguyn l chng minh qui np, khng c b qua bc kim tra P(x) l ng v nu ch c (P(n)P(n+1)) l khng kt lun rng nP(n) l ng. V d : Xt n (n + 3)(n 2) P(n)= i = 0 + 1 + 2 + 3 + ... + n = i =0 2 Gi s P(k) l ng khi n=k. Ta c : K (k + 3)(k 2) i = 0 + 1 + 2 + 3 + ... + k = i =0 2 Cn chng minh: K +1 (k + 3)(k 1) i = 0 + 1 + 2 + 3 + ... + k + (k + 1) = i =0 2 Ta c : K +1 K (k + 3)(k 2) i = i + (k + 1) = i =0 i =0 2 + (k + 1) k 2 2k + 3k 6 + 2k + 2 k 2 + 3k 4 VT = = 2 2 (k 1)(k + 4) VT = = P(k + 1) (pcm) 2 Ta c P(k)P(k+1) l ng. Tuy nhin, khi xt P(0): P(0) = {0 = 3} l mnh sai. Vy nP(n) l sai. Trong trng hp ny ta c th kt lun nh sau : Nu P(k) l ng v nu nk(P(k)P(k+1)) l ng th nk, P(n) l ng. Ch 2 : i khi chng ta cn tnh ton mt biu thc ph thuc vo n, bt u l vic on ra kt qu, cng vic ny c lm bng cch t hay nhiu da vo kinh nghim. Sau , s dng nguyn l chng minh qui np chng minh rng kt qu va tm c l ng. Trang 40

38. Chng 2: Suy lun ton hc & Cc phng php chng minh V d 1: Tnh tng n s l u tin. n S = 1+3+5+7+...+(2n-1) = (2i 1) i =1 Khi n=1 : S = 1 = 12 n=2 : S = 1+ 3 = 22 n=3 : S = 1+3 + 5 = 32 n=4 : S = 1+3+5+7 = 42 n=5 S = 1+3+5+7+9 = 52 ........................................... n Vy c th d on rng S = (2i 1) i =1 = n2 Sau s dng chng minh qui np chng minh kt qu va tm c. n t P(n) = (2i 1) = n 2 i =1 - Khi n=1 : 1 = 1 P(1) l ng - Gi s rng P(k) l ng khi n=k. Ta c : K (2i 1) = k i =1 2 cn chng minh P(k+1) l ng, ngha l : K +1 (2i 1) = (k + 1) i =1 2 K V tri = (2i 1) + (2(k + 1) 1) = k i =1 2 + (2k + 1) = (k + 1) 2 (pcm) Vy nP(n). V d 2: Tng trn c th tnh ton vi mt cch khc nh sau : n n n n(n + 1) S= i =1 (2i 1) = 2 i 1 = 2 i =1 i =1 2 n = n(n + 1) n = n 2 V d 3: Tnh tng Trang 41

39. Chng 2: Suy lun ton hc & Cc phng php chng minh n 1 S= i(i + 1) i =1 1 1 Khi n=1: S = = 2 1+1 1 1 3 +1 2 2 n=2: S = + = = = 2 2.3 2.3 3 2 + 1 2 1 2.4 + 1 3 3 n=3: S = + = = = 3 3.4 3.4 4 3 +1 3 1 3.5 + 1 4 4 n=4: S = + = = = 4 4.5 4.5 5 4 +1 .......................................... n Vy c th d on tng S = n +1 S dng nguyn l qui np chng minh cng thc trn. n 1 n t P(n) = = i =1 i (i + 1) n(n + 1 - Khi n=1 : 1/2 = 1/2 P(1) l ng - Gi s P(k) l ng khi n=k. Ta c K 1 k i(i + 1) = k + 1 i =1 Cn chng minh P(k+1) l ng. Ngha l : K +1 1 k +1 i(i + 1) = k + 2 i =1 (pcm) K +1 1 K 1 1 k 1 V tri = i(i + 1) i =1 i(i + 1) + (k + 1)(k + 2) = k + 1 + (k + 1)(k + 2) i =1 = k ( k + 2) + 1 (k + 1) 2 k +1 = = = (pcm) (k + 1)(k + 2) (k + 1)(k + 2) k + 2 Vy nP(n). Nguyn l chng minh qui np mnh Trang 42

