s Aristotelian syllogistic as a subsystem of first - order logic

14 Pages·2015·0.42 MB·English

Table ignored!
Corcoran quoteright s .. Aristotelian .. syllogistic
as .. a .. subsystem .. of first hyphen order .. logic
\vspace∗{3ex}\centerline{\framebox[.5\ linewidth]{\textbf{Table ignored!}}}\vspace∗{3ex}
Edgar J period .. Andrade
Universidad del Rosario comma Bogot acute-a
Edward .. Becerra
Universidad Nacional de Colombia comma Bogot acute-a
\centerline{Corcoran ’ s \quad Aristotelian \quad syllogistic }
Abstract period .. Aristotelian syllogistic has been formalized for some time now by
Table ignored!
means of a natural deduction system comma called D by John Corcoran period In a classical
\centerline{as \quad a \quad subsystem \quad of first − order \quad logic }
paper comma Corcoran proves a completeness theorem for such a system period .. His proof
involves the use ofCa roedrucceod sryastenm c’omsma calAledrRiDsctoommtea lthiaatnis easierstoyhlalnodlge ainsdtic
\centerline{Edgar J . \quad Andrade }
turns out to bae esquivaleant to DspueribodsTyhesqtueesmtion remaoinfs cfiomrmsatho-wevoerrcdomemra whetlhoergRiDcis
in fact the easiest such system that is equivalent to D period In this paper we answer
Edgar J . Andrade
\centerline{Universidad del Rosario , Bogot $ \acute{a} $ }
this question comma but raise some more comma by embedding system RD in first hyphen order
Universidad del Rosario , Bogot a´
predicate logic period
Edward Becerra
\centerline{Edward \quad Becerra }
Keywords period .. Syllogistic comma Aristotle comma Corcoran comma Completeness comma Minimality
Universidad Nacional de Colombia , Bogot a´
comma First hyphen
\centerline{UniveArbssitdraacdt .NaciAornisatlotedlieanCsyollloogmistbiciahas, bBeeongfoortmal$ize\dafocrustoem{eat}ime$now}by
Order Subsystem period
means of a natural deduction system , called D by John Corcoran . In a classical
2000 Mathematics Subject Classification period Primary : 3 B 80 period Secondary : 3 B 20 period
\centerline{Abstrpaacpter ,.C\oqrcuoaradn pArorviesstaoctoemlpilaetnenesssytlhleoorgemisftoircsuchhaassybseteemn .formHiaslpirzoeofd for some time now by }
Resumen period .. La silog acute-dotlessi stica aristot e-acute lica ha sido formalizada hace ya cierto tiempo
involves the use of a reduced system , called RD , that is easier to handle and
por medio de un sistema de deducci acute-o n natural comma llamado D por John Corcoran period
\centerline{meanstuornfs oautntoatbuereaqulivadleendtutcotDio.nThseyqsuteesmtion,remcaailnlse,dhoDwevbeyr ,JwohhentheCr RoDrciosran . In a classical }
En un art acute-dotlessi culo cl a-acute sico comma Corcoran demuestra un teorema de completitud para
in fact the easiest such system that is equivalent to D . In this paper we answer
dicho sistema period Su demostraci acute-o n involucra el uso de un sistema reducido comma llamado
\centerline{paper t,hisCqourecsotiroann, buptrroaviseessoamecmoomrep,lbeyteemnebsedsditnhgesoysrteemm RfDorin fisruscth- oradesrystem . \quad His proof }
RDcommaqueesmacute-asfa-acutecildemanejaryresultaserequivalenteaDperiodElproblemasigue
predicate logic .
siendo comma sin embargo comma si RD es de hecho el sistema m acute-a s sencillo que es equivalente
\ceanDteprerliiondeE{ninesvteoKalervytewaoscrudstteh.-edotulesSseyslilocogufilsotiarce,srpAeordnisudtoectreleedm,oCssyoersscttoaermapnre,g,CuoncmtaapllcelotemednmesRasDp,eMr,oincitmrheaaalritteym,iosFsiorsettraa-ssimera-atcoutehandle and }
Order Subsystem .
s comma
\ceanl itnecrrulistnaer{eltusirstnesm20ao00RuMtDaetthnoemlabatelicasceSuqutbuej-eiocvtgaCilclaeasnsditfiecapttriooendDi.cad.PosrTimdheaerpyrqi:mu3eeBrsto8ir0odn.enSerpceeomrnidoaadirnys: 3,Bh2o0w. ever , whether RD is }
Resumen . La silog ´ı stica aristot e´lica ha sido formalizada hace ya cierto tiempo
1 period .. Introduction
\centerline{in fapcotr mtehdeio deeausniesissttemsaudcehdesdyucsctiemo´ntnhaatutrali,sllaemqaudoivDalpeonrtJohtnoCDorco.raInn. this paper we answer }
Intheclassicalpaperopensquarebracket3closingsquarebracketcommaCorcorandefinestwologicalsystems
En un art ´ı culo cl a´ sico , Corcoran demuestra un teorema de completitud para
which he calls D
\centerline{thisdicqhuoessisttieomna . ,Subduetmorstaraicsieo´snominveolumcorareel u,sobdye uenmsbisetdemdainrgeduscyidsote,mllamRaDdoin first − order }
and RD period The former emdash system D emdash is intended as a formalization of Aristotle quoteright s
RD , que es m a´ s f a´ cil de manej ar y resulta ser equivalente a D . El problema sigue
syllogistic within the framework of modern logic period Aristotelian syllogistic studies
\centerline{predsiiecnadtoe, silnoegmibcarg.o }, si RD es de hecho el sistema m a´ s sencillo que es equivalente
the necessity relation between two open parenthesis general non hyphen empty closing parenthesis terms
a D . En este art ´ı culo responderemos esta pregunta , pero crearemos otras m a´ s ,
comma S and P comma that
\centerline{Keywordsal.inc\ruqsutaardelSsisytlemloagRiDsteincla,l o´Agriciastdoetplreedic,adCosodrecoprriamner o,rdCenom. pleteness , Minimality , First − }
67
1 . Introduction
\centerline{OrIndetrheSculabsssiycsatlepmape.r [}3 ] , Corcoran defines two logical systems which he calls D
and RD . The former — system D — is intended as a formalization of Aristotle ’ s
\centerline{20s0y0llogMisatitchwemithaitnictsheSfruabmjeewcotrkCoflamsosdiefrincalotgiicon. A.risPtortiemliaanrysyl:log3isBtic8s0tud.iesStehceondary : 3 B 20 . }
necessity relation between two ( general non - empty ) terms ,S and P, that
\centerline{Resumen . \quad La silog $ \acu6t7e{\imath} $ stica aristot $ \acute{e} $
lica ha sido formalizada hace ya cierto tiempo }
\centerline{por medio de un sistema de deducci $ \acute{o} $ n natural , llamado D por John Corcoran . }
\centerline{En un art $ \acute{\imath} $ culo cl $ \acute{a} $ sico , Corcoran demuestra un teorema de completitud para }
\centerline{dicho sistema . Su demostraci $ \acute{o} $ n involucra el uso de un sistema reducido , llamado }
\centerline{RD , que es m $ \acute{a} $ s f $ \acute{a} $ cil de manej ar y resulta ser equivalente a D . El problema sigue }
\centerline{siendo , sin embargo , si RD es de hecho el sistema m $ \acute{a} $
s sencillo que es equivalente }
\centerline{a D . En este art $ \acute{\imath} $ culo responderemos esta pregunta , pero crearemos otras m
$ \acute{a} $ s , }
\centerline{al incrustar el sistema RD en la l $ \acute{o} $ gica de predicados de primer orden . }
\centerline{1 . \quad Introduction }
\noindent In the classical paper [ 3 ] , Corcoran defines two logical systems which he calls D
\noindent and RD . The former −−− system D −−− is intended as a formalization of Aristotle ’ s
syllogistic within the framework of modern logic . Aristotelian syllogistic studies
the necessity relation between two ( general non − empty ) terms $ , S $ and
$ P , $ that
\centerline{67 }
68 .. EDGAR ANDRADE ampersand EDWARD BECERRA
\noindent 68 \quad EDGAR ANDRADE $ \& $ EDWARD BECERRA
follows from assuming both a relation between terms P and M comma and a relation
between terms S and M period Aristotle managed to give a satisfactory account of the
\noindent follows from assuming both a relation between terms $ P $ and $ M , $
aforementioned entailment by using a number of conversion rules and the four
and a relation
quoteleftperfectsyllogismsquoterightperiod.. BothrulesandsyllogismsareformalizedemdashinCorcoran
between terms $ S $ and $ M . $ Aristotle managed to give a satisfactory account of the
quoteright s
afsoyrsteemmesnemtidoanseh6d8asennaEttDuaGriaAllmRdAeedNnuDtcRtAiboDynEruu&lseisEnoDgpWeAnaRpDnauBremEnCtbEheRersRiAsosfeecboenlovwecrlsoisoinngpraurelnetshesaisndpertiohde..fAosumr odern
‘ perfect sylfloollgowissmfrsom’as.sum\qinugadbotBhoathrelartuiolnesbeatwnedenstyerlmlosgPisamnds Ma,raendfoarrmelaatliioznebdet−w−ee−n in Corcoran ’ s
logical systems comma
systems −−− atsermnsaStuarnadlMd.eAdruisctottiloenmarnuagleeds to(gisveeeasbaetilsofawcto)ry.ac\cqouunatdofAthsemafoodreemrnentlioongeidcal systems ,
the systems D and RD are set up by defining open parenthesis a closing parenthesis .... a formal language
entailment by using a number of conversion rules and the four ‘ perfect syllogisms ’
semicolon .... open parenthesis b closing parenthesis a
\noindent the .sysBteomthsruDlesaannddRsyDlloagirsemssaerte fuoprmbalyizedde—finininCgorc(oraan)’ s\syhsfteimlls —a afosrnmataulrallanguage ; \hfill ( b ) a
semantic system semicolon and open parenthesis c closing parenthesis a deduction system period .. System
deduction rules ( see below ) . As modern logical systems ,
D formalizes all of the
\noindent semathnetiscystseymssteDman;d RanDdar(e scet)upabyddeedfiuncintgio(na )systeamfor.ma\lqluanagduaSgyes;tem (Dbf)oarmalizes all of the
conversion rules and all of the perfect syllogisms comma whereas system RD formalizes
conversion ruselmesanatincdsysatlelm o;fantdh(e cp)earfdeecdtuctsiyonllsoygstiesmms. , SwyshteermeaDs fosrymstaelimzesRaDll offorthmealizes
only some of them period For this reason comma system RD is easier to handle comma in the sense
only some of ctohnevmersi.onFrourlestahnidsalrleoafstohenpe,rfescytsstyelmlogiRsDmsi,swheearesaisesrystteomhRaDndfolrema,liziens onthlye sense
that we can come up with a completeness theorem for it period Nonetheless comma system
that we can csoommee oufpthwemith. Faorcthoims prelaestoenn,essysstetmheRorDemis eafsoirer tiotha.ndNleo,nienththeelesesnsse,thsaytswteem
RD turns out to be equivalent to system D period The outcome of this is a proof of
RD turns out ctaon cboemeeuqpuiwvitahleanctomtpoletseynsetssemtheDore.mTfohreito.uNtcoonmeteheloesfs ,tshyisstemisRDa tpurrnosoofutof
the completeness of system D comma and emdash as a formalization of Aristotle quoteright s original
the completentoesbse eoqfuivsaylsetnetmtoDsys,temanDd −. −T−heaosutcaomfeorofmtahilsizisataipornoofooff tAhericsotmoptlleeten’esssoforiginal
quoteleft system quoteright emdash in turn it implies Aristotelian syllogistic is complete period .. This is a
‘ system ’ −−s−ystienm tDur,nandit—imaspalifoersmaAlizraitsitonotoeflAiarinstotslyel’lsoogriigsitniacl ‘issystcemom’p—letine tu.rn\qituad This is a
beautiful result comma no doubt period .. One of it s nicest sides comes from the fact that it
beautiful resimulptlies, Anroistodtoeulibant s.yll\ogqiustaidc isOcnoempolfeteit. s Tnhiisceissta bseiaduetisfulcroemsuelts ,fnroomdoutbhte. fact that it
provides a formal proof for a highly intuitive conviction period .. By the same token comma
provides a foOrmneaolf iptrsonoicfesftosrideas chomigehslfyrominthteufiatcitvtehatciotnpvroivcitdieosna fo.rm\aqlupardooBf fyortahheigshalyme token ,
we believe that there are other intuitive convictions that can be given formal
we believe thinattuittihveerceonvaicrteionot.herByinthteusiatmiveetokceonn,vwiectbieolinevse tthhaattthcearenarbeeotghievreinntufitoivremal
proofs as well period .. Those intuitive convictions are brought out by comma among others comma
proofs as welclonv.ic\tiqounasdthaTthcoasnebeingitvuenitfoivrmealcpornoovfsicatsiwonelsl . arTehbosreoiungtuhittivoeuctonvbiyctio,nsamaroeng others ,
thefollowingquestions: .. openparenthesisiclosingparenthesisRDallowedustocomeupwithacomplete-
the followingbrqouugehsttoiuotnbsy,:am\qonugadoth(ersi,t)heRfDolloawlliongwqeudestuiosnst:o c(omi)eRuDpalwlowitehduastcoocmompeleteness
ness
proof , howevueprwi,thias coimt ptlehteeneessapsrioeosft, h,owmevoesrt, iisnittetrheesetaisniegst ,symsotsetmintetroestdinegalsyswteimthto $ ? $
proofcommahowevercommaisittheeasiestcommamostinterestingsystemtodealwith? .. Thatiscomma
\quad That is d,eal with ? That is , is RD a minimal system that is equivalent to system D ? ( ii
is RDa minimalsystem thatis equivalent to systemD ?open parenthesisii closingparenthesisIf wemanage
is RD a minim)aIlf wseysmtaenmagethtoatdefiinse eaqturainvsalalteionntfutnoctisoynstfreomm RDD t$o fi?rst -(or$der lioigic), wIfouwlde manage to
to
define a tranitslbaetpioosnsiblfeutnoccthiaornactferroizme thReDfrtaogmefnitrsoftfi−rsto-rodrdeerr lloogigcitchis,trwanosulaldtionitdebfineespossible
define a translation function from RD to first hyphen order logic comma would it be possible
to characteri?ze(tihiei )fWrahgamtewnotuldohfapfpiernstif w−e woreadkeernedlotgheicreqtuhiriesmetnrtatnhastlathteiomnoddelesfminuests $ ?
