Moorean Phenomena in Epistemic Logic
Wesley H. Holliday and Thomas F. Icard, III
Department of Philosophy
Stanford University
Stanford, California, USA
Abstract
A well-known open problem in epistemic logic is to give a syntactic characterization of the successful
formulas. Semantically, a formula is successful if and only if for any pointed model where it is true,
it remains true after deleting all points where the formula was false. The classic example of a formula
that is not successful in this sense is the “Moore sentence” p ∧ ¬p, read as “p is true but you do not
know p.” Not only is the Moore sentence unsuccessful, it is self-refuting, for it never remains true as
described. We show that in logics of knowledge and belief for a single agent (extended by S5), Moorean
phenomena are the source of all self-refutation; moreover, in logics for an introspective agent (extending
KD45), Moorean phenomena are the source of all unsuccessfulness as well. This is a distinctive feature
of such logics, for with a non-introspective agent or multiple agents, non-Moorean unsuccessful formulas
appear. We also consider how successful and self-refuting formulas relate to the Cartesian and learnable
formulas, which have been discussed in connection with Fitch’s “paradox of knowability.” We show that
the Cartesian formulas are exactly the formulas that are not eventually self-refuting and that not all
learnable formulas are successful. In an appendix, we give syntactic characterizations of the successful
and the self-refuting formulas.
Keywords: epistemic logic, successful formulas, Moore sentence
1
Introduction
According to the epistemic interpretation of modal logic, the points in a modal model
represent ways the world might be, consistent with an agent’s information. In this
context, “learning” a formula amounts to eliminating those points in the model where
the formula is false. The resulting submodel represents the agent’s information state
after learning has occurred. Some formulas—though not all—remain true whenever
they are learned. A well-known open problem [12,4,5,6,7,2] in epistemic logic is to
give a syntactic characterization of these successful formulas. Partial results have been
obtained (Section 2), but a full solution has proven elusive.
The classic example of an unsuccessful formula is the Moore sentence p ∧ ¬p, read
as “p is true but you do not know p.” This example is a second-person variation of
Wesley H. Holliday and Thomas F. Icard, III
179
G.E. Moore’s famous puzzle [16] involving the paradoxical first-person assertion, “p is
true but I do not believe p.” Hintikka devoted a chapter of his seminal 1962 monograph
Knowledge and Belief [13] to an analysis of such sentences, including the second-person
Moore sentence. Hintikka observed its unsuccessfulness as follows: “You may come to
know that what I said was true, but saying it in so many words has the effect of making
what is being said false” [13, p. 69]. Yet the formal question of unsuccessfulness did
not arise for Hintikka. Only with the advent of Dynamic Epistemic Logic (see, e.g.,
[7]) and the idea of learning as model reduction have the Moorean phenomena and
unsuccessfulness been formally related.
The reason the Moore sentence is unsuccessful is that it can only be true in a model
if p is true at some point and false at some other point accessible from the first. When
the agent learns the sentence, all points where p is false are eliminated from the model,
including all witnesses for ¬p, so the sentence becomes false. This shows that the
Moore sentence is not only unsuccessful, it is self-refuting, for it always becomes false
when learned. Related to the self-refuting property of the Moore sentence is the fact
that the sentence cannot be known. Indeed, the Moore sentence is at the root of Fitch’s
famous “paradox of knowability” [11,3,9]: if there is an unknown truth, then there is an
unknowable truth. For if p is true but unknown, then the Moore sentence p ∧ ¬p is
true, but the Moore sentence cannot be known, because (p ∧ ¬p) is inconsistent with
standard assumptions about knowledge.
While the Moore sentence is conspicuously unsuccessful, other unsuccessful formulas
are less conspicuous. The formula ¬ (p ∨ q) ∨ (p ∧ (p ∨ ♦q)) is also unsuccessful, as
we show in Example 5.12 below, but is the reason Moorean? While on the surface this
formula looks unlike a Moore sentence, it is in fact possible to transform the formula to
reveal its Moorean character. Indeed, we will prove that for a wide range of logics, such a
transformation is possible for every unsuccessful formula. However well-disguised, their
nature is always Moorean.
In Section 2 we establish notation, give the definitions of successful, self-refuting, etc.,
and review what is already known in the literature on successful formulas. In Section 3 we
show that in logics of knowledge and belief for a single agent (extended by S5), Moorean
phenomena are the source of all self-refutation; moreover, in logics for an introspective
agent (extending KD45), Moorean phenomena are the source of all unsuccessfulness as
well. This is a distinctive feature of such logics, for as we show in Section 4, in logics for a
non-introspective agent or multiple agents, non-Moorean unsuccessful formulas appear.
Finally, in Section 5 we relate successful and self-refuting formulas to the Cartesian
and learnable formulas, which have been discussed in connection with Fitch’s paradox,
and to the informative, eventually self-refuting, super-successful formulas, which we
introduce here. In Appendix A we give syntactic characterizations of the successful and
the self-refuting formulas.
2
Preliminaries and Previous Results
Throughout we work with a fixed, unimodal language with and its dual ♦, and an
infinite set Prop of propositional variables. The expression ♦+ ϕ abbreviates ♦ϕ ∧ ϕ. We
180
Moorean Phenomena in Epistemic Logic
use the standard semantics of modal logic, where a model is a triple hW, R, V i with W
any set of points, R ⊆ W × W any relation, and V : Prop → ℘(W ) a valuation function.
A pointed model is a pair M, w with M a model and w ∈ W . The satisfaction relation
between pointed models and formulas is defined as usual.
It is typical to take KD45 as a logic of belief and S5 as a logic of knowledge. Apart
from the assumptions that one does not believe or know anything inconsistent, and that
what is known is true, these logics assume positive (axiom 4) and negative (axiom 5)
introspection: if one believes something, then one believes that one believes it; if one
does not believe something, then one believes that one does not believe it; and mutatis
mutandis for knowledge. Most work on successful formulas has assumed S5. Since our
results apply to a wider range of logics, we assume through Section 3 that we are working
with at least KD45, so all of our models will be serial, transitive, and Euclidean. We
call such models quasi-partitions.
Where M0 is a submodel (not necessarily proper) of M, we write M0 ⊆ M. Rather
than studying formulas preserved under arbitrary submodels, we will study formulas
preserved under a special way of taking submodels, as in the following.
Definition 2.1 Given a model M = hW, R, V i, the relativization of M to ϕ is the
(possibly empty) submodel M|ϕ = hW|ϕ , R|ϕ , V|ϕ i of M, where W|ϕ = {w ∈ W :
M, w ϕ}, R|ϕ is R restricted to W|ϕ , and V|ϕ (p) = V (p) ∩ W|ϕ .
Definition 2.2 A formula ϕ is successful (in logic L) iff for every pointed model (of L),
M, w ♦+ ϕ implies M|ϕ , w ϕ. A formula is unsuccessful (in L) iff it is not successful.
A formula is self-refuting (in L) iff for every pointed model (of L), M, w ♦+ ϕ implies
M|ϕ , w 2 ϕ. 1
In the standard definitions of successful and self-refuting formulas, where L is assumed to be S5, the precondition only requires that ϕ be true at w. Since we are also
working with KD45, we additionally require that ϕ be true at an accessible point, so
that M|ϕ is a quasi-partition provided M is. Our definition reduces to the standard one
in the case of S5. In either case, unsatisfiable formulas are self-refuting and successful.
In the case of KD45, a satisfiable formula such as p ∧ ¬p, read as “p is true but you
believe ¬p,” is self-refuting and successful, since ♦+ (p ∧ ¬p) is unsatisfiable. While
it may be more intuitive to require the satisfiability of ♦+ ϕ for a successful ϕ, we will
follow the standard definition in not requiring satisfiability.
The following lemma relates success and self-refutation across different logics.
Lemma 2.3 Let L be a sublogic of L0 . If ϕ is unsuccessful in L0 , then ϕ is unsuccessful
in L, and if ϕ is self-refuting in L, then ϕ is self-refuting in L0 .
Proof. Immediate from Definition 2.2, given that models of L0 are models of L.
2
An obstacle to giving a simple syntactic characterization of the set of successful formulas is its lack of closure properties. Successful formulas are not closed under negation
1
The term ‘successful’ is used by Gerbrandy [12], by analogy with the success postulate of belief
revision, while the term ‘self-refuting’ is used by van Benthem [3]. Self-refuting formulas have also been
called strongly unsuccessful [2].
N and every set N
Γ and
∈ A,every
thereset
is aΓliteral
∈ A, there
l! ∈ Γisthat
a literal
is notl! the
∈ Γquazi-negation
that is not theoq
l.
l.
Wesley H. Holliday and Thomas F. Icard, III
181
Definition
6. Given
Definition
specification
6. N
Given
"C,
a specification
N,l!set
A#
formula
"C,
N, A#
and
DNF,
formula
N and ϕevery
set
Γ ∈a A,
there
isand
a literal
∈ ΓΓand
that
isthere
not
the
quazi-negation
every
∈ A,
isϕ ain
literal
l! define
∈ Γ ϕt
ϕ
!
! !
!
!
"C,
A#every
"C,
"C,
NΓ N,
,∈A#
A#where
= "C,
N
, the
A# where
set
of⊥)
Nthat
literals
is the
set
l ∈the
ofNall
forliterals
which lthere
∈N
Nl.N,
and
set
A,
there
is N
aisliteral
l →∈
Γall
is
quazi-negation
l.
(take ¬p ∨ p), conjunction
(take
p=
and
¬p),
or
implication
(use
ϕ
[6].
Wenot
is
a
disjunct
d
in
is
ϕ
a
such
disjunct
that
d
d
in
does
ϕ
such
not
contain
that
d
does
∼
l
and
not
there
contain
is
a
∼
selection
l
and
there
Σ
ois
l. they are also not closed under disjunction. Conversely,
show in Proposition 5.11 that
literals,
one
from
literals,
each
Γ
one
in
A,
from
such
each
that
Γ
Σ
in
does
A,
such
not
that
contain
Σ
does
the
quazi-negation
not
contain
th
Definition
6.
Given
a
specification
"C,
N,
A#
and
formula
ϕ
in
DNF,
defi
Definition
6.
Given
a
specification
"C,
N,
A#
if a negated formula is successful, the
unnegated formula may be unsuccessful
(take
ϕ
ϕ
! a specification
!
! literals
! in
! defin
6.not
Given
"C,
N,
A#
formula
DNF,
ofDefinition
l and
dA#does
of
lN
contain
and
dthedoes
the
quazi-negation
not
contain
the
of
quazi-negation
any
Σ.
any
literal
"C,
=
"C,
,ofA#
where
N
isA#
the
set
of
all
for
which
that
"C,
N,
=
"C,
Nand
, A#
wherelin
Nϕof
is
the
N
and
every
set
Γ
∈
A,literal
there
is∈
aNliteral
lset
∈ of
Γin
¬ (p ∧ ♦¬p)), if a conjunction
isN,
successful,
some
conjuncts
may
be
unsuccessful
ϕ
!
!
A# = "C,
Nϕ ,such
A# where
is the
of ϕ
all
l d∈does
Nisfor
which
the
disjunct
disin
that
not
contain
∼literals
l and
anot
selection
Σ
is
adoes
disjunct
in
such
thatthere
contain
(take (p ∧ ♦¬p) ∧ ¬p), and "C,
ifisa aN,
disjunction
successful,
some
or
even
all set
ofd the
disjuncts
l.dN
is
a
disjunct
d
in
ϕ
such
that
d
does
not
contain
∼
l
and
there
is
a
selection
Σ
literals,
one
from
each
Γ
in
A,
such
that
Σ
does
not
contain
the
quazi-negat
literals,
one
from
each
Γ
in
A,
such
that
Σ
does
may be unsuccessful (Proposition 5.9 and Example 5.2).
one
from
Γconjunction
in A,
such
that
Σ6.
does
not
the
quazi-negatio
Definition
Given
acontain
specification
"C, N,
Ifliterals,
ϕ
a conjunction
If not
ϕeach
is
incontain
anormal
form
and
in
L
(ϕ)
$=
form
∅,
then
and
ϕ
Lthe
(ϕ)
is successful.
$= Σ.
∅,
then
ϕ A#
is os
of islpreserved
and
d does
the
of
any
literal
in
of
lquazi-negation
and
dnormal
does
not
contain
quazi-negation
By contrast, the formulas
under
arbitrary
submodels
are
well-behaved.
ϕ
!
!
of
l
and
d
does
not
contain
the
quazi-negation
of
any
literal
in
Σ.
"C,
N,
A#
=
"C,
N
,
A#
where
N
is
the
set
of a
The following result was proved independently by van Benthem and Visser [20,8]. A
If ϕ contains
Proof.
only
Ifonly
ϕ
formulas
contains
of
the
form
formulas
of
and
¬Kα
form
Kα
propositional)
and"
¬Kα
is a disjunct
d .
inKα
ϕ such
d (α
does
not
contain
formula is universal iff it Proof.
can be constructed
using
literals,
∧,only
∨, and
!
! the
"that
!M
!M(
M
M
M
M|ϕ
|∈
!
then
forisany
(M,
then
w)
M,
(M,
w If
!is w)
ϕ,
with
va∈
W
M,
| !wR
ϕ,
∈not
=
WA,
v such
∈
| quazi-negation
wR
W
v |Σ
wR
=(ϕ)
v $=
NIf and
every
set
Γ with
∈for
A,
there
literal
l from
∈w(ϕ)
Γ
that
the
literals,
one
each
Γthen
in
that
does
ϕ
a conjunction
inany
normal
and
L
$= in
∅,vis
ϕ
is
successful.
ϕaform
is
conjunction
normal
form
and
L
Theorem 2.4 A formulaHence
is
preserved
under
submodels
(of
all
relational
models)
iff
it
, w !Hence
ϕ.
Mnormal
ϕ.
l. ϕ isM
|ϕ , w !
If
a |ϕ
conjunction
in
L (ϕ)
$= contain
∅, then ϕ
successful.
ofform
l andand
d does
not
theis quazi-negation
o
N and every
set Γ Kα
∈ A,and
there
is a(α
literal
l! ∈ Γ t
is equivalent (in K) to a universal
Proof. formula.
If ϕ contains only formulas
the
form
¬Kα
propositiona
Proof. of
If
ϕ
contains
only
formulas
of
the
form
K
!
" !
! M M M
l.
M Kα
M
Definition
6.(M,
Given
a#
"C,
formula
If#any
ϕ
formulas
of
form
and
¬Kα
(α
propositional
p Proof.
∧then
♦q ∧for
(q contains
→
pp)
∧∧
♦q
♦only
∧
(q
∧specification
(q
p) →then
♦ the
(q
p)A#
w)
with
M,
wp)
!∧
ϕ,
v ∈∧N,
W
|and
wR
vM,
=w! !vϕ
∈in
WvDNF,
wR|
for
any
(M,
w)
with
ϕ,
∈|ϕW| defin
!
"
ϕ model extensions
Similarly, a formula is preserved under
iff it !is equivalent
M|ϕ
M
!
Mto an exisM
"C,
N,
A#
=
"C,
N
,
A#
where
N
is
the
set
of
all
literals
l
∈
N
for
which
the
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
wR
v
=
v
∈
W
|
wR
Hence M|ϕ , w ! ϕ.
Hence
w !Given
ϕ. ina normal
If[8].
ϕ is M
a conjunction
form and
L
(ϕ)
$=
|ϕ ,!6.
Definition
specification
"C,
N,
A#
tential formula, constructed
using
only
literals,
∧,
∨,
and
♦
every, w
set
A, there
a literal
l ∈ Γ that
is not
the is
quazi-negation
Hence
M
!
ϕ.
is aand
disjunct
ϕ∈
such
that disdoes
notϕ contain
∼! l and
there
pN
pΓ
|ϕ d in
! a selection Σ
"C,give
N, A# following.
= "C, N , A# where N is the set of a
Lemma 2.3 and the right-to-left
direction of Theorem 2.4
l.
literals,
one
from
each
such the
that
Σ
does
not
contain
the of
quazi-negatio
Proof.
If #
ϕ (q
contains
only
formulas
the form K
p ∧ ♦q ∧
# (q
→ p)
∧ ♦Γ(qin∧A,
p)p
∧
♦ (q
∧
p)d does
!not
is ∧a ♦q
disjunct
d→
in p)
ϕ∧
such
that
contain
M
of
l
and
d
does
not
contain
the
quazi-negation
of
any
literal
in
Σ.
p
∧
♦q
∧
#
(q
→
p)
∧
♦
(q
∧
p)
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈ lW
!
!
Corollary 2.5 Universal formulas
are
successful
in
any
normal
modal
logic.
N
and
every
set
Γ
∈
A,
there
is
a
literal
∈does
Γ|t
N
and
every
set
Γ
∈
A,
there
is
a
literal
l
∈
Γ
that
is
not
the
quazi-negation
literals, one
from
in A, such
Σ
Definition 6. Given a specification
"C, N,
A# each
and Γformula
ϕ inthat
DNF,
defin
Hence
M|ϕ , w ! ϕ.
q "C,
q N ! , A# wherep
p N, A#ϕ = "C,
l.
l.
of
d does
quazi-negation
Nl ! and
is the
set ofnot
allcontain
literals the
l∈N
for which theo
As noted by van Benthem
p [4], for any model M and formula ϕ, there is a universal
is
a
disjunct
d
in
ϕ
such
that
d
does
not
contain
∼
l
and
there
is
a
selection Σ
2
0.
formula ϕ0 such that M|ϕ =
M
However,
thisinresult
assumes
the
relation
for
is
IfDefinition
ϕ |ϕ
isevery
a conjunction
normal
form
L
(ϕ)
∅,M
then
ϕ
is
successful.
p a∧
♦q
∧and
#
→
p)a$=
∧
♦
(q
∧l! p)
!(q
Definition
6.
Given
a
specification
N,the
A#
6.
Given
a
specification
"C,
N,
A#
and
formula
ϕ
in
DNF,
defi
N
and
set
N
Γ
and
∈
A,
every
there
set
is
Γ
literal
∈
A,
there
l
∈
Γ
is
that
literal
is
not
the
∈
Γ
quazi-negation
that
is"C,
not
oq
literals,
one
from the
each
Γ in A,Figure
such
that
not
contain the
quazi-negatio
ϕ Σ does
ϕ given
at least a quasi-partition. For
example,
1,
where
the
relation
!
!
! model in "C,
!
N,
A#
=
"C,
N
,
A#
where
N
is
the
set
of
"C,
N,
A#
=
"C,
N
,
A#
where
N
is
the
set
of
all
literals
l
∈
N
for
which
th
q
q
l.not contain
If quazi-negation
ϕM
isof
a the
conjunction
in
normal
form
andIfdϕdoes
the
any
literal
in
Σ. and L (ϕ) a$=
is not Euclidean, there isl.of
nol universal
formula
ψ such
that
= form
M|ψ . of
This
is
contains
only
formulas
(α
|♦p
qProof.
pd adoes
!not
"¬Kα
is
disjunct
dM
in ϕKα
such
that
d! does
containΣ
is a disjunct
d in ϕ such
that
contain
∼M
land
and
there
is propositiona
anot
selection
symptomatic of the fact, established
in Section
4,with
thatM,
in logics
both |axioms
4 = v ∈ W M|ϕ | wRM
then
for any
w)each
!such
ϕ,without
vaone
∈specification
W
wR
vcontain
literals,
from
each
Γ
in
A,
such
that
Σdefine
does
Definition
6.(M,
Given
Definition
a specification
Given
"C,
N,
A#
and
formula
"C,
N,
A#
ϕ
in
and
DNF,
formula
ϕ
literals,
one
from
Γ in6.w
A,
that
Σ
does
not
the
quazi-negat
Proof. If ϕ contains only formulas of
K
! the form
and 5, there are non-Moorean
sources
unsuccessfulness.
ϕ |ϕof
ϕ
Hence
M
,
w
!