40. Chng 2: Suy lun ton hc & Cc phng php chng minh Cho P(n) l mt ng thc c cha bin n, nu P(0) l ng v nu (P(0) P(1)P(2)P(3)...P(k)) P(k+1) l ng th P(n) l mnh ng n (vi 0 l phn t u tin). Ch rng, to ra gi thit qui np vi nguyn tc qui np yu, ngi ta ch gi thit rng P(k) l ng ti n=k. Vi nguyn tc qui np mnh, ngi ta ch ra rng gi thit ng cho tt c cc mnh P(0) P(1)P(2)P(3)...P(k). y chnh l s khc bit c bn ca 2 nguyn tc qui np vi gi thit yu v gi thit mnh. V d 1: Chng minh rng tch ca 3 s lin tip lun chia ht cho 6. Gii : t P(n) = {n.(n+1).(n+2) chia ht cho 6} (n nguyn dng) Ta c : P(1) = 1.2.3 chia ht cho 6. Mnh ng. P(2) = 2.3.4 chia ht cho 6. Mnh ng. P(3) = 3.4.5 chia ht cho 6. Mnh ng. ................................ Gi s n k ta c P(k) l ng. Ngha l : k.(k+1).(k+2) chia ht cho 6. Cn chng minh rng P(k+1) l ng. Nhn thy: (k+1)(k+2)(k+3) = k.(k+1).(k+2) + 3.(k+1).(k+2) Trong : k.(k+1).(k+2) chia ht cho 6. V 3.(k+1).(k+2) chia ht cho 6 = 2.3 (v (k+1).(k+2) l tch ca 2 s t nhin lin tip nn chia chn cho 2). V tng ca 2 s chia ht cho 6 s chia ht cho 6 (sinh vin t chng minh), do (k+1).(k+2)(k+3) chia ht cho 6. P(n) ng vi mi n nguyn dng. V d 2: Chng minh rng nu n l mt s nguyn ln hn 1, khi n c th c vit di dng tch ca cc s nguyn t. Gii : t P(n) = { n = a.b...c } (a, b,..,c l cc s nguyn t) Ta c P(2) = { 2= 2.1} P(3) = { 3= 3.1} P(4) = { 4= 2.4} ...................... P(18) = { 6.3= 3.2.3} Trang 43

41. Chng 2: Suy lun ton hc & Cc phng php chng minh .......................... l cc mnh ng. Gi s P(n) ng n 2 ta c P(k) l ng. Cn chng minh rng P(k+1) l ng. Vi n = k+1 ta c 2 trng hp xy ra nh sau: - k+1 l s nguyn t : k+1 = (k+1).1 P(k+1) ng - k+1 khng l s nguyn t (hp s): k+1 = a.b ( a,b, [2,k] ) Theo gi thit qui np mnh, a, b c th l s nguyn t hoc l tch ca cc s nguyn t. Vy nu k+1 l hp s th n cng s c vit di dng tch ca cc s nguyn t. P(n) ng vi mi n 2. V d 3: Chng minh rng mi bu ph bng hay ln hn 12 xu u c th to ra bng cc con tem 4 xu hay 5 xu. Gii : t P(n) = { n = 4 + ...+ 5+....} Ta c : P(12) = { 12 = 4 + 4 + 4} P(13) = { 13 = 4 + 4 + 5} P(14) = { 14 = 4 + 5 + 5} P(15) = { 15 = 5+ 5 + 5} P(16) = { 16 = 4 + 4 + 4 + 4 } P(17) = { 17 = 4 + 4 + 4 + 5 } Gi s n > 15 v P(n) l ng. Nht thy rng to ra bu ph (n+1) xu ta ch cn dng con tem n-3 xu v cng thm mt tem 4 xu. 2.4. Tng kt chng 2 Chng ta m t cc phng php khc nhau chng minh nh l. C th thy rng khng th a ra mt phng php no chng minh cho mt bi ton no. Nm vng cc phng php chng minh l mt chuyn, bit p dng chng chng minh cc bi ton l mt k thut i hi ngi s dng phi thc tp nhiu ln bng cch th cc trng hp khc nhau. 2.5. Bi tp chng 2 1/ Quy tc suy lun no c dng trong mi lp lun sau : Trang 44