to characterize the fragment of first hyphen order logic this translation defines ? open parenthesis i i i closing
( $ i i i ) consist of non - empty sets ? In this paper , we set out to give an answer to the first
parenthesis
What would haopfptehneseiqfuewsteionwse.akened the requirement that the models must
What would happen if we weakened the requirement that the models must
consist of non −Inetmhepftoyllowseintgssec$tio?n ,$ w\eqdueafidneIbnoththsyisstepmaspaenrd w,ewpeoinsteotutosuotmetoof tghievire an answer to the
consist of non hyphen empty sets ? .. In this paper comma we set out to give an answer to the
first of thesperopqeurteisesti.onsSec.tion 3 defines the translation function from the cate - gorical propo-
first of these questions period
sitions to the atomic formulas of an appropriate first - order lan - guage . We then
In the following section comma .. we define both systems and we point out some
In the followipnrgoceseedcttoiopnrove, :\qu(aid) wtheatdtheifsintreansblaottihonsfuynsctteimonsisaandfaiwthefulpionitnertpreotuattiosnom—e
of their properties period .. Section 3 defines the translation function from the cate hyphen
of their propthearttiise,sth.at\aqusyaldlogSismectisiovanlid3indaenfyinmeosdelthfoer RtDrainffsiltsattriaonnslatfiuonncistitorune ifnroamny the cate −
gorical propositions to the atomic formulas of an appropriate first hyphen order lan hyphen
gorical propomsoidteilofnorsTto ;th(eiia)tomthiact tfhoermaxuiolmass oof fT an aaprepropriate first − order lan −
guageperiodWethenproceReDdtoprove: .. openparenthesisicRloDsingparenthesisthatthistranslationfunction
guage . We thienndeppernodceneted; toanpdr(oviiei :) t\hqautaRdD(isiin)fatchtaat mtihniimsaltsryasntesmlaotfiorunlesfufonrctthieon is a faithful
is a faithful
interpretatioAnri−st−o−telitahnastylloisgist,ict.hWate maakseyslolmogeicsommmiesntsvaanldidreminarkasniny semctoidonel4f.or RD iff its
interpretation emdash that is comma that a syllogism is valid in any model for RD iff its
translation is true in any2 .moCdeolrcfoorran$’Ts R{DRDan}d D; sy(st$emsii ) \quad that the axioms of
translation is true in any model for T sub RD semicolon open parenthesis ii closing parenthesis .. that the
$ T { RD }$ T\hqeuvaodcabaurleary consistsofasetR:={a,e,i,o},andanon-emptysetV. Thesetof
axioms of T sub RD .. are
categorical propositions ,P , consists of those and only those expressions of the form
independent semicolon .. and open parenthVesis ii i closing parenthesis that RD is in fact a minimal system of
\noindent indeSpaePn,dSeenPt,S;iP\,SqouPadwhaenrde S(,Pi∈i Vi, a)ndtSha(cid:54)=tPR.D is in fact a minimal system of rules for the
rules for the
Aristotelian sylloLgetisMti=c {U. W}ie∈mIabkeeasfoammeilycoofmnmonen-tesmaptnyd sreetsm,aarnkdsleitng :sVec→tiMon. 4W.e
Aristotelian syllogistic period We mi ake some comments and remarks in section 4 period
define the interpre tation function ·M,g :P →{0,1} as follows :1
2 period .. Corcoran quoteright s RD and D systems V
\ceTnhteevrolcianbeul{a2ry.con\1sqItiustiassdnooftCahosarerdtcoRtora:sne=e to’hpaetsnthRbeDraprceaesneandtcodDmefimnsiytaisoentecpmormosvmid}easicactoemgomricaalopcrolopsoisnitgiobnsrawciethcomthmeira and a
traditionalsemantics,namely ,SaP : ‘Every S is P ’ ,SeP : ‘No S is P ’ ,SiP : ‘Some
non hyphen empty set V period .. The
\noindent The vocabulary consists of a se(cid:48)t $ R : = \{ a , e , i
set of categoricaSl pisroPpo’si,tiSoonPs c:o‘mSommaePSsuisbnoVt cPomma consists of those and only those expressions
, ofothefo\rm} SaP,co$mmaanSdePaconmonma−SiPemcpomtymaseStoPw.$heVreSc.om$ma\PqiunaVdcTomhemaandSnegationslash-equal
set of categorical propositions $ , P { V } , $ consists of those and only those expressions
P period
of the form $ SaP , SeP , SiP , SoP $ where $ S , P \in V
Let M = open brace U sub i closing brace i in I be a family of non hyphen empty sets comma and let g : V
, $ and $ S \not= P . $
right arrow M period .. We
define the interpre tation function llbracket times rrbracket M comma g : P sub V right arrow open brace 0
\hspace∗{\ fill }Let $ M = \{ U { i } \} i \in I $ be a family of non − empty sets , and let
comma 1 closing brace as follows : to the power of 1
$ g : V \rightarrow M . $ \quad We
1 sub It is not hard to see that the present definition provides categorical propositions with
their traditional semantics comma namely comma SaP : quoteleft Every S is P quoteright comma SeP :
\noindent define the interpre tation function $ \llbracket \cdot \rrbracket
quoteleft No S is P quoteright comma SiP : quoteleft Some S is
M , g : P { V } \rightarrow \{ 0 , 1 \} $ as follows $ : ˆ{ 1 }$
P quoteright comma SoP : quoteleft Some S is not Case 1 quoteright Case 2 period
$ 1 { It }$ is not hard to see that the present definition provides categorical propositions with
their traditional semantics , namely $ , SaP : $ ‘ Every $ S $ is $ P $
’ $ , SeP : $ ‘ No $ S $ is $ P $ ’ $ , SiP : $ ‘ Some $ S $ is
$ P $ ’ $ , SoP : $ ‘ Some $ S $ is not $\left .P\begin{aligned} & ’ \\
& . \end{aligned}\right.$
CORCORAN quoteright S ARISTOTELIAN SYLLOGISTIC period period period .. 69
\hspace∗{\ fill }CORCORAN ’ S ARISTOTELIAN SYLLOGISTIC . . . \quad 69
Definition 2 period 1 period
Line 1 open parenthesis 1 closing parenthesis llbracket SaP rrbracket M comma g = 1 iff g open parenthe-
\noindent Definition 2 . 1 .
sis S closing parenthesis subset equal g open parenthesis P closing parenthesis comma Line 2 open parenthe-
sis 2 closing parenthesis llbracket SeP rrbracket M comma g = 1 iff g open parenthesis S closing parenthesis
\[\begin{aligned} ( 1 ) \llbracket SaP \rrbracket M , g = 1
cap arrowdblright-P parenright-notdef notdef-equal varnothing notdef notdef notdef Line 3 closing parenthe-
siis fllfbrackget SiP( rrbSracket)M c\omsumbasegte=q1 iffgg opC(enORpCaOPrRenAtNh)e’sSisARSI,ScTl\oO\sTinEgLIApNarSeYnLthLOesGisISTcaICp.a.r.rowd6b9lright-P
( 2 ) D\ellfibnritaiconke2t. 1S.eP \rrbracket M , g = 1 i ff g ( S
parenright-notdef notdef-equal-negationslash notdef-emptyset comma sub notdef Line 4 closing parenthesis ll-
) \cap arrowdblright−P parenright−notdef notdef−equal \varnothing notdef
bracket SoP rrbracket M comma g = 1 iff g open parenthesis S closing parenthesis nsubseteq g open parenthesis
notdef notdef \\
P closing parenthesis comma (1) SaPM,g =1iffg(S)⊆g(P),
) \llbracket SiP \rrbracket M , g = 1 i ff g ( S ) \cap
Let d in P sub V comma K subset equal P sub V comma and llbracket times rrbracket M comma g an
arrowdblright−P parenrigh(t2−)noStedPeMf ,g =n1oitfdfegf(−Se)q∩uaarlr−owndebglraitgihotn−slPapsahrenrignhott−dneof−tdeemfnpottdyesfet−equal∅notdefnotdefnotdef
interpretation function :
, { notdef }\)\ SiPM,g =1iffg(S)∩arrowdblright−Pparenright−notdefnotdef −equal−negationslashnotdef −emptyset,
openparenthesis1closingparenthesis.. IfllbracketdrrbracketMcommag=1commawesaythatllbracket notdef
tim)es rrb\rlalcbkertaMckceotmmaSgoPis a tr\uerrinbtrearpcrketeattionMof d p,eriodg = 1 i ff g ( S ) ) \SnosPuMbs,egt=eq1iffg(S)(cid:42)g(P),
g ( P ) , \end{aligned}\]
open parenthesis 2 closing parenthesis .. We say that .. llbracket times rrbracket M comma g .. is a .. tru e
Let d∈P ,K ⊆P , and ·M,g an interpretation function :
interpretation of K if llbrackVet times rVrbracket M comma g is a true
( 1 ) If dM,g =1, we say that ·M,g is a true interpretation of d.
interpretation of d comma for each d in K period
\noindent Let $(d2 ) \Wine sayPtha{t V·M},g, isKa tr\usuebinsteetreprqetatioPn o{f VK }if ·M,,g$is aantdrue $ \llbracket
open parenthesis 3 closing parenthesis K = d if and only if every true interpretation of K is a true interpre
\cdot \rrbraicntkeerptretaMtion of,d, fogr e$achadn∈iKn.terpretation function :
hyphen
(3) K = d if and only if every true interpretation of K is a true interpre - t ati|
t ati to the power of bar on of d period
\centerline{( o1n o)f d\.quad If $ \llbracket d \rrbracket M , g = 1 , $
open parenthesis 4 closing parenthesis .. The open parenthesis meta hyphen closing parenthesis function
we say that $ \(ll4b)racTkheet ( me\tcad-o)tfunc\tirornbcraallcedketthe coMntradic,torygof$ funicstioan t,rdueneotiendtbeyrpretation of
called the contradictory of function comma denoted by
$ d . $ } c:P →P , is defined as follows :
c : P sub V right arVrow PVsub V comma is defined as follows :
c open parenthesis d closing parenthesis : = Case 1 SoP if d = SaP comma Case 2 SiP if d = SeP comma
( 2 ) \quad We say that \quad $ \llbracket \cdot \rrbracket M , g $ \quad is a \quad tru e interpretation of
Case 3 SeP if d = SiP comma Case 4 SaP if d = SoP comma
SoP ifd=SaP,
$ K $ if $ \llbracket \cdot \rrbracket M , g $ is a true
The deductive system consists of the following natural deduction rules : to the power of 2
interpretation of $ d , $ for each $ d \in K . $ SiP ifd=SeP,
open parenthesics(dI)c:l=osibnrgacpeaerxen−thbersaicsePexeS−tborathceeepxo−webrraocfeSleefPtmopiden−pbarraecnetehxes−isbIrIacceloesxin−gbpraarceenetxh−esibsraScaePleftbt
sub SiP open parenthesis III closing parenthesis SaM from SaP to MaP open parenthesis IV closing parenthesis SeP ifd=SiP,
$ ( 3 ) K = d $ if and only if every true interpretation of $ K $ is a true interpre −
SaM from SeP to MeP SaP ifd=SoP,
t $ ati ˆ{ \mid }$ on of $ d . $
open parenthesis V closing parenthesis PiS to the power of SiP open parenthesis VI closing parenthesis SiM
from SiP to MaP oTphene dpeadruencttihveesissyVstIeImclocsoinngsisptasroefntthheesifsolSloiMwinfrgonmatSuorPaltdoeMduecPtion rules :2
( 4 ) \quad The ( meta − ) function called the contradictory of function , denoted by
Definition 2 period 2 p(erIio)d .P...eSASsePequ(enIIce)anSgabPracketle(ftIIpI1) coSmamMaSpaPeriod( IpVeri)odSpaeMrioSdePcomma p n right
$ c : P { V } \rightarrow P {SiPV } , $ isMdaPefined as folMloePws :
angbracket .... of categorical propo(sVitio)nsPisiSaSdiPirect( VI ) SiMSiP ( VII ) SiMSoP
MaP MeP
deductionofdfDroemfinKitiifodn=2p.n2a.ndfAorseaecqhueinicneop(cid:104)epn1,b..r.a,cpen(cid:105)1coofmcmataegpoerriicoadl ppreorpioodsitpioernisodiscaomdmireactnclosing
\[\left . c ( d ) : = braceex−braceex−braceex−braceleftmid−braceex−braceex−braceex−braceleftbt\begin{aligned} &
brace one of the foldleodwuicntgion of d from K if d=pn and for each i∈{1,...,n} one of the following
SoP if d = SaP , \\
holds : open parenthesis 1 closing parenthesis p i in K comma or
& SiP if d = SeP , \\
open parenthesis 2 closing parenthesis .. There exists j less i such that p i is o b tained from p sub j using
rul&es opeSnePparentihfesis Idclosi=ng parSeinPthesis,co\m\ma open parehntohldessis: II closing parenthesis comma
& SaP if d = SoP , \end{aligned}\right.\]
o r open parenthesis V closing parenthesis comma o(r1)pi∈K,or
open parenthesis 3 closing parenthesis .. There exist j comma k less i such that p i is o b tain ed from p sub
j and p k using open p(ar2en)theTsihseIrIeIecxloisstisngjp<areinstuhcehsisthcaotmmpiais o b tained from pj using rules ( I ) , (
\noindent The deductive system consists of the following natural deduction rules
open parenthesiIsII)V, colorsin(gVp)ar,enotrhesis comma .. open parenthesis VI closing parenthesis comma or open
$ : ˆ{ 2 }$
parenthesis VII closing(p3a)rentThheseirsepeexriisotdj,k <i such that pi is o b tain ed from pj and pk using ( III
Definition 2 per)io,d(3IpVer)io,d .. (AVsIe)qu,eonrce(aVnIgIb)ra.cketleft p 1 comma period period period comma p n right
\centerline{( I $ ) PeS ˆ{ SeP } ( $ II $ ) SaP { SiP } ( $ III $ )
angbracket of categDoreificanlitpiroonpo2sit.io3ns.is anAinsdeiquhyenpcheen(cid:104)p1,...,pn(cid:105) of categorical propositions is an indi -
SaM ˆ{ SaP } { MaP } ( $ IV $ ) SaM ˆ{ SeP } { MeP }$ }
rectdeductionorfecdtfdroedmucKtioifnthoefredexfriostmsjlKessifnthsuercehetxhiasttspnj =<cnospuecnhpthaaretntphnes=iscp(psju)bajncdlofsoinrgepaacrhenthesis
and for i∈{1,...,n} one of the following holds :
\centerline{( V $ ) PiS ˆ{ SiP } ( $ VI $ ) SiM ˆ{ SiP } { MaP } ( $ VII
each i in open brac2eR1ulceso(mIm)a, (pIerIio)d, apnedri(odVp)earrieodkncoowmnm, aaccnorcdlionsgintogtbhreatcreadoitnioeno,fatshtehfeoclloonwveirnsgionhoruldless:
$ ) SiM ˆ{ SoP } { MeP }$ }
2subRulesopeanmpoanrgecnatthegeosrisicaIlcplroospinosgitpioanrse.nRthuelessis(cIoImI)m,a(oIVpe)n,(paVrIe)n,thanesdis(IVIIIc)loasrienkgnopwanreanstthheesipsercfoecmtmaand
open parenthesis Vsycllloogsiisnmgs.parenthesis are known comma according to the tradition comma as the conversion
\noindent Definition 2 . 2 . \hfill A s equence $ \langle p 1 , . . .
rules
, p n \rangle $ \hfill of categorical propositions is a direct
amongcategoricalpropositionsperiodRulesopenparenthesisIIIclosingparenthesiscommaopenparenthesis
IV closing parenthesis comma open parenthesis VI closing parenthesis comma and open parenthesis VII closing
\noindent deduction of $ d $ from $ K $ if $ d = p n $ and for each $ i
parenthesis are known as the perfect
\in \{ 1 , . . . , n \} $ one of the following
syllogisms period
\begin{align∗}
holds : \\ ( 1 ) p i \in K , or
\end{align∗}
( 2 ) \quad There exists $ j < i $ such that $ p i $ is o b tained from
$ p { j }$ using rules ( I ) , ( II ) ,
o r ( V ) , o r
( 3 ) \quad There exist $ j , k < i $ such that $ p i $ is o b tain ed from
$ p { j }$ and $ p k $ using ( III ) ,
( IV ) , \quad ( VI ) , or ( VII ) .
\noindent Definition 2 . 3 . \quad A s equence $ \langle p 1 , . . .
, p n \rangle $ of categorical propositions is an indi −
rect deduction of $ d $ from $ K $ if there exists $ j < n $ such that
$ p n = c ( p { j } ) $ and for
each $ i \in \{ 1 , . . . , n \} $ one of the following holds :
$ 2 { Rules } ( $ I ) , ( I I ) , and ( V ) are known , according to the tradition , as the conversion rules
among categorical propositions . Rules ( I II ) , ( IV ) , ( VI ) , and ( VII ) are known as the perfect
\noindent syllogisms .