ϕ.
!
! !
!
M
of
l
and
d
does
not
contain
the
quazi-negation
o
"C,
N,
A#
=
"C,
"C,
N
N,
,
A#
A#
where
=
"C,
N
N
is
,
the
A#
where
set
of
all
N
literals
is
the
set
l
∈
of
N
all
for
literals
which
l
there
∈
N
of
l
and
d
does
not
contain
the
quazi-negation
of
any
literal
in
Σ.
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
If ϕ is a conjunction in normal form and L (ϕ) $= ∅, then ϕ is successful.
qdoes
is a disjunct d inisϕasuch
disjunct
that dd in
ϕ such
not
that
d
does
∼
l
and
not
contain
there
is
∼
a
selection
l
and
there
Σ
o
i
Hence
Mcontain
,
w
!
ϕ.
|ϕ
p ∧ ♦q ∧ # (q
→literals,
p) ∧ ♦Γone
(q ∧A,
p) such
!
literals,
each
from
each
that
Γ
Σ
in
does
A,
such
not
that
contain
Σ
does
the
quazi-negation
not
contain
th
Proof. one
If ϕ from
contains
onlyinformulas
of
the
form
Kα
and
¬Kα
(α
propositional
N and every
l ∈ Γ that
! set ΓM∈ A, there
" is a! literal M
M
Mliteral
|ϕ L
iscontain
avand
conjunction
in
and
(ϕ)in
$=
ofthen
does
not
of w)
l contain
and
does
thewpIfnot
quazi-negation
the
of
any
of
Σ.
any
literal
Ifl and
ϕ for
is daany
conjunction
ind normal
(ϕ)
$=
then
successful.
∧ϕform
♦q
∧
#
→
p)
∧ ∅,
♦
(q
∧=p)ϕinvisform
(M,
with
M,
!
ϕ,
∈ (q
WL
|quazi-negation
wR
vnormal
∈
W
| wR
l.
p
Hence M|ϕ , w ! ϕ.
If the
ϕ contains
of
the
form
K
Proof. If ϕ Fig.
contains
only Definition
formulas
form Kα
andformulas
6.
Given
a only
specification
N, A#
and
!propositiona
!
"¬Kα
! (α"C,
1.
pProof.ϕof
M M
M
M M, w! ! ϕ, vM
|ϕ
!
then
for
any
(M,
w)
with
∈
W
|
any
!A#
ϕ,and
vnormal
∈
| ∅,
wR
vϕ
=(ϕ)
v$=∈
| all
wR
"C,
N,
"C,
N $=
,form
A#
where
is
the
set ofϕ
Ifpthen
ϕ∧ is
ϕw)
conjunction
form
in=
LW
(ϕ)
then
and
LN
is successful.
∅,Wthen
is ls
♦qafor
∧conjunction
#
(q (M,
→If p)
∧isinwith
♦anormal
(q
∧M,
p) w
Hence
, wϕ!such
ϕ.different
Hence M|ϕsimilar
, w ! ϕ.
|ϕslightly
is a2.5,
disjunct
in
that d does not contain ∼ l
Gerbrandy [12] proved aqproposition
to Corollary
but M
in ad
contains
Proof.
only
If ϕ
formulas
contains
only
thefrom
formulas
KαΓof
and
¬Kα
form
propositional)
and
¬Kα
formal context and with ♦Proof.
operator.
He
showed
ifform
all each
instances
of
literals,
one
such
that
Σ "does
no(
!that
!inthe
"A,
!M(αKα
!M
q of
ponly asIfaϕdefined
M
M
M
M|ϕ
|
p
∧
♦q
∧
#
(q
→
p)
∧
♦
(q
∧
p)
then
for
any
(M,
then
w)
with
for
any
M,
(M,
w
!
w)
ϕ,
with
v
∈
M,
W
w
|
!
wR
ϕ,
v
v
∈
=
W
v
∈
|
wR
W
v
|
=
wR
v
∈
p
∧
♦q
∧
#
(q
→
p)
∧
♦
(q
∧
p)
are within the scope of an even number of negations,ofthen
thedformula
is successful.
l and
does not
contain the quazi-negation of an
M|ϕ2.5
, w does
!Hence
ϕ. not M
! ϕ.in the case of formulas
However, the converse ofHence
Corollary
hold,
|ϕ , weven
p
containing no nested modalqpoperators. For example, the formula
♦p ∨ p is successful in
K, but it is not equivalentpto∧any
Yet
other
connections
♦q universal
∧ # (q →formula.
pp)∧∧♦q♦ ∧
(q
#∧(q
p)
→
p)
∧
♦ (q ∧ with
p) universal
If ϕ is a conjunction
in normal form and L (ϕ) $= ∅
formulas do hold. Qian [18] showed that a formula containing no nested modal operators
is successful in K iff it can
into a universal
Proof.
If ϕ contains
only formulas of
Kα
q algorithm
p qbe transformedp using a certain
! the form
M
formula; moreover, a “homogeneous” formula, containingthen
no nested
modal
and w ! ϕ, v ∈ W
for any
(M,operators
w) with M,
| wR
no propositional variables outside the scope of modal operators,
is|ϕsuccessful
Hence M
, w ! ϕ. in K iff it
is equivalent to a universalq formula. The restriction
to K is essential, for the formula ♦p
q
2
p ∧ ♦q ∧ # (q → p) ∧ ♦ (q ∧ p)
The result in [4] is given for the case of multi-agent epistemic logic with a common knowledge
operator, assuming M is finite. The proof for the single-agent case is trivial and requires no assumption
of finiteness.
p
q
182
Moorean Phenomena in Epistemic Logic
is successful in extensions of KD45 (not in K), but it is not equivalent to any universal
formula.
As far as we know, there are no other published results on the syntax of successful
formulas in the basic modal language. However, successful formulas are often studied
in the more general context of Dynamic Epistemic Logic (DEL) [1,12,7], where model
changing operations are formalized in the logic itself by adding dynamic operators to the
language. In the simplest fragment of DEL extending S5, known as Public Announcement Logic (PAL), there are sentences of the form [ϕ]ψ, read as “after the announcement
of ϕ, ψ is true,” for which satisfaction is defined:
M, w [ϕ]ψ iff M, w ϕ implies M|ϕ , w ψ.
An advantage of the PAL setting is that it allows for simple definitions of success and
self-refutation [6]. Successful formulas are those for which [ϕ]ϕ is valid (in S5), and
self-refuting formulas are those for which [ϕ]¬ϕ is valid (in S5). While we will not deal
directly with this language, our work will apply indirectly to PAL given the following
result, due to Plaza [17], which relates PAL to the basic modal language.
Theorem 2.6 Every formula in the language of PAL is equivalent (over partitions) to
a formula in the basic modal language.
Direct results on successful formulas in PAL with multiple knowledge modalities and
the common knowledge operator have been obtained by van Ditmarsch and Kooi in [6],
which also presents an analysis of the role of (un)successful formulas in a variety of
scenarios involving information change.
Another benefit of the PAL definitions of successful and self-refuting is that they
give an upper bound on the complexity of checking a formula for these properties. The
following result including a lower bound for the success problem is due to Johan van
Benthem in correspondence.
Theorem 2.7 The success problem for S5 is coNP-complete.
Proof. For an upper bound, ϕ is successful iff [ϕ] ϕ is valid, and the validity problem
for single-agent PAL is coNP-complete [14]. For a lower bound, we use the following
reduction of the validity problem for S5, which is coNP-complete, to the success problem
for S5: ϕ is valid iff for a new variable p, ψ ≡ p ∧ (♦¬p ∨ ϕ) is successful. From left to
right, if M, w ψ then M|ψ , w p, and since ϕ is valid, M|ψ , w ϕ, so M|ψ , w ψ.
From right to left, take any pointed model M, w. Extend the language with a new
variable p and extend M to M0 with one new point v, related to all other points, with
the same valuation as w for all variables of the old language. Make p true everywhere
except at v. Then since M0 , w p ∧ ♦¬p, we have M0 , w ψ, and since ψ is successful,
we have M0|ψ , w ψ. But v is eliminated in M0|ψ , so M0|ψ , w 2 ♦¬p. Hence M0|ψ , w ϕ.
We conclude that M, w ϕ, for M0|ψ and M differ only with respect to the valuation
of p, which ϕ does not contain.
2
A similar argument shows that ϕ is valid iff for a new variable p, p ∧ (♦¬p ∨ ¬ϕ)
is self-refuting, so the self-refutation problem is coNP-complete given the upper bound
Wesley H. Holliday and Thomas F. Icard, III
183
from PAL. From left to right the reduction is obvious, while from right to left the same
extension of M to M0 works. Similar arguments also show that the success and selfrefutation problems are PSPACE-complete for multimodal S5.
3
The Moorean Source of Unsuccessfulness
To show that Moorean phenomena are the source of all unsuccessfulness in logics for an
introspective agent, we proceed via normal forms for such logics. The following normal
form derives from Carnap [10], and Proposition 3.2 is standard.
Definition 3.1 A formula is in normal form iff it is a disjunction of conjunctions of the
form δ ≡ α ∧ β1 ∧ ... ∧ βn ∧ ♦γ1 ∧ ... ∧ ♦γm where α and each γi are conjunctions of
literals and each βi is a disjunction of literals.
Proposition 3.2 For every formula ϕ, there is a formula ϕ0 in normal form such that
ϕ and ϕ0 are equivalent in K45.
Theorem 1.7.6.4 of [15] gives the analogue of Proposition 3.2 for S5. Inspection of the
proof shows that the necessary equivalences hold in K45.
We use the following notation and terminology in this section and Appendix A.
Definition 3.3 Given δ ≡ α ∧ β1 ∧ ... ∧ βn ∧ ♦γ1 ∧ ... ∧ ♦γm in normal form, we
define δ α ≡ α, δ α ≡ α ∧ β1 ∧ ... ∧ βn , and similarly for δ α♦ , δ , δ ♦ , and δ ♦ .
Definition 3.4 Where χ is a conjunction or disjunction of literals, let L (χ) be the set
of literals in χ. A set of literals is open iff no literal in the set is the negation of any of
the others.
Definition 3.5 A conjunction δ ≡ α ∧ β1 ∧ ... ∧ βn ∧ ♦γ1 ∧ ... ∧ ♦γm in normal form
is KD45-clear iff: (i) L(α) is open; (ii) there is an open set of literals {l1 , ..., ln } with
li ∈ L (βi ); and (iii) for every γk there is a set of literals {l1 , ..., ln } with li ∈ L (βi ) such
that {l1 , ..., ln } ∪ L (γk ) is open. A disjunction in normal form is KD45-clear iff at least
one of its disjuncts is KD45-clear.
In the following, by “clear” and “satisfiable” we mean KD45-clear and satisfiable in
a quasi-partition, respectively. In the case of S5-clear, we must require in clause (ii) of
Definition 3.5 that {l1 , ..., ln } ∪ L(α) is open, in which case (i) is unnecessary. In both
cases, the following lemma holds with the appropriate definition of clarity.
Lemma 3.6 A formula in normal form is satisfiable iff it is clear.
Proof. [Sketch] Suppose ϕ in normal form is satisfiable. Then there is some disjunct
δ of ϕ satisfied at a pointed model. Read off the appropriate open sets of literals from
the current point for clarity condition (i), from some accessible point for (ii), and from
witnesses for each ♦γk in δ for (iii). In the other direction, suppose there are open sets
of literals as described. Construct a model where δ α is true at the root point w, which
is possible by clarity condition (i). For each ♦γk , add a point v accessible from w and
extend the valuation such that all conjuncts of γk and some disjunct of each βi are true
184
Moorean Phenomena in Epistemic Logic
at v, which is possible by (iii). If there are no ♦γk formulas in δ, add an accessible point
v and extend the valuation such that some disjunct of each βi is true at v, which is
possible by (ii). Then ϕ is true at w.
2
The following definition fixes the basic class of Moore conjunctions, as well as a wider
class of Moorean conjunctions. The intuition is that a Moore conjunction simultaneously asserts a lack of information about another fact being asserted, and a Moorean
conjunction is one that behaves like a Moore conjunction in some context.
We write ∼ ϕ for the negation of ϕ in negation normal form, with ¬ applying only
to literals.
Definition 3.7 Let δ be a conjunction in normal form.
(i) δ is a Moore conjunction iff δ ∧ ♦δ α is not clear or there is a ♦γk conjunct in δ such
that δ ∧ ♦ (δ α ∧ γk ) is not clear.
(ii) δ is a Moorean conjunction iff there is a ♦γk conjunct in δ such that δ ∧ ♦δ α ∧
(∼ δ α ∨ ∼ γk ) is clear.
Assuming S5, δ ∧ ♦δ α may be replaced by δ in both (i) and (ii).
The definition of a Moore conjunction generalizes from the paradigmatic case of
p∧♦¬p to include formulas such as p∧♦q ∧ (q → ¬p). The formulas p∧¬p and p∧¬p
are also Moore conjunctions, since for these δ the formula δ ∧♦δ α is not clear, but for the
same reason they are not Moorean conjunctions. An example of a Moorean conjunction
that is not a Moore conjunction is p ∧ ♦q. We consider this formula Moorean because
in a context where the agent knows that q implies ¬p, so (q → ¬p) holds, learning
p ∧ ♦q has the same effect as learning p ∧ ♦¬p. By contrast, the formula p ∧ ♦q ∧ ♦(p ∧ q)
rules out the Moorean context with its last conjunct, and it is not Moorean according
to Definition 3.7(ii).
In the proofs of Lemma 3.9 and Theorem 3.13 below, we will assume without loss of
generality that all models considered are connected (i.e., ∀w, v ∈ W : wRv ∨ vRw), in
which case the following basic facts hold.
Lemma 3.8 Where M is a connected quasi-partition and δ is a conjunction in normal
form:
(i) M, w δ ⇒ M δ ♦ ;
(ii) M, w δ ⇒ M|δ = M|δα and M|δ δ α ;
(iii) M|δ , w δ ⇒ M|δ δ.
We now prove the main lemma used in the proof of Theorem 3.13.
Lemma 3.9 Let δ be a conjunction in normal form. The following hold for both KD45
and S5.
(i) δ is self-refuting if and only if it is a Moore conjunction.
(ii) δ is unsuccessful if and only if it is a Moorean conjunction.
Proof. (⇐ (i)) Suppose δ is a Moore conjunction. Case 1: δ ∧ ♦δ α is not clear. By
Wesley H. Holliday and Thomas F. Icard, III
185
Lemmas 3.6 and 3.8(i), δ ∧ ♦δ α is clear iff ♦+ δ is satisfiable, so in this case ♦+ δ is
unsatisfiable and hence δ is self-refuting. Case 2: δ∧♦δ α is clear, so suppose M, w ♦+ δ.
For the ♦γk in δ such that δ ∧ ♦ (δ α ∧ γk ) is not clear, ¬δ ∨ (¬δ α ∨ ¬γk ) is valid by
Lemma 3.6, so M, w (¬δ α ∨ ¬γk ) given M, w δ. Then since (¬δ α ∨ ¬γk ) is
universal, it is preserved under submodels by Theorem 2.4, so M|δ , w (¬δ α ∨ ¬γk ).
From Lemma 3.8(ii), M|δ , w δ α , so M|δ , w ¬γk . Since ♦γk is a conjunct in δ,
M|δ , w 2 δ. Since M was arbitrary, δ is self-refuting.
((i) ⇒) We prove the contrapositive. 3 Suppose δ is not a Moore conjunction. Then
δ ∧ ♦δ α is clear, and for every ♦γk in δ, δ ∧ ♦ (δ α ∧ γk ) is clear (∗). If there are no
♦γk conjuncts in δ, then δ is a universal formula with ♦+ δ satisfiable, so it is not
self-refuting by Theorem 2.4. Suppose there are ♦γk conjuncts in δ. We claim that
δ 0 ≡ δ ∧ βn+1 ∧ ... ∧ βn+j is clear where {βn+1 , ..., βn+j } = L (δ α ). Given assumption
(∗) and clarity condition (iii) for each δ ∧ ♦ (δ α ∧ γk ), we have that for all ♦γk in δ there
is a set {l1 , ..., ln } with li ∈ L (βi ) such that {l1 , ..., ln } ∪ L (δ α ∧ γk ) is open. Taking
{ln+1 , ..., ln+j } = L (δ α ), {l1 , ..., ln+j } ∪ L (γk ) = {l1 , ..., ln } ∪ L (δ α ∧ γk ) is open, which
gives clarity conditions (i), (ii) and (iii) for δ 0 . Since δ 0 is clear, suppose N , w δ 0 . From
the fact that δ 0 ↔ (δ ∧ δ α ) we have N , w δ ∧ δ α . Given N , w δ α and the
assumption that N is connected, N δ α ; given N , w δ and Lemma 3.8(i), N δ ♦ .
Hence N δ, in which case N , w ♦+ δ and N|δ = N . It follows that N|δ , w δ, so δ
is not self-refuting.
((ii) ⇐) Suppose δ is a Moorean conjunction. Since for some ♦γk in δ, χ ≡ δ ∧
♦δ α ∧ (∼ δ α ∨ ∼ γk ) is clear, there is a model with M, w χ. Given M, w δ ∧ ♦δ α ,
we have M, w ♦+ δ by Lemma 3.8(i). Given M, w (∼ δ α ∨ ∼ γk ), by the same
reasoning as in Case 2 of ((i) ⇐), M|δ , w 2 δ. Therefore δ is unsuccessful.
((ii) ⇒) We prove the contrapositive. Suppose M, w ♦+ δ and δ is not a Moorean
conjunction. Then for all ♦γk in δ, χk ≡ δ ∧ ♦δ α ∧ (∼ δ α ∨ ∼ γk ) is not clear. To show
M|δ , w δ, it suffices to show M|δ , w δ ♦ , since δ α is universal and therefore preserved
under submodels. Consider some ♦γk conjunct in δ. It follows from our assumption that
χk is unsatisfiable, whence (δ ∧ ♦δ α ) → ♦ (δ α ∧ γk ) is valid. Then from M, w ♦+ δ
we obtain M, w ♦ (δ α ∧ γk ), so there is a v with wRv and M, v (δ α ∧ γk ). By
Lemma 3.8(ii), v is retained in M|δ . Since γk is propositional, M|δ , v γk and hence
M|δ , w ♦γk . Since ♦γk was arbitrary, M|δ , w δ ♦ and hence M|δ , w δ. Since M
was arbitrary, δ is successful.
2
Lemma 3.9 gives necessary and sufficient conditions for the successfulness of a conjunction in normal form. We now introduce an apparently stronger notion.
Definition 3.10 A formula ϕ is super-successful (in L) iff for every pointed model (of
L), M, w ♦+ ϕ implies M0 , w ϕ for every M0 such that M|ϕ ⊆ M0 ⊆ M.
If ϕ is super-successful and M, w ϕ, then as points that are not in M|ϕ are eliminated
from M, ϕ remains true at w. Since we take the elimination of points as an agent’s
acquisition of new information, this means that ϕ remains true as the agent approaches,
3
The following argument establishes something stronger than we need for Lemma 3.9, but we use it to
establish Corollary 5.3 below. Compare the (⇒) direction of the proof of Theorem A.3 in Appendix A.
186
Moorean Phenomena in Epistemic Logic
by way of the incremental acquisition of new information, the epistemic state of M|ϕ
wherein the agent knows ϕ. Intuitively, we can say that a super-successful formula
remains true while an agent is “on the way” to learning it.
We will use the next lemma in the proof of Theorem 3.13.
Lemma 3.11 If δ is a successful conjunction in normal form, δ is super-successful.
Proof. Suppose δ is not super-successful, so there is a pointed model such that M, w ♦+ δ and an M0 such that M|ϕ ⊆ M0 ⊆ M and M0 , w 2 δ. Since M0 ⊆ M and
δ α is preserved under submodels, we must have M0 , w 2 δ ♦ . But then M|δ , w 2 δ ♦
given that M|δ ⊆ M0 and δ ♦ is preserved under extensions. Hence M|δ , w 2 δ, so δ is
unsuccessful.
