TON RI RC

TON RI RCNguyn Thnh Nhthttps://sites.google.com/site/nhutclass/toanroiracLOGO1Ni dung: gm 5 phn C s logic Quan h Php m Hm Bool th2Thang imSa bi tp 30%Kim tra gia k 20%Thi cui k 50%Ti liuSlides bi ging.Gio trnh:Ton ri rc, Nguyn Hu Anh.Ton ri rc nng cao, Trn Ngc Danh.Discrete Mathematics and its applications, Kenneth H. Rosen.Chng I: C s logicNi dung:Mnh Dng mnh Qui tc suy dinV t, lng tQui np ton hcI. Mnh 1. nh ngha: Mnh l mt khng nh c gi tr chn l xc nh, ng hoc sai. Cu hi, cu cm thn, mnh lnh khng l mnh .

V d: Mt tri quay quanh tri t.1+1 = 2.Hm nay tri p qu! (khng l mnh )Hc bi i! (khng l mnh )3 l s chn phi khng? (khng l mnh ) I. Mnh K hiu: ngi ta dng cc k hiu P, Q, R ch mnh .

Chn tr ca mnh : Mt mnh ch c th ng hoc sai, khng th ng thi va ng va sai. Khi mnh P ng ta ni P c chn tr ng, ngc li ta ni P c chn tr sai. Chn tr ng v chn tr sai s c k hiu ln lt l 1 (hay ,T) v 0 (hay S,F)

Bi tp lm ngayKim tra cc khng nh sau c phi l mnh khng?

Paris l thnh ph ca M.n l s t nhin.con nh ai m xinh th!3 l s nguyn t.Ton ri rc l mn bt buc ca ngnh Tin hc.Bn c khe khng?x2 +1 lun dng.

I. Mnh 2. Phn loi: gm 2 loia. Mnh s cp (nguyn thy): thng l mt mnh khng nh n.b. Mnh phc hp: l mnh c xy dng t cc mnh s cp nh lin kt bng cc lin t (v, hay, khi v ch khi,) hoc trng t khng.V d: 2 khng l s nguyn t 2 l s nguyn t (s cp) Nu 3>4 th tri ma An ang xem phim hay An ang hc bi Hm nay tri p v 1 +1 =3

I. Mnh 3. Cc php ton: c 5 php ton a. Php ph nh: ph nh ca mnh P c k hiu l P hay (c l khng P hay ph nh ca P). Bng chn tr :

V d :+ 2 l s nguyn t Ph nh: 2 khng l s nguyn t+ 1 > 2 Ph nh : 1 2

I. Mnh b. Php ni lin (hi, giao): ca hai mnh P, Q c k hiu bi P Q (c l P v Q), l mnh c nh bi : P Q ng khi v ch khi P v Q ng thi ng.

Bng chn tr

V d: - 3 > 4 v Trn Hng o l mt v tng - 2 l s nguyn t v 2 l s chn - An ang ht v ung nc I. Mnh c. Php ni ri (tuyn, hp): ca hai mnh P, Q c k hiu bi P Q (c l P hay Q), l mnh c nh bi : P Q sai khi v ch khi P v Q ng thi sai.

Bng chn tr

V d: - p > 4 hay p > 5 - 2 l s nguyn t hay 2 l s chnI. Mnh V dHm nay, An gip m lau nh v ra chnHm nay, c y p v thng minh Ba ang c bo hay xem phim

I. Mnh d. Php ko theo: Mnh P ko theo Q ca hai mnh P v Q, k hiu bi P Q (c l P ko theo Q hay Nu P th Q hay P l iu kin ca Q hay Q l iu kin cn ca P) l mnh c nh bi: P Q sai khi v ch khi P ng m Q sai. Bng chn tr

I. Mnh V d: Nu 1 = 2 th Lenin l ngi Vit Nam Nu tri t quay quanh mt tri th 1 + 3 = 5 p > 4 ko theo 5 > 6 p < 4 th tri ma Nu 2 + 1 = 0 th ti l ch tch nc I. Mnh e. Php ko theo hai chiu: Mnh P ko theo Q v ngc li ca hai mnh P v Q, k hiu bi P Q (c l P nu v ch nu Q hay P khi v ch khi Q hay P l iu kin cn v ca Q hay P tng ng vi Q), l mnh xc nh bi: P Q ng khi v ch khi P v Q c cng chn tr Bng chn tr