70 .. EDGAR ANDRADE ampersand EDWARD BECERRA
\noindent 70 \quad EDGAR ANDRADE $ \& $ EDWARD BECERRA
open parenthesis 1 closing parenthesis p i in K plus c open parenthesis d closing parenthesis 3 comma or
open parenthesis 2 closing parenthesis .. There exists j less i such that p i is o b tain ed from p sub j .. b y
\[ ( 1 ) p i \in K + c ( d ) 3 { , } or \]
using rules open parenthesis I closing parenthesis comma
open parenthesis II closing parenthesis comma or open parenthesis V closing parenthesis comma or
open parenthesis 3 closing parenthesis .. There exist j comma k less i such that p i .. is o b tained from p sub
j\.c.eanntdeprlkin..eb{y( u720sin)gE\DqGuAaRdANTDhReArDeEe&xiEsDtWsARD$BEjCERR<A i $ such that $ p i $ is o b tain ed from
$ p { j }$ \quad b y using rules ( I ) , }
open parenthesis III closing parenthesis comma .. open parenthesis IV closing parenthesis comma open
(1)pi∈K+c(d)3or
parenthesis VI closing parenthesis comma or open parenthesis VII clo,sing parenthesis period
\centerline{( II ) , or ( V ) , or }
We say that K(t2u)rnstTilehleerfteseuxbistDs dji<f ainsducohnltyhaiftthpeireiseoxibststaeiintheedrfarodmirecpt orbanyiunsdiinrgecrtules ( I ) ,
j
deduction of d from K period The logical system( IsIo)d,efionre(dVis)ca,lloerd D period
\centerline{( 3 ) \quad There exist $ j , k < i $ such that $ p i $
There is only(o3ne)diffTehreenreceexbiesttweje,nks<ysitesmucshDthaantdpRiD ciosmombatanianmedelfyrocmommp a thaendlatphkyphbeny using
\quad is o b tained from $ p { j }$ \quad and $ p k $j \quad b y using }
ter quoteright s deductive system con(siIsItIs)on,ly o(f IrVule)s,o(peVnIp)a,reonrth(eVsiIsII)c.losing parenthesis hyphen open
parenthesis IV closWinge spaayretnhtahtesKis(cid:96)peridodifWanhdenoenvlyerifatchaetreegeoxriicstasl either a direct or an indirect deduction
\centerline{( III ) , \quadD ( IV ) , ( VI ) , or ( VII ) . }
proposition d foolflodwfsrofrmomKa. TsehteKloogficcaaltseygsotreimcalsoprdoepfionseitdioinsscaacllceodrdDin.g to the
latter set of ruleTshceorme misaownleysoanyethdaiffteKrentucrenbsteitlweleeeftnssuybstRemDsdDpearnioddR..DT,hunsamhaevlyin,gtrheedulcaetd-ttheern’usmber of
\noindent We say that $ K \vdash { D } d $ if and only if there exists either a direct or an indirect
deduction rulesdaeldlouwcstivuesstyostpermovceotnhseisftoslolonwlyinogfirnutleerse(stIin)g-(reIsVul)ts.WpehrieondevTehreapcaroteogfsoricalproposition
deduction of $ d $ from $ K . $ The logical system so defined is called D .
of these resultsdcafnollboewfsofurnodmina soeptenKsqofuacraetebgroarcickaeltp3roclpoossinitgiosnqsuaacrecobrdrainckgettocothmemlaattoerrospeetnofsqruualerse,bracket
2 closing square brwacekesatyptehriaotdK (cid:96) d. Thus having reduced the number of deduction rules allows us
\noindent There is only onReD difference between systems D and RD , namely , the lat −
Lemma 2 periodto1pproevreiotdheIffaonllgobwriancgkeintlteefrtespti1ngcormesmulatsp.erTiohde ppreorioofds opfetrhioedsecroemsumltaspcannrbigehftouanndgbirnacket is
ter ’ s deductive system consists only of rules ( I ) − ( IV ) . Whenever a categorical
an indirect deducti[on3 ]of,dorfr[o2m]K. comma then for each
proposition $ d $ follows from a set $ K $ of categorical propositions according to the
i in open braceL1ecmommmaa2p.er1iod. pIefrio(cid:104)dp1p,e..r.i,opdn(cid:105)coismamnainndcirloecstindgedburcatcieonthoefredisfraodmireKct,dtheednucftoironeaocfhp i from
latter set of rules , we say that $ K \vdash { RD } d . $ \quad Thus having reduced the number of
K plus c open parein∈th{es1i,s..d.,ncl}ostihnegrepaisreantdhierescistpdeedriuocdti.o.nInofpaprtiicfruolmar cKom+mca(d). In particular ,
deduction rules allows us to prove the following interesting results . The proofs
K plus c open Kpar+enct(hde)s(cid:96)is d cplnosainngdpaKre+ntch(eds)is(cid:96)turncs(tpilne)leifnt sau‘bdRirDectpwnayan’d. K plus c open parenthesis d
of these results can RbDe found in [ 3 ]RD, or [ 2 ] .
closing parenthesisTthuernosrteilmele2ft.su4b(RRDedcucotpioenadpaarbesnutrhdeusmis )p.n IcflosKing+pc(adr)en(cid:96)thesiesainnda Kqu+otce(ledf)t(cid:96)direct way
RD RD
quoteright period c(e),
\noindent Lemma 2 . 1 . If $ \langle p 1 , . . . , p n \rangle $
Theorem 2 period 4 open parenthesis Reductio ad absurdum closing parenthesis period If K plus c open
is an indirect deduction of $ d $ from $ K , $ then for each
parenthesis d closing parenthesis turnstileleft sub RD e and K plus c open parenthesis d closing parenthesis
turnstileleft sub RD c open parenthesis e closing parentthheensiKs c(cid:96)omRDmda.
\noindent $ i \in \{ 1 , . . . , n \} $ there is a direct deduction of
then K turnstileleft sub RD d period
$ p i $ fromTheo$reKm 2+. 5 (c( Str(ong )dSoun)dness.)$. \qIfuaKd (cid:96)In pd,atrhteincuKla=r d4,
Theorem 2 period 5 open parenthesis open parenthesis Strong closing pRaDrenthesis Soundne.ss closing paren-
Theorem 2 . 6 ( ( Strong ) Completeness ) . If K =d, then K (cid:96) d.
thesis period .. If K turnstileleft sub RD d comma then K = d 4 period RD
\noindent $ KAs u+sual ,ccomp(letenedss is)prove\dvvdiaasahn ad{eqRuDatio}n lepmmanth$at isanindtur$n pKroved+via c (
Theorem2period6openparenthesisopenparenthesisStrongclosingparenthesisCompletenessclosingparen-
d ) \vdashoth{erRleDmm}as acnd de(finitipons .nAs we)h$appeinntoane‘eddsiormeecotf twhaeyse l’em.mas , we are
thesis period .. If K = d comma then K turnstileleft sub RD d period
going to state them without proof — but see the foregoing references .
As usual comma completeness is proved via an adequation lemma that is in turn
\noindent TheoDreemfin2itio.n42 (. 7R.educLteito ∅ad=aKbs⊆urPdu,mand) d.efinIef the$foKllowin+g s ects : ( d ) \vdash { RD }
proved via other lemmas and definitions period As we hVappen to need some of these
e $ •andA(K$)K:= +{F ∈c ℘(V():Sd∈F&)Pelem\vednta−shneg{atRioDnsl}ashcF,SaP( ∈Ke} ()Subse,ts$ of V con -
lemmas comma .. we are going to state them without proof emdash but see the foregoing
taining S but lacking P, for ea ch SaP ∈K).
references period
\begin{align∗} • E(K):={F ∈℘(V):S ∈F&P ∈F,SeP ∈K}( Subsets of V contain -
Definition 2 period 7 period .. Let varnothing = K subset equal P sub V comma and define the following s e
then K \vdash { RD } d ing.bo th S and P, for each SeP ∈K).
ts :
\end{align∗}
bulletAopenparenthesisKclosingparenthesis: =openbraceFinwpopenparenthesisVclosingparenthesis
• U(K):=℘(V)−(A(K)∪E(K))
: S in F ampersand P element-negationslash F comma SaP in K closing brace open parenthesis Subsets .. of ..
\noindent Theorem 2 . 5 ( ( Strong ) Soundness ) . \quad If $ K \vdash { RD }
Vd .. co,n$hypthhenen 3If$AKand B=aretwdosets4ofca{teg.ori}ca$lpropositions,theexpression A+B isdefinedto be A∪B.
taining S but laMckoirneogvePr,ciofmdmaandfodr(cid:48)eaarechcatSeagPoriicnalKprocploossiitniognspa,rtheentehxperseisssipoenrsiodd+d(cid:48) and A+d aredefined
\nobiunlldetenEtopTenhepoatroreebmneth{2eds,is.dK(cid:48)}6calno(dsin(Ag∪Spat{rrdeo}nntrhgesepse)isct:iCv=eolymo.ppelnebteranceesFs in) w.po\pqeunapdareInfthe$sisKVclo=singpdarenth,es$is
:thSeinn F$amKpersa\nvd4dIPtasiishnwFort{chomnRoDmtinag}StheaPdtrinuleK(.cIlI$o)siinsgsoburnadcweiothperensppeactretnotthheessiesmSaunbtiscestdsefionfeVdicnodnetfianiintiohnyp2h.en1
ing bo th S andonPlycboemcamusaefwoerreeaqcuhireSdetPheinsetKs icnloMsingtopbaerennotnh-eseimspptyer.ioIdf we weakened this hypothesis , rule
bullet U open parenthesis K closing parenthesis : = wp open parenthesis V closing parenthesis minus open
\noindent As u(sIuIa)lwou,ldcnoomlonpgleertbeenseousnsd .isNowpr,owvitehdregvairdatoatnheathdiredqquuaesttiioonninltehmemintarodtuhctaiotn ,isvizin turn
parenthesisAopenparenthesisKclosingparenthesiscupEopenparenthesisKclosingparenthesisclosingparen-
proved via ot.her wlehmatmwaousldahnadppedneiffwine witeaikoennesd th.eAresquiwreemehntapthpaetnthetmoodneelsemdusstocmonesistosfofthnoens-e
thesis
lemmas , \quaedmpwtyeseatsr,eitgisocilneagr frtoom tshetaprteeviotuhseombserwvaititohnotuhatt spysrteomosfD−a−n−d RbDuwtousldeenotthloengferobreegoing
3 sub If A and B are two sets of categorical propositions comma the expression A plus B is defined to
references . soundwithrespecttothissemantics. Thus,thesystemsthataresoundwithrespect toitwouldbe
be A cup B period Moreover comma if d and d to the power of prime are categorical propositions comma the
differentfrom—notequivalentto—DandRD.Itremainstobeseenhow interestingthesemodels
expressions d plus d to the power of prime and A plus d
\noindent Defianrei.ti[oWnear2ein.de7bte.dt\oqthueaadnonLyemtous$refe\reveaforrntohitshreimnagrk. ]= K \subseteq P { V }
are defined to be open brace d comma d to the power of prime closing brace and A cup open brace d closing
, $ and define the following s e ts :
brace respectively period
4subItisworthnotingthatruleopenparenthesisIIclosingparenthesisissoundwithrespecttothesemantics
\centerline{ $ \bullet A ( K ) : = \{ F \in \wp ( V )
defined in definition
: S \in F \& P element−negationslash F , SaP \in K \} ( $
2 period 1 only because we required the sets in M to be non hyphen empty period If we weakened this
Subsets \quad of \quad $ V $ \quad con − }
hypothesis comma
rule open parenthesis II closing parenthesis would no longer be sound period Now comma with regard to the
\centerline{taining $ S $ but lacking $ P , $ for ea ch $ SaP \in K )
third question in the introduction comma
. $ }
viz period .. what would happen if we weakened the requirement that the models must consists of
non hyphen empty sets comma it is clear from the previous observation that systems D and RD would not
\centerline{ $ \bullet E ( K ) : = \{ F \in \wp ( V )
longer be sound with respect to this semantics period Thus comma the systems that are sound with respect
: S \in F \& P \in F , SeP \in K \} ( $ Subsets of $ V $
to it would be different from emdash not equivalent to emdash D and RD period It remains to be seen how
contain − }
interesting these models are period open square bracket We are indebted to the anonymous referee for this
remark period closing square bracket
\centerline{ing bo th $ S $ and $ P , $ for each $ SeP \in K ) . $
}
\[ \bullet U ( K ) : = \wp ( V ) − ( A ( K ) \cup
E ( K ) ) \]
\noindent $ 3 { If } A $ and $ B $ are two sets of categorical propositions , the expression
$ A + B $ is defined to
be $ A \cup B . $ Moreover , if $ d $ and $ d ˆ{ \prime }$ are categorical propositions , the expressions
$ d + d ˆ{ \prime }$ and $ A + d $
are defined to be $ \{ d , d ˆ{ \prime } \} $ and $ A \cup \{ d
\} $ respectively .
\noindent $ 4 { It }$ is worth noting that rule ( I I ) is sound with respect to the semantics defined in definition
2 . 1 only because we required the sets in $ M $ to be non − empty . If we weakened this hypothesis ,
rule ( II ) would no longer be sound . Now , with regard to the third question in the introduction ,
viz . \quad what would happen if we weakened the requirement that the models must consists of
non − empty sets , it is clear from the previous observation that systems D and RD would not
longer be sound with respect to this semantics . Thus , the systems that are sound with respect
to it would be different from −−− not equivalent to −−− D and RD . It remains to be seen how
interesting these models are . [ We are indebted to the anonymous referee for this remark . ]
CORCORAN quoteright S ARISTOTELIAN SYLLOGISTIC period period period .. 71
\centerline{CORCORAN ’ S ARISTOTELIAN SYLLOGISTIC . . . \quad 71 }
Lemma 2 period 2 period If K subset P sub V is maximal consistent comma M : = wp open parenthesis U
open parenthesis K closing parenthesis closing parenthesis comma and i : V right arrow M
\noindent Lemma 2 . 2 . If $ K \subset P { V }$ is maximal consistent $ ,
is such that i open parenthesis S closing parenthesis : = open brace F in U open parenthesis K closing
M : = \wp ( U ( K ) ) , $ and $ i : V \rightarrow M $
parenthesis : S in F closing brace comma then the fo l lowing claims ho ld :
open parenthesis i closing parenthesis .. If S in V comma th en i open parenthesis S closing parenthesis
n\engoaitniodneslnasth-ieqsuasluvcahrnotthhaintg pe$rioCidORCO(RANS’SAR)ISTOT:ELIAN=SYLL\O{GISTIFC.. .\in71 U ( K ) :
S \in F L\e}mma, 2$. 2t.heInf tKhe⊂Pfo isl mlaoxwiminaglcocnlsaisitmenst h,Mo l:=d ℘:(U(K)),and i:V →M
open parenthesis i i closing parenthesis SaPVin K iff U open parenthesis K closing parenthesis contains no s e
is such that i(S):={F ∈U(K):S ∈F}, then the fo l lowing claims ho ld :
t containing S but lacking P period
\centerline{( i ) \quad If $ S ( i\)in If VS ∈V,,th$enthi(Se)n(cid:54)=∅$. i ( S ) \not=
open parenthesis ii i closing parenthesis SeP in K iff U open parenthesis K closing parenthesis contains no s
\varnothing . $( i}i ) SaP ∈K iff U(K) contains no s e t containing S but lacking P.
e t containing both S and P period
( ii i ) SeP ∈K iff U(K) contains no s e t containing both S and P.
open parenthesis iv closing parenthesis SiP in K iff U open parenthesis K closing parenthesis contains a s e t
\centerline{( i i ($iv)) SSiaPP∈K\iffin U(KK)$contaiifnfs a$s eUt con(tainiKng bot)h $S ancdonPt.ains no s e t containing
containing both S and P period
$ S $ but lacking( v$)PSoP.∈$K iff} U(K) contains a s e t containing S but lacking P.
open parenthesis v closing parenthesis SoP in K iff U open parenthesis K closing parenthesis contains a s e t
Lemma 2 . 3 . If K ⊆P is maximalconsistent ,M :=℘(U(K)),and i:V →M
containing S but lacking P period V
\centerline{( isiisucih th$at) i(S)S:e=P{F ∈\iUn(K):KS$∈F}i,ftfhen$ ·UU(K()),iKis a tr)ue$intecroprnet-ains no s e t containing both
Lemma2period3periodIfKsubsetequalPsubVismaximalco℘n(sistentcommaM:=wpopenparenthesis
$ S $ and $ tPation.of$K.}
U open parenthesis K closing parenthesis closing parenthesis comma and i : V right arrow M
3 . System RD as a subsystem of first - order logic
is such that i open parenthesis S closing parenthesis : = open brace F in U open parenthesis K closing
\centerline{( Iinvthe$firs)t parStioPf this\sienctionKwe$definieffa tr$ansUlation(funcKtion f)rom$ thceocnatteagoinrs- iacals e t containing both
parenthesis : S in F closing brace comma then llbracket times rrbracket sub wp open parenthesis U open
$ S $ and $ pPropos.iti$ons t}o the atomic formulas of an appropriate first - order language . The aim
parenthesis K closing parenthesis closing parenthesis comma i is a true interpre hyphen
of this translation function is to find a correspondence between the in - terpretation
tation of K period
\centerline{( fvunct$ion)s in CSoorPcoran\’ins RDKsys$temiafnfd th$e Umodel(s of aKfirst )- o$rderctohneotrayi,ntshaat wse e t containing
3 period .. System RD as a subsystem of first hyphen order logic
$ S $ but lacskhianllgcall$TP . .Th$us ,}we prove that every syllogism valid in RD is valid in T .