2
We now lift the definition of Moore and Moorean to arbitrary formulas.
Definition 3.12 Let ϕ be an arbitrary formula.
(i) ϕ is a Moore sentence iff any normal form of ϕ is a disjunction of Moore conjunctions.
(ii) ϕ is a Moorean sentence iff any normal form of ϕ contains a Moorean conjunction
as a disjunct.
The following theorem gives necessary conditions for self-refuting and unsuccessful
formulas. In Appendix A we strengthen this result with conditions that are sufficient as
well as necessary.
Theorem 3.13 Let ϕ be an arbitrary formula.
(i) If ϕ is self-refuting in any sublogic of S5, then ϕ is a Moore sentence.
(ii) If ϕ is unsuccessful in any extension of KD45, then ϕ is a Moorean sentence.
Proof. By Lemma 2.3 it suffices to show the consequent of (i) for ϕ that is self-refuting
in S5 and the consequent of (ii) for ϕ that is unsuccessful in KD45. Since ϕ is selfrefuting (resp. unsuccessful) iff any equivalent normal form of ϕ is self-refuting (resp.
unsuccessful), let us assume that ϕ is already in normal form.
(i) Suppose ϕ is not a Moore sentence, so by Definition 3.12 there is a disjunct δ of
ϕ that is not a Moore conjunction. Then by Lemma 3.9, δ is not self-refuting, so there
is a pointed model with M, w ♦+ δ and M|δ , w δ. By Lemma 3.8(iii), M|δ δ. It
follows that M|δ |ϕ = M|δ , since all points in M|δ satisfy one of the disjuncts of ϕ,
namely δ. Then given M|δ , w ♦+ δ, we have M|δ , w ♦+ ϕ and hence M|δ |ϕ , w ϕ.
Therefore ϕ is not self-refuting.
(ii) We prove something stronger. Suppose ϕ is not a Moorean sentence, so by
Definition 3.12 no disjunct of ϕ is a Moorean conjunction. Then each disjunct of ϕ is
successful by Lemma 3.9. Consider a pointed model such that M, w ♦+ ϕ, so for some
disjunct δ of ϕ, we have M, w δ. Since ϕ is a disjunction, M|δ ⊆ M|ϕ . By Lemma
3.11, δ is super-successful, so for any M0 with M|ϕ ⊆ M0 ⊆ M, we have M0 , w δ and
hence M0 , w ϕ. Since M was arbitrary, ϕ is super-successful.
2
Definition 6. Given a specification "C, N, A# and formula ϕ in DN
ϕ 2 , so
χ2 , so
definition
strictly
the
of clarity
definition
doesset
ofnot
clarity
apply.
does
However,
not
apply.
it w
i
!
"C,strictly
N, A# χthe
= "C,
N
, A# where
N ! is the
of all
literals
l∈N
for
�♦
�♦
into is
normal
form
into
normal
theform
sameusing
used
same
withtrick
σ̂
in Definition
withisσ̂a sele
6.
in
a disjunct
d using
in
ϕ such
that
d trick
does the
not
contain
∼ l used
and
there
Holliday
and
Icard
!
!
caseevery
the set
necessary
case
the
modifications
necessary
modifications
add
four
existential
add
quantifiers
existential
literals,
each
in A,
such
that
not
contain
the
N and
Γone
∈ from
A, there
N
is Γaand
literal
every
l ∈
setΓ
Γthat
∈ΣA,
isdoes
there
not
the
isfour
aquazi-negation
literal
l ∈quaz
Γtoto
of
l
and
d
does
not
contain
the
quazi-negation
of
any
literal
in
Σ.
for
simplicity
for
we
simplicity
do
not
write
we
them
do
not
out.
write
When
them
we
out.
say
When
that
χ
we
is
sa
cl
l.
l.
χ2 ,the
somean
strictly
offormula
clarity is
does
not apply. Howe
mean that
modified
thatthe
formula
thedefinition
modified
is clear.
clear.
Definition 6. Given a specification
Definition
"C, N,
6. A#
Given
and aformula
specification
ϕ in DNF,
"C, N,defin
A#
into normal
form using
the
same trick
used with
σ̂ �♦
in Defi
Holliday
Icard
ϕ
ϕ
!
!
!
! and
"C, N,
A#If =
"C,a6.5
N
, A#
"C,
N,
isϕ
A#
the form
set
=
"C,
of N
all
,literals
A#(ϕ)
where
lany
∈N
Nof
is
forϕ
the
which
set
of
ther
ao
Theorem
Theorem
ϕNiswhere
unsuccessful
6.5
unsuccessful
normal
iff
form
normal
is
Moorea
form
ϕ is
conjunction
inN
normal
L
∅,
then
isthat
success
and
every
setisΓiff
∈any
A,and
there
is four
a$=
literal
l! ϕ
∈Γ
is n
case the
necessary
modifications
add
existential
quant
Holliday
Icard
is a pensation.
disjunct d in ϕ pensation.
such that disdoes
a disjunct
not
contain
d in ϕ∼and
such
l and
that
there
d does
is a not
selection
contain
Σ o∼
l.
forIfsimplicity
we
dothe
not
write
them
out.
we! say
tha
Proof.
ϕ
only
formulas
of
form
and
(α
prop
literals, one
from
each
Γ in
A,
literals,
such
that
one
Σfrom
does
each
not
Γclarity
contain
inKα
A,When
such
the"¬Kα
quazi-negatio
that
Σ
does
χ2contains
, so
strictly
definition
of
does
not
apply.
! the
M
M
M
fornot
any
(M,
w)
with
M,
w
!
ϕ,the
vof
∈same
Wclear.
| wR
vΣ.
=
∈
Wfor
mean
that
modified
iscontain
Wesley of
H. lHolliday
and
Thomas
F.the
Icard,
187
and
does
contain
the
of quazi-negation
lIII
and
d formula
does
any
literal
theused
inquazi-negation
ois
bdthen
bnormal
into
form
using
trick
with
σ̂ �♦
in
Definition
6.
Given
anot
specification
"C,
N,
A#v and
χ
strictly
the
definition
of
clarity
does
not
apply.
However,
it
2 , so
Holliday
and
Icard
Holliday
and
Ica
ϕ
Hence
M
,
w
!
ϕ.
!
!
!
|ϕ set
Nnormal
every
ΓbN,
∈ A,
there
is trick
a literal
l ∈with
ΓNthat
isfour
not
the
quazi-ne
A#
=unsuccessful
"C,
N
, A#used
where
is�♦
the
set
of all
bandTheorem
a,
case"C,
the
modifications
add
existential
intoa,
form
using
the
σ̂normal
in
Definition
6.5necessary
ϕ issame
iff any
form
oflitera
ϕ6.is
l.
is
a
disjunct
d
in
ϕ
such
that
d
does
not
contain
∼
l
4 Unsuccessfulness
in
Other
Logics
we
do add
not Holliday
writeexistential
them
out.
When weand
sa
case
necessary
modifications
four
pensation.
pthe
∧the
♦q
∧ #for
(qinsimplicity
→
p)of,(⇒)
∧ϕclarity
♦one
(q
∧from
p)
and
Icard
If ϕ
is a conjunction
normal
If
form
ϕstrictly
is
aand
conjunction
Lnot
(ϕ)
∅,A,
inthen
normal
ϕthat
isform
successful.
and
Lnot
(ϕ)
$=
Proof.
(⇒)
Suppose
Proof.
is
Suppose
unsuccessful,
ϕ definition
is$=
unsuccessful,
so
there
is
aquantifiers
so
model
there
with
isto
aa
χ2 , so
strictly
definition
χ
so
does
the
apply.
However,
of
clarity
itdoes
does
is easy
not
to
literals,
each
Γ
in
such
Σ
co
2
mean
that
the
modified
formula
is
clear.
for
simplicity
we
do
not
write
them
out.
When
we
say
that
χ
is
cle
Definition
6.
Given
a
specification
"C,
N,
A#
and
formula
ϕ
in
DN
We now consider theinto
sources
of
unsuccessfulness
in
logics
for
an
agent
without
intro�♦
of
l
and
d
does
not
contain
the
quazi-negation
of
any
li
M
,
w
�
ϕ.
Given
M
,
w
M,
�
w
ϕ.
�
Given
ϕ,
we
M,
can
w
read
�
ϕ,
off
we
from
can
w
read
the
off
disjunct
from
w
δ
normal|ϕ
form
using
the
into
normal
trickofused
form
with
using
σ̂ only
theinsame
Definition
trick
used
6.2.form
Since
withKt
|ϕ same
bϕ in=modified
! Proof.
Proof.
contains
only
Ifagents.
the
ϕis! contains
form
formulas
¬Kα
(α of
propositional
pϕthat
"C,
N,
A#the
"C, Nformulas
, A#
where
N
is theKα
setand
of "
all
literals
l ∈the
N for wh
spection (logics without axioms
4 If
and
5)
and
logics
for
multiple
mean
formula
clear.
!
!
!
6.5
ϕ
is∈Γassociated
unsuccessful
iff
any
normal
form
χ(M,
strictly
the
ofM
does
not
apply.
Howev
ofany
disjuncts
Swith
ofNdisjuncts
associated
conjuncts
and
conjuncts
disjuncts
and
such
disjuncts
M,
! M
case then
theSfor
necessary
modifications
case
the
add
necessary
four
existential
modifications
four
defi
M
M
2 , so
and
every
set
∈
A,clarity
there
a∼=
literal
lthere
Γthat
that
isM
with
M,
wthen
!definition
ϕ,with
for
W
(M,
|w)
wR
with
vquantifiers
M,
v!add
∈
ϕ,
W∈vto
∈|ϕisthe
W
|aexiste
wR
| onw
aw)Theorem
is
disjunct
d in
ϕ
such
that
dvany
does
not
contain
lwand
selec
From an epistemic perspective,
thea most
interesting
(normal)
proper
sublogics
of is
S5
�♦
pensation.
into
normal
form
using
the
trick
used
with
Now
suppose
Now
for
contradiction
suppose
that
that
a form
θnot
∈
there
Tof
isclear,
that
θDefin
∈
fo
T
for simplicity
we
do
for
them
simplicity
out.
we
When
do
not
we
say
write
that
them
χ σ̂such
is
When
stri
Theorem
6.5
ϕwrite
unsuccessful
iff
any
normal
ϕout.
isain
Moorea
l.is
Hence
!one
ϕ.not
Hence
M
,same
w
!there
ϕ.
literals,
from
Γ
infor
A,
such
that
Σ is
does
contain
quazi
are obtained by dropping
axiom
5M
and
adding
something
in contradiction
its
place.
Indeed,
|ϕ , w
|ϕ
If ϕeach
is aweaker
conjunction
in
normal
form
and
L (ϕ)the
$= ∅,
the
α
α
Proof.
(⇒)
Suppose
ϕ
is
unsuccessful,
so
there
is
a
mod
the
necessary
modifications
add
four
existential
quantifi
M,
wq4,l�case
χ3 .dhave
Hence
M,
w
for
�proposed
χall
Hence
♦γ
inquazi-negation
for
θ,modified
M,
all
♦γ
w
�
in
♦
(σ
θ,any
M,
∧isγliteral
w
) �for♦insome
(σ
mean
the
modified
formula
mean
clear.
3 .isthat
θ the
θ formula
θclear.
logics such as S4, S4.x
for xthat
=pensation.
2, 3,
etc.,
been
as
logics
of
knowledge.
of
and
does
not
contain
the
of
Σ. ∧ σγ
bDefinition
α ∧
αM
6.witness
Given
abe
specification
"C,
N,
A#
and
0 we
p ∧ ♦q
#is(q
→
p)
∧
♦
(q
∧
p)
p
∧
♦q
∧
#
(q
→
p)
∧
♦
(q
∧
p)
,
w
�
ϕ.
Given
M,
w
�
ϕ,
we
can
read
off
from
w
the
di
for
simplicity
do
not
write
them
out.
When
we
say
one
♦γ
one
and
such
let
v
♦γ
be
a
and
let
v
with
a
M,
witness
v
�
σ
with
γ
M,
.
v
Given
�that
σfor
Call logics L and L0 comparable
if∧Lsuch
aM
sublogic
of
L
or
vice
versa.
Proof. If ϕϕ
of
the form
and
|ϕθ
θ contains! only formulas
θ Kα
!
!
!
"f
N
and
every
set
Γ
∈
A,
there
is
a
literal
l
∈
Γ
that
is
not
the
quazi-ne
Theorem 6.5
ϕS isofunsuccessful
any
6.5
normal
isconjuncts
unsuccessful
form
of
iff
Moorean
any
"C,
N, A#iff =
"C,
Nϕ
, A#
where
N ϕisisthe
set
ofnormal
allwithou
litera
a,
bTheorem
�♦
M
M t
disjuncts
with
associated
disjuncts
such
mean
that
the
modified
isM,
clear.
the bfact
that
the
σ ∈
fact
S,
we
that
have
σ(M,
∈formula
M,
S,
we�
have
σ �♦
w v�since
σW
.the
relati
then
for
any
w)w
with
w. M,
!Then
ϕ,and
∈
|Then
wR
vsi
Proposition 4.1 Forpensation.
any normal,
proper
sublogic
L
of
S5,
comparable
to
S4.4,
there
l.
is apensation.
disjunct
d in form
ϕ such
that
d does
contain
∼ l and
p equivalence
p normal
aIf
ϕisisNow
a conjunction
in
and
$= ∅,not
then
ϕθ is∈
successfu
Hence
M
, wnot
! ϕ.
4L (ϕ)
suppose
contradiction
that
there
is
T such
relation
equivalence
(a(⇒)
quasi-partition
relation
(aϕquasi-partition
would
suffice)
would
weaΣ
have
suffice)
M,
vw
�
|ϕ
is a formula (consistent with S5)
that
unsuccessful
in
Lfor
but
is
Moorean.
Proof.
Suppose
is
unsuccessful,
so
there
literals,
one
from
each
Γ
in
A,
such
that
does
Theorem 6.5 ϕ is unsuccessful iff any normal form
ofnot
ϕisiscoaM
α
Definition
6.
Given
a
specification
"C,
N,
A#
and
formula
ϕ
in
DN
M,
wM
�follows
χof
.l�and
Hence
for
in
θ,
M,
�read
(σ
)ϕw
for
It follows
that
It
M,
σonly
that
∧dγGiven
M,
andall
vtherefore
�♦γ
σwθ∧the
γθϕ,
and
M,
vquazi-negation
therefore
�wϕand
∧a♦
γθ¬Kα
,M,
in ∧which
v(α
∧ca
3vw
θany
θdoes
b ϕProof.
bϕ.
not
contain
the
ofγ�with
liγ
�
M,
we
can
off
from
t
If ϕϕ
contains
formulas
form
Kα
propo
(⇒)
unsuccessful,
so
there
model
and is
Icard
Holliday
and
Icard
Proof. First, we claim that
≡ Proof.
♦p ∧pensation.
♦¬p
isSuppose
unsuccessful
in
S4.4
and
hence
in
any
! ϕ
! ∧of
p,♦γ
∧N♦q
∧ is
#
(q
p)
♦the
(q�Holliday
∧set
p)M
!
"and
!∈therefore
α for
θ|ϕ
θ→
"C,
N,
A#
=
"C,
,
A#
where
N
is
of
all
literals
l
N
wh
one
such
and
let
v
be
a
witness
with
M,
v
�
σ
∧
γ
.
M
M
in
M
.
Since
in
γ
M
is
propositional,
.
Since
γ
is
propositional,
we
have
M
,
we
v
�
have
γ
M
,
v
�
γ
|ϕ
θinwith
θ
θ
θ
|ϕ
|ϕ
|ϕ
|ϕ
Holliday
and
Icard
then
for
any
(M,
w)
M,
w
!
ϕ,
v
∈
W
|
wR
v
=
v
∈
W
a,q b2.3.
a,
b
S
of
disjuncts
with
associated
conjuncts
and
disjuncts
sublogic of S4.4 by Lemma
In
the
S4.4
model
M
Figure
2,
ϕ
is
true
at
the
left
q � ϕ, we can read off from�♦
M|ϕis, w
� ϕ. Given
M,
w
w♦the
disjunct
δ
aγthe
disjunct
d of
in
ϕ
such
dwe
does
not
contain
lthat
and
there
is
a�
selec
the
fact
that
σwas
∈that
S,
have
M,
w
�w∼σto
. .χThen
since
χclarity
, θbecomes
so
strictly
the
definition
of
clarity
does
not
apply.
However,
it
is♦te.
χ2 , so
strictly
definition
does
not
apply.
However,
it
is|ϕeasy
put
Since
was
arbitrary,
Since
we
conclude
arbitrary,
that
we
conclude
M
,
�
θ
M
Since
,
w
θ
∈
θ
T
2γ
b
Hence
M
,
w
!
ϕ.
point, but in M|ϕ , the right
point
is
eliminated,
so
ϕ
false
at
the
left
point.
θ
|ϕ
|ϕ
Now
suppose
for
contradiction
that
there
is
a
θ
∈
Tw
p into
S ofliterals,
disjuncts
with
associated
conjuncts
and
disjuncts
such
that
M,
�♦
one
from
each
Γconjunction
in
A,
such
that
Σ
does
not
contain
quazi
normal
form
using
same
trick
used
with
σ̂ �♦
in (ϕ)
Definition
6.2.
into normal
form
using
theϕ
same
trick
used
with
σ̂
inthe
Definition
6.2.
Since
inand
this
χequivalence
, so
strictly
the
definition
of
clarity
does
not
apply.
However,
itwith
is
easy
to
put
χ�
αform
αthe
2M,
If
ϕ
is
a
in
normal
L
=
$
∅,
the
(⇒)
Suppose
is
Proof.
unsuccessful,
(⇒)
Suppose
so
there
ϕ
is
is
a
unsuccessful,
model
M,
so
w
ther
relation
(a
quasi-partition
would
suffice)
we
hav
Note that the formulaProof.
is already
inhave
normal
form
and
is
not
Moorean.
α
we
w
we
�
have
χ
,
which
M,
w
gives
�
χ
,
M,
which
w
�
gives
θ
and
M,
hence
w
�
θ
M
and
,
w
he
1
1
|ϕ
anormal
�♦
M,
wcontradiction
�add
χ
.the
Hence
forquantifiers
all
♦γ
θ,
M,
w6.2.
�quantifiers
♦in(σ
∧
γth�
case
the
necessary
modifications
add
four
tofo
case Now
the necessary
existential
the
definition,
forform
that
there
isin
aof
θexistential
∈Icard
Tliteral
such
that
3four
θ
intomodifications
using
same
trick
used
with
σ̂θto
in
Definition
Since
in this
ofsuppose
l and
d does
not
contain
the�
quazi-negation
any
Σ.
Holliday
and
Holliday
and
Icard
�
M|ϕ ,forwsimplicity
�
ϕ.
Given
M,
w
�
M
ϕ,
we
,
w
can
�
ϕ.
read
Given
off
from
M,
w
w
�
the
ϕ,
disjunct
we
can
read
δ
and
off
sets
fro
p
∧
♦q
∧
#
(q
→
p)
∧
♦
(q
∧
p)
It
follows
that
M,
v
�
σ
∧
γ
and
therefore
M,
v
�
ϕ
∧
γ
,
in
α
|ϕ
for
simplicity
we
do
not
write
them
out.
When
we
say
that
χ
is
clear
wecase
do
not
write
them
out.
When
we
say
that
χ
is
clear,
strictly
we
have
M
,
w
have
�
θ
.
M
For
,
w
suppose
�
θ
.
there
For
suppose
is
some
there
�β
in
is
θ
some
and
�β
v
in
in
θ
θ
α
the
necessary
modifications
add
four
existential
quantifiers
to
the
definition,
|ϕ
|ϕ♦γ
one Proof.
such
and
letθ, vM,
bewIcard
aformulas
M,
vKα
� and
σσ "
Ifθ θϕ♦γ
contains
only
the
M, w �Proof.
