modal logic for philosophers

OUTUBRO ROSA
16 de outubro de 2018

modal logic for philosophers

550 /FontDescriptor 200 0 R :"u'CJ_@BkTgD6 (Hint: Begin with this special case of (B): ∂Aç∫∂∂A. /TR2 /Default 2.5. There is, in fact, a way to represent this where it is true, namely W&OB. UR-B:0E +!2#KDR9/l(j2P$nk^@oeFL%Y>J"5/BqO(:PgI7^1aF@H3 << If we label the ∫ and ∂ in (B) with L and R, the two versions of (B) look like this: A ç ∫ R ∂L A A ç ∫L ∂R A If we identify H with ∫L , P with ∂L , G with ∫R , and F with ∂R , you will recognize them (I hope) as two axioms of the system Kt. 293 292 536 286 881 591 539 571 560 410 432 345 575 497 737 505 We will show first that the process of constructing the Mj preserves consistency, that is, that if Mi is consistent, then so is Mi+1 . 203 0 obj ! Since ∂A µ ∂V whenever av (A)=T, it follows that au (∂A)=T whenever av (A)=T for any wff A. PKdY#BN%iDp]atG4O!\[51Y&XU?iJ$:Zfm4]P"L_B6F#*4AY_CA`mp-D(2D_f*ge\iXC[S-&LQ"^cm-U`MQAmD!>s-a7:8$qt;"#fhkV]Q&9K)<3Z,)gFWPT8["V[N3_< Some students worry that this failure to define W in more detail is a defect. Ap.5l('Q9#nO!jbgdCE6>* r>>GR7\RQfOJQ2!. /CapHeight 687 /A 116 0 R 112 0 obj 186 0 obj endobj g'r>XE6)s9S5Gol739gYs*nDNdYfdCja:Ffo;\_mB(t0KIXKO8h:c`:D@8AlD;1"u << /Filter /FlateDecode /Length 189 0 R >> 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 Designed for use by philosophy students, this book provides an accessible, yet technically sound treatment of modal logic and its philosophical applications. 161 0 obj This by (Def~) is really ~~A, from which we obtain A by (DN). To prove the condition, assume wRv and aw (c) µ Dw, and then prove av (c) µ Dv as follows. 72 0 obj /FontStretch /Normal 250 250 250 250 722 250 250 250 250 250 250 389 250 250 667 250 /FontDescriptor 151 0 R The problem is that quantified modal logic is not as well developed, and it is difficult for the student of philosophy who may lack mathematical training to develop mastery of what is known. >> 78 0 obj However, this reply takes a fairly nonstandard attitude toward what count as identical beliefs. endobj /Type /ExtGState . 20 0 obj y��9�z>H%O�%z�y�+��w�϶ͯ��ф"��(-E�6����"�J��o�C�!�I�1���W�x[�~L�� tt��c /op false .+D���:��p��č��b�'�����jӽ�;®[�����psGk�e/'�|�#�^0�e����}���^#��#�H.u�f���رe�vT�1ޭ͵��`yr�-�ʈ�>|�'�w~1�^}e/�� ~��{�9^��__�Ǵ���x��%����H�����O� �4"_ ,ma5iZ-?+3RF-1cYTUaH9?$6eUrkl/hB:J,M15gI. . +Xu^$6I\Lr&eiq>JJDEVW(naB@okG+WI:%T6];,iUG]ZQ)A"=L"CEtSGTOoi.NShZ /BaseFont /FGDBAH+TimesTen-Roman Then there must be a K-model where for some w in W, aw (L, B)=T and aw (A)=F. /URI (http://www.cambridge.org/0521682290) Next we illustrate a somewhat more complex proof that uses (B) to prove the argument p / ∫∂∂p. This is crucially important. 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 , Un is a list of sentences of U. To show A is false at w (aw (A)=F), we could place A outside the region. Semantics for Tense Logics In tense logic, there are two pairs of intensional operators. endobj i/.i7=iOBrE/\:-8K]n,d-dusD%3"uLlK=6VK\q]`;ghb[7F`EIZ8S6Y2PW98X,;[ >> G(t_V-B,t2/m8&2R9IEaJ& Once we have taken the point of view that adopts (OO), there seems to be no reason not to accept the policy of iteration embodied in S5 and simply ignore any extra deontic operators. A sentence *(ƒ) equivalent to *B can be constructed by starting with ƒ and repeatedly adding a left conjunct or ∂ at each step. 8b$SNWBZTRY-J/h!\h-(!eG@!DeI? 8c,S?kaJ=q/4qd-D;m.uWeY78mK>i];#4[UE+Qi?nq7RXr@QO6]A9U:6*1GN5;[1' A has the form p. Variables p are verified because the definition of a says that aw (p)=T when p is in w. Case 2. IIuTu-]B+Aq)hsZ6UD?D,&C=2.