Section5.3:Disproofs,AlgebraicProofsandBooleanAlgebrasInthissectionwes - PDF document

Section5.3:Disproofs,AlgebraicProofsandBooleanAlgebrasInthissectionweshallconsiderhowtoproveanddisprovecertainstatementsaboutsetsusingalgebraicstyleproofsanddirectproofs.InadditionweshallintroducetheideaofaBooleanalgebra.Specically,theoperationswehavedenedonsetsareinmanyrespectsverysim-ilartothelogicalconnectiveoperations.Itturnsoutthatthereisacommonunderlyingstructuretobothoftheseideas(whichisalsoanunderlyingstructuretomanyotherconceptswealreadyknow),andwecallthisstructureaBooleanalgebra.1.DisprovingSetPropertiesNotethatasetpropertyisastatementwhichisclaimedtobetrueofallsets.Aswediscussedearlier,toshowthatsuchaclaimisnottrue,weneedtoexhibitasinglecounterexample.Therefore,toshowthatagivensetpropertyisnottrue,weneedtobuildaset(s)whichdonotsatisfythestatedproperty.Weillustratewithanexample.Example1.1.Showthat(A\B)[C=A\(B[C)isnotasetidentity.Toshowthatthisisnotasetidentity,weneedtoprovideacounterex-amplei.e.constructsetsA,BandCwhichdonotsatisfythisidentity.NotethatA\(B[C)willbeasubsetofAwhereas(A\B)[CwillhaveCasasubset.Therefore,ifwechooseCtocontainanyelementswhichAdoesnotcontain,thenthesewillbedierentsets.Specically,LetA=f1;2;3g,B=f2;3;4gandC=f4;5g.Then(A\B)[C=f2;3;4;5gandA\(B[C)=f2;3g:AnalternativewaytoprovethisisthroughtheuseofVenndiagrams.Specically,weshadeintheregionscorrespondingtobothsidesoftheidentityandcheckwhetherwehavethesameVenndiagramineachcase: C AB C AB (A\B)[CA\(B[C)1 22.SetProblemSolvingIngeneral,aswithanyprobleminmathematics,thereareacoupleofgeneralstepsonewouldtaketosolveit.Theseare:(i)Askyourselfiftheproblemstatementgivenseemstrue-testwithacoupleofexamplestogetagutfeelingorattemptaVennDiagramifthenumberofsetsissmall.Ifthereisanydoubt,thenstartlookingforacounterexample.(ii)Ifyouaresurethestatementistrue,thenstarttoconstructasetproof-eitherdirectly,indirectlyorwithoneoftheothermethodsofproofwediscussed.Wehavealreadyseenanexampleofhowtodisproveasetidentity,soweshallinsteadconsidersomeexamplesofhowtoprovesetidenti-ties.First,aswedidintheprevioussection,wecanusestandardsetidentitiestoderivenewsetidentities.Example2.1.ShowthatforallsetsA,BandC,A[(BA)=A[B.WehaveA[(BA)=A[(B\Ac)=(A[B)\(A[Ac)bythesetdierenceandDeMorganslaw.Next,usingthecomplementlawsandtheidentitylaws,wehave(A[B)\(A[Ac)=(A[B)\(U)=A[Bprovingtheidentity.Example2.2.ShowthatforallsetsA,BandC,(AB)\(CB)=A(B[Cc).Wehave(AB)\(CB)=(A\Bc)\(C\Bc)bythesetdierencelaw.Next,usingthedistributiveandidempotentlaws,wehave(A\Bc)\(C\Bc)=(A\C)\(Bc\Bc)=(A\C)\Bc=A\(C\Bc):Finally,usingthesetdierencelaw,DeMorganslawandthedoublecomplementlaw,wehaveA\(C\Bc)=A(C\Bc)c=A(Cc[B)=A(B[Cc):Inadditiontothesealgebraicstyleproofs,wecanuseothermethodsofprooftoprovefactsaboutsets.Weillustratewithaclassicalresultfromsettheory.Theorem2.3.Forallintegersn�0,ifasetXhasnelements,thenP(X)has2nelements. 