χ3 . Hence
for
all
inis
�witness
♦ (σsoof!∧with
γθ )form
for
some
and
θϕHolliday
(⇒)
unsuccessful,
there
is�strictly
aγM
mode
mean
that
thepropositional,
modified
formula
iswe
clear.
that thewith
modified
formula
isSuppose
clear.
for
simplicity
we
do
not
write
them
out.
When
we
say
that
χαwe
is clear,
S of χmean
disjuncts
associated
S
of
conjuncts
disjuncts
and
with
disjuncts
associated
such
conjuncts
that
M,
and
w
�
disj
δM
∧
χ
in
M
.
Since
γ
is
have
M
,
v
and
M
q
�♦
M
,
v
�
¬β.
M
Then
,
v
�
since
¬β.
v
Then
was
retained
since
v
was
in
M
retained
,
have
in
M,
,we
vsi
θ
|ϕ
|ϕ
χ
,
so
strictly
the
definition
of
clarity
does
not
apply.
However,
it
is
et
,
so
strictly
the
definition
of
clarity
does
not
apply.
However,
it
is
easy
to
put
χ
|ϕ
|ϕ
|ϕ
|ϕ
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
wR
vw
2
2 one such ♦γ the
fact
∈isS,
wewith
haveM,
M,
σ∧ γθ.. Then
and
let
vthat
beformula
aσ witness
vw
�off
σ� from
Given
M
mean, θthat
the
modified
clear.
p
♦
�♦
�♦
M
w
�
ϕ.
Given
M,
w
�
ϕ,
we
can
read
w
the
disj
Now into
suppose
Now
that
suppose
for
is
contradiction
θ�♦
∈6.2.
T
such
that
for
isSinc
♦γ
aχ a
Theorem
6.5
iswe
unsuccessful
iff
any
normal
form
of
ϕχ
is
Moorean
Theorem
6.5
ϕ
is
unsuccessful
iffM,
any
form
of
ϕ
isanot
Moorean
without
cominto
normal
using
the
same
trick
used
with
σ̂that
in
Definition
6.2.
normal
form
using
the
same
trick
used
with
σ̂
in
Definition
Since
inthen
this
|ϕ
soconjunction
strictly
the
definition
of
clarity
does
apply.
However,
it
is
easy
was
arbitrary,
conclude
that
M
,M
w
�
θ1, successf
.all
2which
ϕfor
isχSince
a,contradiction
in
normal
form
and
L
(ϕ)
∅,
ϕthere
is
σ ∈Iffact
S,
in
σσγθ∈
case
S,
in
which
vnormal
�form
case
given
M,
M,
�
w
β�
χ
.$=
Hence
M,
w
�
.vto
Hence
�put
β,
|ϕ
Hence
M
,β
wϕthere
!
ϕ.(a
1Then
|ϕ
|ϕ
equivalence
relation
quasi-partition
would
suffice)
w
the
that
∈form
S,
we
have
M,
w
�vwith
σ
.given
since
the
relati
�♦
αadd
pensation.
pensation.
case
the
necessary
modifications
four
existential
quantifiers
to
th
the
modifications
add
four
existential
quantifiers
to
the
definition,
6.5
ϕ
is
unsuccessful
iff
any
normal
form
of
ϕ
is
Moorean
without
cominto
normal
using
the
same
trick
used
σ̂
in
Definition
6.2.
Since
in
this
α
STheorem
of
disjuncts
with
associated
conjuncts
and
disjuncts
such
th
M, wcase
� We
χ3 .necessary
Hence
for
all
♦γ
M,
in
w
�
θ,
χ
M,
.
w
Hence
�
♦
(σ
for
all
∧
γ
♦γ
)
for
in
some
θ,
M,
σ
w
∈
�
S.
♦
C
(σ
♦
�
♦
�
Fig.
2.have M, θw � χ1 ,3 which
we
gives
M,
w
�
θ
and
hence
θ
θ
have
shown
We
M
have
,simplicity
w
shown
� When
α M,
∧M
θdo
∧
, four
θw
�
,would
i.e.,
α
∧clear,
M
θout.
∧strictly
, θw ,�we
i.e.,
θ,tosay
in
M
which
�ca
It
follows
that
v
�
σ
∧
γ
therefore
M,
v
�
ϕ
∧
γM
for
we
not
write
them
When
we
that
χ, w
is
clear
for simplicity
wepensation.
do
not
write
them
out.
we
say
that
χ
isand
we
|ϕ(a
|ϕ
|ϕ
|ϕ
case
the
necessary
modifications
add
existential
quantifiers
the
definition,
equivalence
relation
quasi-partition
suffice)
have
M,
v
�
θ
α
Proof.
If
ϕ
contains
only
formulas
of
the
form
Kα
and
¬Kα
(α
propo
bfor
b ♦γ and
�
Now
suppose
contradiction
that
there
is
a
θ
∈
T
such
one such
let
v
be
a
one
witness
such
♦γ
with
and
M,
let
v
�
v
σ
be
∧
a
γ
witness
.
Given
with
M,
M,
w
�
v
p
∧
♦q
∧
#
(q
→
p)
∧
♦
(q
∧
p)
!
"
!
θFor
mean
that
the
formula
is there
clear.
meananother
thatqθthe modified
formula
is.do
have
M
,that
w
�contradiction.
θ write
. modified
suppose
isθϕθM
some
�β
in
an
for
simplicity
we
not
them
out.
When
we
saywe
χ∈isfor
clear,
strictly
we
contradiction.
another
It
follows
that
It∈for
follows
every
that
TαM
every
there
θθM
∈
aθ|ϕ
|ϕ
M
in
M
Since
isθ!extension
propositional,
have
,∈�♦
vW
�is
γca
follows
that
M,
�a,clear.
σ
therefore
M,
v| w
�that
∧
in
Next, we claim that M is a,
aIt
offor
any
logic
Lv
of
S4.4
bis∧aγproper
bmodel
bany
|ϕ
|ϕ
then
(M,
w)
with
M,
w
ϕ,
vin
W
wR
vγ(σ
vwhich
θ γand
θ ,=
�♦
M,
w
�
χ
.
Hence
for
all
♦γ
θ,
M,
�
♦
∧
γ
)
for
mean
that
the
modified
formula
is
clear.
the
fact
that
σ
∈
S,
we
have
the
M,
fact
w
that
�
σ
σ
∈
.
Then
S,
we
since
have
the
M,
w
relation
�
σ
in
.
T
M
3
θ
θ
♦s
M
,Since
v,γwθ�!is
¬β.
Then
vϕwith
retained
M
,Lemma
we� θhav
6.5
isofunsuccessful
iff
any
normal
form
ofby
ϕ ,is
Moorean
Theorem
ϕχ
is
iff
any
form
ofis
iswas
Moorean
without
comM,M
w6.5
�.not,
,unsuccessful
whence
M,
wTheorem
�
M,
,normal
whence
�
χϕϕsince
so
χM,
w
clear
�conclude
χM
by
so|ϕ
Lemma
χ
clear
3.5.
|ϕ
|ϕ
a
and a proper sublogic of S5.
Suppose
so
there
is
theorem
L
M
2
ϕ.
3M
3w
γpropositional,
was
arbitrary,
we
that
M
.
Hence
ϕ.
in
Since
we
have
,wso
v�is�
γin
therefore
θa χ
|ϕ
αw
|ϕ
θvisand
|ϕ
�♦
Proof.
(⇒)
Suppose
ϕ
is
unsuccessful,
there
a
model
with
Proof.
(⇒)
Suppose
ϕ
is
unsuccessful,
so
there
is
a
model
with
M,
ϕ
but
p
one
such
♦γ
and
let
v
be
a
witness
with
M,
�
σ
∧
γ
.
pensation.
pensation.
equivalence
relation
(a
quasi-partition
equivalence
would
relation
suffice)
(athan
quasi-partition
we
have
M,
van
would
�
σ com∧M
σG
Theorem
6.5which
is unsuccessful
iffvany
normal
form
of ϕ
is�
Moorean
without
θϕ
θsuffi
α
σ
∈
S,
in
case
M,
�
β
given
M,
w
χ
.
Hence
M
Change ϕ to ϕ0 by substituting
∧
¬p)
for
any
propositional
variable
q
other
p.
♦
1
|ϕ
Suppose
(⇐)
ϕ
is
Moorean
Suppose
ϕ
without
is
Moorean
compensation,
without
compensation,
so
approp
M
,
w
�
ϕ.
Given
M,
w
�
ϕ,
we
can
read
off
from
w
the
disjunct
δ
an
M|ϕ ,Since
w(p
� (⇐)
ϕ.
Given
M,
w
�
ϕ,
we
can
read
off
from
w
the
disjunct
δ
and
sets
T
and
we
have
M,
w
�
χ
,
which
gives
M,
w
�
θ
and
he
(⇒) Suppose
ϕ isconclude
unsuccessful,
so there
is ,aw
model
with
M, w �θ ϕ∈but
γthe
was
arbitrary,
we
that
M
�
θ
.
Since
T
b
|ϕ
1
�♦
θ Proof.
|ϕ
pensation.
0
fact
that
∈wM
we
have
w
��
.which
Then
since
the
♦from
It therefore
follows
that
M,
vthe
�
σmodel
γconjuncts
follows
therefore
that
vM,
�
σ
ϕsuch
∧
∧
γ
and
in
therefore
case
M,
is
vwrM
�
Since L is normal and
closed
under
substitution,
ϕS,
is
also
aM,
theorem
of
bσ
θIt
θand
θσ
Sp)
ofand
disjuncts
with
associated
conjuncts
disjuncts
such
that
S ofb disjuncts
associated
and
disjuncts
such
that
M,
w
�the
δ,i.e.,
∧disjunct
χ
χ3.5
M
,#
w
�
ϕ.∧
M,
�∧
ϕ,
we
can
read
off
w
sets
TvM,
and
1 ∧M
2 . δ and
αγL
�
pM
∧ with
♦q
∧
(q
→Given
∧
♦
(q
p)
We
have
shown
,
w
�
α
∧
θ
∧
θ
,
,
w
�
θ,
in
w
|ϕ
Let
be
Let
M
be
given
the
by
model
Lemma
given
3.5
by
Lemma
that
M,
such
w
�
χ.
that
For
|ϕ
|ϕ
we
have
M,
w
�
χ
,
which
gives
M,
w
�
θ
and
hence
M
,
w
�
have
M
,
w
�
θ
.
For
suppose
there
is
some
�β
in
1 a,|ϕ
|ϕhave
θS contradiction
θM
bthere
a,
bSince
Now
suppose
for
that
there
isthat
atherefore
θ M,
∈ Twhave
such
that
for
ball
suppose
for
that
isSince
aconjuncts
θcontradiction
∈ p,
TγM
such
that
for
all
♦γ
in
θ,
and therefore of S5.inMoreover,
since
for
variables
q
other
than
¬q,
the
equivalence
relation
(a
quasi-partition
would
suffice)
we
of
disjuncts
with
associated
and
disjuncts
such
�
δ
∧
χ
∧
χ
.
θ
1
2
MNow
.
γ
is
propositional,
in
M
we
.
have
is
,
propositional,
v
�
γ
and
we
M
M
,
w
,
θ for
|ϕ assume M
|ϕ, wthere
|ϕ athere
|ϕ|ϕ
another
Itγthere
that
every
θ such
∈
T
th
�
α∧
α for
,all
w
assume
�contradiction.
M
Then
�that
ϕ.
exists
disjunct
exists
θin
in
aγθθ )disjunct
ϕ
q M,
0Then
0 TS.
w
�¬β.
χ|ϕ
♦γ
in
wsuch
�
♦ (σ
for
some
σtha
∈M
M,
� χ3 . M
Hence
♦γ
in
M,
w
♦Then
(σ
∧
)follows
σ
∈
Consider
M
v θ,contradiction
�M,
since
was
retained
in
M
w
|ϕsuppose
aw
have
,for
�
θunsuccessful,
.ϕ.
For
suppose
is
�β
and
Now
there
is
avθ,
θsome
that
for
all
♦γ
in,in
θ,θ
3 . �Hence
θsome
θ ,for
θall
θv
|ϕ
|ϕ
♦
substitution of (p ∧ ¬p)
forProof.
qγ w
preserves
(un)satisfiability
in
M,
so
M
2
.for
Hence
ϕ∈M,
isconclude
|ϕfollows
It
that
v
σ
∧
γ
therefore
M,
v
�
ϕ
∧
γ
,
in
w
Proof.
(⇒)
Suppose
is
unsuccessful,
so
there
is
a
model
with
M
(⇒)
Suppose
ϕ is we
soθ�
there
is
aϕϕand
model
with
M,
w
�
ϕ
but
Since
was
arbitrary,
conclude
Since
γ
that
was
arbitrary,
M
,
w
�
we
θ
.
Since
θ
∈
that
T
by
M
assum
,
w
p
θ
θ
α
α
α
θ
|ϕ
|ϕ
one
such
♦γ
and
let
v
be
a
witness
with
M,
v
�
σ
∧
γ
.
Given
M
one such
♦γwe
let
v� be
a
witness
with
M,
v
�
σ
∧
γ
.
Given
M,
w
�
χ
and
M,
w
�
χ
,
whence
M,
w
�
χ
so
χ
is
clear
by
Lemma
3.5.
M,
w
χ
.
Hence
for
all
♦γ
in
θ,
M,
w
�
♦
(σ
∧
γ
)
for
some
σ
∈
S.
Consider
2
θ
θ
θ and
θ
3
and
must
have
and
we
θ
∈
must
T
.
For
have
if
θ
θ
∈
∈
/
T
T
.
,
then
For
if
given
θ
∈
/
T
M,
,
then
w
�
given
χ
we
M
h
θformula
θM M,
σ.(⇒)
in
case
vcontaining
�
βcan
given
w
�the
.�1and
Hence
not a theorem of S4.4. But
result
of¬β.
Zeman
[21],
for
,w
vGiven
�
Then
vGiven
was
retained
,Then
we
have
M,
M|ϕ
wθwhich
�ϕany
ϕ.
M,
wψ,
�disjunct
ϕ,
wewe
read
off
from
δvth
an
M|ϕby
,M
wa
�|ϕϕ.
M,
w
�3∈
ϕ,S,
we
can
read
off
from
w M,
the
δin
and
sets
T
and
α
α
Proof.
Suppose
ispropositional,
unsuccessful,
so
there
is
with
M,
w1
ϕαθ.but
�♦
�♦
|ϕ
M
Since
γ,since
have
M
,wM,
vsince
�αχww
γdisjunct
the
fact
that
∈w
S,wsince
we
have
M,
�αa∧
σinmodel
. Given
the
relation
the fact
that
σin
∈
S,
we
have
M,
w
�
σbe
Then
the
relation
isα♦
an
we
have
M,
�
χ
,
which
we
gives
have
M,
�
�
θ
χ
and
,
hence
M
,
w
�
θ
a
one
such
♦γ
let
vis
a. σM,
witness
with
M,
vwhich
�wσ
γM
M,
χ
θ� �
|ϕ
|ϕ
α♦
α♦
α♦
2 and W
1
1
θ and
θgives
|ϕ
Suppose
is
Moorean
without
so
an
�
S�
of
disjuncts
with
associated
conjuncts
disjuncts
such
that
M,
w♦
disjuncts
associated
conjuncts
disjuncts
such
that
M,
w
�the
δcompensation,
∧w
χ
χi.e.,
M,
�with
θMwhich
in
M,
which
w
θtheorem
case
,ϕ�
in
M
which
,S4.4,
w
�have
case
θ�♦
M
or
M,
,to
w
�
�∧,∧relation
(σ
M,
∧
¬β)
w
�
exactly one variable, that S
is ofa σ
theorem
of
S5
but
not
aM,
of
adding
ψ
,,(⇐)
wquasi-partition
� ϕ.
Given
M,
�relation
ϕ,
we
can
read
off
w♦
disjunct
δor
and
Tw
and
1θ
2 .,M
α
|ϕ
|ϕ
|ϕ
∈w
S,
in
case
vwand
β
given
M,
w,from
�
χ
.and
Hence
,sets
v.M
�
aθ
We
have
M
,
�
α
∧
θ
∧
θ
M
,
�
(a
quasi-partition
would
have
M,
�foσ
equivalence
relation
(a
would
suffice)
v that
�
σsince
∧
σsuffice)
γ♦
1�♦
the
fact
that
σequivalence
∈arbitrary,
S,shown
we
have
M,
wwe
�w
σ�
.M,
Then
the
in♦
isvβ,
an
θ . we
|ϕ
�
|ϕ
|ϕ
Since
γ
was
we
conclude
M
,
w
�
θ
Since
θ
Now
suppose
for
that
isand
a|ϕ
∈vTwthat
such
for
Now
suppose
forθ
contradiction
that
there
is|ϕ
a,given
θcontradiction
∈
that
forthere
all
♦γ
inθ M,
θ,
Mand
,w
. disjuncts
For
suppose
have
M
there
w
is�Tσγto
some
θsuch
.and
For
�β
suppose
in
θM,
there
in
is
suc
S
associated
disjuncts
that
δM
∧ that
χαsome
∧χ
θv
S4.4 gives S5. Hencehave
L is S5,
a|ϕ
contradiction.
Since
models
any
logic
S4.4
1
2.
�♦
|ϕ∧
Let
the
by
Lemma
3.5
such
w
ItM
follows
that
M,
γand
therefore
�ααin
ϕ
γis
,�
in
It follows
that
vof�in
σM
∧and
γθbe
and
therefore
M,
vconjuncts
�αv♦
ϕ�∧case
,between
in
which
vsuch
is
retained
�
equivalence
relation
(amodel
quasi-partition
would
suffice)
we
have
M,
v∧�M
σ
∧which
σM,
.i
θIt
θevery
θ∧
σ �∈M,
S,
which
σwith
∈
case
S,
ain
witness
which
¬�β
aissome
witness
iscase
retained
to
¬�β
retained
,some
whenc
|ϕ
another
contradiction.
follows
that
for
θαγcase
∈
shown
M
,
∧
θ
θ
,
i.e.,
,
w
�
θ,
which
M,
ww
��
χw
for
♦γ
in
θ,
M,
wsuch
��
♦θ
(σ
∧and
γin
) all
for
σθca
∈
M, wWe
� χ3qhave
. Hence
for
all
♦γ
in
θ,
M,
�Hence
♦ which
(σ
∧∧
γθall
) gives
for
σM
∈T
S.
Consider
|ϕ
|ϕ
Now
suppose
for
contradiction
that
there
a
θ
∈
that
for
♦γ
in
θ,
3 .α
θ
θ
θ
θ
θ
we
have
M,
w
�
χ
,
M,
w
hence
M
θ
1
and S5, ϕ is unsuccessful
in
these
logics.
2
in
M
.
Since
γ
is
propositional,
we
have
M
,
v
�
γ
and
therefore
M
Since γThen
is
propositional,
we
have
M
,
v
�
γ
and
therefore
M
,
w
�
♦γ
.
M|ϕ ,invM�
since
M
v
was
,
v
retained
�
¬β.
Then
in
M
since
,
we
v
have
was
M,
retained
v
�
σ
in
fo
M
It
follows
that
M,
v
�
σ
∧
γ
and
therefore
M,
v
�
ϕ
∧
γ
,
in
which
case
v
is
retained
θ
θ
θ
|ϕ
|ϕ
|ϕ . ¬β.
|ϕ
|ϕ
|ϕ
|ϕ
θ
θ
α
α
assume
M
,
w
�
ϕ.
Then
there
a
disjunct
θ
in
ϕ
αexists
In
either
case,
In
M
either
,
w
�
case,
θ,
a
M
contradiction.
