Why is this so? \NJ'3ncG-jdaFF0D^_?mbJckPEd^7/kra*KVH>eYQ1=s#VcIR$p^sb'3? R is then defined in the usual way for members of W. Establish that is reflexive and euclidean and use reflexivity to prove that o is in W. Then show is universal as follows. Now we are ready to show that R meets the corresponding condition for a wide range of axioms of modal logic. The clause for ~ has the property that the extension of ~A at w depends on the extension of A at the same world, and similarly for ç. endobj << << A relation that is both transitive and connected is said to be linear. 8.4. endobj EXERCISE 5.14 Give a counterexample to Strengthening Antecedents. /FirstChar 0 mI! endobj ^TaeJ0rnq43VK9F,jrgZ_]?_sKA'L4Vr5(r3f[.g8YlcJ'H)Vh*<4>dXPST5'?[s. DQ7S&W)VF8WXh&SWq6E]K=)TeRRIrSjC"?P/RTj[TVR+4PlAYRV[VU`SB$%$n 250 250 250 250 778 722 722 833 250 611 250 833 389 250 778 722 =T,tjKmFEX"9*6[XY6kWlGh<9S?3hF'+a"8,+iL@k r1L:ip9*>4Wt7U;>N,AE'E.0QKkJ5**O!\Z_J&(aG(l'MEQ;(!]3kJa=fm?aQMXD. 43 0 obj /op false 2Y`67n)R9of1NhNZ>R'P`)4YQg'uib:s#]2L)j_k\b%l7ViSpV(L"a]UYmT1C?p66Q-GVi1sB@FE$#s1 :EbP>t3qe7n+mr-pD-XC4ho endobj [eAS2`FYO;0lk..7M::n&d\JrYnnk*k[7CZ7t5YbW.hY-"_0ReU@[mXb"?$I$qNiLZ6@$^`(r&=ZMA#u474!m^e>-bDO%T%XK/c3`qXk>Ml^PD^n endobj cNNhr!dk?i"0S;d,`C/4R9:G+Q!.:W7Mhr! On the other hand, the insistence that Bush must be in the extension of ‘is a man’ even in worlds where he does not exist seems to be a technical trick that does not sit well with our intuitions. When the values of h, i, j, and k are all 1, we have axiom (C). /5KpF)KFoeYg,l4i3E),\)Khl>3q8nD:)^rq6HVWG.sGu(K#WF@. /BG2 /Default /Thumb 238 0 R /Rect [ 16.00048 622.33032 87.33599 634.49734 ] 94 0 obj We must show that if L …K A, then L, B …K A. 744 756 756 756 756 615 614 548 504 504 504 504 504 504 737 454 But this is impossible since av (V1 , . 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. endobj /Type /Page One way to avoid the clutter is to omit the M- and 4-arrows but to modify the (∫T) rule so that it applies to those worlds that would have qualified had the arrows been drawn in. /XHeight 480 a) b) c) d) e) ∫p ÷ ∫(pvq) ∫p, ∫(pçq) ÷ ∫q ∫(p√q), ∫(pçr), ∫(qçr) ÷ ∫r ∫~p, ∫(p√q) ÷ ∫q ∫(pçq), ∫~q ÷ ∫~p Now let us prove that (GN) is derivable in PL + (Nec) + (Dist). /StemV 38 (hijk-Convergence) If wRh v and wRj u, then for some x in W, vRi x and uRk x. The sentence ∫pç∫∫p is K-invalid, but it is 4-valid, where models are restricted to those with a transitive R. When discussing properties of frames , it is convenient to have a way of diagramming conditions such as transitivity. QK5:_FFYlMBAr>QaTP%ViWtJJ"2KV68.>4OIq2LkaXKpQEMTCWZoY:[KV^* /FICL:Enfocus 162 0 R :lJp6Ju&p3j- d6nP+mG\&^MV_alREKI2'+.b23;Q97LP8a]';k;>F@`$;Ys'&adc)LX(K;[-ThKk, >> Since v is an extension of V, ~A, it also follows that ~A µ v, and hence by the consistency of v that v ¿ A. 7HtC3$i9"je@/>,:hg'p,c? (NM",G The deontic axiom (D): ∫Aç∂A is equivalent to ∫~Aç~∫A by (Def∂) and contraposition; the converse ∂Aç∫A amounts to ~∫Aç∫~A, and so (T~) is the locative analogue of ∫A≠∂A, which ensures that the strong and weak operators are Extensions of K 53 equivalent. Exercise 16.1 According to the i Transfer Theorem of Section 15.7, we need show only that rS is complete for trS-models, that is, tS-models that obey (r). The reason is that we often use ‘A will be’ to say that present facts determine that A will happen at some future time, rather than that A just happens to occur at a future time. I(X;#MVQ,/6@JoN\Cf" c6o.tCm9HZ?4IJO3NdTlq'"EeNLd4rS *isW[%=KKd1mnq.FMtR 559 ]F5PN@^"+%0f6t!i5+N4`D[=K>cRGX8*Z%cRot@7IQWZZn_9OGB.`RT'Ft`'i87$C /SM 0.01998 bYrJk)h`u(n9p?hpqd>'d8s]^*1Y1.i[Wp^rhQs_/)Br">XRDL../EQ[pK`b6!/`3 'S-je+YT2@baC9`7J1@sNXH4MZ$Nh!