Conditionally knowing what

Abstract

Classic epistemic logic focuses on propositional knowledge expressed by " knowing that " operators. However, there are various types of knowledge used in natural language , in terms of " knowing how " , " knowing whether " , " knowing what " , and so on. In [10], Plaza proposed an intuitive know-what operator which was generalized in [16] by introducing a condition. The latter know-what operator can express natural conditional knowledge such as " I know what your password is, if it is 4-digits " , which is not simply a material implication. Essentially this know-what operator packages a first-order quantifier and an S5-modality together in a non-trivial way, thus making it hard to axiomatize. In [16] an axiomatization is given for the single-agent epistemic logic with both know-that and know-what operators, while leaving axiom-atizing the multi-agent case open due to various technical difficulties. In this paper, we solve this open problem. The completeness proof is highly non-trivial, compared to the single-agent case, which requires different techniques inspired by first-order intensional logic.

Full text

Conditionally knowing what
Yanjing Wang Jie Fan
Department of Philosophy
Peking University
{y.wang,fanjie}@pku.edu.cn
Abstract
Classic epistemic logic focuses on propositional knowledge expressed by “knowing
that” operators. However, there are various types of knowledge used in natural lan-
guage, in terms of “knowing how”, “knowing whether”, “knowing what”, and so on.
In [10], Plaza proposed an intuitive know-what operator which was generalized in
[16] by introducing a condition. The latter know-what operator can express natural
conditional knowledge such as “I know what your password is, if it is 4-digits”, which
is not simply a material implication. Essentially this know-what operator packages
a first-order quantifier and an S5-modality together in a non-trivial way, thus mak-
ing it hard to axiomatize. In [16] an axiomatization is given for the single-agent
epistemic logic with both know-that and know-what operators, while leaving axiom-
atizing the multi-agent case open due to various technical difficulties. In this paper,
we solve this open problem. The completeness proof is highly non-trivial, compared
to the single-agent case, which requires different techniques inspired by first-order
intensional logic.
Keywords: knowing what, first-order intensional logic, epistemic logic, conditional
knowledge
1 Introduction
Epistemic Logic (EL), since its birth, has been mainly focusing on reasoning
about propositional knowledge, the knowledge expressed by “iknows that φ”.
However, besides “knowing that”, there are apparently various types of knowl-
edge used in everyday life, expressed by “knowing what” (iknows what dis),
“knowing how” (iknows how to do φ), “knowing whether”(iknows whether φ)
and so on. A natural question which keeps philosophers busy is to ask whether
these types of knowledge can be reduced to propositional knowledge. For ex-
ample, there is a long-lasting debate in philosophy on whether “knowing how”
can be reduced to “knowing that” ever since the seminal work of Ryle [12].
Compared to this heated discussion, “knowing what” received relatively little
attention despite efforts trying to unify “knowing wh-” (what, where, which,
who, why) and “knowing how” in terms of “knowing that” (e.g., [13]).
On the other hand, in computer science and AI, “knowing what” plays an
important role, as argued by McCarthy [9]. In particular, in a security setting,

2 Conditionally knowing what
we need to express that “He knows that she knows [what] her own private key
[is], but he does not know what exactly the key is.” The literal translation
of this sentence in terms of the usual know-that modal operator Kidoes not
work since K1K2p∧ ¬K1pis not consistent in the standard epistemic logic
with Taxiom. This fact has lead a number of authors to propose suitable new
knowledge operators (e.g., [11,6]). In [10], one of the defining works of dynamic
epistemic logic, Plaza introduced a very natural modal operator Kvito express
knowing what in a dynamic epistemic setting. Kvidexpresses exactly that “i
knows what dis”. As for the semantics, Kvidis true on a pointed epistemic
model with world-dependent assignments for diff dhas the same value on all
the epistemically accessible worlds for i. In this setting, it is perfectly possible
that K1Kv2d∧ ¬Kv1dsince two kinds of knowledge are treated differently.
In [16], we generalize the Kvioperator to a conditional one and ob-
tain a complete axiomatization of the single-agent public announcement logic
with both know-that and know-what operators. 1The resulting new formula
Kvi(φ, d) expresses that “agent iknows what dis, given φ.” For example, as it
happens a lot in this internet age, I may forget my own login password for some
website, but I know if the password is 4-digit then it must be 1234, since I have
never used another 4-digit password (though I have several 6-digit passwords).
In such a case people often say “I know my password if it is 4-digit”. This can
be expressed as Kvi(p, d) where pdenotes the proposition that the password is
4-digit. Note that this conditional knowledge is not an implication p→Kvid
nor Ki(p→Kvid). The difference is that, according to the semantics, Kvi(φ, d)
essentially expresses what iwould know if he were informed that φ. This dis-
tinction will become clear when we define the operator formally. In this light,
there is clearly a connection to Public Announcement Logic [10]: to know what
dis given φis similar to knowing what dis after the announcement φ.2
This kind of conditional knowledge has a philosophical connection to the
phenomenon of elusive knowledge studied by Lewis [8]: “Maybe we do know
a lot in daily life; but maybe when we look hard at our knowledge, it goes
away.” One explanation is that what we claim to know is mainly conditional
knowledge where the conditions are often implicit, e.g., “I know I have hands”
can be viewed as an abbreviation of “I know I have hands, given that I am not
a brain in the vat”. This holds for all kinds of common sense “knowledge” that
we have in every day life, which invites systematic logical study.
Coming back to the technical storyline, note that the original Kvioperator
is a special case of the conditional one since Kvidis simply Kvi(>, d). Then
it is natural to ask whether the epistemic logic extended with conditional Kvi
(call it ELKvr) is more expressive than the epistemic logic with the standard
Kvioperator (call it ELKv). In [16] we show that ELKvris indeed strictly
1We called the conditional version of Kvithe relativized Kvioperator in [16], due to the
similarity between it and the relativized common knowledge operator introduced in [15]
2Though there is still a difference: we do not require φto be truthful in Kvi(φ, d): φcan
be just hypothetical.

