ics6bwin11_1.3 - v.” 1 ans roundatlona We.“ SECTION 1.3...

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Unformatted text preview: v...” 1 ans roundatlona: We.“ SECTION 1.3 Meats and Qualifiers 2.a)Thiaiatme,aiocetharelaanalnomuge. b)’l‘hlsiafalae,docatheraianoalnlanou. c)Thioiafaiae.ainoetbemiaooainm. d)“iahtrm,alnoothexeisanainfahe. 4. a) Hm:laatlllcqualto0,oincetlnconditionlaiahe. b) Herez'uatillequalto i,ainoeihemndition'lfalae. c) Thhtimexiaequalto l attheeud,aincetheoonditioniaMaotheatatemmt 1R] lamented. a) Sorneaudem In theachool haavlaited North Dakota. (Alternatiwly. therearxhiaaatudmt in theachoa who has visited North Dakota.) b) Every student in the school has visited North Dakota. (Alternatively. all students in the school have vhlted d) Some Rodent in the school has not visited North Dakota. (Alternatively. there exists a student in the school who has not visitor! North Dakota.) e) Thlslsthenegationofpart (b): Itiamtmiethatm'yatodentintheachoolhuvisitedNorthDahota. (Alternatively, not all students in the admol haw visited North Dakota) 8. Notethat part (b) and part (c) arcnot ibeaoruoflhingaona would nonnallyaay. a) If an animal h a rabbit. then that animal hops. (Alternatively, every rabbit hops.) b) Bvary animal is a rabbit and hope. c) There axbts an animal and; that if it is a rabbit. then it hops. (Note that this is trivially true, satisfied. forexample, by lions, soil is not theaort oithingonowould my.) d) There exists an animal that is a rabbit and hops. (Altamatiwly. some rabbits hop. Alternatiwly, me hopping animals are rabbits.) b) V1:(C(z) V 0(2) V F(.t:)) c) 3.310(2) A Hz) A wD(x)) d) 111': is the negation of part (a): wane“) A 0(2) A F(2)). a) Hate the was; oithme pets can be dlfl’aoent: (320(3))M3: DillMiil? 1:13)). Theme is no harm in using the same dummy variable, but this could also be written, for mole. an (310(2)) A (an 000) A (El: F(z)). 14. I!) Since (—1)3 = —1, this In true. b) Since Q)‘ < G)’. this is hue. c) Since (-3)2 = ((-11?)2 - (—l)’x’ - 3’. we know dint V:I:((—x)2 = 1:2) is true. d) 'I‘wiccupooitivunumber’ulugarthanthenumber, buuhlsinequnlityisnounniornegatlnnumben or 0. Thumb“ V1121 > 1') is fake. 1o. a)true(:r=)/§J b)fnllc(\/-_fisnotamlnumbet) c)tm(the|ofl-hmdsideiulwmuleut2) d)fabe(noumefora:-lorz=0) 18. Bdsteutlal quantifier; are llloe (Injunctions. and animal quantifiers are lllne conjuncflm. Soc ExampIes II and 16. a) Wemttom'tthat ”3:1meth :- inthedomalnmodther P(-2) istrucot P(—l) ismle or P(0) innwot P(l) in true or P(2] istrue. Thusthe answer Is P(—2)vP(-1)VP(0)VP(1]VP(2). The otherpaflsofthisexemisemsimflu. Notethm by Dehhrgau'shmtheupmaioninpm (c) ialogically equlvalemtotheexprendon In part (f),aud Unexpresioninpan (d) ialogicnflyequlvalem loflweupmuion in part. (a). b) P(-2) A P(-l) A P(0) A P(l) A P(2) c) wPI-2) V -nP(—l] v -uP(0) v -P(l) v -P(2] d) fiP(-2) A WP(-l) A HP(0) A HP“) A HP”) a) This is Just. the negation of put. (a): -v(P(—2] v P(-l) v P(0) v P[l) v P(2)) 0 Th isjmt the negation of part (b): -‘(P(-2) A P(—l] A P10) A P[1)A H2» 10. Will quantifies am like dbjunctlom, and universal quantifum am like conjunctions. See Example 11 and m. a) Wow-m 50mm that Pu) lemme forsomc J: inthcdomnin. aoeithet P(-5] Isuueox P(—3) lstmeoc 9(-1) :- we or 9(1) 5 true orP(3) is two or 9(5) "a true. Thus the answer no 9(-5) v 9(—3) v 9(-1) v 9(1)v 9(3)v 9(5). - b) 9(-5) A 9(-3) A 9(-1) A 9(1) A 9(a) A 9(5) "g c) The bum: trunlation a. a Follows: ((—5 ¢ 1) -. 