A Textbook of Belief Dynamics: Solutions to exercises

69 Pages·1999·2.67 MB·English

by  Sven Ove Hansson (auth.)

ATextbookofBeliefDynamics
A Textbook of
Belief Dynamics
Solutions to exercises
by
SVEN OVE HANSSON
Department ofP hilosophy,
Uppsala University
SPRINGER-SCIENCE+BUSINESS MEDIA, B.V.
A C.I.P. 'Catalogue record for this book is available from the Library of Congress.
ISBN 978-0-7923-5329-4 ISBN 978-94-007-0814-3 (eBook)
DOI 10.1007/978-94-007-0814-3
Printed on acid-free paper
AlI Rights Reserved
© 1999 Springer Science+Business Media Dordrecht
Originally published by Kluwer Academic Publishers in 1999
No part of the material protected by this copyright notice may be reproduced or
utilized in any form or by any means, electronic or mechanical,
including photocopying, recording or by any information storage and
retrieval system, without written permission from the copyright owner
CONTENTS
SOLUTIONSFORCHAPTER1+ ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••.1
SOLUTIONSFOR CHAPTER2+ 15
SOLUTIONS FOR CHAPTER3+ 43
SOLUTIONS FOR CHAPTER4+ 57
SOLUTIONS FORCHAPTER5+ 65
SOLUTIONSFORCHAPTER1+
SOLUTIONS FOR CHAPTER 1+
1. No. For a counterexample, let a and ~ belogically independent sentences,
and letA = {av~}.
2. We need to show that forall ~, ~ E Cn((av~a}) holds if and only if ~ E
Cn(0).
Bydeduction, ~ E Cn({av~a}) holdsifandonlyif av~a~ ~ E Cn(0),
(since av~a~ ~ isequivalent with ~) ifandonlyif~E Cn(0).
3.Since a~~ isequivalent with~av~, wehaveCn((a~~}) = Cn({~av~}).
It therefore follows from Observation 1.17 that Cn({~a v ~}) = Cn({,a}) n
Cn({~}).
4. Cn({a~~}) n Cn({~~a}) =
=Cn({,av~}) n Cn({-,~va})
=
Cn({-,a}) n Cn({~}) n Cn({-,~}) n Cn({a})
= Cn({a}) n Co({-,a}) n Cn({~})n Cn({~~})
= Cn({av~a}) n Cn({~v~~}) (Observation 1.17)
=Cn(0)n Cn(0) (Exercise2)
=Cn(0)
S. Suppose that A ~ D c B c Cn(A). It follows by monotony from D c B
thatCn(D) ~Cn(B).
It also follows bymonotony fromB c Cn(A) thatCn(B) ~ Cn(Cn(A». By
=
iteration, Cn(Cn(A» Cn(A), sothatCn(B) ~ Cn(A). Itfollows by monotony
from A ~D that Cn(A) c Cn(D). From Cn(B) ~ Cn(A) and Cn(A) c Cn(D)
weobtainCn(B) c Cn(D).
From Cn(D) ~ Cn(B) andCn(B) ~ Cn(D) wemayconclude that Cn(D) =
Cn(B).
6. Suppose that A ~ Cn(B). It followsfrom inclusion thatB ~ Cn(B), so that
AuB C Cn(B). By monotony, Cn(AuB) c Cn(Cn(B». By iteration,
Cn(Cn(B» =Cn(B), sothatCn(AuB) c Cn(B).
7. For one direction, let Cn(B) = Cn(D). We then have B c Cn(B) = Cn(D)
andD c Cn(D)=Cn(B).
For the other direction, let B c Cn(D) and D ~ Cn(B). It follows by
monotony that Cn(B) c Cn(Cn(D» and byiteration that Cn(Cn(D» =Cn(D),
so that Cn(B) ~ Cn(D) can be concluded. In the same way it follows by
2 ATEXTBOOKOFBEUEFDYNAMICS
monotony that Cn(D) c Cn(Cn(B» and by iteration that Cn(Cn(B» = Cn(B),
so thatCn(D) c Cn(B) canbeconcluded.