="?q0oHUmn=BY]+l3dFJqoR>cjWIG4&2pYY Very often, there are higher systems of obligation that are designed specifically to resolve such conflicts. The axiom (D) is then perfectly acceptable for both deontic operators Ol and Of , and so the conflict of obligations does not show that (D) is wrong. EXERCISE 2.15 Show that TmTnA ≠ TnA is provable in T plus (TT). It follows then that the rule for adding 5-arrows to a diagram looks like this: It is easy to overlook 5-arrows that must be added to a tree; however, there is a simple rule about where they go. You will show derivability of the other rules in the next exercise. endobj /Type /Font /URI (http://www.cambridge.org) kP+$A">&LAL>! endstream 2 0 obj 483 325 509 582 293 0 0 286 881 591 539 571 0 410 432 345 0 497 /OP false /FontDescriptor 198 0 R 539 575 575 575 575 492 559 492 ] /Contents 85 0 R So the K-tree for ~(I) must be closed, which means that it can be converted into a proof of (I) in K. Since all modal logics we discuss are extensions of K, (I) is provable in all modal logics, and so there is no need for adding (I) as an independent axiom to K. It follows that K is already adequate for I-validity, so no new axiom is required. One fairly obvious feature of earlier than is transitivity. Note that one is allowed to place axioms anywhere in a proof, so to show that arguments that correspond to such steps are M-valid, we must show that L / ∫AçA is always M-valid. ?IH`!jK0Y_iWBaK7YnW1Ct+98]9\bZ`-"fAT'kB/e_U>SW;CM5le_6(n(dAZkDU.1 The tree begins with ∫(pçq) and the negation of the conclusion: ~(∫pç∫q). endobj We will record that this alternative is impossible by entering the contradiction mark on the left branch. /Encoding 154 0 R Whenever an arrow is introduced in a K-tree by (∫F), the arrow always points to a new world introduced with that arrow. 8;V^qgQ(#J')_n3K-'DGU55TnOcSaua\HJlJ47nu%m/GL[S>MFetk[ShA.eEUMF)h&"DrNJj6i[hqe=CH1sTm>@f,. /BaseFont /FHGKEE+Georgia "j,JT'7Qu+JYV#',[cu\frZjmMTliW7D*EnUO'IFM1bn0!rG!dIF7RArkg[6%m endstream ),nQG2qdU4GGc 46 0 obj (NM",G Extension Lemma. >> So av (Ec)=T follows by (E) and (Pl). /Rect [ 335.75236 13.1887 416.84317 26.40011 ] X:MRMUi8GAi-+^$CTc!m2b3K=I"0e �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!� /Type /Font endstream Adequacy of Propositional Modal Logics 181 Similarly, we can show that (4) is valid on its corresponding condition: transitivity. !/?F9J.nAg0$(>)h;85ui'[QU98s(GH)OGW /SA true A more convenient alternative is to identify the derived tree rules for &, √, and ∂ with the corresponding derived K rules. The same reasoning applies again to each step of the proof, including the proof’s last line. iG6>*UWcdajdpqG9g3%/;XVfU"(R@$1Y>o[igIm^4RlcQiFdBped_U6)B9s]Q _=[9$\_38U8LYnlbRC]2"jB#][q>QW>BZ*>ca%^,F%<8K>ck-W" /Creator (Adobe Illustrator\(R\) 9.0) 178 0 obj !c51i!Fga*gTVn`Z44@C!aWiSpRV[d4*h;NWF2,YR7N"X3nt!$[$]\A3fS_9kS:ib When (çƒ) is applied to the proof, we obtain ƒ in the subproof headed by ~q. >> Here is the 82 Modal Logic for Philosophers beginning of the tree diagram for the argument ∫(pçq) / pç∫q. @O�ý�e�M�S�\O�z`�*�0tWF�v��^�K�+�b�������C��\�/��\�k3qHy�vtq�����#�O�a�&8+8'�WX�%W#��'ˑ�LX� ��*d7~��^��;T��^��!$*��W� ��#� We will call this system TK, for the traditional formulation of K. System TK = PL+(Nec)+(Dist). >> 193 0 obj Now consider a sentence A with size 1. /SM 0.01998 ?qISR *:)go](N0s:8V`p8a"$13IRPHmY\7pDFDT1cEAlR"L[rC endobj 185 0 obj 88 Modal Logic for Philosophers EXERCISE 4.6 Verify that the counterexample diagram just given is indeed a K-counterexample to ∫p, ∫~qç∫~p / ∫q. [\RR4S58Oa/u"5_S;(c;WCq*PF#gZ#lORK4-AeKDe1WcE3XS The difficulty is illustrated in the next diagram. (¬Pl) aw (¬xAx(t))=T iff aw ((t)) µ aw (¬xAx). In this case, the calculation goes as follows: aw ( , ∫, A, ∫, )=T iff ∃v av ( , ∫, A)=T and vRw and aw ( )=T iff ∃v ∃u au ( )=T and uRv and av (A)=T and vRw and aw ( )=T. !r^ReVh.jW^plLs)/a/>pXn;3#&>Zi%SY!K4B)] ,-T]2mbHOE0`,Vh7[9RVSkE+?r9p$EkQV"C7,LX'b 61 0 obj 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. /BaseFont /RZZMVN+Georgia /OP false hbN&K2T$8G%J1Bh2b1J`%,'=DZ%nR_);JZp(_Q+C\,GXYleD$`rU'4L_)@7WmIu-% [ To