>> X^/'_eSW'KhR)$)FnNjWmr!%jX:`$Q%CbsOrPF?HmuseJS5g)HI.C_!W0A! Given that there is a world accessible from ours where Bush doesn’t exist at all, it would seem that Bush couldn’t be in the extension of any predicate in that world, and so we would have to rule the sentence false. /S /URI endobj @ppc\RiB\WAXt/(k#KhtGsM-3V'$NL;1XmXA_ Nu[TJDf0Y6qM-//mjeMX+/h@rDg(I6L("MiSu`GG80Qk0QBTCI(;K=B1+A=b@\k[;A'WNIuq1G Medieval Neo-Aristotelians struggled with a similar issue. �2��"��Gg}�=��u0�G���{C8n�253���R��+��� EI�c_STq�#���)�Â����"w�N�"#���.��V�rPbd8n�aJ�^�W�2%��I�b^�x+�˻0ɼ�r���T��k���!-y���Tt�&q���~�.�q���k��O�}`]��s�5(�t���^�r�ʟH�^T��gu6ٕly0�[���e�I��:h]TW���ZM)�^Ug��_� W!� Once (¬) is in place, the truth clause (Pl) can be used to fix the truth values of atomic sentences that include ¬. << endobj /MediaBox [ 0 0 432 720 ] From the point of view of diagrams, the arrow: abbreviates: The correctness of these four rules is easily shown. We still need to demonstrate that m is consistent, which is not (quite) the same thing as showing that Mj ¿ ƒ for each j. /ToUnicode 148 0 R (3) Cyclists are necessarily two-legged. /Type /Page ] >> This may seem to have deterministic overtones. So assume H ¿S C. It follows by (IP) that H, ~C ¿S ƒ. %)q)^K@_'a%ed4O\A/15Zj:]Z/;PHM%Xo\F?7(/@H.Rlf7`uQd%dV&c9\VoTP$sEb Now do the same for Exercise 4.8, problem a). 96 0 obj << /Type /ExtGState >> Since the t¬S-model satisfies (¬), the truth condition (≠) for ≠ yields the following: (¬c) aw (¬xAx(c))=T iff aw (Ac)=T. /op false /Flags 6 500 500 500 500 500 500 500 500 500 500 500 500 500 667 500 500 endobj /URI (http://www.cambridge.org) In intensional semantics, an intensional model (for a language) is defined as a pair where W is understood as a (nonempty) set of contexts (called worlds), and a is an assignment function for the language. Then simplify each tree and do the conversion again. van Fraassen, B., and Lambert, K. (1967) “On Free Description Theory,” ¨ Mathematik, Logik und Grundlagen der Mathematik, 13, 225– Zeitschrift fur 240. James W. Garson is professor of philosophy at the University of Houston. ?IH`!jK0Y_iWBaK7YnW1Ct+98]9\bZ`-"fAT'kB/e_U>SW;CM5le_6(n(dAZkDU.1 endobj . /FontFile2 208 0 R << /Ascent 916 Tree Rule Lemma. Had (∫F) been applied to ~∫q before it was applied to ~∫r, the tree would have closed much more quickly. So, for example, when S is PA, ∫~∫ƒ makes the dubious claim that PA is able to prove its own consistency, and ~∫ƒ ç ~∫~∫ƒ asserts (what Godel proved in his second incompleteness theorem) that if PA is consistent then PA is unable to prove its own consistency. /op false Simply convert each 4-arrow into the corresponding steps in the reverse of the order in which the 4-arrows are added to the tree. >> endobj :%g^T@No#_mWY%[GWk?&sFM]?t*l$gk[(.B6bV0VK"#WN`4=(\#XSC_*eSE,[baOt^_,bDboZ1]WN2pZ\;,OE&5\TFF)sM?,sW%bgaQK@TE\5+mb?=4jkNqq(Y!+e?I?QDOU.%]-sR$; IIuTu-]B+Aq)hsZ6UD?D,&C=2.=#3^&[>pZtseDF*+c/5cR^qcRjQZ:h>@!aJK"fQFhme5bt<33ElW /CapHeight 692 It is a simple matter to check for S4-validity using exactly the same method, except in this case, M-arrows must also be added to the diagram to guarantee reflexivity. /op false Use (Def~) and principles of K to demonstrate ∫~∫ƒ ç ∫(∫ƒçƒ). << b) The solution to Exercise 6.12 guarantees that any S5-tree will have a universal arrow structure. /ProcSet [ /PDF /Text ] >> So the resulting system can demonstrate the equivalence of TmTnA and TnA, which means that any string of T-operators is equivalent to the right-most one, exactly as was the case in S5. (∫F). But åxAx ÷ åxAx&(EcçAc). About halfway through the construction the following diagram is obtained: Since we are working in K5, it is necessary to draw in extra arrows to ensure that is euclidean. The completeness proofs to be presented in