3Proof.Weshallprovethisbyinduction.n=0(basecase):ifXhas0elements,thenitistheemptyset.SinceP(X)isthepowersetofX,itconsistsofallthesubsetsofX.Theemptysetisalwaysasubsetofanyset.Therearenoothersubsets(sincetherearenoelementstoformsubsetswith).HenceP(X)has1elementand1=20,sothebasecaseholds.(Inductionstep)Weassumetheresultholdsforannelementsetandproveitforasetwithnelements.SupposethatXhasn+1elementsandz2X.ObservethatwecanbreakupP(X)intotwodistinctsubsets-allsubsetsincludingzandallsubsetswhichdonotincludez.NoticealsothatthisdividesthesetP(X)exactlyinto2.Toseethis,weobservethatwecansetupaone-to-onecorrespondencebetweensuchsets-specically,wecanassociateanygivensubsetwhichdoesnotcontainztothesubsetwiththesameelementstogetherwithzi.e.fx1;x2;:::;xrg$fx1;x2;:::;xr;zgTherefore,thesizeofP(X)willbeequaltotwicethatofthenumberofsubsetsofXi.e.thesizeofthesetP(Xfzg).However,thesetXfzghassizen,sobytheinductionhypothesis,P(Xfzg)has2nelements.ItfollowsthatP(X)has2
2n=2n+1elements.3.BooleanAlgebrasAsobservedpreviously,operationsonsetsareverysimilartologicaloperators.Thisisbecausetheabstractunderlyingstructureofthesemathematicalobjectsarethesame.Inthissection,weshallexplorethisabstractunderlyingstructureandshowthattheresultswehaveprovedwillgeneralizetoallmathematicalobjectswiththisunderlyingstructure.Westartwithaformaldenition.Denition3.1.ABooleanalgebraisasetBtogetherwithtwoop-erations,usuallydenoted+and
suchthatforalla;b2B,a+banda
bareinB,andthefollowingpropertieshold:(i)(CommutativeLaws)Foralla;b2Bwehave(a)a+b=b+a(b)a
b=b
a(ii)(AssociativeLaws)Foralla;b;c2Bwehave(a)(a+b)+c=a+(b+c)(b)(a
b)
c=a
(b
c)(iii)(DistributiveLaws)Foralla;b;c2Bwehave(a)a+(b
c)=(a+b)
(a+c)(b)a
(b+c)=(a
b)+(a
c)(iv)(IdentityLaws)Thereexistselements0and1inBsuchthatforalla2B: 4(a)a+0=a(b)a
1=a(v)(ComplementLaws)Foreacha2B,thereexistsanelementa2Bcalledthecomplement,ornegationofasuchthat:(a)a+a=0(b)a
a=1Justaswesaywithbothlogicaloperatorsandsetoperations,therearemanyfurtherlawswhichcanbederivedsuchasDeMorganslaws,andcomplementlaws(seethetext,Theorem5.3.2onpage288).WenishbyillustratinghowtoprovesuchlawsforanabstractBooleanalgebra.Theorem3.2.(AdditiveIdempotentLaw)ForallainaBooleanal-gebraB,a+a=aProof.Wehavea=a+0=a+a
a=(a+a)
(a+a)=(a+a)
1=a+aTheorem3.3.(AdditiveAbsorptionLaw)ForallaandbinaBooleanalgebraB,(a+b)
a=aProof.Wehave(a+b)
a=(a+b)
(a+0)=(a+b)
(a+bb)=a+b
(b
b)=a+(b
b)
b=a+b
b=a+0=aHomework(i)Fromthebook,pages290-293:Questions:2,6,7,13,17,19,20,23,28,31,36,39,43,49,51,52