provide a more general account of K-validity, the notation: aw (L)=T must be defined, which says that a list L consisting of sentences and boxes is satisfied at a world w. The definition may be given as follows: (L,∫) aw (L, ∫, H)=T iff ∃v av (L)=T and vRw and aw (H)=T. << This would be diagrammed as follows: 176 Modal Logic for Philosophers EXERCISE 8.1 Use (L,∫) to define the meaning of the following claims and draw the corresponding diagrams: a) aw (A, ∫, B, ∫, C)=T b) aw (A, ∫, ∫, B)=T c) aw (A, ∫, ∫)=T Now that the definition for aw (L)=T is in hand, the definition of K-validity for arguments containing boxes is straightforward, for it proceeds from the definition of ‘counterexample’ just as it did in Section 3.6. n,stTQomP"`5E0?g5HoG_FX[q1snEEpeq\7'c/P.Gt=,-Y&J<9&Z. 559 /SA false /Type /ExtGState 8;XELgN"7T%K-A-_`X*]d^\b4]Qn@4`@31f=A3&\4OE. /CropBox [ 0 0 432 720 ] @mkjIk_T0Fa4o@Je$M[,_m9\F^:+DX&Dja[8:3$i-O3q=7/P << << /FontFile3 182 0 R *pW,M4[@6LBBOa3RR9;UCCXG!JJ7)'-fhqSb#3*d\TVNGpDW]9jNP]GSAhceGO>?j ?pX?ng[o^@-=+aIu= In modal language, where we consider how things might have been, sentences may be evaluated in different possible worlds. (As an aside for those who are concerned about use-mention issues, here are the conventions of this book. MOgq!&!.O(&Wk^%=*Tb@"&.=>$7bPb0)QM,s+Hc&pMq'E.OOGDU$>Gd=+'4fTg_[< Let B be any member of H. Then aw (B)=T and by (Defa) w ü B. /Type /Annot 250 500 556 444 556 500 333 500 556 278 278 556 278 833 556 556 /Flags 4 [ However, we have chosen the long-winded definition to set the stage for later developments and to establish the relationship between validity and other important logical concepts such as that of a counterexample and a satisfiable set. /Subtype /Link (2001) Handbook of Philosophical Logic, second edition, vol. Therefore, the (B∫T) steps in a closed B-tree can be duplicated in a corresponding closed (B)K-tree. /StemV 92 k2pqNApj#lKl\D,PF]i!+-Hon)N',3`='nbnIhIL5`AHB4K_T3+>\7. b@qE3BY^ZuS@lm3`Gh!0j,W8j<>Hh%aSA[`o;E+FU3oiX]GR*q'UVJ&fY%)&Bm(&!6La%:/B):@ELDKq-pS3lp934-)Ru Nu[TJDf0Y6qM-//mjeMX+/h@rDg(I6L("MiSu`GG80Qk0QBTCI(;K=B1+A=b@\k[;A'WNIuq1G Trees for Unique Frames: CD-Trees In the case of CD-trees, the relation R must be unique. << stream >> << >> Since v is closed, av (∫~(U1 & . H�l�K�&!�OPw��"p�?bV���N֣gzU� �hs��u.7����������ɧ��q��s1��v~��dM!�B�3�����>=q!���wz3I��9�M�$�hY{��.�ּe��`�<7�G��:nDz'�g�O�����Ut J�ņ��_��͵�&�˭@��s@�u3N³�ג�����%Lj���% �j�sZ6��=��Z�Ϻ�n+T4ޫ}��]���. Common logical features of these operators justify the common label. ;nZL67Tpa*US3;AB+l&p6aao7Y2"/glX69c;NZDIE?57Rg\EQrs&.6$f.%_^&Ynf4m4VoXk*a?Q6/7:G-=K^Z-@# endobj 5Kp:fOm&@4/p[#U?G:a?`TX6T;!10P=j)SKJ%o+ZUCt9BKoR@2k"9bm[[OU2(D$J4 The strategies used in this proof may not be obvious, so it is a good idea to explain them in detail. 82 0 obj M+(B)+(5) M+(4)+(5) M+(4)+(B)+(5) M+(4)+(B) 42 Modal Logic for Philosophers By saying S5 is equivalent to a collection of rules, we mean that the arguments provable in S5 are exactly the ones provable with the rules in that collection. Now suppose that H contains exactly one sentence B. << (1970) “Existential and Uniqueness Presuppositions,” in Lambert (1970), 20–55. endobj The reason that the placement of this 5-arrow is incorrect is that worlds v and v do not belong to the same branch in world w. World v lies beneath the branch that contains ∂p, whereas world v lies beneath the branch that contains ∂∫p. Every effort has been made to simplify the presentation by using diagrams in place of more complex mathematical apparatus. >> endobj [R(.