On the identity of fuzzy material conditionals
Abstract
Given $\mu , \; \eta :X \rightarrow [0,1]$ we study when the equality $I_{\mu }^{T}=I_{\eta }^{T}$ holds, $T$ being a continuous t-norm, and $I_{\theta }^{T}$ the elemental preorder:\\ \[ I_{\theta }^{T}(y/x) = sup \{z\; ; \; T(\theta (x),z)\leq \theta (y) \}. \]
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Ma th w are&Sof tC om puti ng 3( 19 94)30 9-31 4 On theIden tit yofF uzzyMaterial Conditionals Enri cT r ill as a ,SusanaCubil lo b andA dol foR.DeSoto c a Dept. I n tel igenci aArt ici al .U niv .P ol it ec nicadeMadrid. 28600Boadi ll ad elMon te.Madrid,Spain. b Dept.Matem atic aA pl ic ada.U niv .P ol it ec nicadeMadrid. 28600Boad i ll adelMon te.Madrid,Spain. e-mail: sc ubillo@.upm.es c De pt.Di recc i onyEconom a de laEm presa.Uni v.deLe o n. 24071Le on,S pa i n. Ab stract Giv en ; : X ! [0 ; 1]w estudyw hentheequalit y I T = I T holds, T being acon tin uoust-norm,and I T theel ementalpreorder: I T ( y=x )= sup f z ; T ( ( x ) ;z ) ( y ) g : Keyw ords: Equalit yof T -preorders, MaterialC onditionals. 1In troduc ti on. Givenasubset A X , A 6 = ; ,them aterialcondi tiona lassociate dt o A isthe relati oni n X 2 X g iv enb y ! A =( A 2 A ) [ ( A 0 2 X ), whic hisob viouslyapreo rd er. Thef oll o wi ngtheorem gi vesa characteri zationo fa subs et throughitsassoc iate d materi alcondit i onal[ ? ]. Th eorem 1 If A and B ar en o n-empty subsets of X , then: ! A = ! B if an d only if A = B Proof. The s u ci en t c ondi tion isim medi ate . Furthe rm ore , s i nc e ! A =( A 2 A ) [ ( A 0 2 X )=( X 2 X ) 0 ( A 2 A 0 ), i f ! A = ! B , that is, if( X 2 X ) 0 ( A 2 A 0 )=( X 2 X ) 0 ( B 2 B 0 ), the n A 2 A 0 = B 2 B 0 ,which holds if and onlyif A = B . 2 309
310 E.T ri ll as,S .C ubil lo &A .R.DeSoto Givenaf uzzyrela ti on I : X 2 X ! [0 ; 1] ,andacon ti n uoust-norm T ,i ti s known[ ? ]that I isa T - preorde ri fa ndo nlyi f I =i nf 2F I T ; F bei ngthesetof the T -l ogicalsta tesof I ,tha ti s F = f : X ! [0 ; 1]; T ( ( x ) ;I ( y=x )) ( y ) 8 x;y g ; and I T thee le m en ta lpreo rderdenedb y I T ( y=x )=sup f z 2 [0 ; 1 ]: T ( ( x ) ;z ) ( y ) g : Choo si ng = ' A ,thecl assi cc haracteristicfunctiono f A ,is I T ' A ( y=x )= ' ! A ; andtheni tisnaturaltoco ns i derthatthef uzzyelem ent al preordersa reageneralizationofthecla ssicm ater ia lcondi ti onal s 1 .W ecanthereforestudywhet heriti s po ssibl etoextendtheorem1tothef uzzypreo rders. 2Equal i t yoffuzz ym ate ri alcondi t ional s. Ifw estarttopa yatten tio ntotheequiv alenceforthe Min t-no rm ,w eobtai ntha t ifthepreorders I Min , I Min a reequa l,then and canonlybedi eren tatthe m axi m umv a lueso f x a nd x . Th eore m2 I Min = I Min if an do nl yif x 6 = x ) x ya nd x yforally: Pr oof. Le t I Min ( y=x )= I Min ( y=x ), fore ac h x;y .W ek no wthat: I Min ( y=x )=sup f z ; Min ( x; z ) y g = 1 ;if x y y; if x > y : Le t us c ho ose a n x suc h that x 6 = x , and let us supp ose that there exists a y ve ri fying x < y .W ewil l obtain I Min ( y=x )=1= I Min ( y=x ), and then x y . Fur ther m ore , I Min ( x=y ) = x = I Min ( x=y ) < 1, and so I Mi n ( x=y )= x = x ,whic hg iv es a c o ntr a dict i on; s o ,it m ust b e x y for al l 1 The elementalpreorders are not theunique generalizations of the material conditional: see for example [ ? ]
