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Unformatted text preview: 1. (10 marks) True or false? /' @ Circle ’T' if the. statemenf 1.5 true.
Circle ‘1‘“ if the statement is falsp;
For this question. you. should (LSS'H'J'TLH that g: and w are WFFs of SL. @ F If 2,1! is inconsistent, then (,9 is not at tautology. T @ If 1;: is inconsistent then 9* does not entail Q. T ® "Kant. believed that" is a truth functional connective.
T F “33:(Fm —+ 03)" is a valid MPL WFF. T F If X is an inconsistent set of MPL WFFS,
then no member of X entails a member of X. T ® If to entails w then so is consistent.
T ® If so is a tautoiogieal conditional, then the consequent of (p is not inconsistent.
T ® If X is an inconsistent, set. of MPL WFFs. then some member of X is not valid. @ F The following argument can he shown to be valid in SL: “If everyone is
hungry, then Slim is not hungry. Everyone is hungry. So, Slim is not hungry." (I) F q: is not. a “(PP of MP],. /10 [\J 2. (15 marks) Suppose that a new (meplaice mnneri ive ‘53". and a new two—place connective ‘#' are
added to SL. You are informed that: ‘((A#B) —> @A)‘ is a tautology @A' does not. entail '(A#B)‘ “(A#B)' does not. entail “1198‘ ‘©A' is contingent ‘@A’ does not entail '(A V A)‘ Fill in the truth tables for '#' and Circie ‘T’ if the stainingmt 1.5: tT‘ur. (Timute ‘F‘ if the statement is false: ‘(@A&B)’ is contingent
“El/i. (A —’ B) l: B‘ is a valid sequent.
‘(B#A)' is logically equivalent to ‘( B&:A}'. ‘(@A#'[email protected])‘ does not entail ‘(B V A)‘.
[15 3. (10 marks}
For each of the foiiomsing:
Circle “histology” if it "is a. [rVFF .0th that 13 a mummy?)
C’ircie "contingent" if it is rt contingent WFF of SL.
Circle "inconsistent" if it is rm. inconsistent [II/"FF of SL.
()thef'wise‘ don't circte anything. (A <—> —> A};
tautology r nntingis 1t inconsistent ({A v (Beam) —» (A v 0)) [email protected]' contingent inconsistent ((A8; at B) v (AS.an tautology [email protected] inconsistent
((A& ~ A) V (B V mBD
tat®y contingent inconsistent
((24 H t”) —> (51 —' 0))
tau®gy contingent. inconsistent.
\ ( B —, (A —s 8))
[email protected] contingent inconsistent
(A H {CM/1 V B)»
tautology cit inconsistent
(NF H AVA”
tautology cc ingc it inconsistent E
((A —> B) V (B a A))
ty contingent inconsistent ({A <—> B) ——> {(A V (3') <—r (A&("))) tautology con inconsistent I miss??? odd/mark. _, “i
“ I m3; mes/r») f —]
[email protected] j” 4. (15 marks) Tr‘amiate the foiiowrng statements and arguments into MPL.
Preserve as much structure as possibirr.
Use the follown9 tmn.sia.tion scheme: a: Aaron 1'): Barbara Fx: x is friendly
Gx: x is greedy (a) If Aaron and Berberc1. are bot I1 greedy, then someone is friendly. __ Dorinah 3 All Atrium ,
(( Gar2x65» 9m) (b) If everyone is greedy, then sorrieone is. DMWX : A” AﬂMm (iix For *—> 3x5 X) (0) Aaron and Barbara are both greedy only if at least one of the two is not friendly. ((6s?~65)~—‘>(4F1\/AB)) (d) If everyone is greedy then Barbara is greedy. If someone is greedy, then he is
friendly. But. Aaron is neither greedy nor friendly. So either every greedy person is not greedy or Aaron is friendly be mag ,' Air/nan,» ,
(Mr A; an, seam, (A 6’02; Ara)
tag/46m as) v H) (e) Barbara is greedy but. not friendly. Aaron is both friendly and greedy. Therefore,
not everyone who is greedy is friendiy. Démwjﬂ U9” Aﬂ/Wf I (St 2 As) ,( row: A M We) /15 :3. (10 marks)
{3 there on internvetotion under urninn att the fottoioing MPL WFFs are true? If yes,
then gioe one such. intRiprotation. If no. exptoin why there is no and; interpretation. ~EI:I:{A:L'&B.1;} Mir N (Am —> NEH) {4" Mink74 367$ 4; X ’59me VCJT(~BJ: V ~03?) N 33:01: '0 Al 0ft 5 _ PI £551; If M3930 two/to) i *3?! /10
6. (10 marks)
Write down 4 MPL WFFs which. are each consistent. but where there is no interpreta tzon under ‘tﬂt';;tl‘??10?‘€ it}: I : :e WFFS is true. 3 f6 M a,” a” U j!» tom/e14 21:? {AS/— mt
(Fa ﬁst—i ) 4m Mam / 3 7m
ﬂﬁﬁﬁ)
( 42% 3 AH) /10 If new}: minnow).
7". (10 1'narks) Give a consistent MPL L'LWF witmo. 't.‘~' foise under every interpretation which contains
fess than 4' elements in its domain. E“ x a a
a {Ext—(<23) 21 9x( 5x3'4/LX'»EJX(TExX/ﬁx))
7‘ 9X (“taxiAg”) /10 6 8. ([0 marks) S “i
Gi‘h‘f? an rnﬁm'premﬁxJon nn.rﬂ:—:r whrlr'h. " 'v’a:(H 1:84: hr) " is fnise
and iV$HJII —> Gm] " is ﬁrm:
I \ b3 mg“) .L ﬂﬂf‘fl‘UL/S
HAL )t f3 (3.. Auman Em '« X ’3 a nme/ /10
9. (10 marks) :2 @ GILye nn MPL WFF' that as Iogirmfﬁy equivalent to each, of the following WFFs. Your
answer must wincindr: an r:.z:rsr‘.ennm£ qnnnnﬁer if the origrnnﬂ WFF contains a universal quantiﬁer, and m versa.
(MPL WFF .,9 r5 iogtcnﬁﬂy Frrg'nr'vanrnt {.0 MPL WFF 1,6: if and onﬂy ﬁftp entails 111 and 1p
entails $0.) (a) ~Vm(H$\/G:r:)
27x 4 (Hm/6x) M 37x (“Mg/ﬁx) 5" g‘AKAHy—yﬁx) (h) ~33:(~H:r:&t ~ Gm) ﬂéx) or «3x .«(rmzxm v (rt/Mm)
0r ranAm a ﬂax) 22 (An/am)
(d) 3:3{Hx&(?rc)
n VX A (ng‘th) m. ,f/X (AHX Vﬂéx)
or "VX (Hx '3' Aéx)
(9) HynyéLc ~ Fm 7/},«7 NM a}? 7M :3 07 (“mﬁVM/‘K
/10 ...
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 Spring '08
 NoprofOnlinecourse
 Material conditional, Propositional calculus, MPL WFF, valid MPL WFF, mpl wffs

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