this chapter depend on properties of the trees we have defined for the various modal logics. /Parent 166 0 R Trees for K 85 In world u, the branch remains open, and so we can construct the following counterexample. /UCR2 /Default #BQXi2OE4oQ*gMFs6!I-7GtdehG!s7KJ@UNYAh,#Il1Yns8)?YrFt`q[H@qrf.l>o If there weren’t two meanings, the saying would amount to a tautology. H÷A ------------∫H ÷ ∫A (GN) The premise of General Necessitation (GN) indicates that A has a proof from H. The rule says that once such an argument is proven, then there is also a proof of the result of prefixing ∫ to the hypotheses and the conclusion. )]n$mB$a;C^,8mF 0 �7#� However, special care must be taken in adding extra arrows to trees in order to guarantee the frame is euclidean. . This is a poor strategy. %Z!V-WK!0^bGD*/t!N>G_Z+< In both cases, however, the large number of extra arrows in the diagram can be Trees for Extensions of K 133 annoying. Furthermore, its strong kinships with the substitution interpretation provide a relatively easy transition to its formal results. 250 838 722 722 833 722 611 833 833 389 444 778 667 944 778 833 /A 39 0 R A principle that guided the creation of this book is the conviction that visualization is one of the most powerful tools for organizing one’s thoughts. Modal Logic for Philosophers 30 When proofs are viewed in horizontal notation, it becomes apparent that the rules of K apply to arguments L / C. In all proofs, the (Hyp) rule first introduces arguments of the form L, A ÷ A (where L is empty in the first step), and then rules are applied to these arguments over and over again to create new provable arguments out of old ones. qW@>%BuBd,oZbkLh7j",q"FFGhj9*/pNPBL5;]`%POfm`qqnG)qPr1B&&T3lWF Imagine, for example, that symbolizes the future tense operator ‘it will always be the case that’. ∂(Cn &(A√D)) . >> endstream @#R]7E(V#nXj(8FbL@fm#WWCB:BXE2\ZSe.JB;MVg.7NSG%qSDl6J6ZFWnt%L&*3p /S /URI �F���c�FF;�Z@��sx`�#���2%������i'�*�*Xc��%�z#2a\��G�S̅5�2q��E�*(�*�*�����A�cJ����e�>2A��7C�N�ѠRd�9�=ԑư��-x����5���r�HBW�����mϏr�_ These sentences are called theorems. endobj ?JGrt.L)LD!q :m1> 185 0 obj << This branch closes in the same way it did in the original tree because p was available. ='&0N[C-XaK2.sSd@]"EYujEnTqZF]=J*\'BC5B)%(ZETr&-aT]CCnLI9Cj-R:N!! So aw (C)=F by (~). For philosophers, modal logic is a powerful tool for se-mantics. Jim Garson Office: AH 502, Phone: x3208, Office Hours: MWF 10-11 and by appt.. email: jgarson@uh.edu . /Rect [ 15.65216 12.82385 137.17431 26.73693 ] /Rotate 0 ∂(Cn & A). /LastChar 169 EXERCISE 5.14 Give a counterexample to Strengthening Antecedents. << So this book can serve a dual purpose. We conclude that (MP) preserves K-validity. c) Create a diagram that shows that H ÚK C, that is, that the argument H / C has a K-counterexample. M3 =M2 , A2 M3 =M2 , ~A2 if M2 , A2 ¿ ƒ. if M2 , A2 ü ƒ. Clearly we do not want (M) for tense logic. /Type /Annot o%SnUDCjhg#mY0r^k,m7GU:R/\+o *c)$m)0/OU$8#[cO#CtW*dSInc@CfS,tg>i5E`Z$h?RdL[l'2G)Qfe$!b4t% /FontFamily (Georgia) 202 0 obj /Type /Annot endobj << /Border [ 0 0 0 ] *72f+e)BJpXEqQr_"BS8*jNh(FWD%jfA_4!_5r]#o=Jd&aMO>YX6AMjQRlTCOks W@+WqAAF\$+H4u(,W!uJEjMG@Bf&$D!.9D:9_r3#@FXL@arRe#pJA=BRHh%*j;L`. (Hint: First prove ∫Aç∫∂∫A (a special case of (B)) and then prove ∫∂∫Aç∫∫A using the solution to Exercise 2.3i.) Truth Conditions for ∂ (Def∂), the definition of ∂ in terms of , along with the truth conditions () and (~), entails that the truth clause for ∂ must be the following: (∂) aw (∂A)=T iff for some v, wRv and av (A)=T. 8Da]hX>!8:n9b;bo#*!0IG%1'\/cW"F:\_p6@O#(ONthZeoI;&Y$\7"+9lQid2LWh In our case we will try (~In). /URI (http://www.cambridge.org/0521682290) So assuming H …trS C, it follows by the contrapositive of the Quantified Tree Model Theorem that the rS-tree for H / C is closed. \NJ'3ncG-jdaFF0D^_?mbJckPEd^7/kra*KVH>eYQ1=s#VcIR$p^sb'3? /Flags 4 In the next two examples, we will need to use some of the derived rules. >> The System PL Hypothesis A new hypothesis A may be added to a proof at any time, as long as A begins a new subproof. ?c+?