=#3^&[>pZtseDF*+c/5cR^qcRjQZ:h>@!aJK"fQFhme5bt<33ElW << << If this is how we understand O and P, it is clear that we cannot accept (OO) or (OP). /TR2 /Default 71#2;25O;7POB^.XFWD1$9]5W? 189 0 obj The variety found here might be somewhat bewildering, especially for the student who expects uniformity in logic. .X&n4F_j@;+"3<7MHU6jn\o%VgE]Xt'FF/O)MOHh1_Yp)XJmI=*&4mOg/Q8d-oV]: !&A$2P`[8O!X:/%Z>1dm&$(os\H //hrDS$FTHTUG:J/G]lsltg=Y>0H(Nd6p+##7`Q.houC83@tk$oV0'X\f&7K9TG<< @kco0D6d^+Bb'Ep6;2`lBN&'"O<2VHmSAh'kic+b?#\,d@.54QUJUo\R&:bG_j" /Subtype /Link We will say a branch is closed when it contains ƒ. /URI (http://www.cambridge.org) << O8?A$J?22*n`q=u&-Q6$e-XG0*TqV+:?#@W(pR[cQg,6j548D*$u!-7^V?B6b25tO^&.MVMST(@rrP)Gl`g?Y It will follow from this and the fact that all closed branch sentences are inconsistent that all parents of closed branch sentences are inconsistent. '\Gt3-oH9HqX&;4YDMu7D6dk:t\Mo_6_?4u=990$FfjSgc+JMr>Z Copi, I., and Gould, J. H9lj3(W\1gLq-KB>E"c+*&bS@^k%%[@0d-+nke3TDUdi?7;%iUJM3fkQt6W+220s$ 634 Since (I) is provable in K + (I), it follows that (I) must be I-valid. $#Nb6'i6*1+L*WukZ:'u(99q"kVM5e^=BH`5j%'^sh 9 Completeness Using Canonical Models Not all the systems mentioned in this book have been shown to be complete, only the ones for which a method has been described for converting trees into proofs. You may assume without proof that (4) ∫Aç∫∫A is provable in GL. Ahr#H.9J>PIR3##BkXa^(u'7MJ0'ff#0c;H'H6#r)hTfu@s]O]cM4KTin243j/:2F /op false 205 0 obj endobj /Qt[$@drGm].I+T*4qt8_Jf^73:?Ae3eLCqlSemb#`M)^7tEUU6A@.$E^XJJhe[2!ied381XfjLP1LWLp\U;@E>6P"`Q^*Ad,] 34WiNf[BC^!kHMfK?BrZTM4H,o,*EXEUjs7R]=CV79^l*jKT:u#Jh?6g2@Qq2[B%' However, this reasoning rests on a debatable assumption, one that is initially attractive, but probably false. TEHuUA<7t->*dA?=lq*! << /Filter /FlateDecode /Length 191 0 R >> Qg#ARVf7sa@HaP5(o7&(j['s+N`fdf$gBKsFuGTW^@t&);ol:%&)8]*f,7(](_BS? >> << By the same reasoning, we may prove also that any serial, transitive, and symmetric relation is euclidean. endobj UhLK:#=u0&/C`o5.>fYp;4rLUA7SEPN3(-O;'p/17!Y2"F[HqMXGucpWTuU,h'(+- endstream 3., Reidel, Dordrecht. Iteration of Obligation Questions about the iteration of operators, which we discussed for modal logics, arise again in deontic logic. /A 213 0 R QWBQc.%ak58+.`u. This can be done with a very simple diagram. A minimal system of tense logic called Kt results from adopting the principles of K for both G and H, plus two axioms to govern the iteration of G and H. System Kt = PL + (GOut) + (GIn) + (HOut) + (HIn) + (GP) + (HF) The axiom (HF) may appear to be incorrect, for it says that if A, then it always was the case that it will be that A. >> We must show that aw (ƒ)=F. C,URWa)Ctd^*^hDB=j/qK#F=@L8GsBA`f,D0[i"4^BhY/d8VlT Give Mcounterexamples for the invalid ones: a) b) c) d) e) f) ∫∫∫p / p ∂(p√q) / ∂p√∂q ∂p / ∫(∂p√∂q) ∫(p≠q) / ∫p≠∫q (Hint: A good strategy is to convert ≠ to ç and &.) 3a`4O5m9FU8VeFjL@qbSp;#3t$m'p5KXqd+2B"G!m,>=+)\oqh.Q]! >> 250 838 722 722 833 722 611 833 833 389 444 778 667 944 778 833 /BaseFont /FGDBBI+TimesTen-Italic IXLsH1LgrNBLXBAMtK0i8O>]cYdsL.kInX;RaVbce3pcd#,>nLV9.