I. Mnh V d: 2=4 khi v ch khi 2+1=0 6 chia ht cho 3 khi v chi khi 6 chia ht cho 2 London l thnh ph nc Anh nu v ch nu thnh ph HCM l th ca VN p>4 l iu kin cn v ca 5>6 Bi tpTi lp: 1, 2, 4ab, 5

V nh: 3, 4cde, 6, 7, 8, 9II. Dng mnh 1. nh ngha: Dng mnh l mt biu thc c cu to t: - Cc hng mnh - Cc bin mnh p, q, r, , tc l cc bin ly gi tr l cc mnh no - Cc php ton , , , , v du ng m ngoc ().Dng mnh c gi l hng ng nu n lun ly gi tr 1

Dng mnh gi l hng sai (hay mu thun) nu n lun ly gi tr 0.V d: E(p,q) = (p q) F(p,q,r) = (p q) (q r) II. Dng mnh Bng chn tr ca dng mnh E(p,q,r): l bng ghi tt c cc trng hp chn tr c th xy ra i vi dng mnh E theo chn tr ca cc bin mnh p, q, r. Nu c n bin, bng ny s c 2n dng, cha k dng tiu .V d: E(p,q,r) =(p q) r . Ta c bng chn tr sauII. Dng mnh pqrpq(p q) r0000100101010100111110010101111101011111Mnh E(p,q,r) =(p q) r theo 3 bin p,q,r c bng chn tr sau II. Dng mnh Bi tp: Lp bng chn tr ca nhng dng mnh sau

E(p,q) = (p q) p

F(p,q,r) = p (q r) q

u tin cc php tonNgoc ()Ph nhVHayKo theo Ko theo hai chiuV d:p q rhiu l (p q) r p (q r) q hiu l (p (q r)) (q)

Bi tpTi lp: 11ab, 12ab, 13abc

V nh: 10, 11, 12, 13Tng ng logicII.2 Tng ng logicnh ngha: Hai dng mnh E v F c gi l tng ng logic nu chng c cng bng chn tr.K hiu E F (hay E F). V d (p q) p q

nh l: Hai dng mnh E v F tng ng vi nhau khi v ch khi EF l hng ng.Tng ng logic2. Lut De Morgan (p q) p q (p q) p q

3. Lut giao hon p q q p p q q p

4. Lut kt hp (p q) r p (q r)(p q) r p (q r)

Cc lut logicPh nh ca ph nh p p

Tng ng logic5. Lut phn phi p (q r) (p q) (p r)p (q r) (p q) (p r)

6. Lut ly ng p p p p p p

7. Lut trung ha p 0 p p 1 p

Tng ng logic8. Lut v phn t b p p 0 p p 1

9. Lut thng tr p 0 0 p 1 1

10. Lut hp th p (p q) p p (p q) p

30Tng ng logic11. Lut v php ko theo: p q p q q p

V d: Nu tri ma th ng trn nu ng khng trn th tri khng ma Bi tp: Cho p, q, r l cc bin mnh . Chng minh rng:

(p r) (q r) (p q) r 31Gii (p r) (q r) ( p r ) ( q r) (lut 11. v php ko theo) ( p q ) r (lut phn phi) ( p q ) r (De Morgan) ( p q ) r (lut 11. v php ko theo) ( p q ) r (lut 11. v php ko theo)32Php chng minh o ng dng lut v php ko theop q q p CM p q ng, ta CM q p ng.V d:Cho n l s t nhin. CM nu n2 l s chn th n l s chn.Ta CM nu n l s l th n2 l s l.

Php chng minh phn v dng dng lut v php ko theo kt hp lut De Morganp q p q (p q) p q. CM p q sai, ta CM p ng, q sai.Phn v d = trng hp lm M saiV d:Cho n l s t nhin. Nu n2 chia ht cho 4 th n cng chia ht cho 4. CM pht biu trn sai ta tm 1 s n no khng tho. (chng hn n = 6).

Php chng minh phn chng CM p ng ta CM nu p sai th suy ra iu v l hay mu thun.VD:CM cn bc hai ca 2 l s v t.Gii:Gi s cn 2 l s hu t, tc l 21/2 = m/n (dng ti gin) vi m,n l cc s nguyn v UCLN(m,n)=1.(m/n)2 = 2. Hay m2 = 2n2. Nn m chnKhi m=2k. Suy ra n2 = 2k2. Nn n cng chn.Nh vy UCLN(m,n)>1 (mu thun).

Bi tpTi lp: 14a, 15a, 16ab

V nh: 14b

Bi tp v nhc li slide bi ging v chng lin quan trong gio trnh [1], [3].Lm bi tp lin quan cn li trong gio trnhP(P

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