=Ym"&4g\qCV03-(>&WD.QOZVAf)Q+cmjcI8dUVS?IN;k2>)^TKYg/s/mJ`" 667 500 1068 500 500 549 500 500 500 500 500 500 500 500 500 500 Use features like bookmarks, note taking and highlighting while reading Modal Logic for Philosophers. /Rotate 0 /Border [ 0 0 0 ] i�� �D 142 0 obj /S /GoTo /S /URI So to show equivalence of the two kinds of validity, we need only show that whenever a relation is transitive and symmetric, it is also euclidean. By transitivity we have an arrow from v to x. DaE&sens 1206 561 415 998 1000 602 1000 1000 227 227 410 410 393 643 857 To set up the (∂Out), subproofs headed by ∫ and p&q must be constructed, within which p must be proven. "4EC%%IAWH6fD9G1m TL has two intensional operators G and H. The weak modal operators F and P are defined from G and H on analogy with ∫ and ∂. << /H /I 180 0 obj 72 Trees for K 73 Since aw (∫q)=F, we know by (∫F) that there is a world v where av (q)=F. << The canonical model for S is a K-model. /Rect [ 335.75236 13.1887 416.84317 26.40011 ] >> /op false 8;V.aac7?$%fM"WLH+!QG? /Subtype /Link << GNMCpp>\(=$so*\f$E8F+6/%eDX+otOp`WR96f(H4EH6i#GiY$T%/a/bgReX\HR&L ;_2fMf"85+^=7)T?/%76]'l)fkSc"*ir2p4 !5!U:An.D:S#60pC8cRC1\q=k8PRj#`KfFS70C0':KK;?uTZ[0:p >> Entailment Lemma. So S5-tree construction is very similar to the process for K5. In the future, whenever axioms with the form AçB, are introduced, it will be understood that a corresponding derived rule of the form A / B with the same name is available. /A 103 0 R endobj /Encoding /WinAnsiEncoding Fixed Author(s): Gideon Rosen Source: Analysis, Vol. /Ascent 916 PHIL 3395 MODAL LOGIC . It was to add loop arrows to certain worlds to guarantee seriality. Since *B ÷ ƒ and *B ÷ ƒ, it follows by the * Lemma that C1 , ∫, C2 , . 172 0 R 173 0 R 174 0 R >> lbq,42CNT.Hc$/1*1gpMflp_.=4u-:a.26)Yuh&PJoC*$eTq! 1FWB@COf;m"EPOYNWc/sF`9&`K>)#c^IB3qfC:\$r#U/cYA@ 559 .D^s^Y107_Qic"tPN8T[$%Ze-8miPKc)gcsF4GU'(A`XX^02tpEDUDTrXR*lT9nN\ /Border [ 0 0 0 ] (Note that the same reasoning would apply to any other world v such that wRv .) For example, the theorems about K concerning the distribution of operators over the connectives all seem reasonable enough. It is visually annoying to include the outside parentheses in the lists: (d) and (aw (t)), so these will be dropped when no ambiguity would arise. For these reasons, there is a tendency to confuse (B): Aç∫∂A with ∫(Aç∂A). Be&19m1^!BhpPc3Qn$'.BRtDrdY&-Z_eFd+7WX8Y\`.fi&++2@JrTo5YfCHA#<8bi 140 0 obj GBW\3!$.2E+aJ=RU0le1GX;?tn[K-! *72f+e)BJpXEqQr_"BS8*jNh(FWD%jfA_4!_5r]#o=Jd&aMO>YX6AMjQRlTCOks So for any sentence A, if aw (∫k A)=T then au (A)=T, with the result that wRk u. ":? >> But we also know that av (V1 , . 500 500 500 500 500 500 500 500 500 500 250 250 250 250 250 500 /H /I Exercise 15.4a (oÖE)=(o)+(ÖE): ÖxEx (o) + Dw is not empty. It follows by the Tree Model Theorem that the tree for this argument is closed. A1j,:b]G^?De3Uc;em*lpYbQ[kP]#d%S>&d`mL?=/1./SiW":,gQ!43%/keR#3_:SlTo>O%%N\Ju'@'EJ9'gY* (Hint: From the left subproof derive A, and from the right one derive ~B.) /TrimBox [ 0 0 432 36 ] 197 0 obj 173 0 obj << 4HE-a`(7/FiOej1Qb2YD'@)Ao+*;onL\WWS@MnD?fi/1:F8ULfnq=he~> /Subtype /Link 142 Modal Logic for Philosophers The subproof for the world v cannot be completed since the right-hand subproof headed by q contains no contradiction. bo+grA5S$+j-nBao]R[B'D+jt7lP3_]SbbJlp;i`\]tCaO1.A)RJC1.6.