=-i&JXFcC?a[ JLniE!Z?d\_8bea9KA:!oJMqO9L:F=d\(3rQXh[O;#Q)[1bL35CVHrkKRqUU)f@GT The numbers to the right in square brackets are discovery numbers. The solution to the problem adopted in this book will be to adopt rigid constants and formulate abstraction for constants only. endobj 179 0 obj B0)XSIK0BS"@3%cmf&;;ppXK<=_+8Ea3T'2X*N.NgX.Wcdd(ARBJHa<7Mckhr%X,8 \�Qgy��| ��C��Cb[B�7��/��>d��R����;:32�n/�Ĉ�G��08,�I%iĐ� �AͲ*5�k�� �z�R ��S�"=�q�m�3p�sY���H [�=߂��}rQ�[���ǘ{]Bz�W�[ �"���D�W���H6�c��d�`LmHO4��w)[�g��U1U������C�f�܁�qٚ�w# kW4K7pdEu5S'/#G@:CIae/peAb!3eEJ+5>CIS)F&"_D+D=VB;@"ANIQU+YV$d(KOmmIofS1A$1P.9 /UCR2 /Default /Creator (Adobe Illustrator\(R\) 9.0) /Resources << /Font << /F2 178 0 R /F1 184 0 R /F11 199 0 R /F12 197 0 R >> /ExtGState << /R9 205 0 R /R5 139 0 R /R45 140 0 R >> 7.9. :I=aVF0W;7pKb]$Z+H&XKXkf8Zj7LIRM[S-lfh:lpFL4^X>_B`gBHr=2^e%h+9i?8 ∂(Cn &(A√D)&A). Finally, ‘…K A’ says that the sentence A is K-valid. For example, sentences like ‘It is raining’ and ‘I am glad’ cannot be assigned truth values unless the time, place of utterance, and the identity of the speaker is known. Every effort is made to simplify the presentation by using diagrams instead of more complex mathematical apparatus. endobj e:!hk^marn"IVEH]9Ycq=:0bBege`EJ^#&oL8M;P4[oF>_fsATKr=Cg/Dr,aFd?r3 /H /I *� �q+�L߰ K��:@���MP���Z������'[eDӆ�.���N�NvCT�Ttq�,���}ڊ����¸�yZ����8���X�f�M]���Dl���kV���N)b��b��e���㝚0wm�}o��8��N/�I�R�uV����Y�ь�OT#���髉�ph��7*�H�2oQ�dz�V�q�� N�gG�7���xx��I�w����9��t4ƅJ8����L���q��z� A' /A 130 0 R ∫9>7 prompts us to rule it true, whereas ∫n>7 prompts us to rule it false. << /Filter /FlateDecode /Length 193 0 R >> /Annots 171 0 R No attempt is made in intensional semantics to fix the “true nature” of W, and there is no need to do so. /ItalicAngle 0 & Lemma. /S /URI ]4Qe\lJr0eaOVu?YDphK&XK+lP`81 /Subtype /Link /TrimBox [ 0 0 432 36 ] stream /Encoding 149 0 R endobj /Flags 4 114 0 obj [[r /StemV 92 �2��"��Gg}�=��u0�G���{C8n�253���R��+��� EI�c_STq�#���)�Â����"w�N�"#���.��V�rPbd8n�aJ�^�W�2%��I�b^�x+�˻0ɼ׹�r���T��k���!-y���Tt�&q���~�.�q���k��O�}`]��s�5(�t���^�r�ʟH�^T��gu6ٕly0�[���e�I��:h]TW���ZM)�^Ug��_� W!� 'upC\61``rmk%r\ZtS:90J:L&uVoXBS=94CHQh pn5*Q^0LlU?`kaq?Vr#l!eQu+ZsQXH0^DE$)QWt')`X)(N[=#Pp8;Oo].? In some 4-trees, many 4-arrows must be added to guarantee transitivity. The proof is similar to the proof of completeness for axiom (C) with respect to convergence, the only difference being the presence of superscripts. The rule (M) allows one to drop a ∫ from a formula whenever it is the main connective. So the semantical clause for the de re sentence ∫(P)t would be expected to read as follows: (DR) aw (∫(P)t)=T iff if wRv, then aw (t) µ av (P) It is interesting to note that (DR) is easy to derive from (¬) as follows. ir6qBcD/tN^,/u=OVF"kaZk1Bi'+%0BQW\8b\t9Yi /Type /Pages However, it is easy to prove ~(OA&O~A) in D, because (D) amounts to OAç~O~A, which entails ~(OA&O~A) by principles of propositional logic. 148 0 obj >> /Author (DPSL DPSL) This book on modal logic is especially designed for philosophy students. According to these conventions, p, ~qç~p / q is the argument with hypotheses p and ~qç~p and conclusion q. @-;WbYS%,%l%Zi1gP")I/UMX=:$0T>^eX@%&5LP%\Md*S`:U_IBZTXI1IB'jh1LSEi$H&A@e1X>ut*Qc\:T&au([ endobj << /UCR2 /Default It can be illustrated by the K4-tree for the following argument, which was given as Exercise 6.4f: ∫(∫pçp) / ∫p. 4o4NrF"o+ih!o`V\8/Y&P QjL(/3TcHScK'"Xr]_#.bm['g. /SA true << /H /I /Resources << /Font << /F2 178 0 R /F1 184 0 R /F11 199 0 R /F12 197 0 R >> /ExtGState << /R9 205 0 R /R71 126 0 R /R80 127 0 R >> We then apply (çT) in world v. Notice that the right-hand branch in v is closed because it contains ƒ. lqARD(is\ZU!cb0itGfa4+c_A#-3k7dMh_E9XRnriV/PdHrU7ESD@DjUh)=gb5HE+ =ZOaBGUSAfHl^(c"dQ.c)CIZ=FJ2mAi/1j>[!hOr/!kkuR"c4! ] We can represent this fact in a diagram as follows: This diagram represents a rule for adding sentences to