42. Chng 2: Suy lun ton hc & Cc phng php chng minh a. Nhng con kanguroo sng Australia l loi th c ti. Do , kanguroo l loi th c ti. b. Hoc hm nay tri nng trn 100 hoc l s nhim l nguy hi. Hm nay nhit ngoi tri thp hn 100 . Do , nhim l nguy hi. c. Steve s lm vic mt cng ty tin hc vo ma h ny. Do , ma h ny anh ta s lm vic mt cng ty tin hc hoc l mt k lang thang ngoi b bi. d. Nu ti lm bi tp ny c m th ti c th tr li c tt c bi tp. Nu ti tr li c tt c bi tp th ti s hiu c ti liu ny. Do , nu ti lm bi tp ny c m th ti s hiu c ti liu ny 2/ Xc nh xem cc suy lun sau l c c s khng. Nu mt suy lun l c c s th n dng qui tc suy lun no. Nu khng hy ch ra ngy bin no c s dng. a. Nu n l mt s thc ln hn 1 khi n2 > 1. Gi s n2 > 1. Khi n > 1. b. Nu n l mt s thc v n > 3, khi n2 > 9. Gi s n2 9. Khi , n 3. c. Mt s nguyn dng hoc l s chnh phng hoc c mt s chn cc c nguyn dng. Gi s, n l mt s nguyn dng c mt s l cc c nguyn dng. Khi , n l s chnh phng. 3/ Chng minh rng bnh phng ca mt s chn l mt s chn bng : a. Chng minh trc tip b. Chng minh gin tip c. Chng minh phn chng 4/ Chng minh rng tch ca 2 s hu t l mt s hu t. 5/ Chng minh rng mt s nguyn khng chia ht cho 5 th bnh phng ca n khi chia cho 5 s d 1 hoc 4. 6/ Chng minh rng nu n l s nguyn dng khi n l l nu v ch nu 5n + 6 l l. 7/ C 2 gi thit - Mn logic l kh hoc khng c nhiu sinh vin thch mn logic. - Nu mn ton l d thi logic l khng kh. Bng cch chuyn cc gi thit trn thnh cc mnh cha cc bin v cc ton t logic. Hy xc nh xem mi mt trong cc khng nh sau l cc kt lun c c s ca cc gi thit cho khng : Trang 45

43. Chng 2: Suy lun ton hc & Cc phng php chng minh a/ Mn ton l khng d nu nhiu sinh vin thch mn logic. b/ Khng c nhiu sinh vin thch mn logic nu mn ton l khng d. c/ Mn ton l d hoc mn logic l kh. d/ Mn logic l khng kh hoc mn ton l khng d. e/ Nu khng c nhiu sinh vin thch mn logic khi hoc l mn ton khng d hoc l logic khng kh. 8/ Dng nguyn l qui np yu, chng minh cc biu thc tng sau : n n(n + 1)(n + 2) a. i i =1 2 = 6 n n(n + 1)(n + 2)(n + 3) b. i(i + 1)(i + 2) = i =1 4 n c. i(i)!= (n + 1)! - 1 i =1 n i 1 d. (i + 1) = 1 (n + 1)! i =1 n 1 n(n + 3) e. (i + 1)(i + 2) = 4(n + 1)(n + 2) i =1 n f. i.2 i =1 i = 2 + (n 1).2 n+1 n g. 2.3 i =1 i 1 = 3n 1 n n(n + 1)(2n + 7) h. i(i + 2) = i =1 6 9. Tm cng thc tnh cc tng sau v s dng nguyn l qui np chng minh cng thc va tm c n a. (2i 1) i =1 n b. 2 i =1 i 1 n c. i(3i 1) i =1 Trang 46