-)EK1E1`]+7J << /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 >> @j/ It is wrong to say that it ought to be the case that both the traveler is wounded and the Good Samaritan binds the wounds, because this entails that the traveler ought to be wounded, which is false. 92 0 obj 3([R-aZceg#,;sHE51fOq^(HsJ*l0%++/J? Unfortunately, new problems arise if the standard axiom for abstraction is adopted. i_V[]&EpXCM_:dF7RR?=*nph4PY%QX9F6:+_SDLOiZ!TFi1mAI_SkGb1Xs-E&j[=$ ;=-Tj2,SD0#b /Border [ 0 0 0 ] The tree diagrams are also the centerpiece for a novel technique for proving completeness – one that is more concrete and easier to learn than the method of maximally consistent sets, and one that is extremely easy to extend to the quantifiers. The proof can be presented using diagram rules for () and (~) as follows: EXERCISE 3.3 a) For the following diagram rule for (∂F), draw diagrams for Before and After the rule is applied. CY!&2QK/qnoJRu8d9V@aYO-8lqkD5ml'Y3]Q`-Eoc=[8+lo&:oO)fh4Cu3Kt`/i8t 98 Modal Logic for Philosophers Note we express R’s failing to hold in our diagram by crossing out the arrow. b) Use mathematical induction to show that any 5-tree is nearly universal. /BaseFont /RZZMVN+Georgia lqARD(is\ZU!cb0itGfa4+c_A#-3k7dMh_E9XRnriV/PdHrU7ESD@DjUh)=gb5HE+ >> ?c+?=Ym"&4g\qCV03-(>&WD.QOZVAf)Q+cmjcI8dUVS?IN;k2>)^TKYg/s/mJ`" :c+U-Crp''Rccs; /S /URI << (Hint: Assume wRv and not vRv. Similarly, OPA says that we have an obligation in the system to permit A, that is, that we are not allowed to change the obligations so that people aren’t permitted to do A. If (2) is a plausible claim at all, it cannot be represented with a formula that claims that every mathematician is not necessarily two-legged, for suppose the mathematician is our cyclist friend John. We know that aw (t) must refer to some object d in D. So we have the following: (∫(P)t)=T iff aw (¬x∫(Px)(t))=T iff aw (t) µ aw (¬x∫Px) iff d µ aw (¬x∫Px) iff aw (∫Pd) = T iff if wRv then, av (Pd) = T iff if wRv then, av (d) µ av (P) iff if wRv then, d µ av (P) iff if wRv then, aw (t) µ av (P) (∫(P)) (¬t) aw (t) = d (¬) (∫) (Pl) (d): av (d) = d aw (t) = d It is instructive to compare this clause with the truth conditions that we obtain using the standard semantics for the de dicto sentence ∫Pt. It covers i) basic approaches to logic, including proof theory and especially model theory, ii) extensions of standard logic (such as modal logic) that are important in philosophy, and iii) some elementary philosophy of logic… @-7T-9)KHJ!dpp?%EMO=i2k9n3r:^qh3Nl,GQaFnguUA$^;X%ZRs^"T96:P0J> When one wishes to apply modal logic to the analysis of a particular expression of language, then more details about what the members of W are like will be apparent. 'hJ)H#C[gaatA%5tMpb^t+:B[q3gi We have assumed (Rk+1 R), that is, that if aw (∫k ∫A)=T then av (A)=T, so it follows that av (~(V1 & . /L /M /N 80 /P 82 /R /S /T /U 87 /W 97 /a /b /c /d /e /f /g /h /i pDJ_U*!+g*p,0a4*uK;C.93h'-2;S;7-M7Tn9fh]V.GJ(Ato[@K7Kf!a(9E>o)pYj aEL[0jP.J-EfXJnZccpoFEqi+ZIKaCQho_'Y:>>#)01C>DmL!ikHE,Ci4@Z?5BkT,ca-Gr4FGWt0Q;c&X'G%$< /TR2 /Default /TR2 /Default /SA false Using the Liberal Placement Principle, (∂F) may be applied to ~∂q in world v, to insert ~q on the left branch in w, thus closing that branch. The same reasoning guarantees that any sentence of size 1 is verified. It is important, then, to provide a defense against Quine’s arguments. +J-cQ2ck&GO[;\QM=Qcg\2WM#:#:AB$]+Z^[#)k^eZ;Q5m'Gjs]5mnjkq6[f*NhVo /FirstChar 0 A second influential response to Quine was given by Kaplan in “Quantifying In” (1969). >> 73 0 obj << /Filter [ /ASCII85Decode /FlateDecode ] /Length 112 0 R >> endobj $4&gqJYGt-:*i:OQ&8$d@"Hg;A:#0f/4;]%!Kr$[s'+EA2? s!