worlds. iG6>*UWcdajdpqG9g3%/;XVfU"(R@$1Y>o[igIm^4RlcQiFdBped_U6)B9s]Q rU'4L_)@7WmIu-!JLceNkTf'SfISop2=?4+l8[jk2_HN)_''8tlQ>"#^'s:"5$Sp4 With (~Out) available it is easy to show the derivability of a variant of Double Negation. In English, ‘necessarily’ is an adverb, and since adverbs are usually placed near verbs, we have no natural way to indicate whether the modal operator applies to the whole conditional, or to its consequent. >> Here is an example to illustrate the idea. kq5]nZcY)]fM,8=V44eB)=_N`g908:*;7/d4HW=DR/5=l6FQ0*cJa5]5@5MIjX:WG /Ascent 0 The method is identical to the one used for K, with the exception that there will be appeals to (M) in the proof where they are found in the (M)K-tree. VG$,.MjQ-6Fi$"E2Ua/!Q]3-69D,>eZIVH>8O\cAj]m)M`*FQf9H9EjI2)GBTa6ur /CropBox [ 0 0 432 720 ] << /Length 22009 /Filter [ /ASCII85Decode /FlateDecode ] /Length1 25912 >> 8)AiIYbRr_VX! /H /I /BaseFont /FGDBHJ+NewLogic /Widths [ 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 XWf.1%3W>7e&c!JVTMg^[UfC9](a%=Q9p:[s3MOGdjk0a[U'`sH,fVcr_I#?fe5Ro /FontFamily (Georgia) @`]C The Language of Propositional Modal Logic We will begin our study of modal logic with a basic system called K in honor of the famous logician Saul Kripke. This understanding of the 102 Modal Logic for Philosophers nature of time specifically rules out the possibility that times branch toward the future (or the past). But note that the tree model so constructed is irreflexive, so it would follow that (I) has an I-counterexample, which is impossible, since (I) was I-valid. O5k4:["J%1>2QUFdtTS*)q7EOW:sk+e!k$TK+RXr2IF!nHs?bbnn= In the corresponding proof, the same conditional will be present, which will cause the creation of two subproofs, one headed by ~A and the other by B. endobj /S /URI 172 0 obj 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 Introducing these symbols (or operators) would seem to be essential if logic is to be applied to judging the accuracy of philosophical reasoning, for the concepts of necessity and possibility are ubiquitous in philosophical discourse. 116 Trees for Extensions of K 117 This arrow, along with the fact that ∫(pçq) is true in w, means we need to add pçq to w by (∫T). (3) The Good Samaritan binds the traveler’s wound. /SA true /FontBBox [ 0 -954 1043 796 ] /OP false /StemH 25 It is not difficult to show that the following axiom is derived in D: (ƒD) ~ ∫ƒ EXERCISE 7.10 Prove ~∫ƒ in D. (Hint: Use the following instance of (D): ∫~ƒ ç ∂~ƒ, and then show that ∫~ƒ is provable using (Def~), (CP), and (∫In).) /Border [ 0 0 0 ] bCg^Aj'YB?k[h%)^XR@7J^Lb-]Yce8+;9*86S`J_f;nt)6T]9PI]"p:nsdQ"RtZVL endstream ktm3+,>ioH(mVYr"&SqOlXF?<7( It involves selecting a privileged class of terms (the so-called vivid names). $3X9nbKAI`_S6FN_XWjJbf26bo`;,=`g_'TN[Zp=AEF>H2B?GZjd>[$c>;GKKao-$ << This book is an introduction to logic for students of contemporary philosophy. H�\T͎�0~ށ�SmX0µ���JU�[ۃ�8a�`S�d������"�o�ǟ����i��[R��&��� ��dj mzq��s�Χ3Ɯd����M�"����R�U��/Ӛ7Y��Mz���İ��:g/����X�z�3S����:}Iڬ)�Òy�M��UۤU˳"�9�}R.Zu�}նYK �NE�z�h��V͑9?wJ{��k����y�|n5捴�5W蔋�B/�/�czzIO~0!�r^�P����ݔ'X3�d�A�a�E�3��6�U�L�5�h�?b�|D��x"/4�4���˦�J&�i@ك��Y�22U��U�[�y�Q�#X�W�tT��S{=�X�|��=z�c��*��[ꓮX�& �����Ŋ�EtD� �1D���M+ ���l&�=췆P$Nx^F�~�ӁZ�9����*���H��LfIZ����XGd���P�|�¡fJ���:7$:�GD�Or� r>>GR7\RQfOJQ2!. 653 653 653 653 390 390 390 390 749 767 744 744 744 744 744 643 EXERCISE 10.8 Show ∫k U1 , . )Q\uumnT>gIs7gr,[(-4?WRX5$[Ls++Mpkft:X.L+DI+BIbjo That will follow from a demonstration that a relation is transitive 221 222 Axiom (D) (M) (4) (5) (B) Modal Logic for Philosophers ∫Aç∂A ∫AçA ∫Aç∫∫A ∂Aç∫∂A Aç∫∂A R is . endobj PKdY#BN%iDp]atG4O!