44. Chng 2: Suy lun ton hc & Cc phng php chng minh n 1 d. i(i + 1) i =1 n e. (2i 1) i =1 2 n f. i(i + 1) i =1 n g. x i =1 i 10. Dng nguyn l qui np mnh, chng minh cc bt ng thc sau: a. n > 3 : 2n < n! b. n > 4 : n2 < 2n c. n > 9 : n2 < 2n d. n >= 6 : 4n < n2 - 7 e. n > 10 : n - 2 < (n2 - n)/12 Trang 47

45. Chng 2: Suy lun ton hc & Cc phng php chng minh CHNG 2 : SUY LUN TON HC &..................................................................28 CC PHNG PHP CHNG MINH.......................................................................28 2.1. Tng quan .......................................................................................................28 2.2. Suy lun ton hc............................................................................................29 2.2.1. Khi nim ................................................................................................29 2.2.2. Cc qui tc suy lun ................................................................................29 2.3. Cc phng php chng minh........................................................................31 2.3.1. Chng minh rng ( P l sai) ....................................................................32 2.3.2. Chng minh tm thng (Q l ng) ......................................................33 2.3.3. Chng minh trc tip ..............................................................................33 2.3.4. Chng minh gin tip ..............................................................................34 2.3.5. Chng minh phn chng .........................................................................36 2.3.6. Chng minh qui np ................................................................................37 2.4. Tng kt chng 2 ..........................................................................................44 2.5. Bi tp chng 2.............................................................................................44 Trang 48

46. Chng 3: V t v lng t CHNG 3 : V T V LNG T 3.1. Tng quan Mc tiu ca chng 3 Hc xong chng ny, sinh vin phi nm bt c cc vn sau: - Th no l v t, khng gian ca v t, trng lng ca v t. - Th no l lng t, lng t tn ti, lng t vi mi. - Cch biu din mt cu thng thng thnh biu thc logic. Kin thc c bn cn thit Cc kin thc c bn trong chng ny bao gm: - Cc php ton i s, hnh hc c bn xc nh c gi tr ng, sai ca cc pht biu. - C kh nng suy lun. - Nm vng cc php ton logic trong chng 1. Ti liu tham kho Phm vn Thiu, ng Hu Thnh. Ton ri rc ng dng trong tin hc. Nh xut bn Khoa hc v K thut, H Ni - 1997 (chng 1.3, trang 32 - 52). Ni dung ct li - nh ngha v t, khng gian ca v t, trng lng ca v t. - nh ngha lng t, lng t vi mi, lng t tn ti. - Dch cc cu thng thng thnh biu thc logic. 3.2. Cc nh ngha Trong ton hc hay trong chng trnh ca my tnh, chng ta thng gp nhng cu c cha cc bin nh sau : "x>3", "x=y+3", "x+y=z"... Cc cu ny khng ng cng khng sai v cc bin cha c gn cho nhng gi tr xc nh. Trong chng ny, chng ta s xem xt cch to ra nhng mnh t nhng cu nh vy. Trang: 48