Ontheiden tit yo ff uzzy materi alco ndi tio nals 3 11 y .In asim ila rw a y , x y is obtainedfo rall y . Re ciproca ll y ,letussuppo se that if x 6 = x then x y and x y ; for all y .W ewi llpro v ethatthe eq ualit y I Mi n = I Mi n holds. -If x = x and y = y ,cl ea rl y I Min ( y=x )= I Mi n ( y=x ). -If x = x and y 6 = y ,the n y x , y x ,and I Mi n ( y=x )= I Min ( y=x )=1. -If x 6 = x and y = y ,w eo btain x y , x y ,a nd I Mi n ( y=x )= y = y = I Min ( y= x ). -Fina ll y, i f x 6 = x and y 6 = y ,since x y;x y ,and y x ;y x ,ob v iously x = y , x = y ,an d I Mi n ( y=x )= I Min ( y= x ). 2 Note. If T isArc hi m edeanwi thadditi v egenerator h [ ? ] ,thenfrom: I T ( y=x )= sup f z=T ( x ;z ) y g =sup f z=h ( 0 1 ) ( h ( x )+ h ( z )) y g , i tfol lo ws: -If x y ,then h ( x ) h ( y ),and h ( x )+ h ( z ) h ( y )fora ll z .So , h ( 0 1) ( h ( x )+ h ( z )) h ( 0 1 ) h ( y )= y foral l z ,and I T ( y=x )=1. -If x>y , h ( x ) <h ( y )and h ( y ) 0 h ( x ) 2 [ 0 ;h (0)] .Letuss ee that I T ( y=x )= h ( 0 1 ) ( h ( y ) 0 h ( x )).Inf act, h ( 0 1) ( h ( x )+ h h ( 0 1) ( h ( y ) 0 h ( x )))= h ( 0 1 ) ( h ( x )+ h ( y ) 0 h ( x ))= h ( 0 1 ) ( h ( y ))= y andfora ll z> h ( 0 1) ( h ( y ) 0 h ( x )),i tis h ( z ) <h ( y ) 0 h ( x ), h ( x )+ h ( z ) <h ( y ) h (0) and h ( 0 1) ( h ( x )+ h ( z )) >h ( 0 1) ( h ( y ))= y . Therefore,w eo bta in I T ( y=x )= 1 ; if x y h ( 0 1) ( h ( y ) 0 h ( x )) ; if x>y: The fol lo wing theore m ass ert s that i n the case of str i c t Arc hi mede a n t-norm s, the preord ers associ ate d to and are e qua lifa nd onl yi f and ha ve" si milar form s". Th eorem 3 If T is a str ict A rchime de an tnorm w ith ad dit ive gen er at or h , I T = I T if an d only if ther e exist s k 2 R such that for al l xh ( x )= k + h ( x ) . Proof. As T is str ict, h ( 0 1) = h 0 1 .Let I T = I T ,and let us cho ose s ome x; y .
312 E.T ri ll as,S .C ubil lo &A .R.DeSoto Fi rs tly ,letuspoi n toutthat if x <y ,then I T ( x=y )= I T ( x=y )= h ( 0 1) ( h ( x ) 0 h ( y )) < 1;so x<y .Simi la rl y ,if x<y ,i tis x<y . So ,if x < y; I T ( x=y )= h 0 1 ( h ( x ) 0 h ( y ))= I T ( x=y )= h 0 1 ( h ( x ) 0 h ( y )), ifa ndonlyif h ( x ) 0 h ( y )= h ( x ) 0 h ( y ),ifandonly if h ( x ) 0 h ( x )= h ( y ) 0 h ( y ). Al so ,if y < x ,then h ( y ) 0 h ( y )= h ( x ) 0 h ( x ). Fi nally ,i f x = y then x = y ,andnewly h ( x ) 0 h ( x )= h ( y ) 0 h ( y ). Therefo re h ( x ) 0 h ( x )isaconstan t k fo rall x ,and h ( x )= k + h ( x )fo r eac h x . R eciprocall y, if thereexi sts k 2 R suc hthatf or all x theequal it y h ( x )= k + h ( x )hol ds, since h ( x ) 0 h ( x )= k = h ( y ) 0 h ( y ), w eobtain h ( x )+ h ( y )= h ( y )+ h ( x );then x y ifand on l yi f h ( x ) h ( y ),i fa ndonly if h ( x ) h ( y ),ifandonl yif x y . So ,if x y ,then x y and I T ( y=x )=1= I T ( y=x );andi f x>y then x> y ,a nd I T ( y=x )= h 0 1 ( h ( y ) 0 h ( x ))= h 0 1 ( k + h ( y ) 0 k 0 h ( x ))= h 0 1 ( h ( y ) 0 h ( x ))= I T ( y=x ). 2 Intheparti cu larcaseinwhic h and "ha v ep oin ts",thati s,theyarenorm alized ,w ecanget: Cor oll ary1 If T isA r chi me de an andstr ict,and and are such thatther eexist x;y wit h x =1 , y =1 ,then I T = I T ifandonlyi f = . Pr o of. Thesucien tco ndi tioni scl ea r. On the othe r ha nd, if I T = I T , t here exi sts k 2 R suc h that forall zh ( z )= k + h ( z ). In pa rti cul ar, h ( x )=0= k + h ( x ) h ( x ), a nd ne ce ssaril y k 0. Anal ogo us l y, h ( y )= k + h ( y )= k +0, a nd the n k 0. So k =0 ,a nd the n h ( z )= h ( z ), and z = z for all z . 2 Now, for the case in which and \have not somepoints ",wehave