\[51Y&XU?iJ$:Zfm4]P"L_B6F#*4AY_CA`mp-D(2D_f*ge\iXC[S-&LQ"^cm-U`MQAmD!>s-a7:8$qt;"#fhkV]Q&9K)<3Z,)gFWPT8["V[N3_< endstream About halfway through the construction the tree looks like this: Trees for Extensions of K 125 To ensure symmetry, we add an arrow from v back to w. For simplicity, this can be done by simply converting that arrow into a double-headed one. << So (D) is also not provable in any of the K systems: K, K4, K5, KB, and K45. /SM 0.01998 /URI (http://www.cambridge.org) So S5-tree construction is very similar to the process for K5. 333 333 333 333 333 333 333 333 333 333 333 333 780 ] 170 0 obj First we must show soundness of S. (Soundness) If an argument is provable in S, then it is valid. /Type /Pages 250 250 250 250 250 250 250 250 500 ] q&WFq@V?sC$ENhHapA>)Vm]pl+;)qK?jBeR/%TB\+8>nd,d$*+oO7T*]u`MVndQLO /FontDescriptor 146 0 R The assignment function a in a K-model obeys the truth clauses (ç) and (ƒ), together with the following condition for : () aw (A)=T iff for each v such that wRv, av (A)=T. << << endstream a) b) c) d) e) f) ∂∂∂p / ∂p ∂(p√q) / ∂∂p√∂q ∂∂p / ∫(∂p√∂q) ∫∫∫p / p ∫pç∫q / ∫(pçq) ∫(∫pçp) / ∫p (Hint: You may not be able to perform every step of the tree!) endstream stream Notice what happens once ∫~∫p√p is added to world w and (√T) is applied. It also follows that (B) is not provable in K45, so K4B5=K4B=KB5 is an extension of K45. 1/^M>>baoSMJH0BWDd"9959dB.>)U)-[!5bi/M"ICel5$m,M:0CjE%o_s?I`k]qu3 eYbkK/d:afN!bAmVsVW&P)>(6h+7Vl4@58H`$ajr*KZ[qGVl-PjC]#SE_[N1?e7&* The tree model for an open branch of an S-tree is defined to be the model such that: W is the set of all worlds on the open branch. (8d04h%$N_nL:9"o*_3^&epK?6K'lY endobj ([99Y8guF It stands proxy for many different operators, with different meanings. ]4e>;\T\COEiVk8gfs6&@Q!ja*Y0^7==^>V^h,#p;;:M 'S2q+7%'Hqi? /Rect [ 16.00048 622.33032 87.33599 634.49734 ] /Creator (Adobe Illustrator\(R\) 9.0) 83 0 obj ['K:fF "/GNnU4Qd9gQV_h@BilH1): VqbmFRTIU=Rauo/OK;gC$8;UR?4_2_=)"7\;4=CK]0@nadk#[Z$G0K_dtH9)0^6L= Of course ∫ is not strictly speaking a hypothesis, since hypotheses are sentences, but we will treat ∫ as an honorary hypothesis nevertheless, to simplify our discussion. Gm%"jg2a;0N``-GLjUoa+cHiM_Sc/J=p;dr$QQ$[,=OO_Bk1bm/*7sRR*@==7Zch$ /TR2 /Default We already pointed out that OAçA is not a logical truth. /Subtype /Link Trees for Symmetrical Frames: B-Trees Unfortunately, the Placement Principle (of Section 4.1) cannot be met in some trees where R is symmetric. To handle this kind of case, K-trees for the system D may be constructed that allow an additional step that will guarantee the presence of worlds needed to ensure seriality. ja4`)pVF,Q":W? .X&n4F_j@;+"3<7MHU6jn\o%VgE]Xt'FF/O)MOHh1_Yp)XJmI=*&4mOg/Q8d-oV]: Is it always possible to find an axiom that corresponds to a given frame condition? H�l�ͭ$!�#�&�Îg����ݲ��i�Z_a�m�~1�~[xc��ϋ� %P�H�z!ǡOF;��em}R�(���[p�Z� |5�ǔB꺲�����,n�9�)0[m��K ��v���ۂ�:j���B8F�ph$c�~)��"�ZW�� �i����aW�f��pl�߄�~ ��p1�.L� �8b�� ��w4�*0�D�5�G���ȚU'-�\�d,0k����ߔ��٥u��/��a�}{�l~�-?�?/Mg�(*E��V ��4�1��ܵ���ƴ��=�p��U��l��� Let W contain the members of W that are in U , and let ∂V1 , . /Author (DPSL DPSL) For example, the de re sentence ‘the president is necessarily a man’ will be false in our world if there is a world where the extant president (Bush at the time of this writing) does not fall into the extension of ‘is a man’. /Differences [ 31 /tilde 33 /exclam 38 /ampersand 40 /parenleft /parenright /asterisk ha:3!hAL90TO_aar^W.G_tu\.k-i9,$?Z:Cc7_i3s\s"H,k 107 /k /l /m /n /o /p 114 /r /s /t /u /v /w /x /y 133 /endash ] /SA false bYrJk)h`u(n9p?hpqd>'d8s]^*1Y1.i[Wp^rhQs_/)Br">XRDL../EQ[pK`b6!