47. Chng 3: V t v lng t 3.2.1. nh ngha v t (Prdicat) Mt v t l mt khng nh P(x,y,...) trong c cha mt s bin x,y,... ly gi tr trong nhng tp hp A,B,... cho trc, sao cho : - Bn thn P(x,y,...) khng phi l mnh . - Nu thay x, y ,... bng nhng gi tr c th thuc tp hp A, B,... cho trc ta s c mt mnh P(x, y, ...), ngha l khi chn tr ca P(x, y,...) hon ton xc nh. Cc bin x, y,... c gi l cc bin t do ca v t. V d 1: Cc cu c lin quan n cc bin nh: "x>3", "x + y = 5" rt thng gp trong ton hc v trong cc chng trnh ca my tnh. Cc cu ny khng ng cng khng sai v cc bin cha c cho nhng gi tr xc nh. Ni cch khc, v t c th xem l mt hm mnh c nhiu bin hoc khng c bin no, n c th ng hoc sai ty thuc vo gi tr ca bin v lp lun ca v t. V d 2: Cu {n l chn} l mt v t. Nhng, khi cho n l mt s c th l chn hay l l ta c mt mnh : n = 2 :{2 l chn}: mnh ng. n = 5 :{5 l chn}: mnh sai. V t {n l chn} c 2 phn. Phn th nht l bin x l ch ng ca cu. Phn th hai "l chn" cng c gi l v t, n cho bit tnh cht m ch ng c th c. K hiu: P(n) = {n l chn} Tng qut, ngi ta ni P(n) l gi tr ca hm mnh P ti n. Mt khi bin n c gn tr th P(n) l mt mnh . V d 3: Cho v t P(x) = {x>3}. Xc nh chn tr ca P(4) v P(2). Gii: P(4) = {4>3} : mnh ng. P(2) = {2>3} : mnh sai. 3.2.2. Khng gian ca v t (Prdi cat) Ngi ta c th xem v t nh l mt nh x P, vi mi phn t x thuc tp hp E ta c mt nh P(x){, 1}. Tp hp E ny c gi l khng gian ca v t. Khng gian ny s ch r cc gi tr kh d ca bin x lm cho P(x) tr thnh mnh ng hoc sai. Trang: 49

48. Chng 3: V t v lng t 3.2.3. Trng lng ca v t (Prdi cat) Chng ta cng thng gp nhng cu c nhiu bin hn. V t xut hin cng nh mt hm nhiu bin, khi s bin c gi l trng lng ca v t. V d 1: V t P(a,b) = {a + b = 5} l mt v t 2 bin trn khng gian N. Ta ni P c trong lng 2. Trong mt v t P(x1, x2, ..., xn) c trng lng l n. Nu gn gi tr xc nh cho mt bin trong nhiu bin th ta c mt v t mi Q(x1, x2, ... xn) c trng lng l (n-1). Qui lut ny c p dng cho n khi n=1 th ta c mt mnh . Vy, thc cht mnh l mt v t c trng lng l . V d 2: Cho v t P(x, y, z ) = {x + y = z}. Cho x=: Q(y,z) = P(, y, z) = { + y = z} y=: R(z) = Q(, z) = P(, , z) = { + = z} z=: T = P(, , 1) = { + = 1} mnh sai. Cu c dng P(x1, x2, ..., xn) c gi l gi tr ca hm mnh P ti (x1, x2, ..., xn) v P cng c gi l v t. 3.2.4. Php ton v t Php ton v t s dng cc php ton logic mnh v l s m rng ca php ton mnh th hin r hn cc tri thc. V d 1: Cn vit cu "nu hai ngi thch mt ngi th h khng thch nhau" di dng logic v t. Trc khi vit cu trn ta hy tm hiu cc cu n gin c vit nh sau: "Nam thch Mai" c vit theo php ton v t l: thch (Nam, Mai). "ng thch Mai" c vit theo php ton v t l: thch (ng, Mai). Tng qut khng nh trn c vit nh sau: Thch (X, Z) AND thch (Y, Z) NOT thch (X, Y) (Thch (X, Z) thch (Y, Z) thch (X, Y) V d 2: Cho v t "Qu bng mu xanh". Php ton v t cho php m t theo quan h tri thc theo dng: (qu bng, xanh). Cch th hin ny thun tin i vi vic dng bin v hm trong x l tri thc. Trong lnh vc tr tu nhn to, lp trnh trn cc v t ngi ta s dng ngn ng Trang: 50