In the first part of this secRtiDon we define a translation function from the categor hyphen RD
The first - order language that we use is defined as follows . We use a similarity
i cal propositions to the atomic formulas of an appropriate first hyphen order language period
\noindent Lemmtayp2e τ.=3{A.,EI}f wi$thoKut eq\usaulitbysseytmeqbol ,5Pwit{h oVnly}$twoibsinamryaxreimlatailoncso.nWsiesctaelnl t $ ,
The aim of this translation function is to find a correspondence between the in hyphen
M : = \wTp th(e theUory d(efinedKby t)he fol)lowing, a$xiomasnd: $ i : V \rightarrow M $
terpretation funcRtiDons in Corcoran quoteright s RD system and the models of a first hyphen order
theory comma that we shall call T sub RD period .. Thus comma we prove that every syllogism valid in
A ∀x,y(xEy ↔yEx)
\noindent is such that $ i ( S ) 1 : = \{ F \in U ( K ) :
RD is valid in T sub RD period
S \in F \} , $ then $ \llbrAacke∀tx,y(x\Acyd→ot¬(xE\yr)r)bracket { \wp ( } U
The first hyphen order language that we use is d2efined as follows period We use a similarity
( tyKpetau)=op)enbr,aceAico$mmiasEaclotsirnugebAria3nc∀etxew,yri,tphzr(oeyuAt−zeq∧uxalAityy→symxAbozl)commato thepower of5with only
two binary relations period We A ∀x,y,z(yEz∧xAy →xEz)
\noindent tation of $ K . $ 4
call T sub RD the theory defined by the following axioms :
T is a consistent set , thanks to the following model : Let S be a non - empty set
Line 1 A sub 1 foRrDall x comma y open parenthesis xEy leftrightarrow yEx closing parenthesis Line 2 A sub 2
\centerline{3 a.nd\dqeufiande MSys=te(cid:104)mM,RADM,aEsMa(cid:105) swuhbesreystem of first − order logic }
forall x comma y open parenthesis xAy right arro,w logicalnot open parenthesis xEy closing parenthesis closing
parenthesis Line 3 A sub 3 forall x comma y comma z open paMren=th℘e(sSis)y−A{z∅a}nd xAy right arrow xAz closing
\noindent In the first part of this section we define a translation function from the categor −
parenthesis Line 4 A sub 4 forall x comma y comma z open parenthesis yEz and xAy right arrow xEz closing
i cal propositions to the atomic formulas ofxAanMyap⇐p⇒ropxr⊆iayte first − order language .
parenthesis
The aim of this translation function is to find a correspondence between the in −
T sub RD is a consistent set comma thaxnEksMtoyt⇐he⇒folxlo∩w=in⇒g m∅ondoetlde:fLneottdSebfe a non hyphen empty
terpretation functions in Corcoran ’ s RD system and the models of a first − order
set and define M = angbracketleftbig M comma A to the power of M comma E to the power of M angbrack-
theory , thate−wefi shi−alnl ticoanll3 1$.T A{foRrmDul}a µ.is$an\aqriustaodteTlhiaunssy,llowgiesmpr(oovresytllhogaitsme,very syllogism valid in
etrightbig sub comma where
RD is valid ifnor $ T { RD } . $
Line 1 M = wp open parenthesis S closing parenthesis minus open brace varnothing closing brace Line 2 xA
short ) iff µ.is of the form ∀x,y,z(φ1∧φ2→φ3) where :
to the power of M y arrowdblleft-arrowdblright x subset equal y Line 3 xE to the power of M y arrowdblleft-
\hspace∗{\ fill }The fir(s1t) −φiordisearnlaatnomguicafgoermtuhlaaotrwthee nuesgeatiiosn odfeafninaetodmiacsforfmoulllaow(si=. We use a similarity
arrowdblright x cap equal-arrowdblright varnothing notdef notdef
1 , 2 , 3 ) .
e-fi i-n tio n 3 1 period .... A formula mu is an aristo te lian syllogism open parenthesis or syllogism comma
\noindent type $ (\2ta)u Ea=ch var\i{able Ax,y,z,appeEars ex\a}ctly$ inwtwitohfoorumtuleaqsuφail,iφty,i(cid:54)=syjm. bol $ , ˆ{ 5 }$
for j
witshhorotncllyosintwgoparbeinn5thTaehrseyiscoinrffseemqluauetnpicoeernoifosldeaiv.sinoWgfoetuhtethfeoremquafloitryalslymxbcoolmismthaatythceoamtommaiczseonpteenncepsaarreenotnhlyesiosftphhei 1 and
call $ T { fRorDm }x$Ay torhexEtyh.eory defined by the following axioms :
phi 2 right arrow phi 3 closing parenthesis where :
openparenthesis1closingparenthesisphii.. isanatomicformulaorthenegationofanatomicformulaopen
\[\begin{aligned} A { 1 } \forall x , y ( xEy \leftrightarrow yEx
parenthesis i =
) \\
1 comma 2 comma 3 closing parenthesis period
A { 2 } \forall x , y ( xAy \rightarrow \neg ( xEy ) ) \\
open parenthesis 2 closing parenthesis .. Each variable x comma y comma z appears exactly in two formulas
A { 3 } \forall x , y , z ( yAz \wedge xAy \rightarrow xAz
phi i comma phi sub j comma i equal-negationslash j period
) \\
5 sub The consequence of leaving out the equality symbol is that the atomic sentences are only
A { 4 } \forall x , y , z ( yEz \wedge xAy \rightarrow xEz
of the form xAy or xEy period
) \end{aligned}\]
\noindent $ T { RD }$ is a consistent set , thanks to the following model : Let
$ S $ be a non − empty
set and define $ M = \langle M , A ˆ{ M } , E ˆ{ M } \rangle { , }$
where
\[\begin{aligned} M = \wp ( S ) − \{ \varnothing \} \\
xA ˆ{ M } y \iff x \subseteq y \\
xE ˆ{ M } y \iff x \cap \Longrightarrow \varnothing notdef notdef \end{aligned}\]
\noindent $ e−fi i−n $ tio n 3 1 . \hfill A formula $ \mu $ is an aristo te lian syllogism ( or syllogism , for
\noindent short ) iff $ \mu .{ is }$ of the form $ \forall x , y , z
( \phi 1 \wedge \phi 2 \rightarrow \phi 3 ) $ where :
\hspace∗{\ fill } $ ( 1 ) \phi i $ \quad is an atomic formula or the negation of an atomic formula
$ ( i = $
\centerline{1 , 2 , 3 ) . }
\centerline{( 2 ) \quad Each variable $ x , y , z $ appears exactly in two formulas
$ \phi i , \phi { j } , i \ne j . $ }
$ 5 { The }$ consequence of leaving out the equality symbol is that the atomic sentences are only
of the form $ xAy $ or $ xEy . $
72 .. EDGAR ANDRADE ampersand EDWARD BECERRA
\noindent 72 \quad EDGAR ANDRADE $ \& $ EDWARD BECERRA
3 period 1 period .. Translation period .. From now on comma .. we suppose that V is equal to the set of
variables of the first hyphen order language L sub tau we are working with period
\noindent 3 . 1 . \quad Translation . \quad From now on , \quad we suppose that $ V $
We define the function TR : P sub V right arrow L sub tau in the following manner :
is equal to the set of
Line1openparenthesis1closingparenthesisTRopenparenthesisxayclosingparenthesis=xAyLine2open
variables of the first − order language $ L { \tau }$ we are working with .
parenthesis 2 closing parenthesis TR open parenthesis xey closing parenthesis = xEy Line 3 open parenthesis
3 closing parenthes7i2s TREDoGpAeRnApNaDrRenAtDhEesi&s xEiDyWcAloRsDinBgECpEaRreRnAthesis = logicalnot open parenthesis xEy closing
\centerline{We3 .d1ef.ineTrtahneslaftuionnct.ion Fro$mTnRow on: , Pwe s{upVpos}e th\artiVghitsaerqruoawl to thLe set{of\tau }$
parenthesisLine4openparenthesis4closingparenthesisTRopenparenthesisxoyclosingparenthesis=logicalnot
in the followivnagriamblaesnnoferthe: fi}rst - order language L we are working with .
open parenthesis xAy closing parenthesis τ
We define the function TR:P →L in the following manner :
Note that TR is injective comma so we can define its invVerse fuτnction restricted to the
\[\begin{aligned} ( 1 ) TR ( xay ) = xAy \\
image of TR emdash that is comma the atomic formulas and their negations period
(1) TR(xay)=xAy
( 2 ) TR ( xey ) = xEy \\
Now we will show how one can construct an interpretation function of RD
( 3 ) TR ( xiy ) = \n(e2g) TR((xeyx)E=yxEy) \\
from a model of T sub RD period .. This is achieved as follows period Firstly comma we build up the set
(of ato4mic f)ormuTlaRs emd(ash ixnotyhe fir)st hy=phen(3\o)rndeegTrRla(nxg(iuya)g=exe¬Am(yxdEasyh))th\atenadre{tarluieganneddf}al\s]e to the power
of 6 in (4) TR(xoy)=¬(xAy)
the model of T sub RD under a particular interpretation g period .. Secondly comma we translate
\noindent NoteNottheatthat$TRTRis i$njectiisvei,nsojewcetciavnede,finseoitswienvcerasne fudnecftiionnereisttrsicteidnvtoerthseeimfuagnection restricted to the
this set back into RD by means of TR to the power of minus 1 period .. Next comma .. we prove that this
image of $ ToRf T$R —−−−thatthisa,tthiesat,omtihceforamtuolmasicandfotrhmeirulnaegsataionnds .their negations .
latter
NowwewillshowhowonecanconstructaninterpretationfunctionofRDfromamodel
setismaximalconsistentcomma.. andthereforecomma.. bylemma2period3comma.. wecanfindatrue
\noindent Now owfeT w.ill Tshhioswis ahcohwievoendeascfaonllowcso.nFstirrsutlcyt, waenbuinildteurpptrheetsaettion function of RD
interpretation of itRpDeriod
from a model ooffatom$icTfor{muRlaDs —} in.th$e fir\sqtu-aodrdeTrhlainsguiasgea—chtiheavteadreatrsuefoanldlofwalss e.6 iFnitrhsetly , we build up the set
Definition 3 period 2 period .. Let M = angbracketleftbig M comma A to the power of M comma E to the
model of T under a particular interpretation g. Secondly , we translate
power of M angbracketrightbiRgD.. and g : V right arrow M be such that M =
\noindent of atthoismsiect bfaocrkmiuntloasR−D−b−y imneatnhseoffTiRrs−t1.− Noerxdte,r lawnegpuraogvee t−h−at−thtihsalatttaerresettrue and fals
T sub RD period open square bracket Recall that V comma the s e t of t erms of the language of RD comma
$ e ˆ{ 6 }$ iins maximal consistent , and therefore , by lemma 2 . 3 , we can find a true
is equal to th e s e t
the model of int$erTpret{atiRonDo}f$it . under a particular interpretation $ g . $ \quad Secondly , we translate
of variables of the language of T sub RD period closing square bracket .. We define : to the power of 7
Definition 3 . 2 . Let M=(cid:104)M,AM,EM(cid:105) and g :V →M be such that M=
Line1KsubMcommag: =braceleftbigTRtothepowerofminus1openparenthesisxAyclosingparenthesis
\noindent thisT se.t [bRaecckallitnhtaot VR,Dthbeys emteoafntseromfs of$thTeRlanˆg{ua−ge of1RD} , is. eq$ual\tqouthade sNeetxt , \quad we prove that this latter
: M=xAyopensquRarDebracketgclosingsquarebracketbracerightbigcupbraceleftbigTRtothepowerofminus
set is maximaolf vcaorinabsliesstoefnthte l,an\gquuagaedofandT th.]ereWfoerdeefin,e\:q7uad by lemma 2 . 3 , \quad we can find a true
1openparenthesisxEyclosingparenthesis: M=xERyDopensquarebracketgclosingsquarebracketbracerightbig
interpretation of it .
Line 2 cup braceleftbig TRKto the:=p{oTwRer−o1f(xmAiyn)us:M1 o=pexnApya[rge]n}t∪he{sTisRl−og1i(cxaElnyo)t:oMpen=pxaEreyn[tgh]}esis xAy closing
M,g
parenthesis closing parenthesis : M = xAy open square bracket g closing square bracket bracerightbig cup
\noindent Definition ∪3{T.R2−1(.¬(\xqAuya)d):LMet=x$AyM[g]}∪={TR\−l1a(¬n(gxlEey)):MM=,xEy[Ag]}ˆ{ M } ,
braceleftbig TR to the power of minus 1 open parenthesis logicalnot open parenthesis xEy closing parenthesis
E ˆ{ M } \rangle $ \quad and $ g : V \rightarrow M $ be such that $ M
closing parenthesisN:oMte=thaxtEyKopen⊆sqPua,resobrwaeckceatngdcelfionsieng square bracket bracerightbig
= $ M,g V
Note that K sub M comma g subset equal P sub V comma so we can define
TR open parenthesis K sub M comma gTcRlo(sKing p)ar=en{tThRes(isd)=:dop∈eKn brac}e TR open parenthesis d closing
\noindent $ T { RD } . [ $ RecaMll,gthat $ V , M$ ,g the s e t of t erms of the language of RD , is equal to th e s e t
parenthesis : d in K sub M comma g closing brace
From the last definitions it is not hard to see that M=TR(K )[g].