8. We need to show that (I) ifCn isaconsequence operator, then the formula
given in theexercise issatisfied, and (II) iftheformula is satisfied,then Cn is
a consequenceoperator.
Part I: It follows from inclusion and iteration that X c Cn(Cn(X».
Furthermore, it follows from iteration that Cn(Cn(X» ~ Cn(X) and from
monotony that Cn(X)c Cn(XuY).
Part1/: We now assume thatthe formula given in theexercise issatisfied.
In order to show that Cn is a consequence operator, we must prove that it
satisfies inclusion, monotony,anditeration.
Inclusion follows directly fromXc Cn(Cn(X» c Cn(X).
For monotony, suppose that X ~ Y. Then Y =XuY, so that Cn(X) c
Cn(Y) follows direclty fromCn(X)~ Cn(XuY).
One direction of iteration, namely Cn(Cn(X» c Cn(X), is directly given.
The other direction follows frominclusion,thathasalready beenobtained.
=
9. One direction of the desired equivalence is trivial, namely that if Cn Cn',
= =
then Cn(0) Cn'(0). For the other direction, suppose that Cn(0) Cn'(0).
=
In order to show that Cn Cn' we need to prove that for all B and a,a E
Cn(B) ifand onlyifa E Cn'(B). Due tosymmetry,it is sufficient toshow that
for allB and a,ifa E Cn(B), thena E Cn'(B).
It follows by compactness from a E Cn(B) that there isa finite subset Bn
of B such that a E Cn(Bn).By repeated use of Observation 1.18, Cn(Bn)=
Cn((&Bn}). We therefore have a E Cn({&Bn}). By deduction, &Bn~a E
=
Cn(0). Since Cn'(0) Cn(0), wehave &Bn~aE Cn'(0).
Now we can perform thesame procedure backwards, but on Cn' instead of
Cn: It follows by deduction that a E Cn'((&Bn}). By repeated use of
=
Observation 1.18, we obtain Cn'(B") Cn'({&Bn}), so that a E Cn'(B"). By
B" c B and monotony, Cn'(B") c Cn'(B), so thata E Cn'(B),as desired.
10.Suppose that Cno(X)~ Cno(Y).
It follows from inclusion for CnothatXc Cno(X). By monotony for Cn,
Cn(X) c Cn(Cno(X».
It follows bymonotony forCn thatCn(Cno(X» ~ Cn(Cno(Y».
It follows by the supraclassicality of Cn that Cnoey) ~ Cn(Y). By
monotony for Cn, Cn(Cno(Y» ~ Cn(Cn(Y» and by iteration Cn(Cn(Y» =
Cn(Y). so that Cn(Cno(Y» ~ Cn(Y).
We now have Cn(X) c Cn(Cno(X», Cn(Cno(X» c Cn(Cno(Y» .
Cn(Cno(Y» .~ Cn(Cn(Y» , and Cn(Cn(Y» c Cn(Y). They combine to Cn(X)
~ Cn(Y), as desired.
SOLUTIONSFORCHAPTER1+ 3
11. In order to prove that CnT is a consequence operator, ·we need to show
thatitsatisfies inclusion, monotony,and iteration.
Inclusion:By the inclusion and monotony properties of Cnn,A c Cno(A)
=
and Cno(A)c Cno(TvA) CnT(A).
Monotony: Suppose thatA c B.Then TvA!;; TvB and, by the monotony
of Cnn,Cno(TvA) c Cno(TvB), i.e., CnT(A)c CnT(B).