49. Chng 3: V t v lng t Prolog. l mt ngn ng cp cao c c im gn vi ngn ng t nhin, do ng C.Cameraller (i hc Marseilles, Php) v nhm ng s cho ra i nm 1973. V d: Ta c tam on lun sau: "Ngi ta ai cng cht Socrates l ngi Vy Socrates phi cht" Trong phn ny chng ta khng i su vo ngn ng Prolog (v s hc k mn ngn ng lp trnh) m ch gii thiu cc khi nim trong lp trnh Prolog c s dng cc v t. a) Hng: L mt gi tr xc nh trong khng gian ca v t. cc hng c k hiu bi cc ch thng dng t tn cc i tng c bit hay thuc tnh. b) Bin: Dng th hin cc lp tng qut ca cc i tng hay cc thuc tnh. Bin c vit bng cc k hiu bt u l ch in hoa. Vy c th dng v t c bin th hin cc v t tng t. V d: V t "Qu bng mu xanh" c th vit li: "X mu Y". Qu bng xanh l cc hng c xc nh trong khng gian ca v t. X, Y l bin. c) Cc v t: Mt s kin hay mnh trong php ton v t c chia thnh phn. V t v tham s. Tham s th hin mt hay nhiu i tng ca mnh , cn v t dng khng nh v i tng. V d: Cu "X thch Y" c dng thch (X, Y). Thch l v t cho bit quan h gia cc i tng trong ngoc. i s l cc k hiu thay cho cc i tng ca bi ton. d) Hm: c th hin bng k hiu, cho bit quan h hm s. V d: Hoa l m ca Mai, ng l cha ca Cc. Hoa v ng l bn ca nhau. Ta co hm s c vit th hin quan h ny. M (Mai) = Hoa Cha (Cc) = ng Trang: 51

50. Chng 3: V t v lng t Bn (Hoa, ng) Cc hm c dng trong v t l: Bn (M (Mai), Cha (Cc) 3.3. Cc lng t Khi tt c cc trong mtk hm mnh iu c gn cho mt gi tr xc nh. Ta c chn tr ca hm mnh . Tuy nhin, cn c mt cch khc bin cc v t thnh mnh m ngi ta gi l s lng ha (hay lng t). 3.3.1. Lng t tn ti ( ) Cu xc nh "Tp hp nhng bin x lm cho P(x) l ng khng l tp hp rng" l mt mnh . Hay "Tn ti t nht mt phn t x trong khng gian sao cho P(x) l ng" l mt mnh c gi l lng t tn ti ca P(x). K hiu: x P(x) . 3.3.2. Lng t vi mi ( ) Cu xc nh "Tp hp nhng x lm cho P(x) ng l tt c tp hp E" l mt mnh . Hay "P(x) ng vi mi gi tr x trong khng gian" cng l mt mnh c gi l lng t vi mi ca P(x). K hiu: xP(x) V d: Cho v t P(x) = {s nguyn t nhin x l s chn}. Xt chn tr ca hai mnh xP(x) v xP(x). Gii: x P(x) = {tt c s nguyn t nhin x l s chn} l mnh sai khi x = 5. x P(x) = {hin hu mt s nguyn t nhin x l s chn} l mnh ng khi x = 10. Ch : Cho P l mt v t c khng gian E. Nu E = {e1, e2, ... en}, mnh xP(x) l ng khi tt c cc mnh P(e1), P(e2), ... P(en) l ng. Ngha l x P(x) P(e1) P(e2) ... P(en) l ng. Tng t xP(x) l ng nu c t nht mt trong nhng mnh P(e1), P(e2), ... P(en) l ng. Ngha l xP(x) P(e1) P(e2) ... P(en) l ng. Trang: 52