,
w
�
θ,
a
contradiction.
So
θ
∈
T
.
Moreover,
So
θ
∈
one
such
♦γ
and
let
v
be
a
witness
with
M,
v
�
σ
∧
γ
.
Given
M
one such
♦γ
and
let
v
be
a
witness
with
M,
v
�
σ
∧
γ
.
Given
M,
w
�
χ
and
|ϕ
M,
w
�
χ
.
Hence
for
all
♦γ
in
θ,
M,
w
�
♦
(σ
∧
γ
)
for
some
σ
∈
S.
Consider
2
♦
♦
θ
θ
θ
θ
|ϕ
3conclude
M,
ww
χpropositional,
whence
w
�
so
χ
is|ϕsome
Lemma
θassumption,
contradiction.
follows
that
θ,clear
there
a..Tf
�
Since
γ3θθ,It
was
arbitrary,
we
conclude
that
M
w∈� θT�β
.by
Since
θ�θ
∈is
Tand
by
Sinceanother
γθ was arbitrary,
that
M
,θ w �|ϕ
θM,
.have
Since
θfor
∈
by
in
M|ϕM
.we
Since
γ θ�
is
we
M
,χ
vT�every
γαθ�
and
therefore
M
,in
wcontrad
♦γ
|ϕ
θ
|ϕ
|ϕ
have
,
�
.
For
suppose
there
is
�♦
�♦
σ
∈
S,
in
which
case
M,
v
�
σ
β
∈
given
S,
in
M,
which
w
�
case
χ
.
M,
Hence
v
M
β
given
,
v
�
M,
β,
a
w
�
χ
|ϕmust
the
fact
σαwitness
∈ofT
S,knowledge
we
have
M,
w
�♦Tσ
..|ϕ
Then
since
the fact
that
σ ∈
S,such
we
have
M,
w
�vthat
σbe
Then
since
the
relation
in
M
is an
1M
1wa
one
♦γ
and
let
with
M,
v,θ
∧
γthen
M,
wthe
� χrelation
α
and
we
have
θwa.θbe
∈
.hence
For
ifthat
∈
/Since
,Since
given
M,
2w
θgives
θWe
Proposition 4.1 shows that
S5
logics
insofar
θ, is
there
must
be
there
aarbitrary,
vtypical
such
must
v,χ
such
vM
�gives
M
�∈compensation,
γ.
retained
Since
in
we
have
M,
�so
which
w
θ,vααvGiven
and
hence
M
wand
��θ
we have
M,
w
�χ
χamong
, whence
M,
wwe
�that
and
w�
�θσwithout
θ.�♦
.�
also
1is
Since
γwhich
wasthe
conclude
that
,γ.
w
�M,
θwas
T
by
assumption,
|ϕ ,v
|ϕ
|ϕ
|ϕ
θθ,
M,
wunique
�M
M,
w
�
χsuffice)
is,have
clear
Lemma
3.5.
(⇐)
Suppose
ϕaχM
Moorean
�♦|ϕ
3 ,1(a
♦
�
♦
�
equivalence
relation
(a
quasi-partition
suffice)
have
M,
v an
�ασ,
equivalence
relation
quasi-partition
would
we
M,
vby
�
σwould
∧in
σ
∧relation
γM
the
fact
that
σ
∈
S,
we
have
M,
w
�
σ
.
Then
since
the
in
M
is
,
v
�
¬β.
Then
since
v
was
retained
in
,
we
have
θ .θwe
�
�
α♦
α♦
α
α
|ϕ
|ϕ
We
have
shown
M
,
w
�
α
We
∧
θ
have
∧
θ
shown
,
i.e.,
M
,
w
�
α
θ,
∧
θ
which
∧
,
case
i.e.,
M
M
as all of its unsuccessful formulas
are
counterexample
weaker
have
wsome
�given
θ �β
.But
For
there
is such
some
�β
and
inθ|ϕ
M
have for
M|ϕ ,some
w � Moorean.
θwe
. have
For
suppose
is,∈
infor
θsuppose
and
in
that
M,θThe
wBut
�,there
χin
,then
gives
M,
w
�the
and
hence
M
,M,
w�in
�χ
.�and
We
also
|ϕθ,v�
|ϕ∈
M,
w
�
which
M
w
�M
,retained
or
w
♦αvhave
(σ
∧
|ϕ
1M
|ϕ
|ϕ
S.
σwhich
given
χ
M,
w
∈
,,θ�♦
|ϕ
3θ
3
Itis
follows
that
M,
vϕ
�∧σγM,
and
therefore
M,
vθ
�3.5
ϕvso
∧�T
γsuch
inwe
which
It follows
that M,
vσ�|ϕ∈
σ Let
∧for
γθ and
therefore
M,S.
v �case
, γthen
inθ w
which
case
vand
equivalence
relation
quasi-partition
would
suffice)
we
have
M,
∧
σthat
∧vcase
γθ�
.M
θσan
θ∧
M
model
given
by
Lemma
αis
Suppose
ϕsome
Moorean
without
compensation,
appropr
�be
M
, come
v(a
�the
¬β.
Then
since
vbeing
was
retained
in
M
,that
we
have
M,
M|ϕ , vcontradiction.
� (⇐)
¬β. σThen
since
vIt
was
in
M
,�
we
have
M,
vtruly
�
σfollows
for
some
∈
S,
in
which
case
M,
v
β
given
M,
w
�
χ
Hence
M
,
v
have
M
,
w
�
θ
.
For
suppose
there
is
some
�β
in
θ
and
v
M
such
that
|ϕretained
|ϕ.in
|ϕ
logics shows that without
negative
introspection,
one
can
to
know
p
by
1
θ
θ
|ϕ
|ϕ
|ϕ
another
follows
another
that
contradiction.
for
every
θ
It
∈
T
there
is
a
for
♦γ
every
in
θM
in
M
. σSince
γ ,, vv
istherefore
have
M
,♦γ
v retained
�θ .case
γthat
therefore
M
in Mwhich
γ It
isfollows
propositional,
we
M
�propositional,
γaθ¬γ,
and M,
therefore
M
w |ϕ
�is
that
M,
vS,|ϕ
�have
∧
γ�
and
v We
�we
ϕM,
∧
γw
in
which
vαis
retained
θ and
and
σassume
∈
S,
in
which
case
witness
to
¬�β
in
|ϕ . Since case
|ϕ
|ϕ
M
which
,
v
�
case
¬γ,
M
a
contradiction.
�
a
contradiction.
conclude
We
conclude
M
,
w
θχ
θ ,,�
|
|ϕ
|ϕ
|ϕ
σ
∈
in
which
case
M,
v
�
β
given
χ
.
Hence
M
,
v
�
β,
a
c
σ
∈
S,
in
which
case
M,
v
�
β
given
M,
w
.
Hence
M
,
v
�
β,
a
contradiction.
M
,
w
�
ϕ.
Then
there
exists
a
disjunct
θ
1
1
M
,
v
�
¬β.
Then
since
v
was
retained
in
M
,
we
have
M,
v
�
σ
for
some
Let
M
be
the
model
given
by
Lemma
3.5
such
that
M,
w
�
χ.
For
|ϕ
|ϕ
♦
�
♦. Since θ ∈ T by
♦. we
|ϕhave
|ϕχ
told, “You do not know
not
p,”
case
ofis
unsuccessfulness.
Since
γM
conclude
that
M
,M
wχ
�|ϕθis
was
arbitrary,
we
conclude
that
M
,3w
�whence
θα
Since
θ∧
∈θ
T|ϕ
by
assumption,
We
shown
,arbitrary,
w
∧
,γθi.e.,
i.e.,
, wM
�∈,which
θ,
w
in
Ma|ϕ
.surprising
γ♦θM,
propositional,
we
have
M
�The
and
therefore
w
♦γθLe
.
θ was
|ϕ
|ϕ
M,whether
wSince
χγ3θunsuccessful.
,or
whence
M,
χishave
so
w
�
χ
clear
,,�
Lemma
M,
w
3.5.
χ|ϕM
|ϕ
♦, v
,βthere
�
θ,
aTαθM
contradiction.
So
θclear
T�by
.in
Mo
We
shown
M
w
�in
∧
θ�,♦�
,w
�aθ,contradiction.
in|ϕ
case
We�have
shown
M
,either
w
�
αw
∧ case,
θ�
∧
θ�
,M
i.e.,
M
w
θ,by
which
case
M
,so
w
ϕ,
σIn
∈|ϕS,
in
which
case
M,
v|ϕ
�have
given
M,
w
�
χθFor
Hence
,αT
v|ϕ�
�
β,
is
isSince
unsuccessful.
αw
αM
|ϕθ,,�hence
|ϕ
1 .,∧
|ϕ
|ϕ
we
have
M,
w
�
χ
which
gives
M,
w
�
θ
and
hence
M|ϕ , w tha
� θ
we have
M, w �
χ
, Proposition
which
gives
M,
w
�
θ
and
M
w
�
θ
.
We
also
and
we
must
∈
.
if
θ
∈
/
,
then
given
M
following proposition, although
weaker
than
4.1,
shows
that
non-Moorean
assume
M
,
w
�
ϕ.
Then
exists
a
disjunct
θ
in
ϕ
such
1
1|ϕ
Since
γ
was
arbitrary,
we
conclude
that
,
w
�
θ
.
Since
θ
∈
T
byθ
assumption,
|ϕ
θ contradiction. ♦It follows
|ϕ that for every
�
another
∈
T
the
another
follows
for
every
θ case
∈ TM
there
aϕ,♦i
another
contradiction.
follows
for
every
θ θ∈ ,TIti.e.,
is, that
a�
♦γθ,
θwithout
such
θso
have
shown
Mthat
,(⇐)
wcontradiction.
�α
∧ such
θθ��β
M
w
in
which
, wvis�in
(⇐)
Suppose
ϕ. logics
is It
Moorean
compensation,
ϕthere
is
Moorean
an
appropriate
compens
χ
�
αv|ϕ
|ϕM
|ϕ
θ,
there
vSuppose
,M
v,hence
γ.
Since
was
reta
α♦
α♦
have
�
.∧
For
suppose
there
some
in
haveifand
M
w � must
θWe
For
suppose
some
inthat
θin
and
in
such
that
we
have
M,
wknowledge
χθ
,without
which
gives
M,
w
�other
θM
and
M
,w
� �
θvαand
.χ
We
also
|ϕ
unsuccessful formulas appear
ofχmust
and
belief
ways
|ϕ
1be
|ϕ ,weaken
|ϕ
|ϕ�β
M,
w
in
which
M
wisT�
�
θLemma
, aw
or
wwe
�M
have
θ��χ�there
∈
.is3,It,a,wFor
if
θ�
∈
/χ
then
given
M,
h♦
M,
wis
�T
χ
whence
M,
wcase
�T
χ,every
so
χis
is
clear
by
3.5.M,
M, w
�we
χ
whence
M,
w
�
so
clear
by
Lemma
3.5.
13.5.
|ϕ
3 ,we
another
contradiction.
follows
that
for
θ
∈
there
is
♦γ
in
θ
such
α
θ
M,
w
�
χ
,
whence
M,
w
so
χ
clear
by
Lemma
3
MLet
� ¬β.
was
retained
in
M
, and
we
have
M,
v ,� w
M|ϕ be
, v �the
¬β. model
Then
since
v was
retained
in Then
M
, since
we
have
M,
v �M,
σw
some
Let M
Lemma
M
be
the
3.5
model
such
that
given
by
w
Lemma
�
χ.
For
contrad
such
have
Mgiven
�
θby
., vS.
For
suppose
isvα♦
some
�β
in
θ for
and
v3
in
M
such
that
|ϕ
|ϕ
|ϕthere
|ϕ , w
|ϕ3.5
for
some
σ
∈
But
then
given
M,
�
χ
θ
∈
T
as well.
α♦
α
(⇐)
Suppose
ϕ
is
Moorean
without
compensation,
so
an
appropria
(⇐)
Suppose
ϕ
is
Moorean
without
compensation,
so
an
appropriate
χ
is
clear.
M,
w
�
χ
,
whence
M,
w
�
χ
so
χ
is
clear
by
Lemma
3.5.
M,
w � case
θM|ϕ(⇐)
which
case
,w
�M
θ|ϕwithout
,�M
or
�. M,
♦ v(σis
∧
¬β)β,an
for
α ,for
and
∈
S,
in
which
case
a, in
witness
to
σThen
∈ S,
in
which
M,
v�
β
w
�wχ1¬�β
Hence
a ci
σ ∈ S,
in which
M,
v 3�
βσ
given
M,
w
�M
.case
vgiven
β,|ϕM,
a,M,
contradiction.
|ϕHence
1was
,,v in
�
¬β.
since
vχMoorean
we
have
�M
σretained
|ϕ v �some
Suppose
ϕ
is
compensation,
assume
M
, wmodel
�
ϕ.
Then
assume
exists
M
,retained
w
aagiven
disjunct
�compensation,
ϕ.
Then
in
ϕWe
such
exists
that
disju
M
be
model
Lemma
3.5
such
that
M,
w χ�aisso
χ.M
For
Let M
be|ϕthe
given
by Let
Lemma
3.5
such
that
M,
wby
� χ.
For�θcontradiction,
(⇐)
Suppose
ϕthere
is�
Moorean
without
sothere
an
appropriate
clear.
|ϕ
|ϕ c,a
which
case
,i.e.,
vvthe
�
¬γ,
contradiction.
conclude
that
♦∧
♦M
|ϕ
We
have
shown
M
,
w
�
α
∧
θ
θ
,
i.e.,
M
,
w
�
θ,
in
which
case
have
shown
M
,
w
�
α
∧
θ
∧
θ
,
M
,
w
�
θ,
in
which
case
M
,
w
�
ϕ,
Proposition 4.2 For anyWe
sublogic
L
of
KTB
or
KD5,
there
is
a
formula
(consistent
σ
∈
S,
in
which
case
M,
�
β
given
M,
w
�
χ
.
Hence
M
,
v
�
β,
a
contradiction.
|ϕ
|ϕ
|ϕ
|ϕ
|ϕ
1
|ϕ
In
either
case,
,θw
�ϕ¬�β
θ, athat
contradiction.
SoM,
θw
∈
T
and
∈ Let
inThen
which
case
aa|ϕwitness
to
isexists
retained
in
M
,such
whenc
|ϕ
assume
M
wM
ϕ.then
Then
a|ϕ
θwe
in
ϕ
that
assume
Mσ
�S,ϕ.M
there
exists
disjunct
in3.5
such
M
, such
w
�
the
model
given
by
such
wdisjunct
�
contradiction,
|ϕ , wLet
M
model
given
by
Lemma
3.5
that
�
and we
must
θ Moorean.
∈be
Tfollows
. the
For
and
if
we
θ� α∈
/,every
must
Tθ�Lemma
given
θthere
∈|ϕis,that
T
.�♦γM,
For
w
�
χθ1θ,
∈
/case
T
,|ϕ
have
then
♦, ∧
another
contradiction.
follows
that
for
θFor
∈
TM
there
is�that
agiv
♦
another
contradiction.
Itbe
that
for
θ have
there
aM,
θifχ.
such
isnot
unsuccessful.
θ inevery
We
have
shown
M
,w
∧given
θ∈�T,TIt
i.e.,
M
w
θ,
which
, wwe
ϕ,
with S5) that is unsuccessful
Lmust
buthave
isassume
|ϕT
|ϕ
and
we
must
have
θ
∈
.
For
if
θ
∈
/
T
,
then
given
M,
w
�
χ
hav
andin
we
have
θ
∈
T
.
For
if
θ
∈
/
,
then
M,
w
�
χ
we
have
that
either
1
1
M
,
w
�
ϕ.
Then
there
exists
a
disjunct
θ
in
ϕ
such
that
M
,
w
�
θ,
θ,
there
must
be
a
v
such
that
M
,
v
�
γ.
Since
v
wa
In
either
case,
M
,
w
�
θ,
a
contradiction.
So
θ
∈
T
.
Moreover,
|ϕχ |ϕ
|ϕ in ϕ sf
α♦
α♦
αdisjunct
α♦
|ϕ
M,
wis �clear
χw
, whence
M,
w
�αM,
χ every
sowexists
χcase
isθ clear
by
Lemma
3.5.
wθ�α♦
χα♦
M, wcontradiction.
�
so
χw
by
Lemma
3.5.
assume
M
,
�
ϕ.
Then
there
a
θ
3It
3,, whence
another
follows
that
for
∈
T
there
is
a
♦γ
in
θ
such
α♦
α♦
α
α♦
M, wM,
�
in
which
case
M
M,
w
,
�
�
θ
θ
,
in
,
or
which
�
♦
M
(σ
∧
,
w
¬β)
�
θ
for
,
some
or
M,
θ
|ϕ
|ϕ
M,
�
,For
inM,
which
case
M
, w for
� θM,
,wor��β
M,
w
♦ (σ that
∧ ¬β)
for �
s
M, w � θ , in which
M|ϕ
, ww
�θ|ϕ
θ∈ θT,. or
∧ |ϕ
¬β)
some
and wecase
must
have
ifwθ �∈
/♦
T(σ
, then
given
χ1in
weθ�have
either
(⇐)
Suppose
ϕM
is isMoorean
without
compensation,
so3anand
appropria
Suppose
ϕ iswthe
Moorean
without
compensation,
so
an�
appropriate
χαis
clear.
θ,
there
must
be
a
v
such
that
,
v
γ.
Since
v
was
retained
some
σ
∈
S.
But
then
given
M,
w
�
χ
θinθ χ
∈M
Proof. For KTB considerand
ϕ (⇐)
≡
♦p
∧
♦q
and
model
M
in
Figure
3.
In
M
,
the
left
M,
�for
χα♦
, whence
M,
w
�∈
χ so
χ
clear
by
Lemma
3.5.
3
�
|ϕ
|ϕ
α♦
and
we
must
have
θ
T
.
For
if
θ
∈
/
T
,
then
given
M,
w
�
σ ¬�β
∈σ
S,∈
in
which
to �a
¬�β
is, w
retained
M|ϕ ,�β
whence
σS,
∈ S,ininwhich
which
case
to
is
retained
in
whence
M
�¬β)
θ whence
.forin¬�β
M, w
� θa witness
, ainand
which
case
M
,in
w case
�which
θisaM
,witness
or
M,
w
♦witness
(σ
∧|ϕ
some
in
|ϕ , case
|ϕ
and
σ
∈
case
witness
and
to
S,
¬�β
retained
in
M
,
to
M
is
reta
,
|ϕ
|ϕ cw
Let
M
be
model
given
Lemma
3.5
that M, w χ� is
χ.clear.
For
Let M so
model(⇐)
given
byat
Lemma
3.5the
such
M,compensation,
wby
� χ.
For contradiction,
Suppose
ϕ
isthen
Moorean
so such
anSo
and right points are eliminated,
ϕthebecomes
false
the
center
point.
forbecase,
some
σ,w
S.
But
given
M,
w
��
χfor
and
θappropriate
T. ,M
we
have
which
case
M
,�without
vthat
¬γ,
contradiction.
conclude
α♦
In
either
case,
M
,�
wTM
�. θ,
aisa
contradiction.
θ∈
∈ We
T
Moreover,
In either
M
w∈∈�
θ,θin
a contradiction.
So
θ|ϕ ∈
Moreover,
every
in
3α♦
|ϕ
and
σ
S,
which
case
a witness
to
¬�β
retained
in
Mdisjunct
,♦γ
whence
, w(σ
�α
θ�for
.
|ϕ
|ϕ
|ϕ
M,
�
,
in
which
case
,
w
θ
,
or
M,
w
�
♦
∧
assume
M
,
w
ϕ.
Then
there
exists
a
θ
in
ϕ
such
that
assume
M
,
w
�
ϕ.