Iteration: One direction of iteration follows directly from inclusion. For
the other direction of this property, suppose that XE CnT(CnT(A».Then XE
Cno(TvCno(TvA». By inclusion and monotony for Cnn,Tc Cno(TvA), so
that TvCno(TvA) = Cno(TvA). We thus have XE Cno(Cno(TvA» and, by
the iteration property for Cnn,XE Cno(TvA), i.e.XE CnT(A).
Supraclassicality: Let XE Cno(A). Then by monotony for Cnn,XE
Cno(TvA), i.e.,XE CnrtA).
Deduction:'I'E CnT(Av{x}) holds if and only if 'I'E Cno(TvAv{X}),
thus (by deduction for Cnn)ifand only if <X~'l') E Cno(TvA), ifand only if
(X~'l') E CnrtA).
Compactness: Suppose that XE CnT(A), i.e., XE CnO(TvA). By the
compactness of Cno, there are finite subsets SofT andA' ofA such that XE
Cno(SvA'). By monotony for Cnn,XE Cno(TvA'), i.e.XE CnT(A').
12.Weneed toshow that Cn'satisfies inclusion, monotony,and iteration.
Inclusion: Let a EA.Thena E Cn'(A) follows from a E Cn({a}).
Monotony: Suppose thatAc B and a E Cn'(A).Then there is some a E A
such that a E Cn({a}). Itfollows fromAc B thata E B, from which we may
conclude that a E Cn'(B).
Iteration: One direction follows directly from inclusion. For the other
re»
direction, let a E Cn'(Cn'(A» . Then there is some ~ such that a E Cn(
and ~ E Cn'(A). It follows from ~ E Cn'(A) that there is some a such that ~
E Cn({a}) and a EA.
From W} c Cn({a}) follows by monotony Cn({~}) c Cn(Cn({a})) and
by iteration Cn({~}) ~ Cn({a}). From this and a E Cn({~}) follows a E
Cn({a}). Since a E A. thisissufficient toprove that a E Cn'(A).
13. a I- ~ if and only if ~ E Cn({a}), by deduction if and only if a~~ E
Cn(0), ifand only ifI- a~~.
14. a I- ~ and ~ I- a
~ E Cn({a}) and aE Cn({~})
{~} c Cn({a})and a E Cn({~})
Cn({~}) C Cn(Cn({a})) and aE Cn({~}) (monotony)
Cn({~}) ~Cn({a}) and aE Cn({~}) (monotony)
4 ATEXTBOOKOFBEUEFDYNAMICS
s
E Cn({a})
al-~
15. Suppose that a I- ~ and ~ f- a, i.e., ~ E Cn((a}) and a E Cn((~}). It
follows from ~ E Cn((a}) that {~} <;;;; Cn((a}), so thatbymonotonyCn({~})
c Cn(Cn({a})). By iteration, Cn(Cn({a})) =Cn({a}), so that Cn({~}) ~
Co({a}).
In the same way we can also prove that Cn({a}) ~ Cn({~}). We may
=
therefore conclude that Cn((ol) Cn((~}).Thus, for all ~, ~ E Cn((a}) if
andonly if~E Cn((~}), thatisa I-~ ifandonlyif~f- ~.
16. Suppose that Xf- a and Xu {a} f- ~, i.e., that a E Cn(X) and ~ E
Cn(X u (a}). It follows by deduction from ~ E Cn(Xu (a}) that a~~ E
Cn(X). Since a and a~~ truth-functionally imply ~, it follows by
supraclassicality that ~E Cn(X),thatisXI- ~.
17. a. Suppose to the contrary that A f-a. ThenXu{~~a}f-a, i.e., by the
deduction property Xf-(~~a)~a, or equivalently Xf-av~, contrary to the
conditions.
b. Directly from {av~, ~~a} I-a
18. a. {pvq,T} b. {p~q,T} c. {p,pvq,q~p, T}d. {T}
19. One direction is trivial: if A is logically closed, then there is some B,
= =
namely B A suchthatA Cn(B).