51. Chng 3: V t v lng t - Nu khng gian E l mt tp trng th xP(x) v xP(x) c chn tr nh th no ? (Sinh vin t gii p). V d: Cho P(a,b) = {cp s nguyn tng ng tha a + b = 5} Hy xc nh chn tr ca cc mnh sau: (a,b) P(a,b) {Tt c cp s nguyn tng ng F (a,b) P(a,b) {Hin hu mt cp s nguyn tng ng (a,b) sao cho a + b V = 5} ba P(a,b) {Hin hu mt cp s nguyn tng ng b sao cho cho mi F s nguyn tng ng a ta c a + b = 5} ab P(a, b) {Mi s nguyn tng ng a, hin hu mt s nguyn tng V ng b sao cho a + b = 5} ab P(a,b) {Hin hu mt cp s nguyn tng ng a sao cho cho mi F s nguyn tng ng b ta c a + b = 5} ba P(a, b) {Mi s nguyn tng ng b, hin hu mt s nguyn tng V ng a sao cho a + b = 5} nh l 1: Cho v t P(a, b) c trng lng l 2. Khi : a) ab P(a,b) v ba P(a, b) l c cng chn tr. Ngha l : ab P(a,b) ba P(a, b) K hiu: (a,b) P(a,b) b) ab P(a,b) v ba P(a, b) l c cng chn tr. Ngha l: ab P(a,b) ba P(a, b) K hiu: (a,b) P(a,b) c) Nu ab P(a,b) l ng th ba P(a,b) cng ng nhng iu ngc li cha ng. Ngha l : ab P(a,b) ba P(a,b) d) Nu ba P(a,b) l ng th ab P(a,b) cng ng nhng iu ngc li cha ng. Ngha l : ba P(a,b) ab P(a,b) Trang: 53

52. Chng 3: V t v lng t nh l 2: 1. ( x P(x)) v x ( P(x) l c cng chn tr. 2. ( x P(x)) v x ( P(x) l c cng chn tr. Gii thch: 1. Ph nh vi x P(x) ni rng tp hp nhng x lm cho P(x) ng khng l tt c tp hp E. Vy ni rng hin hu t nht mt phn t x E m chng P(x) l sai hay ni rng hin hu t nht mt phn t x E m chng P(x) l ng. 2. x P(x) ni rng tp hp nhng x m chng P(x) l ng l tp hp trng. Ngha l, tp hp nhng x m chng P(x) l sai l tp hp E hay khng c phn t no lm P(x) ng. Ta c x ( P(x)). V d: Ph nh ca "Mi s nguyn n l chia chn cho 3" l "Tn ti t nht mt s nguyn n khng chia chn cho 3" - Phng php ng dng. t c ph nh ca mt mnh xy dng bng lin kt ca nhng bin ca vi t vi phng tin nh lng, ngi ta thay th nhng nh lng vi mi bi tn ti , tn ti bi vi vi mi v sau cng thay th v t bng ph nh ca v t . nh l 3: Cho P v Q l hai v t c cng khng gian. 1. Mnh x (P(x) Q(x)) v (x (P(x) x (Q(x)) l c cng chn tr. 2. Nu mnh x (P(x) Q(x)) l ng th ta c mnh : (x P(x)) (xQ(x)) cng ng. 3. Mnh x (P(x) Q(x)) v (xP(x) xQ(x)) l c cng chn tr. 4. Nu mnh x (P(x) Q(x)) l ng th ta c mnh xP(x) xQ(x) l ng, nhng iu ngc li khng lun lun ng. Ch thch: Nu P v Q l hai v t c cng khng gian E. Ta c : - Tp hp A E : Tp hp nhng phn t x thuc E m chng th P(x) l ng. - Tp hp B E: Tp hp nhng phn t x thuc E m chng th Q(x) l ng. Trang: 54