Then
there
exists
a
disjunct
θ
in
ϕ
such
that
M
,
w
�
θ,
|ϕ
In either
case,
M
, wbecase,
�theθ,θ,M
In
a|ϕ ,,contradiction.
either
case,
M
So
w
θ |ϕ
∈
�,∈vM
T
θ,
..γ.,a|ϕ
Moreover,
contradiction.
for ♦γ
every
So
Let
M
model
by
Lemma
3.5
that
M,
w
�vχ.�vσFor
contradiction,
|ϕ
|ϕ , So
there
a vv was
such
that
M
�
Since
was
in
M
θ, there
must
be
a |ϕ
veither
such
that
vwgiven
�must
γ.|ϕ
Since
retained
in
M,
|ϕ
In
M
�then
θ,contradiction.
abe
contradiction.
θ∈
T
Moreover,
forwretained
every
in
|ϕ
which
case
M
,
v
�
¬γ,
a
We
conclude
that
M
,
w
is
unsuccessful.
and
we
must
have
θ
∈
T
.
For
if
θ
/
T
,
then
given
M,
�
χ
we
hav
and
we
must
have
θ
∈
T
.
For
if
θ
∈
/
T
,
given
M,
w
�
χ
we
have
that
either
|ϕ
|ϕ
4 S4.4 is S4 plus ϕ → (♦ϕ →
1
1
assume
,w
�which
ϕ.
Then
there
exists
θthat
in
ϕis¬γ,
such
that
M
�, θ,
|ϕgiven
|ϕ
It α♦
a be
proper
ofθ,
S4.3
and
therefore
also
S4.2
[21].
and
σ
∈M
S,
in
case
aθγ.
to
¬�β
retained
in, whave
M
,
for
some
σ�|ϕ
∈χ
given
w
�
χ
and
θ ,∈
T|ϕ
we
M
forϕ).
some
σis ∈
S.
But
then
M,
w
Ta
,aof
we
have
M,
v retained
�
in
θ, there
must
vextension
such
that
there
M
,must
vand
�But
be
vdisjunct
such
vM,
was
v(σ
in
�α,, M
γ.
M
3M
|ϕ
3S.
θ,a
there
must
be
such
M
,witness
v∈then
�Since
γ.
vSince
�for
σ
α♦
α♦
α
|ϕ
|ϕ
19
19
M,
θTα♦
,that
in
which
case
M
, w for
�v θwas
,wretained
or��β
M,
w
� have
♦M
∧M,
¬β)
s
M, wis� unsuccessful.
θ , in which
case
M|ϕ
, waw
�vθ �
θ∈
, .or
M,
∧Since
¬β)
some
θin
and
we
must
have
For
ifvwθ�|ϕ�∈
/♦
T(σ
,athen
given
M,
χ
we
that
either
|ϕ
1in
which
case
M
,
¬γ,
contradiction.
We
conclude
that
M
,
w
�
which case M|ϕ
,
v
�
¬γ,
a
contradiction.
We
conclude
that
M
,
w
�
ϕ,
whence
ϕ
|ϕ � θ,
|ϕMor
some α♦
σ case,
∈ S.for
But
then
given
w
�But
χ3 |ϕ
and
θ∈
∈ Tgiven
wewe
have
�. ¬γ,
inan
Infor
,M
w
So
θinM,
∈
α♦
for some
S.
But
given
M,
σ|ϕw
∈M,
�
χ
and
then
Tis,,αw
,retained
wv|ϕTM,
�
χ
vin
|ϕin
σM
∈some
S,
which
case
aM
to
¬�β
M
,�β
whence
M
and σ ∈σS,∈
in which
case
witness
to
¬�β
is
retained
in
, M,
whence
M
θ�M,
.have
3contradiction.
3 θ�
M,either
w � θathen
, inand
case
,w
�S.
θa
,witness
or
wθ�
♦ (σ
∧�¬β)
for
some
|ϕ
|ϕ
is,which
unsuccessful.
is unsuccessful. which
case M|ϕ
v � ¬γ, a contradiction.
We conclude that M|ϕ✷
, w � ϕ, whence �ϕ
In
either
case,
M
,
w
�
θ,
a
contradiction.
So
θ
∈
T
.
Moreover,
for
In
either
case,
M
,
w
�
θ,
a
contradiction.
So
θ
∈
T
.
Moreover,
for
every
♦γ
in
be
a
such
that
M
,
v
�
γ.
Since
v
was
retai
|ϕ to
σ∈
S,must
inawhich
case
av
witness
¬�β
is
retained
in
M
,
whence
M
,
w
�
θ
.
which case M|ϕθ,,isand
v|ϕthere
�
¬γ,
contradiction.
which
case
M
We
,
v
conclude
�
¬γ,
a
contradiction.
that
M
,
w
�
We
ϕ,
wh
co
|ϕ
|ϕ
|ϕ
|ϕ
|ϕ
unsuccessful.
✷
a vv was
suchretained
that So
M|ϕ
�
Since
M
θ, there must beIn
a veither
such case,
that θ,
M
,,vw�must
γ.θ,Since
in
, Moreover,
M,
v �vσwas
|ϕ
Mthere
�But
abecontradiction.
θ ,∈vM
T|ϕ.γ.
forretained
every
♦γin
in
|ϕ
for
some
σ
∈
S.
then
given
M,
w
�
χ
and
θ
∈
T
,
w
is unsuccessful.
is
unsuccessful.
3
for aM,
some
But
then
given
M,
w
�
χ
and
θ
∈
T
,
we
have
M
for some σ ∈ S.θ,But
then
given
w σ� ∈
χ3S.and
θ
∈
T
,
we
have
M,
v
�
¬γ,
in
3
there
must
be
v such
that
M
,
v
�
γ.
Since
v
was
retained
in
M
,
M,
v
�
σ
|ϕ
|ϕ
which
,case
vthen
�We
contradiction.
We
conclude
which
M¬γ,
, v �aM,
¬γ,
a �contradiction.
conclude
that
Mthat
which case M|ϕ
,for
v �some
¬γ,case
aσcontradiction.
that
, w θ� ∈
ϕ, TWe
whence
ϕ M,
|ϕconclude
|ϕ , win�M
∈M
S. |ϕ
But
given
w
χM
and
,
we
have
v
�
¬γ,
3 |ϕ
19
is ,unsuccessful.
is unsuccessful.iswhich
case M|ϕ
v � ¬γ, a contradiction. We conclude that M|ϕ✷, w � ϕ, whence ϕ
unsuccessful.
is unsuccessful.
✷
19
19
19
19
literals,
one
from
each
Γformula
in A,
that
Σ does
Definition
specification
"C,l N,
A#
and
ϕquazi-negation
innot
DNF,
defin
and M
every, 6.
setpGiven
Γ ∈ A,a there
is aa disjunct
literal
Γ ϕthat
is
not
thesuch
Hence
is
d∈in
contain
pq N
|ϕ w !
!such that d does
!
Hence
Mϕ|ϕ=
,w
! ϕ.
ϕ
ql. ϕ.
p N, A#
p
!N, A#
"C,
N
,literals
A# where
NN
is
the
set of
l.l.
of
l
and
d
does
not
contain
the
quazi-negation
o
"C,
= "C,
N ! , A# where"C,
N
is
the
set
of
all
l
∈
for
which
theaK
literals,
from
each
Γ
in
A, suchof
that
does
Proof.
Ifone
ϕ (q
contains
only
formulas
the Σ
form
∧
♦q
∧
#
→
p)
∧
♦
(q
∧
p)
pp ∧ ♦q ∧ # (q → p) ∧ ♦ (q ∧ p)p
!
isdoes
a disjunct
d in ϕ∼such
that
d does
not
contain
is
a
disjunct
d
in
ϕ
such
that
d
not
contain
l
and
there
is
a
selection
Σ
M
land
and
d#
does
not
the
quazi-negation
pN
∧ ♦q ∧every
# (q
→
p)Γ∧∈♦
(q
∧ 6.
p)N
then
for∧
any
w)that
with
M,
wthe
!form
ϕ,
vand
∈ lW
|o
!!Ldefine
! A#
IfGiven
a"C,
conjunction
in
normal
pof
∧ϕ
(q
→
p)
∧contain
♦
(q
∧
p)
Definition
6.6.
Given
specification
N,
formula
DNF,
Definition
aone
specification
"C,
N,
and
formula
every
Γ
∈
A,
there
is
literal
∈
Γ
set
there
is
a♦q
l(M,
∈
is
not
quazi-negation
literals,
from
each
Γformula
in
A,A#
such
that
does
Definition
Given
aA,
specification
"C,
N,
and
ϕaain
inquazi-negatio
DNF,
defin
N and
and ϕone
every
set
Γ ∈aA,
there
is
aisliteral
literal
l!set
∈A#ΓΓ
that
is
not
the
quazi-negation
N
and
every
set
Γand
∈
A,
there
isϕ
literal
lΣ
∈(ϕ)
Γ ϕtt$=
literals,
from
each
Γ
in
A,
such
that
Σ
does
not
contain
the
ϕ
! ! ! M|ϕ , w ! ϕ.!
Hence
ϕ "C,q
q"C,
pq N,
p
A#
NN!N,
,!contain
A#
NNquazi-negation
set
all
l the
for
which
there
"C,
A#where
=the
"C,
N
A#
Nall
is
set
ofN
all
literals
l(ϕ)
∈
qIf
l.
of
l isand
dwhere
does
contain
quazi-negation
o$=
"C,
A#d =
= "C,
, A#
where
is, the
setofofnot
literals
l∈∈
N
for
which
theN
l.l N,
l.
ϕ
is
athe
inthe
normal
form
and
L
ofl.
and
does
not
of literals
any
literal
in
Σ.
Proof.
Ifconjunction
ϕ contains
formulas
of
the there
form
p
qaadisjunct
pdoes
isis
not
∼∼l lonly
and
Σ
o
isϕϕ
asuch
disjunct
ϕ such
that
d does
not
contain
l!selection
and
iK
disjunctddinin
suchthat
thatdddin
does
notcontain
contain
andthere
thereisisa∼aselection
Σ
M
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
If
ϕ
is
a
conjunction
in
normal
form
and
L
(ϕ)
$=
p
∧
♦q
∧
#
(q
→
p)
∧
♦
(q
∧
p)
!
!
6.
Given
aacontain
specification
"C,
N,
A#
Definition
6.
aainin
specification
N,
A#
and
formula
ϕ
in
DNF,
defi
literals,
one
from
each
A,
such
does
contain
the
Definition
6.N
Given
specification
"C,
N,
A#
and
formula
ϕquazi-negation
in
DNF,
defi
literals,
from
inΣ
A,
such
that
not
contain
th
Definition
6.
Given
specification
"C,
N,
A#
Nliterals,
and every
set
ΓGiven
and
∈ A,Γevery
there
set
isDefinition
aΓeach
literal
∈that
A,
there
lΣ
∈
Γis
that
anot
literal
is
not
lΣthe
∈does
Γthe
quazi-negation
that
is
not
the
oK
q
Proof.
IfΓϕ"C,
ϕ
contains
only
formulas
of
the
form
one
from
each
Γone
A,
such
that
does
not
quazi-negatio
ϕ
!
!
!
!!
!
ϕ
ϕ
Hence
M
,
w
!
ϕ.
! is
!
!
"C,
N,
A#
=
"C,
N
,
A#
where
N
is
the
set
of
a
|ϕ
"C,
N,
A#
=
"C,
N
,
A#
where
N
the
set
of
all
literals
l
∈
N
for
which
th
M
q
q
of
lϕ
does
contain
the
quazi-negation
ofw)
any
literal
Σ.
"C,
N,adA#
= not
"C,
, A#
where
NN,
isA#
the
set
literals
N
for
which
th
of
lN
and
dnormal
does
not
contain
the
quazi-negation
of
literal
"C,
=
"C,
N
,∅,
A#
where
N
isany
the
set
ofin
l.of
l.
then
any
(M,
with
M,
!form
ϕ,
vthe
∈
W
|aK
If
ϕ
isfor
aIf
conjunction
in
normal
and
L
(ϕ)
$=
land
and
dconjunction
does
not
contain
the
quazi-negation
of
any
literal
in
Σ.
is
in
form
and
L
(ϕ)of
$=all
then
ϕwlin
is∈
successful.
Proof.
ϕ contain
contains
only
formulas
of
form
qIfis
pHence
disjunct
dd in
ϕ
such
that
dd does
not
contain
not
∼
ll and
there
is
aa!
selection
Σ
qdd aadoes
is aa disjunct
disjunct dd in
in ϕ
ϕ such
such that
thatis
does
not
contain
∼
and
there
is
selection
Σ
is
disjunct
in
ϕ
such
that
does
not
contain
M
,
w
!
ϕ.
M
|ϕ (M, w) with M, w ! ϕ, v ∈ W
then
for∧
any
|ϕ
pGiven
∧such
♦q
#
(q
→does
p)
∧not
♦formula
(qcontain
∧N,
p)
literals,
one
from
each
Γ
in
A,
such
that
does
Definition
Given
Definition
a specification
6.A,
"C,
athat
N,
A#
and
"C,
A#
ϕ
in
and
DNF,
formula
define
literals,
from
each
ΓΓ formulas
in
that
Σ
the
quazi-negat
! Σ
Proof.
If
ϕspecification
contains
only
formulas
of
the
form
K
literals,
one
from
each
in
A,
such
ΣΓ
does
not
contain
the
quazi-negat
literals,
one
from
each
Γ
in
A,
such
that
Σ
does
Proof.
Ifϕone
ϕ 6.
contains
only
of
the
form
Kα
and
¬Kα
(α
propositiona
N
and
every
set
∈
A,
there
is
a
literal
l
∈
Γ
that
!
ϕ
"literal
!ofNquazi-negation
Hence
!
! ! !M|ϕ , w ! ϕ.!
M
of
l
and
d
does
not
contain
the
M
M
"C,
N,
A#
=
"C,
"C,
N
N,
,
A#
A#
where
=
"C,
N
N
is
,
the
A#
where
set
of
all
N
literals
is
the
set
l
∈
all
for
literals
which
l
there
∈
N
of
l
and
d
does
not
contain
the
quazi-negation
of
any
in
Σ.
M
M
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
|ϕ
If
ϕ
is
a
conjunction
in
normal
form
and
L
(ϕ)
=
$
∅,
then
ϕ
is
successful.
of
l
and
d
does
not
contain
the
quazi-negation
of
any
literal
in
Σ.
If w)
ϕ is
aEpistemic
conjunction
in
normal
(ϕ)
$= W
∅, then| ϕ
is oos
l ϕ,
and
the
∧
♦q
∧and
(q
→not
p)
∧contain
(q
then
any
(M,
with
M,
vd#
∈does
W
|form
wR
v ∧L
=p)
∈
wR
l. wpof!
If
ϕ isfora Phenomena
conjunction
in
normal
form
L (ϕ)
$=
∅,♦and
then
ϕ
isvquazi-negation
successful.
188
Moorean
in
Logic
pdoes
q
isHence
a disjunct
d
in
is
ϕ
a
such
disjunct
that
d
d
in
ϕ
such
not
contain
that
d
does
∼
l
and
not
contain
there
is
∼
a
selection
l
and
there
Σ
oi
Hence
M
,
w
!
ϕ.
|ϕ
M|ϕ , w ! ϕ.
psuch
∧each
♦q
∧Γ#
(qdoes
→ Kα
p)not
∧of
♦
(q¬Kα
∧Σ
p)does
!
Proof.
If
ϕ
contains
only
formulas
of
the
form
and
(α
propositional)
Proof.
If
ϕ
contains
only
formulas
the
form
Kα
and
¬Kα
(
literals,
one
from
literals,
each
Γ
one
in
A,
from
that
Σ
in
A,
such
that
contain
the
quazi-negation
not
contain
th
Definition
6.
Given
specification
Proof. If ϕ contains only formulas
and
(α "C,
propositional
N
and
setform
Γ ∈Kα
A,a !
there
is!a! literal
l N,
∈"A#
Γ that
!of
""¬Kα
!and
p every
! the
M
M
M
M
M
ϕ
|ϕ|ϕ| wRM
|
M
M
!
!
M
M
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
wR
v
=
v
∈
W
then
any
(M,
with
w(ϕ)
!
ϕ,
∈∧W
|of
wR
vliteral
∈
ϕ
is
aav=
conjunction
in
form
and
L
(ϕ)
ofthen
l∧and
does
not
of
l contain
and
d normal
does
the
not
quazi-negation
the
of
any
literal
Σ.
any
ϕ
is
conjunction
and
$=
then
successful.
∧w)
♦q
∧
#
(q
→
p)
∧ ∅,
♦
"C,
A#
"C,
N
A#
where
Nϕ
the
set
of
allvinl$=
for
w)
with
M,
!
ϕ,
∈M,
WL
|,quazi-negation
wR
vnormal
=p)
∈
W
|=wR
If
ϕ
is∧daaany
conjunction
in
normal
form
and
L
(ϕ)
$=
∅,(q
then
ϕinvis
is
successful.
IfN,
ϕform
iscontain
conjunction
in
normal
form
and
L
(ϕ)
$=
pIf
♦q
#
(q(M,
→
p)
∧for
♦in
(q
∧l.
p)wpIf
p
q
Hence
ϕ.ϕ. M|ϕ ,iswa!disjunct
Hence
ϕ.
HenceM
M|ϕ|ϕ, w
, w!!
d in ϕ such that d does not contain ∼ l
If
ϕ
formulas
of
the
form
Proof. If ϕ contains only Definition
formulas
of
the
form
Kα
and
(α
propositiona
Given
aΓonly
specification
N,does
A#
and
formulas
of
the
form
Kα
and
¬Kα
(α"C,
If
ϕ contains
contains
only
formulas
of
the
form
K
!
!
""¬Kα
!! that
pProof.
from
each
in
A,
such
Σ
noK
! 6.
!proposition
qProof.one
p Proof. If ϕ contains only literals,
M
M
M
M
M
ϕ
|ϕ
!
!
M
M
M
M
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
wR
v
=
v
∈
W
|
wR
|ϕ
"C,
N,
A#
=
"C,
N
,
A#
where
N
is
the
set
of
all
l
pIfp∧then
♦q
∧
#
(q
→
p)
∧
♦
(q
∧
p)
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
wR
v
=
v
∈
W
|
wR
p
∧
♦q
∧
#
(q
→
p)
∧
♦
(q
∧
p)
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
ϕ∧ is
ϕ ∧isin♦anormal
conjunction
indoes
normal
L not
(ϕ) $=
form
∅, then
and
ϕ
L quazi-negation
(ϕ)
is successful.
$= ∅, then ϕofisas
and dand
contain
the
♦qa∧conjunction
# (q →If p)
(q
∧ofp)l form
qdisjunct
M
,, w
ϕ.
Hence
,Fig.
w !3. ϕ.
is aHence
Hence M
M|ϕ
Hence
Md|ϕ
wϕ!
!such
ϕ. that d does not contain ∼ l
|ϕ , w ! ϕ.
|ϕin
Proof.
If
ϕ
contains
Proof.
only
If
ϕ
formulas
contains
of
only
the
form
formulas
Kα
and
¬Kα
form
propositional)
and
¬Kα
literals,
one
such
that
Σ "does
no(
ppq
p
! fromMeach Γof
!inthe
"A,
!M(αKα
!M
q
M
Mthe
M|ϕ
|
For KD5 consider ψ ≡
¬q
∨
(p
∧
♦♦q)
and
the
model
M
in
Figure
4.
Only
∧
(q
→
p)
∧
♦
∧
then
then
w)p)
for
M,
(M,
wp)
ϕ,♦q
with
v∧
∈#
W
w |!
wR
ϕ,
=
Wp)
v∈
| wR
W
v (ϕ)
|=
wR
pp ∧
♦q
∧
(q
∧
(q
∧
lpp!and
dconjunction
contain
quazi-negation
of
∧for
♦qany
∧#
#(M,
(q →
→
p)with
∧♦
♦any
(q of
∧
p)
∧w)
∧does
#M,
(qnot
→
p)
∧normal
♦vv(q
(q∈the
∧
p)
If
ϕ
is
a♦q
in
form
and L
$=van
∅∈
right point is eliminated in
M
,
so
ψ
becomes
false
at
the
left
point.
|ψ M|ϕ , w !Hence
Hence
ϕ.