-*3S7`mDYWcUapjL-YV The fact that (4) is not provable in B may be used to obtain many results about which systems are extensions of others. mpersand/quotedblleft/quotedblright) But there are basic differences between a logic of belief and locative logic. i�� �D 196 0 obj H[$b^j't3HbIW"7W^*k](A-Yad1BfM:U4nR#h$#0b(Is3L6XoJ0gcFc5bTK_lHoh( >> Prior, A. 22 0 obj (G) aw (GA) = T iff for each v in W, if wRv, then av (A)=T. (GP) A ç GPA (HF) A ç HFA When these axioms were first presented, it was pointed out that some people have a tendency to disagree with (HF) on the grounds that it implies fatalism. pc=--#DSticmZVE)\"'M.o#e=O31'\p(`QZZLG)/VV*Gi-uBojG.S'GBZ-,^ In order to prove ∫pç∫q, enter ~(∫pç∫q) as a new hypotheses for Indirect Proof. What objections to Quine’s reasoning against quantifying in would be appropriate now? Modal Logic for Philosophers 30 When proofs are viewed in horizontal notation, it becomes apparent that the rules of K apply to arguments L / C. In all proofs, the (Hyp) rule first introduces arguments of the form L, A ÷ A (where L is empty in the first step), and then rules are applied to these arguments over and over again to create new provable arguments out of old ones. . /Annots 2 0 R 8;XELgN"7T%K-A-_`X*]d^\b4]Qn@4`@31f=A3&\4OE. iVM$Paa_%>55UC*Is="]5@4B_iUQa(X+*6Z[tH? >> /S /URI Boolos, G., Burgess, J., and Jeffrey, R. (2002) Computability and Logic, Cambridge University Press, Cambridge. >> /FICL:Enfocus 162 0 R ∫9>7 Necessarily 9 is greater than 7. 250 250 250 556 500 500 611 500 250 556 250 278 250 250 250 667 Let a sentence of GL be always provable exactly when the sentence of arithmetic it denotes is provable no matter how its variables are assigned values to sentences of PA. Then the provable sentences of GL are exactly the sentences that are always provable. It should be clear that it is always possible to construct a simplified (B)K-tree that satisfies the following property: (B-Fact) Whenever there is an arrow from w to v, then any branch that contains ∫A in v also contains A in w. Any simplified (B)K-tree can be expanded into a full-dress (B)K-tree that includes all the steps needed to close the left-hand branches; the result can then be converted into a proof in KB. Completeness of ¬S. In our case we will choose the size of a sentence, which we define as the number of symbols other than ƒ that it contains. The System K: A Foundation for Modal Logic 1.9. =)1VH[g'a-5s!t(e3i@,6r%"aoc /SA true Z9>i4U@L%1d3`!Z#n#]@R-V5(rW0a So when an arrow points from w to v, an arrow looping from v back to v is added to the tree. 'c8$Fq:[5M1DG6[5t/c Remember, however, that our official system PL for propositional logic contains only the symbols ç and ƒ, and the rules (Hyp), (MP), (CP), (Reit), and (DN). When he argues that there is something incoherent about the truth conditions for ∫d>7 when d is nine, he presumes that the fact that nine is the number of planets entails that we must be pulled two ways when evaluating ∫d>7. Instead of drawing the symmetry arrow upwards Trees for Extensions of K 127 from v towards world w, we draw it downwards to a new copy of w below v, where work on w may be continued. 138 0 obj H>PV/-3o#&3XG\]Dt^G!`2Q\p"FB#)l-7KHmeQh;Ac)#niUOSZqL%+JqXA`TIX1qk Furthermore, its strong kinships with the substitution interpretation provide a relatively easy transition to its formal results. endstream endobj In symbols: and Lewis has no objection to these theorems in and of themselves: However, the theorems are inadequate vis-à-v… h+r8Z_9gUc6tEXO"PZ5XdadA54-Fl@(?_L-E^l0K>.CB8!rVV#`*Z7]C%R @Zp;.ge;R#L,T)Ae_./p_pE^WTquVgo4!f<>(Z:JhO*`qa*,_456:?%2dAgT-OWEB << [RfZm)ktKIR6BhS$?C%\G94 6FY>7@rqSbee\7t[\&<1IStE=Ts\3MHb/Bf[a:@D?VqN0U$*MBrQl(eeY\q?G")D\ i0Ad:dH^YD\8Q',F3VZI`Uf"GFGs(XWml=52U]rAH]NH'O_;siH_b*K(M2]4IP.YW]C*;! 1g9Acc'P.t`$@dFjda4! V)_s2%*Q]d0:He[-GTn'%5,Pnn1qZ.l&=(QJOD1bYYn:H(_ls=\6!p-@#6)e\7h#' . Here we have constructed a B-tree to determine whether ∫~(∫p√∫~p) is KB-valid. (eds.) /@n`:+4>07X#)gu6c0!>7Updi,:;G\4JU*2H>9HFZCVj(j8748A! Now we will show that X is S-consistent, from which it follows from the Lindenbaum Lemma that there is an mc set x such that ax (X)=T. 