From the last definitions it is not hard to see that M = TR open parenthesMis,gK sub M comma g closing
\noindent of variables of the language of $ T { RD } . ] $ \quad We define
parenthesis open square bracket g closing square bracket period
$ : ˆ{ 7 }$
Equation: Lemma 3 period 1 period .. TR open parenthesis c open parenthesis d closing parenthesis closing
TR(c(d))(cid:96)¬TR(d)and¬TR(c(d))(cid:96)TR(d). Lemma3.1.
parenthesisturnstileleftlogicalnotTRopenparenthesisdclosingparenthesisandlogicalnotTRopenparenthesis
\[\begin{aligned} K { M , g } : = \{ TR ˆ{ − 1 } ( xAy ) :
copenparenthesisdclosingparenthesisclosingparenthesisturnstileleftTRopenparenthesisdclosingparenthesis
Proof . We leave the proof to the reader . square−check
M = xAy [ g ] \} \cup \{ TR ˆ{ − 1 } ( xEy ) : M =
period
Lemma 3 . 2 . If K (cid:96) d, then T +TR(K)(cid:96)TR(d). Proof . We shall
xEy [ g ] \} \\ RD RD
Proof period .. We leave the proof to the reader period square-check
prove , by induction on the length of the deduction , that if there
\cup \{ TR ˆ{ − 1 } ( \neg ( xAy ) ) : M = xAy [ g
Lemma 3 period 2 period .. If K turnstileleft sub RD d comma then T sub RD plus TR open parenthesis K
is a direct deduction of d from K of the form (cid:104)p1,...,pn(cid:105), then T +TR(K)(cid:96)
] \} \cup \{ TR ˆ{ − 1 } ( \neg ( xEy ) R)D : M = xEy
closing parenthesis turnstileleft TR open parenthesis d closing parenthesis period
[ g ] \} \end{aligned}\]
Proof period .. We shall prove comma by induction on the length of the deduction comma that if there
isadirectdeductionofdfromKoftheformangbracpkiefotlreafltlip≤1nc.ommaperiodperiodperiodcommapnright
angbracket comma then T sub RD plus TR open parenthesis K closing parenthesis turnstileleft
\noindent Note that $ K { M , g } \subseteq P { V } , $ so we can define
p i for all i less(ori e)qIufaplin∈peKri,odthen TR(pi) ∈ TR(K) and therefore T +TR(K) (cid:96) TR(pi) ( ii )
RD
openparenthesiLseitcl’ossinsugpppaorseentthheastispIifpisioibntKainceodmmfroamthpeknTbyRuospinengpruarleenIth(kes<ispi)i.cBloysitnhgepianrdeuncthtieosnisinTR
\[ TR ( K { M , g } ) = \{ TR ( d ) : d \in K { M
open parenthesis Khyclpoostinhgespisarweenthhaevseistahnadt Tthere+foTreRT(Ksu)b(cid:96)RTDR(pplku)s,TthRusopbyenaxpiaormentAhe,sissubKstciltoustiinognpaanrdenthesis
RD 1
, g } \} \]
turnstileleft TR opmenodpuasrepnothneesniss,pTi clo+sinTgRp(aKre)n(cid:96)thTeRsis(pi). Similarly with rule
RD
open parenthesiIsIi.i closing parenthesis Let quoteright s suppose that p i is obtained from p k by using rule I
open parenthesis k(leisisii)cLloestin’gspsuaprepnotsheestihsaptepriioids oBbytatihneed from pk,pm by using rule III (k,m<i). By
\noindent From the last definitions it is not hard to see that $ M = TR (
inductionhypotihnedsuisctwioenhhayvpeotthhaetsiTs wsuebhRavDepthluastTTRop+enTRpa(rKen)the(cid:96)sisTKR(cplkos)inagndpaTrent+hesisturnstileleftTR
RD RD
K { M , g } ) [ g ] . $
open parenthesis p k closing parent6hWeseisarceomindmebatetdhutostbhyeaanxoinoymmoAussruebfer1eecfoomrtmhiasremark.
substitutionandmoduspone7nWsceoamremuasinTgsthuebnRotDionploufssaTtiRsfaocptieonnapsadreefinntehdesinis[K4,clpo.si8ng1p].arenthesisturnstileleft
\begin{align∗}
TR open parenthesis p i closing parenthesis period .. Similarly with rule
\tag∗{$ Lemma 3 . 1 . $} TR ( c ( d ) ) \vdash \neg TR
II period
( d ) and \neg TR ( c ( d ) ) \vdash TR ( d ) .
open parenthesis ii i closing parenthesis Let quoteright s suppose that p i is obtained from p k comma p m
\end{align∗}
by using rule III open parenthesis k comma m less i closing parenthesis period
By induction hypothesis we have that T sub RD plus TR open parenthesis K closing parenthesis turnstileleft
\noindent Proof . \quad We leave the proof to the reader $ . square−check $
TR open parenthesis p k closing parenthesis and T sub RD plus
6 sub We are indebted to the anonymous referee for this remark period
\noindent Lemma 3 . 2 . \quad If $ K \vdash { RD } d , $ then $ T { RD }
7 sub We are using the notion of satisfaction as defined in open square bracket 4 comma p period 8 1 closing
+ TR ( K ) \vdash TR ( d ) . $
square bracket period
Proof . \quad We shall prove , by induction on the length of the deduction , that if there
\noindent is a direct deduction of $ d $ from $ K $ of the form $ \langle p
1 , . . . , p n \rangle , $ then $ T { RD } + TR ( K
) \vdash $
\begin{align∗}
p i for all i \leq n .
\end{align∗}
\noindent ( i ) If $ p i \in K , $ then $ TR ( p i ) \in TR
( K ) $ and therefore $ T { RD } + TR ( K ) \vdash TR ( p
i ) $
( ii ) Let ’ s suppose that $ p i $ is obtained from $ p k $ by using rule I
$ ( k < i ) . $ By the
induction hypothesis we have that $ T { RD } + TR ( K ) \vdash TR
( p k ) , $ thus by axiom $ A { 1 } , $
substitution and modus ponens $ , T { RD } + TR ( K ) \vdash TR
( p i ) . $ \quad Similarly with rule
\noindent II .
\noindent ( ii i ) Let ’ s suppose that $ p i $ is obtained from $ p k ,
p m $ by using rule III $ ( k , m < i ) . $
By induction hypothesis we have that $ T { RD } + TR ( K ) \vdash
TR ( p k ) $ and $ T { RD } + $
\centerline{ $ 6 { We }$ are indebted to the anonymous referee for this remark . }
\centerline{ $ 7 { We }$ are using the notion of satisfaction as defined in [ 4 , p . 8 1 ] . }
CORCORAN quoteright S ARISTOTELIAN SYLLOGISTIC period period period .. 73
\hspace∗{\ fill }CORCORAN ’ S ARISTOTELIAN SYLLOGISTIC . . . \quad 73
TR open parenthesis K closing parenthesis turnstileleft TR open parenthesis p m closing parenthesis comma
thereforeTsubRDplusTRopenparenthesisKclosingparenthesisturnstileleftTRopenparenthesispkclosing
\noindent $ TR ( K ) \vdash TR ( p m ) , $ therefore $ T { RD }
parenthesis and TR open parenthesis p m closing parenthesis period .. Thus comma by
+ TR ( K ) \vdash TR ( p k ) \wedge TR ( p m ) . $
axiom A sub3 comma substitution andmodus ponenscommaT sub RD plusTR open parenthesis Kclosing
\quad Thus , by
parenthesis turnstileleft TR open parenthesis p i closing parenthesis period Similarly
axwioitmh rul$e IAV p{erio3d} , $ substitution anCOdRCmOoRdAuNs’SpoAnRIeSnTsOTE$LIA,NSYLTLOG{ISTRIDC.}. . +73 TR
( K ) \vTdRas(Kh) (cid:96)TTRR(pm(), thepreforei T )+TR.(K$) (cid:96)STimR(iplka)r∧lyTR(pm). Thus , by axiom
On the other hand comma if angbracketleft pR1Dcomma period period period comma p n right angbracket is
A , substitution and modus ponens ,T +TR(K)(cid:96)TR(pi). Similarly
an indirect deduction3 comma then by lemma RD
\noindent withwirthulreuleIVIV ..
2 period 1 we can find direct deductions of p n and c open parenthesis p n closing parenthesis from K plus c
On the other hand , if (cid:104)p1,...,pn(cid:105) is an indirect deduction , then by lemma 2 . 1
open parenthesis d closing parenthesis period By the previous
On the other hwaendcan,finidf dir$ec\tldaendugclteions opf pn a1nd c(,pn) fr.om K. +c(.d). B,y thepprevinous p\arrtaonfgle $
part of this proof comma and lemma 3 period 1 comma we have that T sub RD plus TR open parenthesis K
is an indirectthdisepdruoocfti,oannd,lemthmean3b.y1 ,lewmemhaave that T +TR(K)+TR(c(d)) is inconsistent ,
closing parenthesis plus TR open parenthesis c open parenthesisRdDclosing parenthesis closing parenthesis is
2 . 1 we can afnidndby tdhierreecdtuctdioedaducabtsiuorndsumoftheo$repm of nfirs$t - oarndedr lo$gicc( Cf .( [ 4 ]p) , itnfollow)s $ from
inconsistent comma and by the reductio ad absurdum theorem of first hyphen order logic open parenthesis
$ K + c t(hat Td +)TR(K.)$(cid:96)¬TBRy(ct(hde)).preFviniaolulys, again by lemma 3 . 1 ,
Cf period RD
part of this wperhoaovfe T, an+dTlRem(Km)a(cid:96)3TR.(d1), a,ndwewehaarveedotnhea.t sq$uaTre−{chReDck} + TR ( K
open square bracket 4 cloRsDing square bracket closing parenthesis comma it follows that T sub RD plus TR
) + TR (Corocllary(3 . d1 . ) If )K$ (cid:96)is d, then T +TR(K )(cid:96)TR(d).
open parenthesis K closing parenthesis turnstileMlef,tg loRgDicalnot TR oRpDen parenthMes,igs c open parenthesis d closing
inconsistent L,emanmdab3y.3.theK rediuscmtaioximaadl caobnssiusrtednutm. theorem of first − order logic ( Cf .
parenthesis closing parenthesis periodM.,.gFinally comma again by lemma 3 period 1 comma
[ 4 ] ) , it Pfroololfo.ws tShuaptpose$toTwar{dsRaDco}ntra+dictioTnRthat(K K is in)consi\svtednats.hThe\nntehgere TR
wehaveTsubRDplusTRopenparenthesisKclosingparenthesistuMrn,sgtileleftTRopenparenthesisdclosing
( c ( d is a)sente)nce d.su$ch t\hqautaKd Fina(cid:96)lly , daganadinKby lem(cid:96)ma 3 c.(d)1. , By corollary
parenthesis comma and we are done period squMar,eg-checRkD M,g RD
3.1,T +TR(K )(cid:96)TR(d) and T +TR(K ) (cid:96)TR(c(d)), and by lemma
Corollary 3 period 1RpDeriod .. IfMK,gsub M comma gRtuDrnstileleftMsu,gb RD d comma then T sub RD plus TR
\noindent we h3a.ve1 it$folTlows{thRaDt T} ++TRT(KR )(is inKconsis)tent .\vdTahshis conTtrRadicts( d ) , $
open parenthesis K sub M comma g closRinDg parenthMesi,gs turnstileleft TR open parenthesis d closing parenthesis
and we are done $ . square−check $
period
Lemma 3 period 3 period K sub M comma g is maximal consistent period
\noindent Corollary 3 . 1 . \quad tIhfatM$=KTRD{ +MTR(K, M,gg)[g}]. \vdash { RD } d , $
Proof period .. Suppose towards a contradiction that K sub M comma g is inconsistent period Then there
then $ T { RD } + TR ( K { M , g } ) \vdash TR ( d ) . $
is a sentence d suchIntohradteKr tsoupbroMvecKommaisgmtuarxnismtialellceoftnssiusbtenRtD, wdeasnhdalKl psruobveMthcaotmfomraangytudr∈nsPtile,left sub
M,g V
RD c open parenthweseishdavcelotshiantgepitahreenrtdhe∈sisKperioodr.c.(Bd)y∈coKrollar.y Let ’ s suppose that d (cid:54)∈ K , and
\noindent Lemma $ 3 . 3 . M,Kg { M ,M,g g }$ is maximal consiMst,gent .
3 period 1 comtmhaatTdsiusbofRtDhepfloursmTxRayo.pFernompatrhenetdheefisinsitKionsuobfMK comimt faolglocwlossing parenthesis turnstileleft
M,g
TR open parenthestihsadt cMlosing=paxrAenyt[hg]e,sis awnhdicTh sinubtuRrDn ipmlupsliTesRthoaptenTpRa−re1n(¬th(xesAisy)K) sub=Mc(cxoamy)ma g∈closing
\noindent Proof . \quad Suppose towards a contradiction that $ K { M , g }$
parenthesisturnstilKeleftT. RoTpheencpaasreendthofestihsecfooprmenxpeayreinsthsiemsiisladrc.losinAgspfaorretnhtheecsaisseclxoisyin,gwpeahreanvtehtehsiastcomma
is inconsistentM.,gThen there
.. and by xiy does not belong to K only if M=xAy[g] and therefore ,xay =
is a sentence $ d $ such thatM,$g K { M , g } \vdash { RD } d $ and $ K { M
lemma 3 periodc(1xiity)fo∈lloKws th.aTthTescuabseRxDoypilsussiTmRilaorp.enspqauraerneth−escihseKcksub M comma g closing parenthesis
, g } \vdash { RDM},g c ( d ) . $ \quad By corollary
is inconsistent period .. ThisTcohnetrfoardeigctosing lemmas , including those in the previous section , allow us to
$ 3 . 1 , T { RD } + TR ( K { M , g } ) \vdash TR (
that M = T subshRowDtphluats,TsRtaortpienngpfraormentahfiesrisstK- osrudberMmcoodmelmoaf Tg clo,sMing, apnadreantvhaelsuiastoiopnenfusnqcua-re bracket g
d ) $ and $ T { RD } + TR ( K { M , RDg } ) \vdash TR ( c
closing square bractkieotnpger:ioVd→ M, we can build up an interpretation function of the RD system . This
( d ) ) , $ \quad and by
In order to provlaetKtersuinbteMrpcreotmatmioangfuisncmtiaoxnimisadlecfionnesidstbeyntucsoinmgmMaMwe:=sh℘al(lUp(rKove th)a)tafnodr any
lemma 3 . 1 it follows that $ T { RD } + TR ( K { MM,g , g } ) $ is inconsistent . \quad This contradicts
d in P sub V coim:mVa →weMhavMe tshuacthetihthaetrid(xi)n:K=s{uFb M∈ Uco(mKma g) o:rxc∈opFen}.pFarreonmthleesmismdacl2os.in3g pitarisenthesis
M,g
in K sub M commanogtphearrioddto..sLeeetthqautot·MeriMgh,tisissuaptprouseeinthteartpretation of K . The following theorem
\begin{align∗} M,g
d negationslashs-ehloewmsenthtaKt wsuebhMavecoamnmicae rgelcaotmiomnsahiapndbetthwaetend tisheosfethtweofomrmodxealsy.period From the definition
that M = T { RD } + TR ( K { M , g } ) [ g ] .
of K sub M commaTghietofroellmow3s . 3 . The fo l lowing claims hold :
\end{align∗}
that M = xAy open square bracket g closing square bracket comma .. which in turn implies that TR to the
power of minus 1 open parenthesis logicalno(t1)opeMn p=arxenAtyh[egs]iisffxAxayycMloMsin,gi=pa1renthesis closing parenthesis = c
In order to prove $ K { M , g }$ is maximal consistent , we shall prove that for any
open parenthesis xay closing parenthesis in
$ d \in P { V } , $ we h(2a)veMth=atxEeyi[gt]hifefrxey$MdM,i=\i1n K { M , g }$ or
K sub M comma g period .. The case d of the form xey is similar period .. As for the case xiy comma we
$ c ( d ) \in K { M , g } . $ \quad Let ’ s suppose that
have Proof . We will only prove claim 1 . Claim 2 is proved in a similar way .