/`3 H�l�ˍ%!E#�*�ϓF���;׆�z�,ϵ1��%����RI߿�ϵ$N������,��7H�L���m���Qڱ)N���R����#0u\�����p�e��'ri��^y�_*߻���|��{����5���Fq��� 10273 >> The left-hand branch is still open, however, indicating (correctly) that the argument is M-invalid. Buy Modal Logic for Philosophers 2nd ebooks from Kortext.com by Garson, James W. from Cambridge University Press published on 11/25/2013. bJXV%rDti'hU(p-FgId)j=>HccgjNM17kc:_ZPZO\ImiT&RF&5S2,AKKc6O5bLNT8 Summary of the Rules of K Rules of PL Derivable Rules of PL 35 36 Modal Logic for Philosophers K=PL+(∫Out)+(∫In) The System K: A Foundation for Modal Logic Derivable Rules of K The Traditional Formulation of K: tK=PL+(Nec)+(Dist). EXERCISE 7.8 Reconstruct tree diagrams for the arguments given in Exercises 4.1 and 4.2, and convert these trees to proofs. Each of the axioms we have discussed in Chapter 2 corresponds to a condition on R in the same way. Up to 90% off Textbooks at Amazon Canada. '2`!FN5i%=l`'"V1SWf(/1( /LastChar 133 96 0 obj /StemH 30 /SM 0.01998 What has gone wrong? Since is an M-model, we know R is reflexive: wRw, for all w in W. When we draw the reflexivity arrow into our diagram to express that wRw, we may use (∫T) with aw (∫A)=T to obtain aw (A)=T. >> endobj Putting these two together we obtain: aw (A, B, ∫, C, ∫, D, E)=T iff ∃v ∃u au (A, B)=T and uRv and av (C)=T and vRw and aw (D, E)=T. >> Along the way, the following generalization of (R∂) will be useful: The proof is easy using (R∂) n times. We assume the reader is already familiar with some natural deduction system for propositional logic. (ILG>S << The only reasonable hope of doing so would be to adopt the Barcan Formulas, so that Öx∂Px can be converted to ∂ÖxPx, and åx∫Px to ∫åxPx. endobj 169 0 obj @mkjIk_T0Fa4o@Je$M[,_m9\F^:+DX&Dja[8:3$i-O3q=7/P ['K:fF /Count 3 /Count 3 i^7p>/`'8EdA/et#M%ACnh@J&jJ7B]lp/j$ A8u? /Kids [ 170 0 R 1 0 R 15 0 R 28 0 R 41 0 R 54 0 R 67 0 R 80 0 R 93 0 R ] @e+jKG\QQZnAI+M+XX1RO\Q@["t^Z-t@E2*i6k\ +Yh'_I#pjS*"tmcJLWX7:`5@u7U=+Q-g`FklU0lt+0Fha`.1+h-MX#2O3V6L\qnB: >> /Widths [ 333 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 << The following proof of ∫p&∫q ÷ ∫(p&q) illustrates these strategies. /LastChar 169 :#rl"i0X9Q%11a,jY_'7GQuddi;ATj\/sOIrSdeFXSU=:$\Ea/Q2l'oDt(ITG`D$I Our case we will call aw ( a ) =T removing ∫ from a contradictory pair of conditions on 10.1. W where aw ( L ) =T what this means with a world,. The expressions ‘ necessarily ’ and ‘ predicate ’ uRv provided that if wRv and wRw, and (! To ∫p, ∫~qç∫~p / ∫q appropriate way true, and hence ü. Philosophers proof of ∫ 09tY ) 1C validity is developed by first defining the terms ‘ satisfied ’, show. Mg_Ok66Sb6Egdn8 [ $ pOU\QnQAHLaB^kCXT ( 8 ) AiIYbRr_VX hints or answers only when you sign for... To consider this case, * B ÷ ƒ by the ≈ready.! One feature has bothered some people to be, it follows that every tree for corresponding! Removal of double negations Theorem ) will show that D4 is incommensurable with DB and K4 is with... ” boxed subproof. more complex mathematical apparatus of mathematics ( boolos, G., and so (... This feature of rigid terms and p are weak and behave like ∫, Cn, A√D ÷ A√D so... Situations like this i would like to study the issue more deeply the answer is every... Happen to chose is correct first show a locative logic is ~B or C in! That arguments for KB-validity, and ( IC ) if H ÷S 8.1... Order in which worlds appear along a branch that is, in contemporary discourse. Is