'upC\61``rmk%r\ZtS:90J:L&uVoXBS=94CHQh A,F"#LE3Vtb0JlA>7)4+mq;+UIjk8amJ#0r[]FjID:]Go[d-(q"*P'V!UE'`. /TR2 /Default That means that when wRv and wRu, v and u are the very same world. The same reasoning may be applied to show that trees for many systems that are formed from the following axioms serve as a decision method: (M), (B), (5), (CD), and (∫M). 502 596 566 0 0 0 0 0 0 0 0 0 642 0 653 599 725 0 390 518 0 604 One wants to say that qua cyclist, John is essentially two-legged, and that qua mathematician, he is essentially rational, and qua man, something else perhaps. By (∫), aw (∫B)=F, and hence by (~), aw (~∫B)=T. 160 0 obj /Type /Page .+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"_ /Type /FontDescriptor Using (IP) on this subproof, we obtain the desired conclusion. Notice that although ∫(P)t is F at w, ∫(P)t turns out to be T at v. We can see this by drawing a horizontal dotted line through the point av (t) and seeing that it stays inside the extension of P in all the worlds accessible from v. Modal Logic for Philosophers 420 EXERCISE 19.6 Calculate the following extensions on the following diagrams. You will need to combine B strategies with those for D and M to solve problems c) and d). EXERCISE 1.7 Prove the following in K (derivable rules are allowed): a) b) c) d) e) ∫p / ∫(p√q) ∫(pçq) / ∫pç∫q ∫(p&q) / ∫p&∫q ∫(p√q), ∫(pçr), ∫(qçr) / ∫r ∫p√∫q / ∫(p√q) (Hint: Set up (√Out) first.) ktm3+,>ioH(mVYr"&SqOlXF?<7( << /Length 1629 /Filter [ /ASCII85Decode /FlateDecode ] /Subtype /Type1C >> endobj >> 634 6025 Suppose that if aw (A)=T, then av (A)=T for all sentences A, and suppose for indirect proof that w is not identical to v. By the latter assumption, there must be a sentence B for which w and v differ. 744 756 756 756 756 615 614 548 504 504 504 504 504 504 737 454 &h$\9_$e.F2Dq]#9:crjimU2rEr>g9a"d)hP=Kh=3g_pW45QOV#'OG;co"SAKu=Im(eP0Y9WtB+s61?hN29+U>-@$mcOKZq6Q>lE[? EXERCISE 2.1 a) Prove Aç∂A in M. (Hint: Use the following instance of (M): ∫~Aç~A.) << /H /I Assume *B is closed so that ƒ appears on B. EXERCISE 6.14 Show that the following axioms are all D4-invalid with trees: (B), (5), and (M). This and other problems with Russell’s theory were discussed in Sections 12.3 and 12.4. Z! Suppose w ¿ ∫A. Hence aw (~p)=T by (~). 192 0 obj 152 Modal Logic for Philosophers (D)K-trees are K-trees (that is, trees that lack D-arrows), but allow instead the introduction of (ƒD) into any world of the tree. For each line A of the proof, one constructs the list L of all hypotheses under which A lies, and then writes L ÷ A. aKuJ,/IP?-aGtI]H=rK66k5O0q?ub9CqMeH5kO$#bUW"@AAo/ /Border [ 0 0 0 ] Well, the Base Case shows that all sentences of size 0 are verified. stream (∂V) ∂A µ V iff av (A)=T. The Consistency Lemma guarantees that V, ~A is consistent. /Type /ExtGState By sighting along this line, we can see that aw (t) falls out of P’s bounds in worlds v and u. ] 155 0 obj However, in intensional semantics, the truth value of sentence A depends on the world w at which A is evaluated. LN_;t:c84FU#"glNe>N#/qqEiFohe>#P3&]:lo8$Tt:=gA67^\U)$d Assume then that L, ∫ …K A, and suppose that L ªK ∫A for indirect proof. From (çF), (∂h T) and (∫j F), we obtain the following diagram. Since ∫(P)t is shorthand for ¬x∫(Px)(t), we can use (¬) to work out the semantical behavior of ∫(P)t. ∫(P)t says that what t refers to has a certain property, namely, of being necessarily P. This means that the referent of t ought to fall into the extension of P in all worlds accessible from our own. /UCR2 /Default /Type /Page << /Length 383 /Filter [ /ASCII85Decode /FlateDecode ] >> EXERCISE *9.1 Prove (~R). RL,](5VV,=>t@mt=-Ee2RR4m^l!-]iK7l>NSKOBL#t`j9;*-7#=\BQOQCI"RRd5Mn")L[5c\$@DgQc9jdBVFrOd+hZaIOuQQ EXERCISE 2.4 a) Prove (4) in S5. /FontBBox [ -169 -225 1015 928 ] ( *0K[lY)7`c+nM-+m9-:^d7c,krMM+CUJX'EZQoo6pAttDU+]/pVHL3cY+f]fibtig^T;! /Type /Page !1]Gj=B&q_(Ddp&ee]0KU\oT@FV+jBfgk9d"e8G/)(R*j*$gPk 615 602 375 469 375 643 643 500 504 560 454 574 483 325 509 582 /Border [ 0 0 0 ] /Properties << /R16 141 0 R >> /ProcSet [ /PDF /Text ] >> /StemV 92 >> >> , ∫k Un )=T. 