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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(theoﬂhmdsideiulwmuleut2) 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) wPI2) 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) Wowm 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.) tunlath): 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. equivlent to 9(1) v 9(3) v9(5).
0) mu thosetxmd put new round the domain: (IP(—5]VIP(3)VP(l)VvP(l)VP(3)VvP(5))A
mn 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.
gﬂAdnminmisungousmodunommmalmumummmcm."ammonium."
‘ A olthe United State. then an. I: «manly false.
jrmdomdnhnﬂnddemoﬂthnitedStmthenthhhlme.Ifthedomainisthemtofpupikina
ff“ chauifnbe.
"'ﬁMmhtsofaﬂtheUniudStaumchuwhonhstmmisBush,thenthestmmtb
ﬁremanmomwmd Staten Pruidentsﬁhenthentahmenthfahe.
r__' mﬂnddmofunUnItedSMtu,chenthislacemlnlytmltmhedomalnmistsof 'hmMMmMmmulduputthawememtobctalscﬂt‘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)AH(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: 3342(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 WmﬂﬂenmtieVzﬂzJ.
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) «Veal1:). 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 3260:). 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
‘ «as5(2), the negation is Bil3(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
monkimamplcs).
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 nitmags.“ 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. Ratherme. therightrhandeidei
also in» (elnce PL!) VA is true for all .2). 0n the other hand. if P(.r) i ﬁalae forsome x. then bothaida
arefalaaHUmbrengainthetwoaldeaarelgicaﬂyequivalenl. .
b)The1earetvmcaeta. 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 lefthand side In true. “inhuman. the the hemhand side In equivalent to V: P(.t), which ’a equivalent to the righthand 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. leahand side is equivalent to irP(an), which i!
equivalent to the righthand side. 52. a)Thielefake.elneedmaremwvalueeofxthetmalnez>ltrue.
h)Thlria£alae.eineethcreuetwovalueaofxthazmabz3=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/mmwifthuauemmieumlnCheatcry.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, mum1y 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'Qttl) 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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 Winter '09
 Meenakshisundaram

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