9(—5))A ((—3 at 1) .5 P(-3))A((-l ,4 1) _. 9(-1))A ((1 151).. P(l))A((3 ,e 1) —. 9(3)) A((5 # 1) —. 9(5)). Hm. aincethe [1pr 3 ¢ 1 is false when _. a: I. l and two when 1 in anythingother than 1, we have momaimply P(—5)AP(-3)AP(-1]AP(3)AP(5). ‘ d) mum.) tun-lath): In as follows: ((-5 z o)A9(-5))v((-3 2 0)AP(-3))v((-12 o)A9(-1))v((1 2 0)AP(1)) v((5 2 o) AP(3)}V ((5 z o) A 9(5)). Since only three ofthe x's In the domain meet. the condition. ‘. 0h m I. equiv-lent to 9(1) v 9(3) v9(5). 0) mu thosetxmd put new round the domain: (-IP(—5]V-IP(-3)V-P(-l)V-vP(l)V-P(3)V-vP(5))A m-n A P(-3) A 9(-5)). m i. equivalent to («9(1) v «9(3) v ~9(5)) A (9(—1) A 9(4) A 9(-5)). ~ mmpodbloineachcue. gflAdnm-inmisungousmodunommmalmumummmcm."ammonium." ‘ A olthe United State. then an. I: «manly false. jrmdomdnhnflnddemoflthnitedStmthenthhhlme.Ifthedomainisthemtofpupikina ff“ chauifnbe. "'fiMmhtsofafltheUniudStaumchuwhonhstmmisBush,thenthestmmtb fireman-momwmd Staten Pruidentsfihenthentahmenthfahe. r__' m-flnddmofunUnItedSMtu,chenthislacemlnlytmltmhedomalnmistsof 'hmMMmMmmuldupu-tthawememtobctalscflt‘anotmdwtht “how” warm/hen”). 24. Inorderwdothetnnelntlontheoeoondwny.mletC(z)hethepmpoeitionnlfunetion “zininyou. . lnallofthmgwewllllet Y(;t) betheproponitionnl functionthnt: isinmnreehoolordmunppmpeinte. Notethat brtheeeeond my. wealwnyownnt toueeconditioml etetementewith univennl queuing. conjunctions with eudetentinl quantifies. Iemnd way. . b) Let F(.r) he “I has seen a [amigo movie." Then we have Elzflx) the first way, or 3110(3) A513” (5.) eemnd way. . n) "we let 0(1) be “a: heavieited Uzbeidetnn.’ then we have 3::U(z) lfthcdomaln isjmt your Idtoolmotee, or 33012:) AU(2)) if the domain in all people. lfwe let V(x,y) mean that penon 1: he. wilted country 3, then we can rewrite this lost one as 32011:) A 312:, Uzbekistan». b) If we let 0(1) and P(x) be the propositional [auctions muting that 3 has studied calculu- end C++. respectively. then we have V1(C(:)AP(;I:)) if the domaln iajmt )our adloolmatas. or V:(Y(.r) —. (C(z)AP(z))] ifthedomnin isnll people. llwolet Shay) menuthet person s haeetudiedeubjeot y. then ween rewrite this last one an V3013) —o (8(1. calculus) A S(.r, C+ +))). c) lfwe let 8(1) and Mt'x) be the propositional functions enacting that :: owns a bicycle and a motorcycle. respectively, then we have VJ(H(B(z-)AM(::))) l! the domain hjust your echoohnetea. or V2013) —. ~(B(::)A M(1)))lfthedomnin In all people. Not. that “no one‘ buxom “tor nu not.” lfwe let 0(z.y) moon that person 3 owns item 1:. then We can rewrite this last one as V1011) —o «[O(:. bicycle) A 0(x.motorcyde))). d) If we let 3(2) be “r 5 happy.“ then we have irwlflx) if the domain is jut your acheolmetu, or 32(Y(1)A-H(z)) if the domain Is all people "we let E(z.y) mean that puma :- is In mental state p, then we can mwrite this Int one a 3:011) A -:E(.t. happy”. 1:) If we let Ttx) be '2' m born in the twentieth century,“ then we have V:T(x) i! the dmnnin is just your admolmntee, or V3013) —o T(z)) if the domain is all people. If we let 8(2. p) men that person 2: was born in the y" century, then we can write this last one as Vx(Y(z-) —’ B(2,2o)). Let R(z) be “.r is In the entrant place”; let. Bu) he “a- is in excellent condition"; let th) be “.t in n [or ’ your] tool”; and let the domain ofdineounie be all things. n)'l1teeeatisteeomething not in themed place: 33-42(3). e) V: (3(1) A 8(3)) d) This '3 saying that anything fails to misty the condition: V: -(R(:r) A 8(3)). 