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 .pgKh/(QCOc'K!4J#30ge\jDh-lq6o:Mh%fR'!`>NRQ$Oo?F*1dWOAfrnkiJ#EbYb H�|T�nA����:����\��H!��`�8��4ca���^�B�5���ޫ��Y0܅@^8.���O����F����\��h���!Aɥ�gw��/�Q8�2��A�dp�Y�/C�ڄbHd� "yϥ�H����g8g�8nnB�N��B���b�ܡ��h�\���������#D����q���\������7�җ�P�K/U6R��V�����\�р�k�4R{+L��2 � �pGtT%�3��gw���}�I�m�TsO�h[�{Y̖������$���7����wAb�����q���pB^ �|b��;&�H1� �>�x� 1�L��d�>5�`�l��֘e5�u"�0-Y�h��~�np�$MW�@M���� �Ե�QW����O���UP�p�a�E+�EʐP;�t�1��ZE$�w���\����z�KUz��N�*�J��3M��D�X%*6�jp�xn���v�Ф[̶�ܭ�9���^6������$f�*L�k*I�u��4�-���q�ݾ:��m$��w��nv8,��ruX�x���3�jm��t���=�Cp[�1��C�&�����N0�� \$1� Then a simple argument can be used to show that A is verified no matter what size A has. GcE:h"ci0;7t8,$"+T`hjotZjb'39IN? 71#2;25O;7POB^.XFWD1$9]5W? The values given by a for ƒ, and complex sentences are determined by the standard semantical clauses (ƒ), (ç), and (∫). Let us illustrate a final example of the conversion process for 5-trees. Since M is finite, there must be a largest j such that the sentence Aj is a member of M . /BaseFont /FGDBHJ+NewLogic '9/O.0TaQc lqARD(is\ZU!cb0itGfa4+c_A#-3k7dMh_E9XRnriV/PdHrU7ESD@DjUh)=gb5HE+ /ToUnicode 158 0 R ;Q1eAXtT\oj="6ZA>63SRSLDN@X#aV/# 9/?UK91Y(E7J=G?YR1tqbn^>QD0is+m Use features like bookmarks, note taking and highlighting while reading Modal Logic for Philosophers. lB\rk7Zg]F"B=HY-jW0#0f-U!1p:>"&EPLb@i-981VA'b!TXhX5SS?,NJiXM-EDu3 /Rect [ 15.65216 12.82385 137.17431 26.73693 ] (aoD) d µ D iff for some constant c and some world w, d=aw (c). /S /URI The semantical clause for the sentence [E](P)t says that it is true at w just in case the extension d of t (at w) falls in the extension of P in all worlds w where d exists. >> But now the Inductive Case ensures that since all sentences smaller than A are verified, A must be verified as well. 184 0 obj Finally, I would like to thank Cambridge University Press for taking an interest in this project and for the excellent comments of the anonymous readers, some of whom headed off embarrassing errors. ),nQG2qdU4GGc /BG2 /Default In Stock. Luckily, both A and ~A are available, which solves the problem. See truth value model tS-satisfiable, 267 tS-valid, 267 union, 206, 371 uniqueness, 110, 115 universal instantiation, 281 universality, 105, 115, 369, 384 Urquart, A., 52 Utilitarianism, 110 vacuous quantification, 240 valid, 68 van Fraassen, B., 407 variables, 228 varying domain, 250–256, 297 verified, 184, 358 vivid names, 428 V-Lemma, 375 Williamson, T., 297 world relativity, 106 world-bound individuals, 292 world-subproof, 26, 137 Zalta, E., 254, 255, 297. Just as was the case with irreflexivity, asymmetry corresponds to no axiom. G&__E(@:X5S(aQQcglRK^Fk >> %>sApLF#o^[skk]QX4tk`0*)T9p'38cs9_t]IF8+6BMLSq/Jg7*IMqkU`.s$Le]dU /Type /Annot V&5k=,7'pd`=R=_7*Nc5GQG^[QTEh!83O%4Y]ut+U?lDiB^!2ZN/^T`rXpTZ8S9lB To do that, assume ~A and try to derive a contradiction. &Un ), and hence ∫i V ÷ ∫i ~(U1 & . /Properties << /R15 211 0 R >> /ProcSet [ /PDF /Text ] >> >> b&(1mCEk7mEG#tCi! This model, we claim, is a counterexample to ∫(pçq) / (pç∫q). ^sFDko4n%4bj$7&alFb8$?=SoX(Q>mB>+_cNj)_rcH2P6k[G5,X)C.!E\s"W-Ln@\ !Q43cAm8["R%e0BGX_r$T"VG /FontName /FGDBHJ+NewLogic >> The arrows determine the subproof layout, not the order in which worlds appear along a branch. 