53. Chng 3: V t v lng t Khi ngi ta lu rng, AB l tp hp nhng x thuc E m chng mnh P(x)Q(x) l ng. Trong khi AB l tp hp nhng x ca E m mnh P(x)Q(x) l ng. 3.4. Dch cc cu thng thng thnh biu thc logic Sau khi c gii thiu v cc lng t, chng ta c th biu din c mt tp hp rng ln cc cu thng thng thnh cc biu thc logic. Vic lm ny nhm mc ch loi i nhng iu cha r rng v ngi ta c th s dng cc cu suy lun ny trong vic lp trnh logic v tr tu nhn to. V d 1: Biu din cu "Mi ngi u c chnh xc mt ngi bn tt nht" thnh mt biu thc logic. Gii: Gi s B(x,y) l cu "y l bn tt ca x". dch cu trong v d cn ch B(x,y) mun ni rng i vi mi c nhn x c mt c nhn khc l y sao cho y l bn tt nht ca x, nu z l mt c nhn khc y th z khng phi l bn tt nht ca x. Do , cu trong v d c th dch thnh: x y z [B(x,y) ((z y) B(x, z))] V d 2: Biu din cu: "Nu mt ngi no l ph n v sinh con, th ngi s l m ca mt ngi no khc" thnh mt biu thc logic: Gii: Gi s F(x) = "x l ph n" P(x) = "x sinh con" v M(x,y) = "x l m ca y" V trong v d p dng cho tt c mi ngi nn ta c th vit n thnh biu thc nh sau: x (F(x) P(x)) y M(x,y) V d 3: Xt cc cu sau. Hai cu u tin l tin v cu ba l kt lun. Ton b tp hp 3 cu ny c gi l mt suy l. "Tt c s t H ng u hung d". "Mt s s t H ng khng ung c ph". "Mt s sinh vt hung d khng ung c ph". Gii: Gi P(x)= {x l s t h ng} Q(x)= {x hung d} R(x)= {x ung c ph} Gi s rng khng gian l tp hp ton b cc sinh vt, ta c cch suy din sau: Trang: 55

54. Chng 3: V t v lng t x ( P(x) Q(x) x ( P(x) R(x)) x ( Q(x) R(x)) 3.5. Tng kt chng 3 C mt s iu cn lu trong vic ph nh cc lng t trong nh l 2. V d : Hy xt ph nh ca cu sau y : "Tt c sinh vin trong lp u hc mn Ton ri rc 2" Cu ny chnh l cu s dng lng t vi mi nh sau: xP(x) Trong P(x) = { x hc mn Ton ri rc 2 }. Ph nh ca cu ny l : " Khng phi tt c cc sinh vin trong lp u hc mn Ton ri rc 2". iu ny c ngha l :" C t nht mt sinh vin lp ny cha hc Ton ri rc 2" . y chnh l lng t tn ti ca ph nh hm mnh ban u c vit nh sau : xP(x). Ta c : xP(x) xP(x) xP(x) xP(x) Php ph nh cc lng t c minh ha r hn trong bng ch thch sau: Ph nh Mnh tng Khi no ph nh l Khi no sai ng ng xP(x) xP(x) P(x) sai vi mi x C mt x P(x) l xP(x) xP(x) C mt x P(x) sai ng P(x) ng vi mi x 3.6. Bi tp chng 3 1. Cho 2 v t P(x) xc nh nh sau: P(x) = {x 3} Q(X) = {x+ 1 l s l} Nu khng gian l tp s nguyn, hy xc nh chn tr ca nhng mnh sau: Trang: 56

55. Chng 3: V t v lng t a) P(1) b) Q(1) c) P(3) d) Q(6) e) P(7)Q(7) f) P(3)Q(4) g) P(4) h) (P(-4)Q(-3) i) P(-4) Q(-3) 2. Cc v t P(x), Q(x) c cho nh bi tp 1.