M|ϕ , w ! ϕ.
pp If ϕ contains only formulas of the form Kα
p
Proof.
qq p
q
!
M
p ∧ ♦q ∧ # (q → pp)∧∧♦q♦ ∧
(q#∧(q
p)
→
p)a∧any
♦ (q(M,
∧ p)w) with
then
M, w !form
ϕ, and
v∈W
| $=
wR
If
ϕ isfor
conjunction
in normal
L (ϕ)
∅
Hence M|ϕ , w ! ϕ.
Proof.
Kα
qq If ϕ contains only formulas of
p qq
p
! the form
M
then
for
any
(M,
w)
with
M,
w
!
ϕ,
v
∈
W
|
wR
p ∧ ♦q ∧ # (q → p) ∧ ♦ (q ∧ p)
Fig. 4.
Hence M|ϕ , w ! ϕ.
q in KD45 to the
q normal form
p ¬q ∨ (p ∧ ♦q), which is not
The formula ψ is equivalent
p
Moorean, since neither disjunct is a Moorean conjunction.∧5 ♦q ∧ # (q → p) ∧ ♦ (q ∧2p)
There are other formulas that one may wish to categorize
as Moorean, perhaps even
pq
as Moore sentences, but which are inconsistent with KD45.
Example 4.3 The formula p∧♦¬p is a kind of “higher order” Moore sentence, which
q
is satisfiable on intransitive frames but is also self-refuting over such frames. Similarly,
the formula ♦p ∧ ♦¬♦p is satisfiable yet self-refuting on non-Euclidean frames. The first
formula says that the agent is not aware of what he believes, while the second says that
he is not aware of what he does not believe. These formulas are the very witnesses of a
failure to validate axioms 4 and 5, respectively.
A natural question is whether there are non-Moorean sources of unsuccessfulness in
languages more expressive than the basic modal language. Consider, for example, a
language with multiple modalities, as in multi-agent epistemic logic. Without giving a
formal definition of Moorean in the multi-agent case, it is nonetheless clear that there
are more ways to be Moorean in the multi-agent case than in the single-agent case,
even assuming quasi-partitions for each relation. For example, there are self-refuting,
indirect Moore sentences such as a p ∧ ♦b ¬p, which imply single-agent Moore sentences
(in this case, p ∧ ♦b ¬p) in multi-agent S5. There are also self-refuting, higher order
Moore sentences such as a p ∧ ♦b ¬a p and ♦a p ∧ ♦b ¬♦a p, which resemble the higher
order, single-agent Moore sentences consistent with logics weaker than KD45, noted in
Example 4.3. However, not all unsuccessful formulas in the multi-agent context have a
Moorean character.
Example 4.4 In the model M in Figure 5, the formula ϕ ≡ ♦a p ∧ ♦a ♦b ¬p is true at
the left point but false at the right point, since ♦a p is false at the right point.
5
Note that while the right disjunct is not a Moorean conjunction, in S5 (but not in KD45) it implies
the Moorean conjunction p ∧ ♦q. This is an instance of the more general fact that in S5 some successful
formulas imply unsuccessful formulas.
a, b
a, b
b
p
p
a, b
b
a, b
Wesley H. Holliday and Thomasp F. Icard, III
a, b
b
189
a, b
p
Fig. 5.
As a result, in the model M|ϕ the right point is eliminated, in which case ϕ becomes
false at the left point because ♦a ♦b ¬p becomes false there.
The formula ♦a p ∧ ♦a ♦b ¬p does not resemble any single-agent Moorean formula,
yet it is nonetheless unsuccessful. Given the connection that we have observed between
introspection and Moorean phenomena in the single-agent case, we can see why there
should be non-Moorean unsuccessful formulas in the multi-agent case: agents do not
have introspective access to each other’s knowledge or beliefs.
5
Related Classes of Formulas
Theorem 3.13 shows that in a wide range of logics, every self-refuting formula is a Moore
sentence (i) and every unsuccessful formula is a Moorean sentence (ii). However, neither
the converse of (i) nor the converse of (ii) holds in general. In this section, we relate
the failures of the converses of (i) and (ii) to other classes of formulas. In Theorems A.3
and A.6 in Appendix A, we overcome these failures and give syntactic characterizations
of the self-refuting and unsuccessful formulas as the strong Moore sentences and strong
Moorean sentences, respectively.
For simplicity we assume in this section that we are working with S5 only.
5.1
Informative, Cartesian, and eventually self-refuting formulas
Definition 5.1 A formula ϕ is (potentially) informative iff there is a pointed model
such that M, w ϕ and M|ϕ 6= M. Otherwise ϕ is uninformative. A formula ϕ is
always informative iff for all pointed models such that M, w ϕ, M|ϕ 6= M.
Note that if a formula is not always informative, then it is not self-refuting, for there is
a model such that M, w ϕ but M|ϕ = M, so M|ϕ , w ϕ. This observation explains
some of the counterexamples to the converse of Theorem 3.13(i), Moore sentences that
are not self-refuting.
Example 5.2 The formula (p ∧ ♦¬p) ∨ (q ∧ ♦¬q) is a Moore sentence, but it is not
always informative (take a model with only two connected points, one of which satisfies
p but not q and the other of which satisfies q but not p) and hence not self-refuting. The
formula (p ∧ ♦¬p) ∨ (¬p ∧ ♦p) is a Moore sentence, but it is uninformative and hence
successful, but not self-refuting. These examples show that neither the self-refuting nor
the unsuccessful formulas are closed under disjunction.
From our previous results, we have the following corollary.
Corollary 5.3 A conjunction in normal form is always informative iff it is self-refuting.
190
Moorean Phenomena in Epistemic Logic
Proof. By inspection of the proof of Lemma 3.9(i).
2
However, a Moore sentence that is always informative may not be self-refuting.
Example 5.4 The formula ϕ ≡ (p ∧ ♦¬p) ∨ (p ∧ q ∧ ♦¬q) is always informative, for if
M, w (p ∧ ♦¬p), then there is a witness to ♦¬p that is eliminated in M|ϕ , and if
M, w (p ∧ q ∧ ♦¬q), then either the witness to ♦¬q does not satisfy (p ∧ ♦¬p), in
which case it does not satisfy ϕ and is eliminated in M|ϕ , or it does satisfy (p ∧ ♦¬p),
in which case there is a witness to ♦¬p that is eliminated in M|ϕ . In either case,
M|ϕ 6= M, so ϕ is always informative. However, ϕ is not self-refuting, as shown by the
partition model M with W = {w, v, u}, V (p) = {w, v}, and V (q) = {w, u}. We have
M, w p ∧ q ∧ ♦¬q, and given W|ϕ = {w, v}, also M|ϕ , w p ∧ q ∧ ♦¬q.
Interestingly, the formula ϕ is self-refuting within two steps; given M, w ϕ and
M|ϕ , w ϕ, we have M|ϕ |ϕ , w 2 ϕ. Let δ1 ≡ (p ∧ ♦¬p) and δ2 ≡ (p ∧ q ∧ ♦¬q)
be the disjuncts of
ϕ. All ¬p-points are eliminated in M|ϕ , so M|ϕ ¬δ1 . Hence
M|ϕ |ϕ = M|ϕ |δ . But since M|ϕ , w δ2 and δ2 is self-refuting, M|ϕ |δ , w 2 δ2 .
2
2
Since δ1 is existential, M|ϕ |δ , w 2 δ1 , so we conclude that M|ϕ |ϕ , w 2 ϕ.
2
Example 5.4 points to the interest of self-refutation “in the long run.”
Definition 5.5 Given
a model M, we define M|n ϕ recursively by M|0 ϕ = M,
M|n+1 ϕ = M|n ϕ |ϕ . A formula ϕ is self-refuting within n steps iff there is an n
such that for all pointed models, if M, w ϕ, then M|m , w 2 ϕ with m ≤ n; ϕ is
eventually self-refuting iff for all pointed models, if M, w ϕ, then there is an n such
that M|n , w 2 ϕ.
Using Definition 5.5, we can generalize Corollary 5.3. First, recall Fitch’s paradox,
mentioned in Section 1, which we can restate as: if all truths are knowable, then all
truths are (already) known. One proposal for avoiding the paradox [19] is to restrict the
claim that all truths are knowable to the claim that all Cartesian truths are knowable,
in which case it does not follow that all truths are known.
Definition 5.6 ϕ is Cartesian iff ϕ is satisfiable. 6
The class of formulas ϕ for which ϕ is satisfiable seems a natural object of study in its
own right. The following proposition establishes a connection between these formulas,
defined “statically,” and the other formulas classes that we have defined “dynamically”
in terms of model transformations. It also generalizes Corollary 5.3.
Proposition 5.7 The following are equivalent:
(i) ϕ is always informative.
(ii) ϕ is not Cartesian.
(iii) ϕ is eventually self-refuting.
6
The term ‘Cartesian’ is due to Tennant [19], though his definition is in terms of consistency rather
than satisfiability. The term ‘knowable’ seems more natural, but ‘knowable’ is used in a different sense
in the literature [9], for what we call ‘learnable’ in Definition 5.13.
Wesley H. Holliday and Thomas F. Icard, III
191
Proof. Suppose ϕ is always informative and M, w ϕ. Without loss of generality,
assume M is connected. Since ϕ is always informative, M|ϕ 6= M. Hence there is a
point v such that wRv and M, v 2 ϕ, in which case M, w 2 ϕ. Since M was arbitrary,
ϕ is unsatisfiable, so ϕ is not Cartesian.
Suppose ϕ is not Cartesian, and without loss of generality, assume ϕ is in normal form. We prove by induction that for all n ≥ 0, if M|n ϕ , w ϕ, then there are
♦γ1 , ..., ♦γn+1 distinct subformulas of ϕ with M|n+1 ϕ ¬♦γ1 ∧ ... ∧ ¬♦γn+1 (∗). It
follows that since ϕ has some finite number n of distinct ♦γk subformulas, we must
have M|m ϕ , w 2 ϕ for some m ≤ n. For the base case, assume M, w ϕ. Since ϕ
is not Cartesian, M|ϕ , w 2 ϕ. Let v be such that wRv and M|ϕ , v 2 ϕ, and note
that M, v ϕ, for otherwise v would not have been retained in M|ϕ . From the fact
that universal formulas are preserved under submodels, it follows that for some ♦γ1
in ϕ, we have M, v ♦γ1 but M|ϕ , v 2 ♦γ1 . Hence M|ϕ ¬♦γ1 by Lemma 3.8(i).
For the inductive step, assume that M|n+1 ϕ , w ϕ. Since w is retained in M|n+1 ϕ ,
we have M|n ϕ , w ϕ, so by the induction hypothesis there are ♦γ1 , ..., ♦γn+1 distinct
subformulas of ϕ for which (∗) holds. By the same reasoning as before, since ϕ is Cartesian, M|n+2 ϕ , w 2 ϕ, so there is some z with wRz and some ♦γn+2 in ϕ such that
M|n+1 ϕ , z ♦γn+2 but M|n+2 ϕ ¬♦γn+2 . Moreover, since ♦γk formulas are preserved
under extensions, M|n+2 ϕ ¬♦γ1 ∧...∧¬♦γn+1 as well. Finally, given M|n+1 ϕ , z ♦γn+2
and (∗), ♦γn+2 is distinct from ♦γ1 , ..., ♦γn+1 .
Suppose ϕ is not always informative. Then there is a model with M, w ϕ and
M|ϕ = M, in which case M|n ϕ = M for all n, so ϕ is not eventually self-refuting. 2
Corollary 5.8 ϕ is eventually self-refuting iff it is self-refuting within n steps, with n
bounded by the number of distinct diamond formulas in a normal form of ϕ.
Proof. By inspection of the proof of the previous proposition.
5.2
2
Successful, super-successful, and learnable formulas
The converse of Theorem 3.13(ii) fails because a formula ϕ in normal form that contains
a Moorean conjunction as a disjunct may be successful. For example, it is easy to
see that if the disjunction of the non-Moorean conjunctions in ϕ is a consequence of
the disjunction of the Moorean conjunctions in ϕ, then ϕ is not only successful but
super-successful, given Lemma 3.11. Examples of such ϕ include (p ∧ ♦¬p) ∨ p and
(p ∧ ♦¬p)∨♦¬p. There are also successful formulas ψ that contain Moorean conjunctions
as disjuncts, but do not meet the condition of ϕ. These formulas nevertheless manage
to be successful by a kind of compensation: when one disjunct of ψ goes from true at
M, w to false at M|ψ , w, another disjunct compensates by going from false at M, w to
true at M|ψ , w. The formula (p ∧ ♦¬p) ∨ p exhibits this kind of compensation against
a Moore conjunction, while the formula in the proof of the following proposition does
so against a Moorean conjunction.
Proposition 5.9 Not all successful formulas are super-successful.
Proof. The formula ϕ ≡ (p ∧ ♦q) ∨ p is successful but not super-successful. Suppose
M, w ϕ, and without loss of generality assume that M is connected. Case 1: M, w 192
Moorean Phenomena in Epistemic Logic
p. Then M|ϕ = M, so M|ϕ , w ϕ. Case 2: M, w 2 p. Then M|ϕ = M|p∧♦q ,
and given M|p∧♦q p, we again have M|ϕ , w ϕ. Hence ϕ is successful. But
ϕ is not super-successful, as shown by the partition model N with W = {w, v, u},
V (p) = {w}, and V (q) = {u}. We begin with N , w ϕ, for while N , w 2 p, it holds
that N , w p ∧ ♦q. Moreover, given W|ϕ = {w}, we still have N|ϕ , w ϕ, for while
N|ϕ , w 2 p ∧ ♦q, it holds that N|ϕ , w p. However, in the extension N 0 of N|ϕ with
W 0 = {w, v}, we now have N 0 , w 2 ϕ. The fact that u ∈
/ W 0 gives N 0 , w 2 p ∧ ♦q, and
0
0
the fact that v ∈ W gives N , w 2 p.
2
Proposition 5.10 A formula ϕ is super-successful iff for a propositional variable p that
does not occur in ϕ, ϕ ∨ p is successful.
Proof. (⇒) Suppose ϕ is super-successful and M, w ϕ ∨ p, where p does not occur in
ϕ. If M, w p, then M|ϕ∨p , w p, and if M, w ϕ, then given M|ϕ ⊆ M|ϕ∨p and the
assumption that ϕ is super-successful, M|ϕ∨p , w ϕ. In either case, M|ϕ∨p , w ϕ ∨ p,
so ϕ ∨ p is successful.
(⇐) Suppose ϕ is not super-successful, so there is an M = hW, R, V i with w ∈ W
such that M, w ϕ, and an M0 = hW 0 , R0 , V 0 i such that M|ϕ ⊆ M0 ⊆ M and
M0 , w 2 ϕ. Let M∗ = hW, R, V ∗ i be the same as M except that V ∗ (p) = W 0 \ W|ϕ .
Then since p does not occur in ϕ, we have M∗ , w ϕ ∨ p given M, w ϕ. Moreover,
∗
since W|ϕ∨p
= W 0 , we have M∗|ϕ∨p , w 2 ϕ given M0 , w 2 ϕ. From the assumption that
M, w ϕ, it follows that w ∈ W|ϕ , in which case w ∈
/ V ∗ (p) and hence M∗|ϕ∨p , w 2 p.
∗
But then M|ϕ∨p , w 2 ϕ ∨ p, so ϕ ∨ p is unsuccessful.
2
By Proposition 5.10, complexity and syntactic characterization results for successful
formulas carry over immediately to super-successful formulas.
From the previous propositions we obtain a surprising failure of closure.
Corollary 5.11 The set of successful formulas is not closed under disjunction.
Proof. Immediate from Propositions 5.9 and 5.10.
2
Example 5.12 The formula ¬p ∧ ¬q is successful, and by the proof of Proposition
5.9, (p ∧ ♦q) ∨ p is also successful. However, the disjunction of these formulas, χ ≡
(¬p ∧ ¬q) ∨ ((p ∧ ♦q) ∨ p) is unsuccessful, as shown by the model N in the proof of
Proposition 5.9. We begin with N , w χ, since while N , w 2 ¬p ∧ ¬q (and hence
N|χ , w 2 ¬p ∧ ¬q), we have already seen that N , w (p ∧ ♦q) ∨ p. However, N|χ = N 0 ,
and we have already seen that N 0 , w 2 (p ∧ ♦q) ∨ p.
By contrast, the set of super-successful formulas is closed under disjunction, since M|ϕ∨ψ
is an extension of M|ϕ and M|ψ .
Another consequence of the previous results concerns the relation between successful
and learnable formulas. Like the Cartesian formulas, the learnable formulas have been
discussed in connection with Fitch’s paradox [3,9].
Definition 5.13 A formula ϕ is (always) learnable iff for all pointed models, if M, w Wesley H. Holliday and Thomas F. Icard, III
193
ϕ, then there is some ψ such that M|ψ , w ϕ. 7
Formulas that are learnable (and satisfiable) are Cartesian according to Definition
5.6, but the converse does not hold [3]. For example, the formula p ∧ ♦q is Cartesian,
since (p ∧ ♦q) is satisfiable, but it is not (always) learnable, for if M, w (p → ¬q)
and M0 ⊆ M, then M0 , w 2 (p ∧ ♦q).
All successful formulas are learnable [9], since if M, w ϕ and ϕ is successful, then
not only M|ϕ , w ϕ but also M|ϕ , w ϕ. For if v is retained in M|ϕ , then M, v ϕ,
in which case M|ϕ , v ϕ by the successfulness of ϕ. Hence for a successful ϕ, we can
always take ψ in Definition 5.13 to be ϕ itself. On the other hand, it is natural to ask
whether some unsuccessful formulas are learnable as well.
Corollary 5.14 Not all learnable formulas are successful.
Proof. Let δ1 ∨ δ2 be an unsuccessful disjunction with δ1 and δ2 successful, as given
by Corollary 5.11. If M, w δ1 ∨ δ2 , then M, w δi for i = 1 or i = 2. Since δi is
successful, we have M|δi , w δi , in which case M|δi , w (δ1 ∨ δ2 ). Since M was
arbitrary, δ1 ∨ δ2 is (always) learnable.
2
6
Conclusion
From a technical point of view, we have studied the question of when a modal formula is
preserved under relativizing models to the formula itself. In the epistemic interpretation
of modal logic, this preservation question takes on a new significance: it concerns whether
an agent retains knowledge of what is learned, which requires that what is learned
remain true. We have shown (Theorem 3.13) that for an introspective agent, the only
true sentences that may become false when learned are variants of the Moore sentence.
For an agent without introspection or multiple agents without introspective access to
each other’s knowledge or beliefs, there are non-Moorean sources of unsuccessfulness
(Propositions 4.1 and 4.2, Example 4.4).
In connection with our study of Moorean phenomena, we have observed a number of
related results. We saw that the sentences that always provide information to an agent,
no matter the agent’s prior epistemic state, are exactly those sentences that cannot be
known—and will eventually become false if repeated enough (Proposition 5.7); we saw
that there are sentences that always remain true when they are learned, but whose truth
value may oscillate while an agent is on the way to learning them (Proposition 5.9); and
we saw that there are sentences that sometimes become false when learned directly, but
which an agent can always come to know indirectly by learning something else (Corollary
5.14).
In Appendix A, we return to the problem with which we began. We give syntactic
characterizations of the self-refuting and unsuccessful formulas as the strong Moore
sentences and strong Moorean sentences, respectively.
7
The term ‘learnable’ is used by van Benthem [3]. Balbiani et al. [9] use the term ‘knowable’.