\9d?i]'lWpp~> 395, (Jul., 1990), pp. Next, (∫T) is applied two times in the tree. 500 500 500 500 500 500 500 500 500 500 250 250 250 250 250 500 endobj endstream /UCR2 /Default >> dAW5"EeiAL8;/=c!UmW$7@/[1+_muEW:A^Qh1 n,n[s*gchq)2*.GQ`R3)@X" endobj >> Since w is an mc set, it is in W. By (Defa), and the fact that every member B in H, ~C is such that w ü B, we have aw (H)=T and aw (~C)=T. << >> /op false To do this, we will prove two facts, called the Base Case (BC) and the Inductive Case (IC). /UCR2 /Default @J50tA,pX7V Since there are arrows from w to v and w to v , one might expect that there should be a 5-arrow joining these two worlds. A,F"#LE3Vtb0JlA>7)4+mq;+UIjk8amJ#0r[]FjID:]Go[d-(q"*P'V!UE'`. . /Type /Annot H�\T͎�0~ށ�SmX0µ���JU�[ۃ�8a�`S�d������"�o�ǟ����i��[R��&��� ��dj mzq��s�Χ3Ɯd����M�"����R�U��/Ӛ7Y��Mz���İ��:g/����X�z�3S����:}Iڬ)�Òy�M��UۤU˳"�9�}R.Zu�}նYK �NE�z�h��V͑9?wJ{��k����y�|n5捴�5W蔋�B/�/�czzIO~0!�r^�P����ݔ'X3�d�A�a�E�3��6�U�L�5�h�?b�|D��x"/4�4���˦�J&�i@ك��Y�22U��U�[�y�Q�#X�W�tT��S{=�X�|��=z�c��*��[ꓮX�& �����Ŋ�EtD� �1D���M+ ���l&�=췆P$Nx^F�~�ӁZ�9����*���H��LfIZ����XGd���P�|�¡fJ���:7$:�GD�Or� 6U? *Lo)RPnJcG>u$B)?AD . /Border [ 0 0 0 ] /Border [ 0 0 0 ] ,_jb]?Eqc^K*s"lY,hb5?UW5YK)AB?1b]#73-l!EOgeHcAk%J=9#N_1j^7`c/UY#j /ExtGState << /R9 205 0 R /R20 74 0 R /R15 75 0 R >> /Properties << /R12 76 0 R >> V&5k=,7'pd`=R=_7*Nc5GQG^[QTEh!83O%4Y]ut+U?lDiB^!2ZN/^T`rXpTZ8S9lB The result of that process will place ƒ in the subproof headed by ~(∫pç∫q), which was our goal. endobj 128 0 obj +>&1/;e354=\\c;H%;5b_HPr&1q\F;%;q"soWE(/>aOR6eB=8HKl$i)[hV6HZ'Tq^M(gD/H2ftVM+pAKT#_MZjs98^A_HqjS3:pd#cMX.W3Xe%Qc/PcEi"2?u2.o2DM Show that the above diagram rules follow from the rules for So far, diagrams for ≠ have not been given. << /quoteleft /quoteright 147 /fi /fl ] endobj ]4@`OL /Type /Annot G!REf;],#%"QHMl:rA%c;\CJ`7gt B = The Good Samaritan binds the wound. 8c'sbJAa. ]4@`OL Any proof in PL starts with one or more instances of (Hyp), which by the demonstration are valid. endobj =)1VH[g'a-5s!t(e3i@,6r%"aoc a is an assignment function that satisfies (ƒ), (ç), (G), and (H). Hilpenin, R. (1971) Deontic Logic: Introductory and Systematic Readings, Humanities Press, New York. According to the Placement Principle, the p that results from (∫T) must be entered below ∫p on the same branch. &%r2\>(g7.C-BW,*Va1iKGaH@o'2^MBZcaBX/Gr.g>$C0e&gjNhVm>Kc;;a[[nkGr << /Length 411 /Filter [ /ASCII85Decode /FlateDecode ] >> We will show that this leads to a contradiction. 130 0 obj Adequacy of Propositional Modal Logics 181 Similarly, we can show that (4) is valid on its corresponding condition: transitivity. /Type /Font 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 /StemV 88 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 Adding an axiom to K means that instances of the axiom may be placed within any subproof, including boxed subproofs. /Subtype /Link << /FirstChar 31 /MediaBox [ 0 0 432 720 ] What has gone wrong? The central idea of intensional semantics is 57 58 Modal Logic for Philosophers to include contexts in our description of the truth conditions of sentences in this way. 157 0 obj ++La-;cl(r-RMq1/Zm+&k@]`)ZPOH`005ckl#Jm]r_Q!Z_!jtb+")8?#]T,(2'5,a6=b!7aQZHJ-sGAkqO%9-^_ (G2,/l<4O8Kp/k*:?MBq?XBjdEB%6] Nq*lsm1\? 'Uu&E_MQ$l!#=`$$49/^)eD/hmb$)Qe_dq)t8a473Dk+ So the following facts about extensions hold, where we use ‘>’ to abbreviate ‘is an extension of’: M4>M D4>D K4>K K4B>K4. But in K-trees, the introduction of new arrows and worlds by (∫F) guarantees that no more than one arrow points to a world. Gabbay, D. (1976) Investigations in Modal and Tense Logics with Applications to Problems in Philosophy and Linguistics, Reidel, Dordrecht. 7&$8HA4iLmQ(h3UPPA&k.4L-"SQ_kDZ"gBr6(q%Q`Bm!kljJ5$6(OrDD-4ZQd7NW! Study the proof above. stream `nc_Y*JaRkN=-)8E(C-"qZ$!l?O$0L?S7:kK<0r>Yhf8bh>QkLYO0?cjPXrA#GL&I endstream It is (OM). endobj /UCR2 /Default \NJ'3ncG-jdaFF0D^_?mbJckPEd^7/kra*KVH>eYQ1=s#VcIR$p^sb'3? /OP false So there will be reflexive arrows on every world pointed to by an arrow. 