=
For the other direction, let B besuch that A Cn(B). By iteration,
Cn(Cn(B)) = Cn(B). BysubstitutingA forCn(B) weobtainCn(A)=A, so that
A islogicallyclosed.
20. Part I:For onedirectionof thedesiredequivalence,weare going toshow
that if AuB is logically closed, then either A~B orB~A. We are going to
prove this implication initsconverseform.Thus,weare goingtoshowthat if
it is not thecase thatA C B orB ~A, thenAuB is not logically closed. For
this purpose, suppose that that A ~Band B ~ A. Our task is to prove that
AuB :F-Cn(AuB).
It follows from A ~ B that there is some a such that a E A and a ~ B.
Similarly, it follows from thatB~ A thatthereissome ~ such that~ E Band
~ e: A. Weare goingtoshow(1) that af-7~ E Cn(AuB),and(2)that af-7~ ~
AuB.
For (I): Since a E A,a E AuB. Similarly. since ~ E B. ~ E AuB. By
truth-functional logic, af-7~ E Cn({a,~}). Since {a,~} ~AuB, monotony for
SOLUTIONSFORCHAPTER1+ 5
Cn yields Cn«(a,~}) ~ Cn(AvB). We therefore have aH~ E Cn(AvB), as
desired.
For (2): In order to show that aH~ ~ AvB it is sufficient to show that
aH~ ~ A and aH~ ~ B. In order to show that aH~ ~ A, suppose to the
contrary that aH~ E A.Since a E A,and ~ follows by truth-functional logic
from a and aH~, it would follow that ~ E Cn(A), and - since A is closed
under logical consequence - that ~ E A, contrary to the conditions. We may
conclude from thiscontradiction that aH~ ~ A.
The proof that aH~ ~ B issimilar.Wenowknow that aH~ ~ AvB.
In summary, we have found a sentence (aH~) that is an element of
Cn(AvB) butnotanelementofAvB. Thisissufficienttoprove thatCn(AvB)
;=AvB, i.e.,thatAvB isnotlogicallyclosed.
Part II: For the other direction of the desired equivalence, suppose that
eitherA~B orBd. IfA~B, thenAvB =B, and the logical closure of AvB
follows from that of B. Similarly if B~A, then AvB =A, and the logical
closure ofAvB followsfromthatofA.
21.Only c,d and eare {p,pvq}-closed.Inaand b,pvq is missing.
22. For one direction, let B1c B2. Then Cn(B1)c Cn(B2) follows directly
from Bl ~B2. Suppose thatCn(BI)= Cn(B2). SinceBl andB2 areA-closed,
we then have BI =AnCn(BI) =AnCn(B2) =B2, contrary to BI C B2. It
follows from this contradiction that Cn(BI) :f;.Cn(B2), and since we have
already proved thatCn(BI) c Cn(B2)wecanconcludethatCn(BI) C Cn(B2).
The other direction followsfromObservation 1.30.
23. Let c(BI) = c(B2). We then have Cn(BI) = Cn(c(BI» = Cn(c(B2» =
Cn(B2). Since BI andB2 areA-closed subsets of A, we can therefore derive:
Bl =AnCn(Bl) =AnCn(B2)·=B2.
24. LetAl and A2 beB-closed subsets of B. Weare going to show thatAl is
A2-closed, i.e., that Cn(AI)nA2 ~AI.
SinceA2~B, wehaveCn(AI)nA2 c Cn(AI)nB. SinceAl isB-closed, we
also have Cn(AI)nB ~A1.It follows that Cn(Al)nA2 c AI, i.e., that Al is
A2-closed. In the same waywecanprove thatA2isAj-closed.It follows that
A1andA2 are mutually closed.
25. Suppose thatA and Bare closed under implication. Weneed toshow that
AnB isalsoclosed underimplication.
Let a and ~ beelements of AnB. Then a E A and ~ E A, and by the
closure under implication of A, a-t~ E A. In the same way it follows that