0) 'I'lm east: It tool with this propetty: 3:: (T(:) A fillet) A 5(3)). . a) 90.3) v P(2,3) v P(3,3) b) m. 1) A P(l,2)/\ P(l,3) c) ~P(2. n) v -P(2. 2) v -.P(2, 3) d) «Pun» A -P(2. 2) A “P61. 2) a) Let 17(3) he “x has flees.’ondletthedomninofdieeounebedon- WmflflenmtieVzflz-J. Immaonie 3raF(1-). lnEngliahthiemds‘Therebndogthatdoeenothevefleu.” )h‘zcanlddu' wherethsdomsinofdhcoumeishom. 11mourorl¢inalstateumnth3rfi(x). ~.. bv‘w“),[amushthbbreaderedmostsimplyss‘l‘iohorsecanaddf 52:10“) he“: ”climb.” andletthedomainol’dbcoursebelmalas. Our originalstatementisVa‘Ck). ~ .3 3:4(3). InEnglishthisreads Misahoalathatoannotclimb.’ 7(3) Mdtmnpmkfienda." andlotthodomainofdiacoursebemnkeys. Ouroriflnalstatcmant ;.-. N3) «Veal-1:). its ligation is 3317(3). In Engl'ahthhnearh ”Hm'uamonheythatcanqiealt I ,5 , 3(,) he N; canswlm’ and let C(z) be ‘1 can catch ihh.‘ whercthedomainoldiscourscispiy. Then ' 01W statement ls 32(S(:i:) A 0(1)). It: negation is V: -(S(z) A 0(2)). which could abo be written , _ '4“) v -.C(1:)) by De Morgan’s law. In English this is “No pig can both swim and catch fish." or “Every j‘ " mkumbbwwimuhmbhwcatchflsh." "“1 us 8(3) be “t obeys the speed limit.” where the domain of discourse '3 driven. The oriy'nai statement ”:3 I 32-60:). the mgntion ’- V:S(z). “All drivers obey the speed limit.” . 1‘.) m 3(3) be ”a: is m“ where the domain of dbcourse a. Swedish mm. The original am. a ._: ”5(1). the negation is 3:-S(:r). “Some Swedish movia are not serious.” 1:) Let S(:) be “a- can ineep a ascret.‘ where the domain of discourse is people. The original statement is ‘ «as-5(2), the negation is Bil-3(a). “Some people can lneep a sweet.” (I) Let A(~.r) be “at has a good attitude.” where the domain of discourse is people in th'n class. The original mm is 3: HA“). the negation in Va- A(:) . “Everyom. in this clam has a good attitude.” as. a) Since 1’ = l.thisstatementhl'abe; .t- l isaoountermmple. Sois .1: n0 (thasearetheonlytwo monk-imamplcs). b) There are mmuMemmpiss: :r: = y/i and z = —\/5. c) There is one countermampls: z = 0. 38. I) Some system is open b) Every system is either malfunctioning or in a diagnostic state. c) Some system is open. or some system is in a diagnostic state. I!) Some system is unavailable. 0) No system is working. (We could also say “My system is not working." as long as we underflood that filth is dilferent from "Notewerysystem hi working") 40. There are many ways to write these. depending on what we use br predicates. a) Let He) he 'I’herehlemthan :- nugalrymfreaonthelmddisk.” with thsdomainofdlscoursebcing positive numbers. and let W(x) he “Dar :1! is sent a warning nit-mags.“ Then we have H30) -o V: W (.t). I») Let 0(1) be “Directory 1: can he opened.” let C(:) be “Fllc a: can becloaed.’ and let 3 bethe proposition “System errors ham been detected.” Then we have E -9 ((Vx ~0(z)) A (VxHC(m))). c) Let. B bethepropmition “'I'hefllesystcsncnn bebacluedup.” andlet L(:r) be“Uset a: icurrentlylonxed on.” Then we have (32 L(.t)) —o *8. d) [at 0(1) be “Product 1 can beddivered.’ andlet Mm be ‘Thercamatleafl x megnhytcsolmem- ory available” and 8(1) be "The connection speed is at least. a: ldlohits per second.“ wine the domain at discourse for tho last two propositional functions are positive numbers. Then we have (1|!(8) A S(56)) _. D(video on demand). 