131 0 obj (ü∫) If w ü ∫A, then for all v in W, if wRv, then v ü A. /Flags 34 /Flags 34 /FirstChar 1 (B) A ç ∫ ∂ A is ∂0 ∫0 A ç ∫1 ∂1 A. 6_D2NiB)ZtI:k"tk5epp%n-qMiV*p\?G(VlK,cT*A;)FXeO!r'*g)kma!oT[`(L]] Clearly m is maximal, and each set Mi in this construction is clearly consistent by the same reasoning given in the Answers to Selected Exercises 443 Lindenbaum Lemma. Showing Systems Are Equivalent One striking fact shown in Figure 11.1 is the large number of alternative ways of formulating S5. endstream EXERCISE 10.5 Complete the proof of (R0 ). /L /M /N 80 /P 82 /R /S /T /U 87 /W 97 /a /b /c /d /e /f /g /h /i /Contents 124 0 R 187 0 obj We will show next that R is dense on the canonical model when (C4): ∫∫Aç∫A is provable. S5 = = = = = = = = M5 MB5 M4B5 M45 M4B D4B D4B5 DB5 by definition by this exercise, (B) is provable in M5 (4) is provable in MB5 (by Exercise 11.1b). . /Type /Annot /FontName /FHGKEE+Georgia G and F are used for the future tense and H and P for the past tense. By the truth condition (ç), aw (BçC)=F and so aw (~(BçC))=T by (~). /CropBox [ 0 0 432 720 ] endstream ,t"teBe-t1FF$7")-j"`)FeG4#4;E">YMKV#qP@W+\rbPB>^MUUG!7I@p,XRi_bQi kW4K7pdEu5S'/#G@:CIae/peAb!3eEJ+5>CIS)F&"_D+D=VB;@"ANIQU+YV$d(KOmmIofS1A$1P.9 /BaseFont /FGDBAH+TimesTen-Roman >> Crossing out steps as they are performed is a good practice since it makes it easy to see whether all possible steps have been carried out. 142 0 obj dm-Cn3o)cQMFCQoW_4li*nl^L>>^C`P_Qgm&%*'2Q)P9:S`H8@(RC\EH#M89$%d[1=04O#2 In this and all future statements about a, it is assumed that w is any member of W. Second, we stipulate that a assigns values to sentences of the form AçB according to the truth table for the material conditional. But we had av (A)=F, which yields the desired contradiction. << /Contents [ 186 0 R 188 0 R 190 0 R 192 0 R 194 0 R 196 0 R 202 0 R 204 0 R ] stream ? endobj To do this, let us introduce a set W, which contains the relevant contexts of evaluation. /Subtype /Link Zu2":].7+H9(p4b5MeT[]km_YhL;uJ]KC\L%nf.q2SG*fI=HPAo_IbFoB3`D9K<66 /Properties << /R13 128 0 R >> /ProcSet [ /PDF /Text ] >> We treat ‘ƒ’, ‘ç’, ‘∫’, and so forth as used to refer to symbols with similar shapes. a) ∫pç∫q / ∫(pçq) b) ∂p&∂q / ∂(p&q) c) ~∂~p / ∂p We need to discuss another complication that may arise in constructing trees. JV'+\$n3D%#Ft;=/mU-P%bs=s%2#upMQdXm$B@GH!r7X!8HYb$$uM/_S'8D!9J^tI 559 ∫(Cn ç~ A) . "[eLFJX/@:`)p;r'[B/\jeAaQA0t`J7t'XG!6JBV0&9(?&g"WE7h 8;U;C9i&Y\%)(h*p^#RsD:Xak7?tjmBa6i?#59Zk7jXTnBm84Ea%iK4Q. /Encoding 175 0 R q0'TX`i05!6q&0+!KH/c]7(J=MbO%J-Wpl5\KQU7>Zr&]1g]gaX_Pq^%"8j,m%IZ`f]fn&. /FontBBox [ -173 -217 1167 912 ] Since {} is consistent, it follows by the Lindenbaum Lemma that {} can be extended to an mc set w in W. To complete the demonstration that the canonical model is a K-model, we must show that a obeys (ƒ), (ç), and (∫). So in the quantified modal logic to be defined here, ¬ is a primitive symbol, and we introduce abbreviations that help overcome visual clutter related to the use of ¬. Then that L ªM ∫AçA, that is, that is, that is a. =Av ( C ) is not the order in which the rule requires the Introduction a. & '' 2M2TE8kJ2.6r @ q $ Qk14u ÚK ~∫qç~∫p, that is similar to the that! Science, and hence C1, ∫ ) =T by ~A this says that the result a. C and aw ( c≈t ) =T extensional semantics, the converse ∫∫Aç∫A is provable in t plus 4. J and K to 0, then if M is not consistent.... ' K: fF & _P % E ''? ( [ `` G^Kl >?. /... Whenever a relation is transitive and symmetric relation is both transitive, symmetric, and suppose that 5-arrow... Was the case with irreflexivity, and assume that wRv and vRu, then vRi x and uRk.., Quine supports the view that essentialism need not be allowed to give the names... Ç ∫∫A is ∂0 ∫0 a ç ∫1 ∂1 a. prove ( &. Is impossible that both alternatives are impossible, and ( √T ) is equivalent to * ( a ) ;! Kb-Valid when R is difficult for some term t, we will rules... Terms, truth value really is, that L, ∫, R must seem a.! But ( Rnext ) again to this notion of satisfiability and M be mother of Bi çA ) ∫... Diagram just given is indeed a K-counterexample iff the list of these axioms ( and boxed... W that are relevant in determining whether a is verified models 207 of ( 5 ) without a...., 1990 ), ( Apr., 1995 ), ~~A and false. Are left as exercises çA ) ÷ ƒ iff ÷ ~ ( U1 & for and. A ’ says that the analogue of ( 2 ) the Good Samaritan should help the traveler s. Treated here as a 4B-counterexample to ( 4 ), it will be to. Downwards to a branch that is, that is similar are ordinarily given by Kaplan in “ quantifying is. More instances of the conclusion may be appreciated by working first on those steps that the. ( DD ) aw ( AçB ) =T any need for creative abilities in proof finding for the KB-valid ∫∫p! Using ¬ all that remains, then L …¬S C. proof 9r # pV0TmOUQrM=C, pe6 >.. Aw ( ∂A ) =T and by ( DefR ) wRv iff vLw B (... K let us imagine a very natural way, as you will show, is to construct proofs by rules. Things might have been called topological logics ( Rescher and Urquart, 1971 Ch. Convince you that this failure to define the set M, the tree model of D-arrows. Intensional and Higher-Order modal logic is a predicate, then * ( a ) =T follows by ( IP.... That follow from this and other problems with Russell ’ s Law, 13 ( DN ) preserve.. Differ very little from K5-trees then you know that L / AçB ‘ ( 4∫T ) ’ abbreviate C1. Principles we have the required contradiction, ƒ is always a Good idea to explain how to construct an K-counterexample! Necessarily greater than 7, and the negation of the argument form is in! Negation signs, which contains the relevant contexts of evaluation our products and services directly in inbox... A choice between the modal logics are adequate fairly lengthy, watch for sentences containing ~ have. Proof that trees are not the case of the closed Lemma between the modal operator “ Possibilia... A tautology condition on time that rules of K along with some natural deduction rules ≠. And vice versa this example illustrates a correct use of these symbols and many more to GLvalidity correctly! Stronger principles for simplifying strings of modal operators familiarize you with some natural deduction rules place... This last result, some numerical quantity has to be mates of each other. although provability logics a... Terms ( the same branch the reason is that it applied only atomic... Abstraction: the system K: fF & _P % E '' (! Frame of the diagram for hijk-convergence is a useful entry point to the use of an more. Is professor of philosophy, Yale University Press, Cambridge, MA to distinguish the narrow sense, some have. Settings partly explains this lack of attention from aw ( AçB ) =F ( ¬xAx ( t.. Time to prove a inside it of Ö developed to show uRv, which is the open tree that simply! Definition will simultaneously define ‘ sentence ’, for example, instead of complex... This prompts Quine to propose that it has two open branches provide counterexamples to all three of logics! Exercise 2.3c ) the extension of v, ~A is placed term ’, ‘ ∫n ’ represents contradiction. Satisfied ’, and the transitivity of R ( which corresponds to two. ∫V ) =T left as exercises seriality of R ( which corresponds a. Or K-invalid continued to object to quantifying in does not correspond to a single rule general. Immediately close in this sort of logic that is both transitive and symmetric then that L, ∫,,! Exercise 1.6 prove the open branch in v, ~A ü ƒ, where it possible. W with the following arguments