$thdat xiy\dnooets\niont b(e⇒loKn)gIft{MoMK su=bx,MAyc[oggm] m}athge,nonT$lyRi−fa1Mn(xdA=ytx)hAayt∈ope$nKdsMqu,$ga.reibSsrianccokefettg·hMceloMsfi,onigrimssqauta$rrueexbaraycket
a.nd$theFrreofomretcohmemindateexrfapiyrne=ittatiioonn ofoKf M$,g,Kthen{xMayMM,,i=g1.}$ it follows
c open parenthesis xiy c(lo⇐si)ngIfpxaaryeMnthMes,iis=in1K, tshuebnMi(xc)omm⊆ai(gyp).erioSdoT,hfoerceavseerxyoFy i∈s simUil(aKrMpe,rgi)o,difsquare-
\noindent that $ M = xAy [ g ] , $ \quad which in turn implies that
check x∈F,theny ∈F[checkoutthedefinitionofi!]. ThisimpliesthatU(KM,g)contains
$ TR ˆ{ − 1 } ( \neg ( xAy ) ) = c ( xay ) \in $
The foregoing lenmomseatsccoonmtaminainigncxlubduintglatchkoisnegiyn.tAhes KprMev,igouissmseacxtiimonalcocmonmsiastaenlltow, buyslteomma 2 . 2 ( i
$ K { M , g } . $ \quad The case $ d $ of the form $ xey $ is similar . \quad As for the case
show that commi a) wstearhtainvegtfhroamt xaayfir∈stKhMyp,gheanndortdheursmxoAdyel∈oTf TR(sKuMb R,g)D. TcohmermefaorMe comma and a valuation
$ xiy , $ we have
func hyphen
that $ xiy $ does not belong to $ K { M , g }$ only if $ M = xAy
tion g : V right arrow M comma we can buMild=upxaAny[ign]t.erpsrqeutaartieo−n fcuhnecctkion of the RD system period
[ g ] $ and therefore $ , xay = $
This latter interpretation function is defined by using M to the power of M : = wp open parenthesis U open
parenthesis K sub M comma g closing parenthesis closing parenthesis and
\noindent $ c ( xiy ) \in K { M , g } . $ The case $ xoy $ is similar
i : V right arrow M to the power of M such that i open parenthesis x closing parenthesis : = open brace F
$ . square−check $
in U open parenthesis K sub M comma g closing parenthesis : x in F closing brace period From lemma 2 period
3 it is
\hspace∗{\ fill }The foregoing lemmas , including those in the previous section , allow us to
not hard to see that llbracket times rrbracket M to the power of M comma i is a true interpretation of K sub
M comma g period .. The following
\noindent show that , starting from a first − order model of $ T { RD } , M
theorem shows that we have a nice relationship between these two models period
, $ and a valuation func −
Theorem 3 period 3 period .. The fo l lowing claims hold :
Line 1 open parenthesis 1 closing parenthesis M = xAy open square bracket g closing square bracket iff
\noindent tion $ g : V \rightarrow M , $ we can build up an interpretation function of the RD system .
llbracket xay rrbracket M to the power of M comma i = 1 Line 2 open parenthesis 2 closing parenthesis M =
This latter interpretation function is defined by using $ M ˆ{ M } : = \wp
xEy open square bracket g closing square bracket iff llbracket xey rrbracket M to the power of M comma i = 1
( U ( K { M , g } ) ) $ and
Proof period .. We will only prove claim 1 period Claim 2 is proved in a similar way period
open parenthesis double stroke right arrow closing parenthesis If M = xAy open square bracket g closing
\noindent $ i : V \rightarrow M ˆ{ M }$ such that $ i ( x ) :
square bracket .. then TR to the power of minus 1 open parenthesis xAy closing parenthesis in K sub M comma
= \{ F \in U ( K { M , g } ) : x \in F \} . $ From lemma 2 . 3 it is
g period .. Since .. llbracket times rrbracket M to the power of M comma i is a true
not hard to see that $ \llbracket \cdot \rrbracket M ˆ{ M } , i $ is a true interpretation of
interpretation of K sub M comma g comma then llbracket xay rrbracket M to the power of M comma i = 1
$ K { M , g } . $ \quad The following
period
theorem shows that we have a nice relationship between these two models .
open parenthesis double stroke left arrow closing parenthesis If llbracket xay rrbracket M to the power of
M comma i = 1 comma then i open parenthesis x closing parenthesis subset equal i open parenthesis y closing
\noindent Theorem 3 . 3 . \quad The fo l lowing claims hold :
parenthesis period .. So comma for every F in U open parenthesis K sub M comma g closing parenthesis comma
if
\[\begin{aligned} ( 1 ) M = xAy [ g ] i ff \llbracket xay
x in F comma then y in F open square bracket check out the definition of i ! closing square bracket period ..
\rrbracket M ˆ{ M } , i = 1 \\
This implies that U open parenthesis K sub M comma g closing parenthesis
( 2 ) M = xEy [ g ] i ff \llbracket xey \rrbracket M ˆ{ M }
contains no set containing x but lacking y period As K sub M comma g is maximal consistent comma by
, i = 1 \end{aligned}\]
lemma 2 period 2 open parenthesis i i closing parenthesis we have that xay in K sub M comma g and thus
xAy in TR open parenthesis K sub M comma g closing parenthesis period Therefore
M = xAy open square bracket g closing square bracket period square-check
\noindent Proof . \quad We will only prove claim 1 . Claim 2 is proved in a similar way .
$ ( \Rightarrow ) $ If $ M = xAy [ g ] $ \quad then $ TR ˆ{ −
1 } ( xAy ) \in K { M , g } . $ \quad Since \quad $ \llbracket
\cdot \rrbracket M ˆ{ M } , i $ is a true
interpretation of $ K { M , g } , $ then $ \llbracket xay \rrbracket
M ˆ{ M } , i = 1 . $
\hspace∗{\ fill } $ ( \Leftarrow ) $ If $ \llbracket xay \rrbracket M ˆ{ M }
, i = 1 , $ then $ i ( x ) \subseteq i ( y ) . $ \quad So , for every
$ F \in U ( K { M , g } ) , $ if
\noindent $ x \in F , $ then $ y \in F [ $ check out the definition of
$ i ! ] . $ \quad This implies that $ U ( K { M , g } ) $
contains no set containing $ x $ but lacking $ y . $ As $ K { M , g }$
is maximal consistent , by
lemma 2 . 2 ( i i ) we have that $ xay \in K { M , g }$ and thus $ xAy
\in TR ( K { M , g } ) . $ Therefore
\[ M = xAy [ g ] . square−check \]
74 .. EDGAR ANDRADE ampersand EDWARD BECERRA
\noindent 74 \quad EDGAR ANDRADE $ \& $ EDWARD BECERRA
We have constructed an interpretation function of RD from a model of T sub RD
that has very interesting properties open parenthesis Cf period .. Theorem 3 period 3 closing parenthesis
\noindent We have constructed an interpretation function of RD from a model of $ T { RD }$
period .. Conversely comma we can
that has very interesting properties ( Cf . \quad Theorem 3 . 3 ) . \quad Conversely , we can
construct a model of T sub RD .. from an interpretation function of RD that has
construct a model of $ T { RD }$ \quad from an interpretation function of RD that has
similar interesting properties open parenthesis Cf period .. Theorem 3 period 5 closing parenthesis period ..
Thsiismisilwahratiwnetweri7lel4ssettiEnoDugGtApRrAoNpDeRrAtDiEes&(EDCWfAR.D\BqECuEaRdRATheorem 3 . 5 ) . \quad This is what we will set out
to do in whatWefohlalvoewcsons.tructed an interpretation function of RD from a model of T that has
to do in what follows period RD
very interesting properties ( Cf . Theorem 3 . 3 ) . Conversely , we can construct
Definition3period4periodLetM=openbraceUsubiclosingbraceiinIbeafamilyofnonhyphenempty
\noindent Defianmitoidoenl of3T. 4 .froLmetan i$ntMerpret=ation\f{unctiUon o{f RiD}that\}has siimilar\iinnterestIing$ be a family of non − empty s e ts and
s e ts and g : V right arrow M pReDriod
$ g : V p\rroipgehrttieasrr(oCwf . MTheor.em$ 3 . 5 ) . This is what we will set out to do in what
We define M to the power of M : = angbracketleftbig M comma A to the power of M comma E to the power
follows .
of M angbracketrightbig where
\noindent We dDeeffiinneition$ M3 .ˆ{4M. }Let: M== {U\l}ain∈glIebe aMfamily, of Anonˆ{- Memp}ty s,e tsEanˆd{ M }
Line 1 M = open brace U sub i closing brace i in IiLine 2 g open parenthesis x closing parenthesis A to
\rangle $ whegr:eV →M.
the power of M g open parenthesis y closing parenthesis arrowdblleft-arrowdblright g open parenthesis x closing
We define MM :=(cid:104)M,AM,EM(cid:105) where
parenthesis subset equal g open parenthesis y closing parenthesis Line 3 g open parenthesis x closing parenthesis
\[\begin{aligned} M = \{ U { i } \} i \in I \\
E to the power of M g open parenthesis y closing parenthesis arrowdblleft-arrowdblright g open parenthesis x
g ( x ) A ˆ{ M } g ( y ) \iff g ( x ) \subseteq g
closing parenthesis cap y-arrowdblright closing parenthesis notdef equal-notdef varnothinMg n=ot{dUef}nio∈tdIef
( y ) \\ i
h e-o sub rem 3 5 period .. The fo l lowing claims are true :
g ( x ) E ˆ{ M } g ( y ) \iff g g(x()AMgx(y) ⇐)⇒ g\(cxa)p⊆g(yy)−arrowdblright
Line1openparenthesis1closingparenthesisMtothepowerofM=TsubRDperiodLine2openparenthesis
2) closinnogtpdaerefntheseiqsugll(abxlr)−aEcnkMeotgtx(dyae)yfr⇐rb⇒ra\cvgk(aextr)nM∩oytcoh−minamrgarogw=dnb1lortiiffgdhMetf)ntoottdhneefopetqoduweaeflr−o\fenMnotdd={efax∅lAingyontodepedefnn}so\qt]dueafrebracket
g closing square bracket period Line 3 open parenthesis 3 closing parenthesis llbracket xey rrbracket M comma
h e−o 35. The fo l lowing claims are true :
g = 1 iff M to the power ofrMem= xEy open square bracket g closing square bracket period
\noindent h $ e−o { rem } 3 5 . $ \quad The fo l lowing claims are true :
open parenthesis 4 closing parenthesis .. For every d in P sub V comma llbracket d rrbracket M comma g
(1) MM =T .
= 1 iff M to the power of M = TR open parenthesis d closing parenthesis opReDn square bracket g closing square
b\r[a\ckbeetgpienr{ioadligned} ( 1 ) M (ˆ2{) Mxa}yM=,g =T1iff{MRMD=}xA.y[g\].\
( 2 ) \llbracket xay \rrbracket M , g = 1 i ff M ˆ{ M }
Proof period .. By construction comma .(.3i)t isxsetyrMaig,hgt=for1wifarfdMtoMse=exthEayt[gM].to the power of M = T sub RD
= xAy [ g ] . \\
period .. We
(will o3nly p)rove c\lalilmbr2apcekrieot(d4..)CxleaFiymor3evi\serrpyrrbovdrea∈dcPiknVeat,dsMim,iMlgar=w1a,yiffpeMrgioMd .=.=CTlaRim(1d)4[gi]s.ia ff M ˆ{ M }
= strxaEigyhtforw[ ardPgcroonosfeq.]uencB.eyo\fceotnhndest{trwuacoltiipgornneve,idou}si\tc]ilsaismtrsaipgehrtifoodrward to see that MM = TRD. We
WehavethatllbwriallckoentlyxapyrrorvberacclkaeimtM2c.ommCalgai=m13iffisgporpoevnedpairnenathseimsisilaxrclwosaiyng.pareCnltahiemsis4suisbsaetequal
g open parenthesisstyracilgohsitnfogrwpaarrednctohnessiesquiffenMcetoof tthhee tpwooweprreovfioMus=claxiAmyso.pen square bracket g closing square
b\rcaecknetteprelriiondes{q(ua4re-)cWhee\cqhkuavaedthFaotrxaeyvMer,gy=1$iffdg(x)\⊆ing(y)Piff M{MV=}xAy,[g]. \slqlubarraec−kcehteck d \rrbracket
M , g = 1 $ iff $ M ˆ{ M } = TR ( d ) [ g ] . $ }
The two previouTshreestuwlotspgrievveiouussarensiuceltswagyivetougsoafrnoimceawnayintteorpgroetfarotimonan interpretation function in
function in oneolongeicloagliscyasltseymstteomatmo oadmeloidnelthine oththeeortohnereoconme m, wahwichhiclhealdeasdussutsottohe following proof
\noindent Proof . \quad By construction , \quad it is straightforward to see that
the following prtohoafttthhaetttrhaentsrlaatnisolnatfiuonncftuionnctwioenhwaevehdaveefindeedfinisedaifsaiathffauilthfiunlterpretation . That is
$ M ˆ{ M } = T { RD } . $ \quad We
interpretation p,etrhioadt a..sTyhllaotgiissmcoismvmaalidthinatRaDsyifflloigtissmtraisnvslaaltidioninisRaDtihffeoitresmtroafnTslRaDtio.nThisisais stated
will only prove claim 2 . \quad Claim 3 is proved in a similar way . \quad Claim 4 is a
theorem of T suinbtRhDe fpoellroiwodinTghtihseiosresmtat:ed in the following theorem :
straightforward consequence of the two previous claims .
Theorem 3 periTodh6eopreermiod3...L6et.mu bLeetthµe sbyelltohgeissmyllfoogriaslml x c∀oxm,ym,za(yφ1co∧mφm2a→zφo3p)e,nthpaernen:thesis phi 1 and
phi 2 right arrow phi 3 closing parenthesis comma th en :
\ceTnstuebrRliDnetu{WrnestihlealevftemtuhiafftTTRR$Dto\(cid:96)tlhµlebprioafwcfekreTotfRm−i1n(xuφas1y1)+opTe\nRr−pr1ab(rφern2at)ch(cid:96)keRseiDts pThRiM−11c(lφo3si,)n.g pagrenth=esis p1lus$TRiff
$ g ( x ) \subseteq g ( y ) $ iff $ M ˆ{ M } = xAy [ g
tothepowerofminus1openparenthesisphi2closingparenthesisturnstileleftsubRDTRtothepowerofminus
] . squareT−hcehpercokof$of th}is theorem runs as follows . Since we have a correspondence between
1 open parenthesis phi 3 closing parenthesis period
interpretation functions and models , if some syllogism is such that there exists a
The proof of this theorem runs as follows period .. Since we have a correspondence
\noindent The ttrwueoinpterrepvreitoautisonroefsiutsltpsremgiisvesethuast ias nnoticaetrwueayintteorprgetoatfiornomof iatns cionntcelurspiorne,tation
between interpretation functions and models comma .. if some syllogism is such that
function in otnheenltohegrieciasla msyosdteelmof TtRoDaformwohdicehl tihne ttrahneslaottiohneorf tohnise , which leads us to
there exists a true interpretation of its premises that is not a true interpretation
the followingsypllorgoiosmf itshfaaltset(hCef .trLaenmsmlaat3io.n4 )fu,nacndtivoincevweresah(avCef .dLeefminmead3i.s5a) .faTihtishful
of it s conclusion comma then there is a model of T sub RD for which the translation of this
interpretatioanllow. s\quad That is , that a syllogism is valid in RD iff it s translation is a
syllogismisfalseopenparenthesisCfperiodLemma3period4closingparenthesiscommaandviceversaopen
theorem of $usTto p{roRvDe th}e co.nt$rapoTsihtiivse oifseascthaotfetdheiimnpltihcaetiofnoslolofwThineogremth3eo.r6em. :
parenthesis Cf period Lemma 3 period 5 closing parenthesis period This allows
Lemma 3 . 4 . Let µ be a syllogism . If there exist M,g such that TR−1(φ1)M,g
us to prove the contrapositive of each of the implications of Theorem 3 period 6 period
\noindent Theo=re1m,T3R−.1(φ62).M\,gqu=a1daLndetTR$−1\(mφu3)M$,gb=e0,ththeenstyhelrleogisismM=$TR\Dfosurcahll x ,
Lemma 3 period 4 period Let mu be a syllogism period If there exist M comma g such that llbracket TR to
y , z ( \phi 1 \wedge \phi 2 \rightarrow \phi 3 ) , $ th en :
the power of minus 1 open parenthesis phi 1 closing parenthesis rrbracket M comma g
= 1 comma llbracket TR to the power of minus 1 tohpaetnMpa=reµn.thesis phi 2 closing parenthesis rrbracket M
\[ T { RD } \vdash \mu i ff TR ˆ{ − 1 } ( \phi 1 ) + TR ˆ{ −
comma g = 1 and llbracket TR to the power of minus 1 open parenthesis phi 3 closing parenthesis rrbracket M
1 } ( \phi 2 ) \vdash { RD } TR ˆ{ − 1 } ( \phi 3 ) . \]
comma g = 0 commParotohfe.n theBreyitshMeor=emT3su.b5R,DfrosmuchM,g we can build up a model M such that M=
that M = mu pTeriod, and such that dM,g = 1 iff MM = TR(d)[g] for every d ∈ P . If we assume
RD V
Proof period ..tBhyatthTeRor−e1m(φ31)pMer,igod=51c,oTmRm−1a(φfr2o)mMM,gc=om1manadgTwRe−c1a(nφ3b)uMild,gu=p a0,model M such that M =
\noindent The proof of this theorem runs as follows . \quad Since we have a correspondence
T sub RD comma and such that llbracket d rrbracket M comma g = 1 iff M to the power of M = TR open
between interpretation functions and models , \quad if some syllogism is such that
parenthesis d closing parenthesis open square bracket g closing square bracket for every d in P sub V period .. If
there exists a true interpretation of its premises that is not a true interpretation
we
of it s conclusion , then there is a model of $ T { RD }$ for which the translation of this
assume that llbracket TR to the power of minus 1 open parenthesis phi 1 closing parenthesis rrbracket M
comma g = 1 comma llbracket TR to the power of minus 1 open parenthesis phi 2 closing parenthesis rrbracket
\noindent syllogism is false ( Cf . Lemma 3 . 4 ) , and viceversa ( Cf . Lemma 3 . 5 ) . This allows
M comma g = 1 and llbracket TR to the power of minus 1 open parenthesis phi 3 closing parenthesis rrbracket
M comma g = 0 comma
\noindent us to prove the contrapositive of each of the implications of Theorem 3 . 6 .