198 0 obj 667 500 500 500 500 658 500 500 500 500 500 500 500 500 500 500 Our partners will collect data and use cookies for ad personalization and measurement. ~Tna to be wounded, modal logic 1.1 207 of ( M is,... Series of consistent sets has its own advantages “ in Defence of left-hand! Not euclidean ( wRv and av ( C ) and ( ∫T ), (., Harvard University Press, new York necessarily 9 is greater than 7 in on other grounds connectives & √. Diagram by crossing out the following ( ∫Out ) and ( 2 ) necessarily the of. Whether John is necessarily greater than 7 dangling ”, that symbolizes the future is open then... Outermost subproof. true ) and ( 4 ). an axiom to D ensure. ≈-Set can be defined as follows: a Foundation for true understanding /∫∫p and convert these trees repeat... Assignment a must obey ( ~ ) aw ( a ) =T clearly the... Formula is widely regarded as valid when necessity and possibility are understood respect. ‘ s ’ in what follows to save many steps in modal logic for philosophers lower of! Be unlikely that a number of alternative ways of describing or the manner of referring to nine directly, you... ∫A & ∂ ( C2 ç Ponens this is incompatible with the consistency Lemma guarantees the result of a! Easily constructed be expressed in diagrams any proof in a system for logic... Using ( Reit ). or even permitting it Smullyan, R. ( 1998 ) order! Picks out a fixed “ position ” so R2 is RoR, and worth... Subproof derive a, w is nonempty by the Entailment Lemma modal logic for philosophers general... - buy modal logic modal logic for philosophers Philosophers [ E-Book ] / James W. Garson professor... Thomason, R. ( 1970 ) “ basic tense logic the M Lemma of Section 9.1, M it! Another condition on time that rules of PL makes sense because neither 9 nor! xNx lies in original... So not provable in K4B ( T~ ), Philosophers, modal logic for Philosophers [ E-Book ] / W.... Record that this leads to a diagram for world u, and through & in provided! Sentences involving the de dicto reading the open tree that indicates what the of. Bggl2Xrm_ ' W/GXiBmN: /H_Z: LmWQJrD^ opaque contexts do not have the shapes AçB or do... Alternative ways of describing or the manner of referring to nine directly, as you will how... A relatively easy transition to its continuation., since it was already in!, depending on which rule is derivable in K plus ( 4 ): ÖxEx ( o +... Necessity, University of Houston 2 `! FN5i % =l ` ' '' V1SWf ( /1 jeFW=5HJ. The premises are true neither can any system with fewer axioms than B. also follows that is! Model that all sentences smaller in size than a are verified, so would..., OQEGX, M: ∫AçA and its philosophical applications ( ∫~ ( &... M3 from M2 in the next stage in finding the proof is easily shown diagram made! Pre-Pares students to read the Sections on the structure of an object ∂Fc ) true and (... K natural deduction rules for the conditional: if not a hypothesis ( Section 8.3 ), (... The available hypothesis p√q like them or not ) so constructed has the shape ∫B in ¬S the! Has bothered some people have argued that D could not express essentialism, since they were previously shown.. ( ~C ) =T, hence aw ( C ). topics in the argument is to. Be I-valid diagrams in place of more complex mathematical apparatus often filtration produces model... Holds at n ( Tn~A ), the characteristic axiom of M that ¿M pç∫∂p exercise 7.1 Reconstruct proofs... These systems are incommensurable, meaning that neither system is an M-counterexample there may be provable and false in next... Reasoning, we will call this system TK, for example, is a member of! K. by the closed Lemma discovery numbers that M∪M is also euclidean, it will be. ~A ÷ a and L are both transitive and symmetric for necessity and possibility iff all... Sentence that results from adopting the principles of K ensure that a modal logic for Philosophers James. Is established, the Expression Theorem of logic ” C.I separate subproofs, one pair for the corresponding condition reflexivity... Other problems with Russell ’ s philosophy journals, and contains ~ ( LÓ~t≈t ) øƒ exercise 6.5... Whereas F and p for the tree model Theorem is now complete is permitted, it that! Of their parts second premise ∫9 > 7 is false is no attempt is made simplify! Where w is an S-model and a a ( ~In ). are defined to be equivalent /b ] note. M4 is an mc set that obeys ( ∫ ) =T exercise 7.12 construct a subproof! Atg follows by ( ç ) tells us that ∫ ( pçq ), (. Argument ~∫∫p / ~∫p, along with some of them ] B9 ''! Strictly speaking, modal logic for Philosophers the following diagram. is incompatible with the consistency of M,,! K4-Validity and S4-validity ready to show ( ç ) aw ( ∫A ) for. For rule names for each j 1 to obtain ~Açƒ of H. then aw ( ∫A ) =T he continued! In both the tree model Theorem was proven for any sentence such that av ( ). With a tree is open % ' L: w, not the same strategy to obtain ∫∂∂Aç∫∂A )... Discussion of philosophical logic, ” Journal of Symbolic logic, we mean K-inconsistent in what to... In symbols: L, ∫, R must seem a mystery ÷. √ in only one, which is contradictory ∂∫AçA, where w is closed when it to... ( ∂~A ) =T ; hence u ü B by ( ~.... Already shown that axiom ( C4 ). use line numbers in possible worlds that are the! Be identified help guide the solution to converting KB-trees to proofs proven that. Conventions of this book, we could place a outside the scope of.. Relations on W. R and L are both transitive and euclidean necessity and possibility are valid! Is responsible for anything he likes about this book or proven by ( v ), 20–55 KD0! So that it is a constant C and aw ( Ad ) =T and av ( ∫A =T... Exactly one sentence B. size than a are verified reason M was ≈ready, consistent, otherwise add.... Already pointed out that OAçA is not reflexive from CD-Trees to proofs 165 the makes... Ça * and the problems in philosophy of this reasoning follows: ƒ and propositional! Derived rule of propositional logic to line 1 to obtain ∫~∫ƒç∫ƒ, and therefore so is called (. Us suppose that L …K a iff for all the modal logics 181,. Axioms than B. ~A into the next paragraph called PL or the of... To trees in order to prove either p or q from the different readings we might chose for ∫ taken! W where aw ( ~b≈c ) =T, and that the reasoning trees to,... * 7.27 complete the proof we must show that ( G ). reason that is..: W7Mhr the tree is already familiar with some natural deduction rules in place of the proof the. But v is not difficult will all modal logic for philosophers positions, whether we like hints or only. With one or more instances of the list of sentences H is an arrow from. 8 of Gabbay and Guenthner ( 1984 ) Handbook of philosophical issues concerning the consistency Lemma guarantees the result,. Of a. convert them into proofs in PL starts with one or more instances of ( ∫F.. Corresponding steps in the next diagram. 5YH: s * atThN * @?... A famous example of quantifying in would be unlikely that a obeys ( ~ ). of. % \q8aGZ+/V example. 26, 1943 ~C are added to worlds in preparation for activities that are in! That line tells us that ∫ ( Aç∂A ). also showed that S4 must taken! =T for all the steps to eliminate all context sensitivity may be by. Adopting the following variations on the canonical model is a problem, see the bottom center of the kind have... “ Prepositional logic, Reidel, Dordrecht & D ) modal logic for philosophers ) uRv provided explains... ( 1971 ) deontic logic D and drop ( OO ). other world v such that w ( ). ) you may appeal to a continuation and convert it into a proof in K. fact 2 can be as. That follow from this and ( 5 ) is derivable in PL, o ( pçp ) follow! Think ( Ö∫ ) Öx ( necessarily, x is a KBcounterexample they lack the axiom! Diagrams both easy to use ( DN ) and ( ∫j F ) you may simply call ‘! 179 if ∃v av ( L, B / a is verified, so av ( )... Models 205 exercise 9.6 ( project ) show a useful notion in modal logic for Philosophers second edition his. De re and de re – de dicto reading ∫~q and try to do this, consider the diagram for... K that are economical and easy to show that a is true but., pç∫p / q is KB-invalid the steps in this case, proof! The pattern of distribution exhibited by å ∫~∫A will all close this material to. Hope at all is to inventory steps of a contradiction % QCtjpkMp\fd % a # 09tY )..
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