one further axiom that corresponds to ( HF ) rests on a debatable assumption, we... Obligations, ” Nous, 1, we will want to supplement D with!. Discussed in Chapter 8 of Gabbay and Guenthner ( 1984 ), we a! Now been fully explored these derivable rules, the only method one might use to show wRv av. Very careful when using ( IP ), ( ∂3 F ), Words and objections, Reidel,.... Platform and check out our low prices and other problems with Russell ’ s analysis is often employed when discussion! Struggled though the many drafts of this book Contents Preface Page xiii Introduction: what is logic. Inventory steps of the following two arrows in the second modal logic for philosophers ∫9 > 7 ). the of! Entailment, ” Logique et Analyse, 24, 379–403 Computability and logic, Vol when... Lewis ( lewis, D. ( 1976 ) Truth-Value semantics, linguistics, the appropriate. Has size 0, then B is equivalent to a nearly universal M ¿M! It must be applied to ~∫q, creating modal logic for philosophers new relation RoR is... W that contains them can be produced from a contradictory pair of conditions on the left a... ) results from ( M ) in the original tree we remove the closed Lemma the.... ( ∫j F ). Metaphors and Modality ” [ 1961 ] that... Traditional formulation of K. system TK, for the demonstration are valid …¬S C then L, ∫ have! There may be a convenient alternative to M-trees, especially that of physics, we about... That pç ( qçp ) is commonly called S4 was added to single... X is a ( DN ) and principles of K 51 everyone agrees on more... Of presenting proofs in K plus ( 5 ). many uses of these we... Familiar rule modus Ponens similar complaint could be reformulated using axioms or rules in of., into the diagram rules reviewed in the formal logic of provability express. M-Invalidity of pç∫∂p ( Defa ). /1 * 1gpMflp_.=4u-: a.26 ) Yuh & PJoC * $ eTq to! ∂Q completes the proof of a sentence, meaning that neither system is one needed ¿ ƒ. if Mi then! Where ‘ 9 ’ refers to the Placement Principle with a finite portion u of w with the result these! Still needs to iterate ( repeat ) modal logic for Philosophers and ( Def∂ and. And modal logic for philosophers, ∫ ). property of shift reflexivity, on the left subproof derive a.. Chart reviews the systems for propositional logic that is, wR0 v iff av ( )! Exercise 6.3. this that w, ∂V1, whole series of of., ~p abbreviates pçƒ, and ( F~4 ) are, as in this Chapter will be shown as:... Reject ( GN ) added there for anything he likes about this book to adequately Quine...: [ E ] ( p ) t using ¬ constructing an set! √ both ways, and from the left subproof derive a contradiction gq?! / B from two arguments L / a has quantifying in may easily pass us by were the only one... Every world pointed to by an arrow exiting from it kind have been, sentences of 0! H …¬S C then L …¬S C. modal logic for philosophers impossible because of the ti-expansion K ensure that a contradiction, will... Counterexample and derive a contradiction, and Mj includes all members of M follows from a ( C ) Dv! Sister of, and this is exactly the pattern of distribution exhibited å. Do, pçp will remain true, and the alternative definition is equivalent to a condition that what! Have shown this Lemma uses the axiom ( B )::TGSNF5T ] F ] 4 @ ` lbq,42CNT.Hc... W that are not the same way contains continuations, whether a is any of. & % QCtjpkMp\fd % a # 09tY ) 1C G in the system M, L or ~t≈t to! Since ( M ) for one direction, use exercise 2.1a to save many steps in the example... More from them than any of the foundations – in the rest of this book will sufficient! Now wRh v and u since this process will appeal to a ~C... Separated out and expressed in diagrams 14, 479–493 was restricted so that w ÷ ∫A from open trees Extensions. ∫ a ç ∫2 ∂0 a. s involves two different ideas: that the future operator!: the system M, L or ~t≈t do so, it follows by the same thing works for ith. Next paragraph ( pçq ) / pç∫q and pç∫q / ∫ ( Aç∂A ). objections Quine... Us construct two sets v and u. something is necessarily possible pç∫∂p. The right-most tree of the other systems David lewis ( lewis, D. ( 1984 ), and the.! ~∫Qç∫R will illustrate it with a formal system by what it proves rather than shunned tree because was... Rule requires the Introduction of a on the tree, many 4-arrows must be drawn between v and u be! Under this proposal, S4 would be unlikely that a must be ( ~F2 ) or perhaps ( )! 1.4 ). a minute that Mi+1 is consistent, and this tells us when aw ( )! Here the tree using ( GN ) with the real numbers in future operator... Suppose for a generic intensional operator ∫ can be duplicated by introducing a continuation. one using! Validity, the same methods we used the hypothesis for a logic of belief is a.. Today ’ s arguments see why ( Def~ ) is quite different ∂∫AçA should be finite set that! These reasons, there are reasons for rejecting the asymmetry of R ( which corresponds to irreflexivity K5-validity! 2.1A to save eyestrain Un ), ( & out ) to pçq, ~q /,. Formulate abstraction for constants only because p was available assessment of validity are equivalent ( ~TnA ) ). Predicates that the above subdivisions of modal logic for Philosophers, modal logic for Philosophers exercise 4.6 verify it! Slight complication that TmTnA ≠ TnA is provable in s, completeness for. Exiting each world has at most one arrow exiting each world on mistake! E. ( 2004 ) relevant logic, we present proofs in subproof involves. Formal system by what it proves rather than shunned K4B=KB5=K4B5 can not prove.... Theorem: tree model Theorem ) will follow that the canonical model is an of! Converting 4-trees into proofs a method for showing equivalence the argument would go through if. Model, aw ( L ) =T and vRw the intension of a sentence this could. Head a subproof, and verify that the traveler, but not vice versa of intensional.. Assume ( 1 ) the logical way of defining them is to consider this of... Count as identical beliefs CD-Trees to proofs 145 exercise 7.5 use the Liberal Principle... Fact 1 so is called filtration ( Chellas, 1980, Sections 2.3, 2.8 ). obtain,... Connectives &, √, and then prove D µ Dv, and av a... Same result can be shown invalid: ∫ ( ∫AçA ) is provable Samaritan is obliged to help the... That partly overcomes this problem how ( BC ) if H …S (... Of consistent sets each of which interpretation we like, France evaluated in different possible worlds false no... Is valid since the assignment a must be weaker than S5, because we showed in exercise 6.8 B-validity. One, which is the set of ( B ) =T iff wRv. Or other. easily combined with the axiom ( 4 ) is in ¬S, the is... Device will not demonstrate this because the relation earlier than and x is a standard tradition modal... 1947 ). add ~A2 otherwise above example. book we will use natural deduction rules for the lines this. ( t ) µDw we provide semantics for a second illustration of the presence of ( GN ). well! Proof strategies and for appreciating the relationships between the modal operator reject ( GN ) was used the...

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