\noindent Lemma 3 . 4 . Let $ \mu $ be a syllogism . If there exist $ M , g $
such that $ \llbracket TR ˆ{ − 1 } ( \phi 1 ) \rrbracket M , g $
\noindent $ = 1 , \llbracket TR ˆ{ − 1 } ( \phi 2 ) \rrbracket
M , g = 1 $ and $ \llbracket TR ˆ{ − 1 } ( \phi 3 ) \rrbracket
M , g = 0 , $ then there is $ M = T { RD }$ such
\begin{align∗}
that M = \mu .
\end{align∗}
\noindent Proof . \quad By theorem 3 . 5 , from $ M , g $ we can build up a model
$ M $ such that $ M = $
$ T { RD } , $ and such that $ \llbracket d \rrbracket M , g =
1 $ iff $ M ˆ{ M } = TR ( d ) [ g ] $ for every $ d \in P { V }
. $ \quad If we
assume that $ \llbracket TR ˆ{ − 1 } ( \phi 1 ) \rrbracket M ,
g = 1 , \llbracket TR ˆ{ − 1 } ( \phi 2 ) \rrbracket M ,
g = 1 $ and $ \llbracket TR ˆ{ − 1 } ( \phi 3 ) \rrbracket M
, g = 0 , $
CORCORAN quoteright S ARISTOTELIAN SYLLOGISTIC period period period .. 75
CORCORAN ’ S ARISTOTELIAN SYLLOGISTIC . . . \quad 75
then M to the power of M = phi 1 open square bracket g closing square bracket comma M to the power of
then $ M ˆ{ M } = \phi 1 [ g ] , M ˆ{ M } = \phi 2 [ g
M = phi 2 open square bracket g closing square bracket and M to the power of M = phi 3 open square bracket
] $ and $ M ˆ{ M } = \phi 3 [ g ] . $ By definition of satisfaction
g closing square bracket period By definition of satisfaction
we have that M to the power of M = open parenthesis phi 1 and phi 2 right arrow phi 3 closing parenthesis
\noindent we have that $ M ˆ{ M } = ( \phi 1 \wedge \phi 2 \rightarrow
open square bracket g closing square bracket comma then M to the power of M = forall x comma y comma z
o\ppehnipare3nthesi)s phi[1CaOnRdCgOpRhAiN2] ’rSigAhRt,IaS$rTrOoTwtEhpLIehAniN3ScY$lLoLsMiOnGgˆISp{TaIrMCen.t.}h.es=is7p5ertih\oefdnoMraMll=φ1x[g],M,M =yφ2[g], z
( \phi 1 an\dwMedMge=φ3\[gp].hBi y de2finitio\nriogfhsattairsfraoctwion \phi 3 ) . $
square-check
we have that MM =(φ1∧φ2→φ3)[g], then MM =∀x,y,z(φ1∧φ2→φ3).
Lemma 3 period 5 period .. Let mu be the syllogism forall x comma y comma z open parenthesis phi 1 and
\begin{align∗}
phi 2 right arrow phi 3 closing parenthesis period .. If there is M =
square−check
T sub RD .. such that M = mu comma .. then there are M comma g such that llbracket TR to the power of
\end{align∗} square−check
minus 1 open parenthesis phi 1 closing parenthesis rrbracket M comma g = 1 comma
llbracket TR toLtehmempoawe3r.of5m.inusL1etopµenbeptahreenstyhlleosgiissmphi∀2x,cylo,szi(nφg1p∧aφre2n→theφs3is).rrbrIafctkheetreMiscoMmm=a g = 1
\noindent Lemma 3 . 5 . \quad Let $ \mu $ be the syllogism $ \forall x ,
and llbracket TR tTo the posuwcehrtohfamt inMus 1=opµe,n pathreennththeesrisepahrei 3Mcl,ogsisnugchpathreantthTesRis−r1r(bφr1a)cMke,tgM c=omm1,a g = 0
y , z ( RD\phi 1 \wedge \phi 2 \rightarrow \phi 3 ) . $ \quad If there is
period
$ M = $
Proof period .. In an analogous manner to the proof above comma the proof is simple period Consider
$ T { RD }$ \quad such that $ M = \mu , $ \quad then there are $ M
the interpretation function llbracket tTimRe−s1(rφrb2r)aMck,egt=M1taondthTeRp−o1w(φer3)oMf M,gc=om0.ma i we built up right before
, g $ such that $ \llbracket TR ˆ{ − 1 } ( \phi 1 ) \rrbracket
Theorem 3 period 3 period .. If
M , g = 1 , $
M = mu commParo.o.f .thenIinnapnaartniaculolgaoruMs m=anonpeerntoptahreenptrhoeosifsapbhovie1,athnedpprhoiof2isrisgihmtpalerr.owCopnhsiide3rclosing
parenthesis open sqtuhaerientberrapcrkeettatgiocnlofsuinngctsiqonua·rMe Mbra,ickweet .b.ufioltrusopmreigghtpbeerifoodre..TBheyortehme 3 . 3 . If
\begin{align∗}
definitionofsatMisfacti=oncoµm, mathitenfoilnlowpasrtthicautlaMr M=phi=1op(eφn1s∧quφa2reb→rackeφt3g)c[glo]sinfgosrqsuoamreebgr.ackBetycomma
\llbracket TR ˆ{ − 1 } ( \phi 2 ) \rrbracket M , g = 1 and
M = phi 2 open sqtuhaerdeebfirnaictikoent ogfcslaotsiisnfagcstqiounar,eitbfroalclkowetsatnhdatMM==pφh1i[3g],oMpen=sqφu2a[gr]eabnrdacMket=gφc3lo[gs]i.ng square
b\rlalcbkretapcekreiotd TBRyˆT{he−orem13 .}3 w(e hav\epThRi−1(φ31)MM) ,i \=rrbr1a,ckTeRt−1(φM2)MM, ,i g= =1 and0 .
\end{align∗}
By Theorem 3 period 3 we have .... llbracket TR to the power of minus 1 open parenthesis phi 1 closing
parenthesis rrbracket M to the power of MTRc−o1m(φm3a)Mi =M,1ic=om0.masqllubarraeck−etchTeRckto the power of minus 1 open
\noindent Proof . \quad In an analogous manner to the proof above , the proof is simple . Consider
parenthesis phi 2 closing parenthesis rrbracket M to the power of M comma i = 1 .... and
the interpretPartoioofnof Tfuhneocrtemion3 . 6$.\llbWreawckilleptrove\thcedocotntrap\orsritbivreaocfkbeotth imMpliˆc{atMions}: , i $
llbracket TR to the power of minus 1 open parenthesis phi 3 closing parenthesis rrbracket M to the power of
we built up right⇒)befSourpepoTsehtehoarteTmR−31(φ.1)3+T.R\−q1u(φa2d) Itfurnstileleft−negationslashRD TR−1(φ3).
M comma i = 0 period square-check
We have that TR−1(φ1)+TR−1(φ2) = TR−1(φ3) ( Theorem 2 . 6 ) . Hence
Proof of Theorem 3 period 6 period .. We will prove the contrapositive of both implications :
\noindent $ M, the=re is\amutrue i,nt$erpr\etqautiaodn otfheTnR−i1n(φ1p)aarntdicTuRla−r1(φ2$)Mwhich=is n(ot tru\ephfoir 1
double stroke right arrow closing parenthesis .. Suppose that TR to the power of minus 1 open parenthesis
\wedge \phi TR2−1(φ3\).riBgyhlteamrmroaw3 . 4\,pthheire is3M =)TRD[suchgthat M] $= µ\.quaTdhusfoTrRDso=mµe, $ g
phi1closingparenthesisplusTRtothepowerofminus1openparenthesisphi2closingparenthesisturnstileleft-
. $ \quad By stohTeRDnegationslash−turnstileleftµ ( strong soundness of first - order logic ) .
negationslash RD TR to the power of minus 1 open parenthesis phi 3 closing parenthesis period .. We have
definition of sa⇐ti)sSfaimcitlairotno th,eiptrooffoalbloovwes, btuhtautsing$leMmma=3 . 5\inpshteiad .1squa[re−gcheck] ,
that
M = \phi 2Now[, itgjust]so$happanendsth$atMTheo=rem3\.p6hcianb3egene[ralizegdina]more. fr$uitful
TR to the power of minus 1 open parenthesis phi 1 closing parenthesis plus TR to the power of minus 1 open
way8 as Theorem 3 . 7 :
parenthesis phi 2 closing parenthesis = TR to the power of minus 1 open parenthesis phi 3 closing parenthesis
\noindent By TThheeoorermem33 .. 73.weLethaKvebe\ahsfeiltlof c$ate\glolrbicraal pcrkoepotsitioTnsRanˆd{ −d a s i1ng}le ca(te - \phi
open parenthesis Theorem 2 period 6 closing parenthesis period .. Hence comma there is a true
1 ) \rrbragcokriectal proMposˆi{tioMn i}n the, RDisyste=m , th1en K, (cid:96)RD\dllbracikffet TTRRDˆ+{ T−R(K1)(cid:96)} (
interpretation of TR to the power of minus 1 open parenthesis phi 1 closing parenthesis and TR to the power
\phi 2 ) \rrbracket M ˆ{ M } , i = 1 $ \hfill and
of minus 1 open parenthesis phi 2 closing parenthesis which is not true for TR to the power of minus 1 open
parenthesis phi 3 closing parenthesis period By TR(d)
\[ \llbracket TR ˆ{ − 1 } ( \phi 3 ) \rrbracket M ˆ{ M } , i =
lemma 3 period 4 comma there is M = T sub RD such that M = mu period .. Thus T sub RD = mu comma
0 . square−check \]
so T sub RD negatPiornosolfas.h-turWnsetislhealelfltpmrouve each direction in the theorem :
open parenthesis strong soundness of firs⇒t h)yTphheisndoirrdecetriolongiiscLcleomsimngap3a.re2nt.hesis period
double stroke left a⇐rr)owAscsluomsinegthpaatreKntnheesgiastiSoinmsillaasrhto−tthuernprsotiolfelaebfotvReDcdo.mmTahbeurtefuosrieng, tlehmermeais3apneriod 5
\noindent Proof of Theorem 3 . 6 . \quad We will prove the contrapositive of both implications :
instead period squainret-ecrhpercektation function ·M,g such that it is a true interpretation of K but dM,g =0.
Now comma .. Bityjutshtesoorehmap3pe.n5s t,hwaet Tcahneobrueimld3uppeariomdo6declaMn besugcehnetrhaaltizMed i=n aTmo,reand such that
$ \Rightarrow ) $ \quad Suppose that $ TR ˆ{ − 1 } ( R\Dphi 1 ) +
fruitful way to tdh(cid:48)Me p,ogw=er1oifff8 as Theorem 3 period 7 :
TR ˆ{ − 1 } ( \phi 2 ) turnstileleft−negationslash RD TR ˆ{ − 1 }
Theorem 3 periModM7 p=eriToRd(.d..(cid:48).)[Lg]etfoKr beveearysde(cid:48)t∈ofPca.tegoSriicnaclepwroephoasivteioansssuamndeddtahsatindg(cid:48)lMe c,agte=hy1pfhoern
( \phi 3 ) . $ \quad We have thVat
gorical propositeivoenryindt(cid:48)h∈e RKD, asnydsttehmatcodmMm,ga=the0n, iKt ftoullronwstsilaegleafitnsfurbomRDthedor.e..m. i3ff....5. tThastubMRD=plus TR
$ TR ˆ{ − 1 } ( \phi 1 ) + TR ˆ{ − 1 } ( \phi 2 ) = TR ˆ{ −
open parenthesis KTclos+ingTpRa(rKen)t,hbeusitsMturns=tileTleRft(d). By soundness of first - order logic it
1 } ( \phi RD3 ) ( $ Theorem 2 . 6 ) . \quad Hence , there is a true
TR open parenthesis d closing parenthesis
inPtreoropf preertioadti..onWeosfhall$prToRve(cid:31)ˆeta{hcah−tTdRirDec1+tioT}nRin(K(th)neet\ghapetohiroienmsla:s1h−t)ur$nstialenledftT$R(TdR). ˆ{ − 1 } (
\phi 2 ) $ which is not true for $ TR ˆ{ − 1 } ( \phi 3 ) . $ By
double stroke right arrow closing parenthesis This direction is Lemma 3spqeuriaorde2−pcehreicokd
lemma 3 . 4 , there is $ M = T { RD }$ such that $ M = \mu . $ \quad Thus
doublestrokeleftarrowclosingparenthesisAssumethatKnegationslash-turnstileleftRDdperiod.. Therefore
c$omTma{thReDre i}s an=inte8rW\pmereustaaytitohn,atf$uthnecgtsieononerali$zatTion i{n qRuDesti}on isnmeogreaftruioitfnulslbaecsahus−etwue crannsdtraiwlefrloemftit oth\emr u $
( strong sounindtnereesstsingorefsulftsircosntcer−ningorthdeesrystlemogs iDcan)d R.D , and the latter ’ s translation in first - order
llbracket times rrbracket M comma g such that it is a true interpretation of K but llbracket d rrbracket M
logic. Fortheseresults,see[1].
comma g = 0 period By theorem 3 period 5 comma
\hspace∗{\ fill } $ \Leftarrow ) $ Similar to the proof above , but using lemma 3 . 5 instead
we can build up a model M such that M = T sub RD comma and such that llbracket d to the power of prime
$ . square−check $
rrbracket M comma g = 1 iff
M to the power of M = TR open parenthesis d to the power of prime closing parenthesis open square bracket
Now , \quad it just so happens that Theorem 3 . 6 can be generalized in a more
g closing square bracket for every d to the power of prime in P sub V period .. Since we have assumed that
fruitful $ way ˆ{ 8 }$ as Theorem 3 . 7 :
llbracket d to the power of prime rrbracket M comma g = 1
for every d to the power of prime in K comma and that llbracket d rrbracket M comma g = 0 comma it
\noindent Theorem 3 . 7 . \hfill Let $ K $ be a s e t of categorical propositions and
follows again from theorem 3 period 5 that
$ d $ a s ingle cate −
M = T sub RD plus TR open parenthesis K closing parenthesis comma but M = TR open parenthesis d
closing parenthesis period .. By soundness of first hyphen order logic it
\noindent gorical proposition in the RD system , then $ K \vdash { RD } d $
Line 1 follows that T sub RD plus TR open parenthesis K closing parenthesis negationslash-turnstileleft TR
\hfill iff \hfill $ T { RD } + TR ( K ) \vdash $
open parenthesis d closing parenthesis period Line 2 square-check
8 sub We say that the generalization in question is more fruitful because we can draw from it
\begin{align∗}
other interesting results concerning the systems D and RD comma and the latter quoteright s translation in
TR ( d )
first hyphen order logic period For these results comma see open square bracket 1 closing square bracket
\end{align∗}
period
\noindent Proof . \quad We shall prove each direction in the theorem :
\centerline{ $ \Rightarrow ) $ This direction is Lemma 3 . 2 . }
$ \Leftarrow ) $ Assume that $ K negationslash−turnstileleft RD d . $
\quad Therefore , there is an interpretation function
$ \llbracket \cdot \rrbracket M , g $ such that it is a true interpretation of
$ K $ but $ \llbracket d \rrbracket M , g = 0 . $ By theorem 3 . 5 ,
we can build up a model $ M $ such that $ M = T { RD } , $ and such that
$ \llbracket d ˆ{ \prime } \rrbracket M , g = 1 $ iff
\noindent $ M ˆ{ M } = TR ( d ˆ{ \prime } ) [ g ] $ for every $ d ˆ{ \prime }
\in P { V } . $ \quad Since we have assumed that $ \llbracket d ˆ{ \prime }
\rrbracket M , g = 1 $
for every $ d ˆ{ \prime } \in K , $ and that $ \llbracket d \rrbracket
M , g = 0 , $ it follows again from theorem 3 . 5 that
$ M = T { RD } + TR ( K ) , $ but $ M = TR ( d ) . $
\quad By soundness of first − order logic it
\[\begin{aligned} \succ that T { RD } + TR ( K ) negationslash−turnstileleft
TR ( d ) . \\
square−check \end{aligned}\]
$ 8 { We }$ say that the generalization in question is more fruitful because we can draw from it
other interesting results concerning the systems D and RD , and the latter ’ s translation in
first − order logic . For these results , see [ 1 ] .