42. Therearemany waystowfitethmdependingonwhat we Ilsei'or predicata. a) Let Mar) he “User 2 has access to an electronic mailbmr.” Then we have V: Apr). b) Let A(.r.y) be “Group member 1: can mean resource 1:.” and let 511,11) be “System z: is in state y.” Then we hate S(iile systemJoclocd) —. V: A(a:,systun mailbox). 10 Chapter 1 The Fbundafiam: Logic Ind d)LetT(1']he‘l‘helhrwshputhatleenxkbpe,‘wherethedomalnofdhcouraelspodtin - let M(::.y) be “Raouree 1 la in mode y." and let 8mm} be “Rainer .r is In state y.” Then We (mm) A «71500) A -M(pmxy unendiamtic» —v 31:8(32, normal). 44. “hmmitioual functlone Panthhatareeometlmea. hutnotalweyedmehothauln that A iel‘ahe. HP“) himebrall 1:.thmtheleflphandeldeiatnle. Rather-me. therightrhandeidei also in» (elnce PL!) VA is true for all .2). 0n the other hand. if P(.r) i fialae for-some x. then bothaida arefalaaHUmbrengainthetwoaldeaarelgicaflyequivalenl. . b)The1earetvmcaet-a. lfA istmeJhen [3rP(x))VA litmeumdeince P(x)VA lstnneforaomeheally all) .z. 31-(P(.r)VA) is also true. Thus both aidenof the logical equivalence aretme (hence equivalent). Now suppose that A Is false. If P(::) is true for at least one 1-. then the left-hand side In true. “inhuman. the the hem-hand side In equivalent to V: P(.t), which ’a equivalent to the right-hand side. b) Therearetwoeeeu. HA isfalae. then botheideeol'theequlvalencearetrue, becaueeacondltlonal mm with a felse hypothesis In true (and we are assuming that. the domain in nonempty). If A In true, then A —. P(;e) 8 equivalent to Plx) for each x. no the. lea-hand side is equivalent to irP(an), which i! equivalent to the right-hand side. 52. a)Thielefake.elneedmaremwvalueeofxthetmalnez>ltrue. h)Thlria£alae.eineethcreuetwovalueaofxthazmabz-3=l true. c)“Isletruemincebyalgebmweaeethatlheunlqueaolutloubotheequationle1=3. d) Thiahhhgeineethereaeenovalueeofxthatmakex=x+l true. 54. “new. are only three can In which Ball’tx) is true. so we tom the disjuncllon oftheae three calm. The answer is thus (PO) A -.P(2) A «P(3))v(~P(1)/\ P(2) A «PG!» v («Pm A -P(2) A P(a)). so. APmbgquezyreturmayes/mmwifthuauemmieumlnCheat-cry.anditretumthemmtm mahethequuy trueifthereere. Section 1.4 Nested Quantifim 17 n) Noneofthcfactawasthat Kcvinmenrollulin 32222 Sothelespomeisno. h) One of the fact: was that Kiko was enrolled in Math 273. So the mponaa 'u you. c) Prolog returns the mum of the comes for which Grumman h the instructor, namely just c3301. d) Prolog returns the name of the instmctor for CS 301. namely groom. e) Prolog returns the nuns of the Instructors teaching any course. that Kevin is enrolled in, mum-1y chm, since Chan is the lnstmctor in Math 273. theonly mum Kevin is enrolled in. Rillowing the iden and syntax of Exmnple 28. we haw the folbwlng rule: “Mantra,” :- tubal-(1.2) . father(2,Y): “theta,” . moms-(2.0. Note that we used the comma to mean “and" and the mloolon to mean “or.” Pb! X to be the yandhthe: of Y, I must be eithet Y’s lnthu‘a fatheror Y's mother’s latlm. - a) w(PU?) -° 0(1)) b) 3103(1) A'Qt-tl) c) 3r(R(l)/\ “Phil (1) Yes. The uusatinhictory excuse guaranteed by part (b) cannot be n clan: explanation by part. (a). a) “(PM -' "-511” 5) “(3(1) -° 3(3)] c) VIIQlt) - PM) 6) V4100?) -‘ “1“!» 0) Yes. If a: '5 one 0! my poultry. then he is a duck (by part (c)). hence not willing to waltz (part (3)). Since oliim me always willing to waltz (put (b)], a‘ is not an oilioer. ...
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