194
Moorean Phenomena in Epistemic Logic
Acknowledgement
We wish to thank Hans van Ditmarsch and the anonymous referees for their helpful
comments on an earlier draft of this paper, and Johan van Benthem for stimulating our
interest in the topic of successful formulas.
References
[1] Baltag, A., L. Moss and S. Solecki, The logic of public announcements, common knowledge and
private suspicions, in: I. Gilboa, editor, Proceedings of the 7th Conference on Theoretical Aspects
of Rationality and Knowledge (TARK 98), Morgan Kaufmann, 1998 pp. 43–56.
[2] Baltag, A., H. van Ditmarsch and L. Moss, Epistemic logic and information update, in: P. Adriaans
and J. van Benthem, editors, Philosophy of Information, North-Holland, 2008 pp. 361–456.
[3] van Benthem, J., What one may come to know, Analysis 64 (2004), pp. 95–105.
[4] van Benthem, J., One is a lonely number: on the logic of communication, in: Z. Chatzidakis,
P. Koepke and W. Pohlers, editors, Logic Colloquium ’02, ASL & A.K. Peters, 2006 pp. 96–129.
[5] van Benthem, J., Open problems in logical dynamics, in: D. Gabbay, S. Goncharov and
M. Zakharyashev, editors, Mathematical Problems from Applied Logic I, Springer, 2006 pp. 137–192.
[6] van Ditmarsch, H. and B. Kooi, The secret of my success, Synthese 151 (2006), pp. 201–232.
[7] van Ditmarsch, H., W. van der Hoek and B. Kooi, “Dynamic Epistemic Logic,” Springer, 2008.
[8] Andréka, H., I. Németi and J. van Benthem, Modal languages and bounded fragments of predicate
logic, Journal of Philosophical Logic 27 (1998), pp. 217–274.
[9] Balbiani, P., A. Baltag, H. van Ditmarsch, A. Herzig, T. Hoshi and T. de Lima, ‘Knowable’ as
‘known after an announcement’, The Review of Symbolic Logic 1 (2008), pp. 305–334.
[10] Carnap, R., Modalities and quantification, The Journal of Symbolic Logic 11 (1946), pp. 33–64.
[11] Fitch, F. B., A logical analysis of some value concepts, The Journal of Symbolic Logic 28 (1963),
pp. 135–142.
[12] Gerbrandy, J., “Bisimulations on Planet Kripke,” Ph.D. thesis, University of Amsterdam (1999),
ILLC Dissertation Series DS-1999-01.
[13] Hintikka, J., “Knowledge and Belief: An Introduction to the Logic of the Two Notions,” College
Publications, 2005.
[14] Lutz, C., Complexity and succinctness of public announcement logic, in: P. Stone and G. Weiss,
editors, Proceedings of the Fifth International Joint Conference on Autonomous Agents and
Multiagent Systems (AAMAS 06) (2006), pp. 137–143.
[15] Meyer, J.-J. Ch. and W. van der Hoek, “Epistemic Logic for AI and Computer Science,” Cambridge
Tracts in Theoretical Computer Science 41, Cambridge University Press, 1995.
[16] Moore, G. E., A reply to my critics, in: P. Schilpp, editor, The Philosophy of G.E. Moore, The
Library of Living Philosophers 4, Northwestern University, 1942 pp. 535–677.
[17] Plaza, J., Logics of public communications, in: M. Emrich, M. Pfeifer, M. Hadzikadic and Z. Ras,
editors, Proceedings of the 4th International Symposium on Methodologies for Intelligent Systems:
Poster Session Program, Oak Ridge National Laboratory, ORNL/DSRD-24, 1989 pp. 201–216.
[18] Qian, L., Sentences true after being announced, in: Proceedings of the Student Session of the
1st North American Summer School in Logic, Language, and Information (NASSLLI), Stanford
University, 2002.
[19] Tennant, N., “The Taming of The True,” Oxford: Clarendon Press, 1997.
Wesley H. Holliday and Thomas F. Icard, III
195
[20] Visser, A., J. van Benthem, D. de Jongh and G. R. R. de Lavalette, NNIL, A study in intuitionistic
propositional logic, Logic Group Preprint Series 111, Philosophical Institute, Utrecht University
(1994).
[21] Zeman, J. J., A study of some systems in the neighborhood of S4.4, Notre Dame Journal of Formal
Logic 12 (1971), pp. 341–357.
A
Appendix
In this appendix, we address the problem of giving syntactic characterizations of the
successful and self-refuting formulas. Given Theorem 2.6, Lemma 3.6, and the PAL definitions of successful and self-refuting, there is a trivial characterization of both classes: ϕ
is successful iff the result of reducing the PAL formula ¬[ϕ]ϕ to normal form in the basic
modal language is not clear; ϕ is self-refuting iff the result of reducing the PAL formula
¬[ϕ]¬ϕ to normal form in the basic modal language is not clear. Although by Proposition 2.7, the PAL definitions of successful and self-refuting lead to optimal methods
for checking for these properties, they do not provide much insight beyond the semantic
definitions of the formula classes. In Theorems A.3 and A.6 below, we give syntactic
characterizations of the self-refuting and unsuccessful formulas that reveal more than
the trivial characterization.
We state the following results for S5 with the standard definitions of successful
(∀M, w : M, w ϕ ⇒ M|ϕ , w ϕ) and self-refuting (∀M, w : M, w ϕ ⇒ M|ϕ , w 2
ϕ). With minor changes the results also hold for KD45, given the modified definitions
using precondition ♦+ ϕ instead of ϕ.
We will continue to use the abbreviations δ α , δ ♦ , etc., from Definition 3.3. We will
also use δ̂ ♦ to denote a conjunct of δ ♦ , i.e., a βi or ♦γk in δ. As before, ∼ ϕ is the
negation of ϕ in negation normal form, with ¬ applying only to literals.
Definition A.1 ϕ is a strong Moore sentence iff for any normal form ϕ∗ of ϕ, no disjunct
of ϕ∗ is compensated in ϕ∗ . A disjunct δ ≡ α ∧ β1 ∧ ... ∧ βn ∧ ♦γ1 ∧ ... ∧ ♦γm of ϕ∗
is compensated in ϕ∗ iff there is a disjunct δ 0 of ϕ∗ , a subset S of the disjuncts of ϕ∗ , a
sequence of disjuncts σγ1 , ..., σγm (not necessarily distinct) from S, and for every σ ∈
/S
a conjunct σ̂ ♦ of σ ♦ , such that χ ≡ δ 0 ∧ δ α ∧ χ ∧ χ♦ is clear, where
χ ≡
^
σ∈S
σ ♦ ∧
^
σ ∈S
/
χ♦ ≡
∼ σ̂ ♦ ∧
^
k≤m
^
σ∈S,j≤n
(∼ σ α ∨ βj ) ;
♦ σγαk ∧ γk .
Note that χ is in normal form, so the definition of clarity properly applies. The
reason for the use of ∼ σ̂ ♦ is that ∼ σ ♦ may be a disjunction, in which case χ would
not be in normal form. Hence we pull the disjunction out of the formula and add an
existential quantifier to the condition, since σ ♦ is false iff there is some conjunct σ̂ ♦
of σ ♦ that is false.
As we will see in the proof of Theorem A.3, the existence of a compensated disjunct
in ϕ is equivalent to there being a model in which ϕ has a “successful update,” in the
196
Moorean Phenomena in Epistemic Logic
sense that M, w ϕ and M|ϕ , w ϕ. For ϕ to have a successful update in M, there
must be a disjunct δ 0 of ϕ true at a point w in M and a disjunct δ of ϕ (possibly distinct
from δ 0 ) true at w in M|ϕ . What is required for δ to be true at w in M|ϕ is that its
propositional part δ α is true at w in M, that its universal part δ is already true at w
in M or becomes true at w in M|ϕ , and that its existential part δ ♦ remains true at w
in M|ϕ . This is exactly what χ captures.
Proposition A.2 Every strong Moore sentence is a Moore sentence.
Proof. By Definitions A.1 and 3.12, it suffices to show that for ϕ in normal form, if
a disjunct δ of ϕ is not compensated in ϕ, then δ is a Moore conjunction. To prove
the contrapositive, suppose δ is not a Moore conjunction. Then by Definition 3.7,
δ ∧ ♦δ α is clear and δ ∧ ♦ (δ α ∧ γk ) is clear for every ♦γk conjunct in δ. It follows that
V
V
δ∧
♦ (δ α ∧ γk ) is clear and hence satisfiable. Given M, w δ ∧
♦ (δ α ∧ γk ), let
k≤m
k≤m
V ♦
V
σ ∧
¬σ ♦ . For every σ ∈
/ S,
S be the set of disjuncts in ϕ such that M, w σ∈S
σ ∈S
/
V ♦
V
pick a conjunct σ̂ ♦ of σ ♦ such that M, w σ
∧
∼ σ̂ ♦ . Since δ ∈ S, let
σ∈S
σ ∈S
/
the sequence σγ1 , ..., σγm of disjuncts from S be such that σγi ≡ δ for 1 ≤ i ≤ m. We
claim that for the χ as in Definition A.1 based on these choices, M, w χ. From the
V
fact that M, w ♦ (δ α ∧ γk ) and our choice of σγ1 , ..., σγm , we have M, w χ♦ .
k≤m
V
Since M, w δ , it is immediate that M, w (∼ σ ∨ βj ). Together with
σ∈S,j≤n
V ♦
V
M, w σ
∧
∼ σ̂ ♦ , this gives M, w χ . Finally, setting δ 0 ≡ δ, we have
σ∈S
σ ∈S
/
M, w χ. Hence χ is clear, so δ is compensated in ϕ.
2
We can now generalize Theorem 3.13(i).
Theorem A.3 ϕ is self-refuting if and only if ϕ is a strong Moore sentence.
Proof. ϕ is self-refuting iff any equivalent normal form of ϕ is self-refuting, so let us
assume ϕ is already in normal form.
(⇐) We prove the contrapositive. Suppose ϕ is not self-refuting, so there is a pointed
model such that M, w ϕ and M|ϕ , w ϕ. Given M|ϕ , w ϕ, there is a disjunct δ
of ϕ such that M|ϕ , w δ. We claim that δ is compensated in ϕ, so ϕ is not a strong
Moore sentence. It suffices to show the satisfiability of an appropriate χ as in Definition
A.1.
Since M, w ϕ, there is a disjunct δ 0 of ϕ such that M, w δ 0 . This gives the first
conjunct of χ. Since M|ϕ , w δ, we have M, w δ α because δ α is propositional and
hence preserved under extensions. This gives the second conjunct of χ.
Next we claim M, w χ . Let S be the set of the disjuncts of ϕ such that M, w V ♦ V
σ ∧
¬σ ♦ , and let σ̂ ♦ be a false conjunct of each σ ♦ for σ ∈
/ S. For reductio,
σ∈S
σ ∈S
/
V
suppose M, w 2
(∼ σ α ∨ βj ), where β1 ∧ ... ∧ βn ≡ δ . Then there is some
σ∈S,j≤n
v with wRv and M, v σ α ∧ ¬βj for some σ ∈ S and j ≤ n. Since σ ∈ S, we have
M, w σ ♦ . It follows by Lemma 3.8(i) that M, v σ ♦ , in which case M, v σ
Wesley H. Holliday and Thomas F. Icard, III
197
given M, v σ α . Hence v is retained in M|ϕ . But then given M|ϕ , v 2 βj , we have
M|ϕ , w 2 βj , which contradicts the assumption that M|ϕ , w δ. We conclude that
M, w χ , which gives the third conjunct of χ.
Finally, we claim M, w χ♦ . Given M|ϕ , w δ ♦ , take an arbitrary ♦γk in δ, and
let v be such that wRv and M|ϕ , v γk . It follows that M, v σ for some disjunct σ
of ϕ, which we label as σγk , for otherwise v would not be retained in M|ϕ. Since γk is
propositional, M, v γk given M|ϕ , v γk . Therefore M, w ♦ σγαk ∧ γk . Then from
the fact that M, v σγ♦
, we have M, w σγ♦
by Lemma 3.8(i), so σγk ∈ S. Since ♦γk
k
k
was arbitrary, M, w χ♦ , which gives the final conjunct of χ.
(⇒) Again we prove the contrapositive. Suppose δ is not a strong Moore sentence,
so there is some δ that is compensated in ϕ, for which an appropriate χ as in Definition
A.1 is clear. Where M, w χ, we claim that M|ϕ , w δ. We will show M|ϕ , w δ α ,
M|ϕ , w δ , and M|ϕ , w δ ♦ separately.
Given M, w δ, we have M|ϕ , w δ α since δ α is propositional.
Next, for any v retained in M|ϕ , we have M, v σ α for some disjunct σ of ϕ. It
must be that σ ∈ S, for otherwise M ¬σ given M, w χ and Lemma 3.8(i). Then
V
from the fact that M, w (∼ σ α ∨ βj ), we have M|ϕ , v βj for all j ≤ n.
σ∈S,j≤n
Since v was arbitrary, M|ϕ , w δ .
Finally, given M, w χ♦ , for any ♦γk in δ we have M, w ♦ (σγk ∧ γk ) with σγk ∈ S.
Let v be such that wRv and M, v σγαk ∧ γk . Since σγk ∈ S, M, w σγ♦
. It follows
k
♦
by Lemma 3.8(i) that M, v σγk , in which case M, v σγk given M, v σγαk . Hence
v is retained in M|ϕ . Since γk is propositional, we have M|ϕ , v γk given M, v γk ,
whence M|ϕ , w ♦γk . Since γk was arbitrary, M|ϕ , w δ ♦ .
We conclude that M|ϕ , w δ, in which case M|ϕ , w ϕ. Given M, w χ, we have
M, w δ and hence M, w ϕ, so ϕ is not self-refuting.
2
Finally, we will prove an analogous generalization of Theorem 3.13(ii).
Definition A.4 ϕ is a strong Moorean sentence iff for any normal form ϕ∗ of ϕ, there
is a disjunct δ and non-empty sets S and T of disjuncts of ϕ∗ , with for every θ ∈ T , a
♦γθ in θ, such that χ ≡ δ ∧ χ1 ∧ χ2 ∧ χ3 is clear, where
t(θ) ≡ θα♦ ∧
χ1 ≡
^
θ∈T
t (θ) ∧
^
θ ∈T
/
χ3 ≡
^
σ∈S, β in θ
(∼ σ α ∨ β) ;
¬t (θ) ; χ2 ≡
^
σ∈S,θ∈T
^
σ∈S
σ ♦ ∧
^
σ ∈S
/
¬σ ♦ ;
(∼ σ α ∨ ∼ γθ ) .
The formula χ is not yet in normal form, due to the ¬t(θ) conjuncts in χ1 and the
¬σ ♦ conjuncts in χ2 , so strictly the definition of clarity does not apply. However, it
is straightforward to put χ into normal form using the same method involving σ̂ ♦ as
in Definition A.1, together with some distribution of ∧ and ∨. Since in this case the
necessary modifications add four existential quantifiers to the definition, for simplicity
198
Moorean Phenomena in Epistemic Logic
we do not write them out. When we say that χ is clear, strictly we mean that the
modified formula is clear.
As we will see in the proof of Theorem A.6, the clarity of χ is equivalent to there
being a model in which ϕ has an unsuccessful update. For ϕ to have an unsuccessful
update in M, there must be some disjunct δ in ϕ such that M, w δ, but no disjunct
δ 0 such that M|ϕ , w δ 0 . To ensure that there are no such δ 0 , we need only keep track
of those disjuncts θ of ϕ whose propositional part θα and existential part θ♦ are true in
M and whose universal part θ was already true in M or becomes true in M|ϕ , since
all other disjuncts will be false in M|ϕ . This is the purpose of χ1 . For each such θ,
we must ensure that there is a diamond formula ♦γθ in θ that becomes false in M|ϕ ,
because none of its witnesses satisfy any of the disjuncts that are satisfied somewhere in
M. This is the purpose of χ3 and χ2 .
Proposition A.5 Every strong Moorean sentence is a Moorean sentence.
Proof. Assume ϕ is a strong Moorean sentence, so an appropriate χ as in Definition
A.4 is clear. Note that the distinguished disjunct δ of ϕ∗ must be a member of both S
and T . Then given that δ ∧ χ3 is clear, we have that δ ∧ (∼ δ α ∨ ∼ γδ ) is clear, where
♦γδ is a conjunct of δ. Hence δ is a Moorean conjunction by Definition 3.7, in which
case ϕ is a Moorean sentence by Definition 3.12.
2
Theorem A.6 ϕ is unsuccessful if and only if ϕ is a strong Moorean sentence.
Proof. As before, let us assume that ϕ is already in normal form.
(⇐) Suppose ϕ is a strong Moorean sentence, so an appropriate χ as in Definition
A.4 is clear. Consider a pointed model such that M, w χ. For reductio, assume
M|ϕ , w ϕ. Then there exists a disjunct θ in ϕ such that M|ϕ , w θ.
We claim that θ ∈ T . For if θ ∈
/ T , then given M, w χ1 there are two cases. Case 1:
M, w 2 θα♦ . Then M|ϕ , w 2 θα♦ since θα♦ is existential. Case 2: M, w ♦ (σ α ∧ ¬β)
for some σ ∈ S and β in θ. Let v be such that wRv and M, v σ α ∧ ¬β. Since σ ∈ S,
we have M, w σ ♦ , in which case M, v σ given Lemma 3.8(i) and M, v σ α . Hence
v is retained in M|ϕ , and since β is propositional, M|ϕ , v 2 β, so M|ϕ , w 2 β and
M|ϕ , w 2 θ . In both cases, M|ϕ , w 2 θ, a contradiction. Therefore θ ∈ T .
Given M|ϕ , w θ, for every ♦γ in θ there is a v with wRv and M|ϕ , v γ. Since
v was retained in M|ϕ , we have M, v σ for some σ ∈ S. Then given M, w χ3 and
θ ∈ T , we have M, v ¬γ. Since γ is propositional, M|ϕ , v ¬γ, a contradiction. We
conclude that M|ϕ , w 2 ϕ, so ϕ is unsuccessful.
(⇒) Suppose ϕ is unsuccessful, so there is a pointed model with M, w ϕ but
M|ϕ , w 2 ϕ. To show that ϕ is a strong Moorean sentence, it suffices to show that
an appropriate χ as in Definition A.4 is satisfiable. Given M, w ϕ, we can read off
from w the disjunct δ and sets S and T such that M, w δ ∧ χ1 ∧ χ2 . It only remains
to show M, w χ3 . For reductio, suppose there is a θ ∈ T such that for all ♦γ in θ,
V
M, w 2
(∼ σ α ∨ ∼ γ), i.e., M, w ♦ (σ α ∧ γ) for some σ ∈ S. Then we claim that
σ∈S
M|ϕ , w θ. As before, we take the three parts of θ separately.
Wesley H. Holliday and Thomas F. Icard, III
199
For θ♦ , consider any ♦γ in θ and let v be such that wRv and M, v σ α ∧ γ. Given
M, w χ2 and the fact that σ ∈ S, we have M, w σ ♦ , so M, v σ by Lemma 3.8(i).
Hence v is retained in M|ϕ . Since M, v γ and γ is propositional, M|ϕ , v γ and
therefore M|ϕ , w ♦γ. Since ♦γ was arbitrary, M|ϕ , w θ♦ .
For θ , take any v retained in M|ϕ such that wRv, and we have M, v σ α for some
σ ∈ S. It follows that for any β in θ, we have M, v β given M, w χ1 . Since β is
propositional, M|ϕ , v β. Since v and β were arbitrary, M|ϕ , w δ .
Finally, for θα , since by assumption θ ∈ T and M, w χ1 , we have M, w θα and
hence M|ϕ , w θα .
We have shown M|ϕ , w θ and hence M|ϕ , w ϕ, which contradicts our initial
assumption. It follows that for every θ ∈ T there is a ♦γθ in θ such that M, w χ3 . 2