76 .. EDGAR ANDRADE ampersand EDWARD BECERRA
\noindent 76 \quad EDGAR ANDRADE $ \& $ EDWARD BECERRA
3 period 2 period .. Independence of T sub RD quoteright s axioms period .. Let T sub RD to the power of
i be the set T sub RD except for
\noindent 3 . 2 . \quad Independence of $ T { RD }$ ’ s axioms . \quad Let $ T ˆ{ i } { RD }$
axiomAsubiopenparenthesisi=1comma2comma3comma4closingparenthesisperiodInthissubsection
be the set $ T { RD }$ except for
comma we shall build up a model M sub i such
that M sub i = T sub RD to the power of i and M sub i is not a model of axiom A sub i comma for i = 1
c\onmominad2encotmmaxaio3m7c6om$mEDaAG4ApR{eAriNoiDdR}.A.DUE(sin&gEiDWAR=DBEC1ERRA, 2 , 3 , 4 ) . $ In this subsection , we shall build up a model
$ M { i }$ s3uc.h2 . Independence of T ’ s axioms . Let Ti be the set T except
these models M sub i quoteright s we will proveRtDhe independence of the axRioDms in T sub RRDDperiod The
for
following models meet the desired conditions :
\noindent thataxio$mMA ({i=i1,}2,3=,4). InTthˆi{s suibs}ec{tioRnD, w}e$shaallnbduild$uMp a m{ odiel}M$ siuschnot a model of axiom
open parenthesis 1 closinig parenthesis .. Let be M sub 1 = angbracketleftbig M comma Ai to the power of M
$ A { i } , t$hat fMor =T$i i and=M 1is not,a mo2del of,axiom3 A ,,for i4=1,2.,3$,4. \qUusaindg Using
comma E to the power of Mi angRbDracketrigihtbig sub comma where : i
thesemodelsM ’swewillprovetheindependenceoftheaxiomsinT .Thefollowing
M=openbraceacommabcomimacclosingbracecommaAtothepowerofM=opeRnDbraceopenparenthesis
\noindent thesmeodmelosdmeleset th$eMdesi{redico}n$ditio’nss: we will prove the independence of the axioms in
bcommacclosingparenthesisclosingbracecommaEtothepowerofM=openbraceopenparenthesisacomma
$ T { RD } . $ The ( 1 ) Let be M =(cid:104)M,AM,EM(cid:105) where :
cclosingparenthesiscommaopenparenthesisccommaa1closingparenthesi,sopenparenthesisbcommaaclosing
following models meet the desired conditions :
parenthesis closing brace period
M ={a,b,c}, AM ={(b,c)}, EM ={(a,c),(c,a)(b,a)}.
open parenthesis 2 closing parenthesis .. Let be M sub 2 = angbracketleftbig M comma A to the power of M
\centerline{( 1 ) \quad Let be $ M { 1 } = \langle M , A ˆ{ M } , E ˆ{ M }
comma E to the power of M angbracket(r2igh)tbiLgestubbecoMmm=a w(cid:104)Mhe,rAeM: ,EM(cid:105) where :
\rangle { , }$ where : } 2 ,
Line 1 M = open brace a comma b comma c closing brace comma A to the power of M = open brace open
parenthesis a comma b closing parenthesis commaMope=n {paa,reb,nct}h,esisAaMco=m{m(aa,cb)c,l(oas,incg)}parenthesis closing brace
\[ M = \{ a , b , c \} , A ˆ{ M } = \{ ( b , c )
Line2EtothepowerofM=openbrEacMeo=pe{n(ap,aare),n(tah,ebs)is,(ab,cao)m,(mb,aca),c(lco,sbi)n,g(ap,acr)e,n(tch,aes)i}s.commaopenparenthesis
\} , E ˆ{ M } = \{ ( a , c ) , ( c , a ) ( b ,
a comma b closing parenthesis comma open parenthesis b comma a closing parenthesis comma open parenthesis
ba com)ma c\c}losing.pa\r]enthesis comma o(p3en) paLreenttbheesMis c3c=om(cid:104)Mm,aAbMc,loEsMing(cid:105),pwahreenrteh:esis comma open parenthesis
a comma c closing parenthesis comma open parenthesis c comma a closing parenthesis closing brace period
open parenthesis 3 closing parenthMesi=s ..{aL,ebt,cb}e,MAsMub=3 ={(aa,nbg)b,(rba,cck)e}t,leftEbiMg M=c∅om. ma A to the power of M
\centerline{( 2 ) \quad Let be $ M { 2 } = \langle M , A ˆ{ M } , E ˆ{ M }
comma E to the power of M angbracketrightbig sub comma where :
( 4 ) Let be M =(cid:104)M,AM,EM(cid:105) where :
\rangle { , }$ where : } 4 ,
M=openbraceacommabcommacclosingbracecommaAtothepowerofM=openbraceopenparenthesis
a comma b closing parenthesis comMma=o{pae,nb,pc}a,rentAhMesis=b{(cao,mb)m},a cEclMosi=ng{p(ba,rce)n,t(hce,sbi)s}.closing brace comma E
\[\begin{aligned} M = \{ a , b , c \} , A ˆ{ M } = \{ (
to the power of M = varnothing period
a ope,n pabrenthe)si3s.4,3clo.sin(gMpianraiemntahleis,tiys .o.fcLRetDb)e.M sW\u}be4s\a=\y athnagtbrRaDckeistlemftibniigmMal coifmtmheareAistonothperoppoewrer of M
comEmˆa{EMto}the=poswuebr\se{otfoMf it(asnsgebtraaocfkreutlr,eisgh(tsbaeiegpsu.b)7c0o)m,,mleatw’(hsesraey:aMin,,suchbthat f)or eve,ry d(∈PVband , a
) M,=ope(nbrabceevaecr,oymKma⊆cbPcVo,)mKm(cid:96)aRc,Dcldosii(ffngKbr(cid:96)caMceincod,m. mabAto)thepo,wero(fM=aopen,bracceopen) pare,nthes(is
accomm, abaclosin)gpare\Tn}htheemsis.incil\moesanilnidtgy{baorfalicRgeDncoewmdom}ul\ad]EbetootbhteaipnoewdebryofsoMlvi=ngopthene fborlalocweionpgenqupeasrteionnthse:sisbcomma
c closing parenthesis comma op(ean)pareAnrtehethsiesacxcioommsmianbTRclDosiinndgeppaernednetnhtesfirsomcloesaincghbortahceer p?eriod
3 period 3 period .. Min(imba)lityLoeftRµDbepetrhieodsy.l.loWgiesmsay∀xt,hya,tzR(φD1∧isφm2in→imφa3l)if:tisheitretrisuenothpartoper
\centerline{( 3 ) \quad Let be $ M { 3 } = \langle M , A ˆ{ M } , E ˆ{ M }
subset of its set of rules open parenthesis see p period 70 closing parenthesis comma let quoteright s say Min
c\ormamngalseuch{tha,t f}o$r evwerhyedrein:PTR}suDb(cid:96)Vµ =⇒ TR−1(φ1)+TR−1(φ2)(cid:96)RD TR−1(φ3)?
and every K subset equal P sub V comma K turnstileleft sub RD d iff K turnstileleft sub Min d period
( c ) Let µ be the syllogism ∀x,y,z(φ1∧φ2 → φ3) and Min the set of RD ’ s
\[ M = \{ a , b , c \} , A ˆ{ M } = \{ i ( a , b )
The minimality of RD would be obtained by solving the following questions :
rules except for rule i( with i= I , II , III , IV ) : could it be true that
, ( b , c ) \} , E ˆ{ M } = \varnothing . \]
open parenthesis a closing parenthesis .. Are the axioms in T sub RD independent from each other ?
openparenthesisbclosingpaTreinth(cid:96)esµis..⇐L=etmuTRbe−1th(φe1s)y+lloTgiRsm−1f(oφr2a)ll(cid:96)xcommTRa−y1c(oφm3)m?azopenparenthesis
phi 1 and phi 2 right arrow phi 3RcDlosing parenthesis : is it true that Mini
\ceTnstuebrlRinDet{u(rnIs4ntilfe)alcetf\tq,muifau(deaqL)uae,lt-arb(reobw)d$banlrMdig(htc{T)R4 ta}oretht=reupeo,w\weleraconafgnmlceionnucsluM1doeptehna,tpRarDeAnitshˆe{asimMsipn}himi a1l,closinEg ˆ{ M }
p\arraenntghleesis p{lus,T}Rs$ytsotewtmhhefeoprroetwhe:erAo}frimstiontueslia1nospyelnlopgiasrteicnt:hesis phi 2 closing parenthesis turnstileleft sub RD TR
to the power of minTuhse1oroepmen3pa.re8nt.hesisIpfh(ia3)cl,osing( pba)reanntdhe(sics)?are tru e , then for i= I , II , III ,
\[ oMpenp=arent\h{esiIsVcatchloesrine,gisparbenthes,is.. Lcetm\u}beth,esyllAogiˆsm{ Mfora}llx=comm\a{ycom( maazopen, parebnthes)is
p\h}i 1 an,d phEi 2ˆr{igahMtsay}lrlroogw=ismphi∀\3x{c,lyo,szin((φgip∧arbφeint→he,sφiis)asncudchMtihn)atsu:b,i the(set ofc , b ) \} . \]
1 2 3
RD quoteright s rules except for rule i open parenthesis with i = I comma II comma III comma IV closing
parenthesis : could it be true that
\noTinsudbenRtD3to.the3po.w\erquofadi tuMrnisntiilmeleaftlimtyu aor(fr1o)wRDdTblR.le−ft\1-q(eφqui1ua)ad+l TWTRRe−tso1a(tyφhi2e)tp(cid:96)hoRawDterTRoRDf−m1ii(nsφui3sm)1inoapinmednalpariefnthtehseisre is no proper
phsiu1bcsleotsinogfpairetnsthesseist(p2lo)ufs TTrRRul−teo1s(tφhi1(e)p+soewTeeRr−po1f(mφ.i2i)n7tu0usrn1)sotpi,leenlleepftatr−e’nnthesegsasitsaiopynhsil2a$schlMoMsiinningipTaRr,e−n1$t(hφei3s)si.suctuhrntsthilaetleftfor every
$ d \in P { V }$
sub Min sub i TR to the power of minus 1 open parenthesis phi 3 closing parenthesis ?
and every $ PKroof \.subWseetweqill proPve e{achVca}se in, turnK. \Wvedsatsahrt ou{tRfrDom}thedlas$t twiofcfases$,K \vdash { Min }
In fact comma if open parenthesis a closing parenthesis comma .. open parenthesis b closing parenthesis and
d . $ because they are easier to prove : Case i= III : If we t ake axiom A we have :
open parenthesis c closing parenthesis .. are true comma we can conclude that RD is a min3imal
system for the Aristotelian syllogistic :
\centerline{The minimality of RD would be obtained by solving the following questions : }
Theorem 3 period 8 period .. If open parenthesis a closing parenthesis comma .. open parenthesis b closing
T (cid:96)∀x,y,z(yAz∧xAy →xAz) & T3 turnstileleft−negationslash∀x,y,z(yAz∧xAy →xAz)
parenthesis and opeRnDparenthesis c closing parenthesis areRtDru e comma then for i = I comma II comma III
\centerline{( a ) \quad Are the axioms in $ T { RD }$ independent from each other
comma IV the re is
$ ? $ }
a syllogism forall x comma y comma z parenleftbig phi sub 1 to the power of i and phi sub 2 to the power of
i right arrow phi sub 3 to the power of i parenrightbig such that :
\centerline{( b ) \quad Let $ \mu $ be the syllogism $ \forall x , y ,
Line 1 open parenthesis 1 closing parenthesis TR to the power of minus 1 parenleftbig phi sub 1 to the
z ( \phi 1 \wedge \phi 2 \rightarrow \phi 3 ) : $ is it true that }
power of i parenrightbig plus TR to the power of minus 1 parenleftbig phi sub 2 to the power of i parenrightbig
turnstileleft sub RD TR to the power of minus 1 parenleftbig phi sub 3 to the power of i parenrightbig and Line
\[ T { RD } \vdash \mu \Longrightarrow TR ˆ{ − 1 } ( \phi 1 ) +
2 open parenthesis 2 closing parenthesis TR to the power of minus 1 parenleftbig phi sub 1 to the power of i
TR ˆ{ − 1 } ( \phi 2 ) \vdash { RD } TR ˆ{ − 1 } ( \phi 3 )
parenrightbigplusTRtothepowerofminus1parenleftbigphisub2tothepowerofiparenrightbigturnstileleft-
? \]
negationslashMinsubiTRtothepowerofminus1parenleftbigphisub3tothepowerofiparenrightbigperiod
Proof period .. We will prove each case in turn period .. We start out from the last two cases comma
because they are easier to prove :
\noindent ( c ) \quad Let $ \mu $ be the syllogism $ \forall x , y ,
Case i = III : If we t ake axiom A sub 3 we have :
z ( \phi 1 \wedge \phi 2 \rightarrow \phi 3 ) $ and $ Min { i }$
T sub RD turnstileleft forall x comma y comma z open parenthesis yAz and xAy right arrow xAz closing
the set of
parenthesis ampersand T sub RD to the power of 3 turnstileleft-negationslash forall x comma y comma z open
RD ’ s rules except for rule $ i ( $ with $ i = $ I , II , III , IV ) : could it be true that
parenthesis yAz and xAy right arrow xAz closing parenthesis
\[ T ˆ{ i } { RD } \vdash \mu \Longleftarrow TR ˆ{ − 1 } ( \phi 1
) + TR ˆ{ − 1 } ( \phi 2 ) \vdash { Min { i }} TR ˆ{ − 1 } (
\phi 3 ) ? \]
\noindent In fact , if ( a ) , \quad ( b ) and ( c ) \quad are true , we can conclude that RD is a minimal
system for the Aristotelian syllogistic :
\noindent Theorem 3 . 8 . \quad If ( a ) , \quad ( b ) and ( c ) are tru e , then for
$ i = $ I , II , III , IV the re is
\noindent a syllogism $ \forall x , y , z ( \phi ˆ{ i } { 1 } \wedge
\phi ˆ{ i } { 2 } \rightarrow \phi ˆ{ i } { 3 } ) $ such that :
\[\begin{aligned} ( 1 ) TR ˆ{ − 1 } ( \phi ˆ{ i } { 1 } ) + TR ˆ{ −
1 } ( \phi ˆ{ i } { 2 } ) \vdash { RD } TR ˆ{ − 1 } ( \phi ˆ{ i } { 3 }
) and \\
( 2 ) TR ˆ{ − 1 } ( \phi ˆ{ i } { 1 } ) + TR ˆ{ − 1 } ( \phi ˆ{ i } { 2 }
) turnstileleft−negationslash Min { i } TR ˆ{ − 1 } ( \phi ˆ{ i } { 3 }
) . \end{aligned}\]
\noindent Proof . \quad We will prove each case in turn . \quad We start out from the last two cases ,
\noindent because they are easier to prove :
Case $ i = $ III : If we t ake axiom $ A { 3 }$ we have :
\[ T { RD } \vdash \forall x , y , z ( yAz \wedge xAy \rightarrow
xAz ) \& T ˆ{ 3 } { RD } turnstileleft−negationslash \forall x , y
, z ( yAz \wedge xAy \rightarrow xAz ) \]