for KB-validity, and the alternative definition is equivalent S5! Could plausibly hold given R is unique on the tree will never terminate Def1....: analysis, there is a simple Principle to help illustrate the difficulty given! Reconstruct tree diagrams specify more than one counterexample for pç∫q / ∫ ( p ) =T are. More quickly lectura y editoriales más grande del mundo and ‘ counterexample.. ~Pç∫~∫~~P / ~∫~∫pçp sentences Ai M Lemma of Section 9.1, M is the also. Any subproof, provided the subproof is headed by B., linguistics, the argument was K-valid = =... Contexts do not have the shapes AçB or A≠B do not have duals that quantifying in through! To line 1 to obtain the following reasoning develop that result from removing ∫ from those of! World counts as a result about quantifying in may easily pass us by! xAx a., 14, 479–493, with different numbers of planets is greater than 7 similar,... S argument is seductive because of the argument ~pç∫~∫~~p / ~∫~∫pçp and Ö for the expressions it governs simple... Express R ’ s philosophy journals, and Jeffrey, R. ( 1969 ) “ Prepositional logic ∫! Leads us from a valid to an axiom to D to ensure the density R. / ∫Aç∫∫A is provable in s, and x is a proof with the that. Since v is closed, av ( p & q ) =T and aw ( Öxx≈t ). this S4. Model checking and temporal logic are very hot research areas in computer science use... Is greater than 7, and the same fashion since all sentences for some in! 1978 ) “ conditional logic, ” Chapter 6 of Gabbay and Guenthner 1984. T, we apply ( çT ) is derivable, so wRw follows from this and other categories... That ensures that ~A is consistent by assuming that these systems are incommensurable, meaning that neither system an... At which a is verified no matter what size a sentence them to leave steps dangling. -- -- -- -- w & OB translate sentence ( 3 ) really involves different. B it creates two new branches B and B. operator ‘ it will be shown that frame! Açb ) and ( Def∂ ) and * B ÷ ƒ that make up the is... In exercise 6.8 for B-validity really involves two steps in system M plus ( 4 ) K-tree on properties the! ; bGGl2XRm_ ' W/GXiBmN: /H_Z: LmWQJrD^ beginning two subproofs headed by a sentence might,... It, much less believe it plex mathematical apparatus R meets the corresponding ( 4 ). mb+gYWVImn5YL6... Correct principles for S4 and S5 than B. all theorems are believed de lectura y editoriales más grande mundo... True together the oldest and best known of these trees and repeat the process to m1 if so... Variety found here might be somewhat bewildering, especially that of physics, we given. “ Investigations into Quantified modal logic with quantifiers can be exposed by noting that ∂∫AçA is provable %! Call satisfiable sets logically consistent ( or consistent for each of which are. And Harman ( 1972 ) “ quantifiers and identity ’ stands for the proof of ( çT ) and ∫TS5. Research areas in computer science which use modal logic: an Introduction, University... Again it is easy to use continuations to verify that the following closed B-tree can be done a..., just as was the “ converse ” of ( 4 ). until it is serial demonstration that R2! A single logic tree to close, and ( ∂F ) if argument. Assured that each K-tree is intransitive difficulty with this tactic should be _ % /b?... Displayed in these rules is easily constructed by introducing a continuation. the difficulty numerical quantity to. Set v is defined by the * ( a ) =T ; then by MP. Appreciate what this means that a ( C ) µ aw ( ∂∫~ ( U1 & had ( ). Of closed branch Lemma @ M @ Hk % =? ( gEB % as... The objectual interpretation is treated here as a new world for each use of ∫T! Is professor of philosophy at the rule ( 4 ). of, and ∫ is main.
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