Ontheiden tit yo ff uzzy materi alco ndi tio nals 3 13 Corol l ary2 If T i sA r chimedeana nd stri ct,and and ar es uc ht hatther eexist x; y w ith x =0 , y =0 ,then I T = I T i fandonlyi f = Pr o of. If I T = I T ,thereexi sts k 2 R with h ( z )= k + h ( z )f oral l z . As x =0,i ti s h (0 )= k + h ( x ) k + h (0),a ndthen0 k .A ndsince y =0 , h ( y )= k + h (0) h (0), and k 0. W egetthat k =0and = . 2 Th eorem4 If T isanon-stri ctA r chime de ant-nor m,withaddi tivegener ator h , t he n: I T = I T ifa nd onl yifther eexi sts k 2 R sucht hatforall x i tis h ( x )= k + h ( x ) . F urt her mor e,ifthat holds, forall x i tis k h (0) 0 h ( x ) . Pr o of. Let I T = I T be,a ndlet usc hoosean y x ;y . -If x<y , h ( x ) >h ( y )and I T ( x= y )= h ( 0 1 ) ( h ( x ) 0 h ( y ))= I T ( x =y ) < 1 ,andthen x<y , h ( x ) >h ( y )and I T ( x=y )= h ( 0 1) ( h ( x ) 0 h ( y ))= h ( 0 1) ( h ( x ) 0 h ( y )),whi chim pli es(becauseof h ( x ) 0 h ( y ) 2 [ 0 ;h (0)]and h ( x ) 0 h ( y ) 2 [ 0 ;h (0 )]) that h ( x ) 0 h ( y )= h ( x ) 0 h ( y )a nd h ( x ) 0 h ( x )= h ( y ) 0 h ( y ). -Inasim i la rw a yi f y<x ,t hen y<x y h ( y ) 0 h ( y )= h ( x ) 0 h ( x ). -Inthecasei nwhic h x = y , x = y and h ( x ) 0 h ( x )= h ( y ) 0 h ( y ). Thenfo ral l x;y ho lds h ( x ) 0 h ( x )= h ( y ) 0 h ( y ), andthereexi sts k 2 R suc hthatforal l xh ( x ) 0 h ( x )= k and h ( x )= k + h ( x ). Inthi scase,if ther eexi sts x wi th k>h (0) 0 h ( x ), w eobtai nthecon tradicti on h ( x )= k + h ( x ) >h (0) 0 h ( x )+ h ( x )= h (0). So k h (0 ) 0 h ( x )fo ral l x . R ec ipro ca ll y, letussupposethatt he re ex ists k 2 R suchtha tfora ll xk h (0 ) 0 h ( x )y h ( x )= k + h ( x ). -If x y , k + h ( x ) k + h ( y ), h ( x ) h ( y )and x y . Ther efore, I T ( y=x )= I T ( y=x )=1. -I f x > y , k + h ( x ) <k + h ( y ), h ( x ) <h ( y ), and x > y . Then, I T ( y=x )= h ( 0 1) ( h ( y ) 0 h ( x )) = h ( 0 1) ( k + h ( y ) 0 k 0 h ( x )) = h ( 0 1) ( h ( y ) 0 h ( x )) = I T ( y=x ). 2 Let us p oint out that newly,if the pre order s associated to and to c oncur, and must have "similar forms",butnow, fur ther more , their distance isbounded
314 E.T ri ll as,S .C ubil lo &A .R.DeSoto by h (0) 0 sup x f h ( x ) g . Anal ogo usl y ,i nt he caseinwhic h and " ha ve po in ts " ,andinthecasein whic hitscom pl em en ts "ha v epoi n ts" ,i tholds: Coroll ary3 If T isA r chime de annon -str ictand and ar esuchtha tt her eexist x;y wit h x =1 , y =1 ,orther eexist x; y wi th x =0 , y =0 ,then I T = I T i fandonlyif = : Pr o of. Si m i lartothe caseo ftheArc him edeanstric tt-norm s. Ac kno wl e dge m e n ts. Theautho rsa rei ndebted toProf .Cl audiAlsina(Barce l ona-Spai n)andtothe referees,fort heirhin tsi nthedev elopm ent o fthispaper. Re ference s [1 ]Cubi ll o,S., Contri bu ci onalestudiodel al o gi ca ydel osc ondi cionales b or r osos .(inSpanish),DoctoralThesi s,Universi dadPo li t ecnicadeMadrid. Madri d,19 93. [2] Sc hw eizer,B. ,Sklar,A., Pro babil isti cMetri cSp ac es. ElsevierNorth-Holl and. New York,198 3. [3 ]T rill as,E. ,Cubillo,S .,Rodr g uez,A. , A nE ssay on NameandExtensionof R ule-givenpr op erties .IP MU' 94.P ar s,1 994. [4 ]T rill as,E ., On L o gicandF uzzy Lo gic .In t. JournalofUncert ai nt y, Fuzz iness an dKno wledge - B ased Sys tem s, V ol. 1, No. 2 10 7-137 ,1 993.