Wang and Fan 3
more expressive than ELKv. More interestingly, we show that adding the
public announcement operators to ELKvrdoes not increase the expressive
power, i.e., ELKvris closed under announcement updates. As in the standard
epistemic logic, this is a good property for a logic as a foundation of epistemic
reasoning (cf. e.g.,[14]).
To really layout the foundation for reasoning about both knowing that and
knowing what, we need to axiomatize ELKvr. In [16], an interesting system is
given to axiomatize the single-agent ELKvr. The completeness proof relies on
a canonical model construction which consists of two copies of each maximal
consistent set. However, such a method does not generalize to the multi-agent
case thus leaving the axiomatization of the multi-agent ELKvropen.
In this paper, we solve this open problem by showing that the multi-agent
version of the system proposed in [16] is indeed complete for the multi-agent
ELKvr. The techniques used here are quite different from the single-agent
completeness proof and are inspired by the following observation: ELKvrcan
be viewed as a fragment of first-order intensional logic (FOIL) proposed and
studied in [4,5]. FOIL features two kinds of variables: the object (rigid) vari-
ables and the intension (non-rigid) variables where the latter variables range
over the functions from the set of possible worlds to the set of objects. A
technique called predicate abstraction is applied to abstract predicates from
formulas. Now consider the following fragment (FOIL−) of FOIL where the
non-rigid variables are not quantified, the only predicates are unary ones over
rigid variables, and the only (implicit) predicate abstraction is applied to equal-
ities between a rigid variable xand a non-rigid variable d:3
φ::= > | P x |d=x| ¬φ|φ∧φ|Kiφ| ∀xφ
ELKvrcan then be viewed as a small fragment of FOIL−by recursively trans-
lating Kvi(φ, d) into ∃xKi(φ0→d=x) where φ0is the FOIL−-translation of φ.
Actually, this first-order formulation is also in accordance with the treatment
of “knowing-wh” in terms of “knowing that” in [7,13]. In this way we can see
clearly that Kvipackages a first-order quantifier and a modality together.
This observation motivates our construction of the canonical model for
multi-agent ELKvr. However, the method for axiomatizing FOIL as in [5]
cannot be applied directly here, due to two reasons: first, our language is much
weaker and we cannot express the desired first-order axioms in ELKvr; sec-
ond, to our knowledge, it is unknown, how to axiomatize FOIL on S5 frames
due to the diffcuities introduced by symmetry property as explained in [5]. In
this work, we found a way to provide just enough extra information in the
states of the canonical model to encode the “omitted” information expressible
by potential FOIL−formulas, while keeping it controlled by purely ELKvr
axioms. We believe that this method can be applied to other similar fragments
of first-order intensional logic (over S5 frames).
3In Fitting’s syntax of FOIL,d=xshould be formalized as hλy.y =xi(d) (cf. [5]). Here,
dmay also be viewed as a constant since it is never quantified in the language.

4 Conditionally knowing what
In the rest of this paper, we first review the syntax and semantics of multi-
agent ELKvrand the proof system ELKVrin Section 2. In Section 3, we prove
our main result that ELKVrcompletely axiomatizes multi-agent ELKvr. We
conclude with future work in Section 4.
2 Preliminaries
Given a countably infinite set of proposition letters P, a countably infinite set
of agent names I, and a countably infinite set of (non-rigid) constant symbols
D, the language of ELKvris defined as follows:
φ::= > | p| ¬φ|(φ∧φ)|Kiφ|Kvi(φ, d)
where p∈P, i ∈I, d ∈D.
Kiφsays that the agent iknows that φ.Kvi(φ, d) says that the agent
iknows what dis, given φ. More precisely, Kvi(φ, d) says that the agent i
would know what dis if he were informed that φ. The original (unconditional)
Kvidformulas proposed in [10] can be viewed as Kvi(>, d). As usual, we
define ⊥,(φ∨ψ),(φ→ψ),(φ↔ψ),ˆ
Kiφas the abbreviations of, respectively,
¬>,¬(¬φ∧ ¬ψ),(¬φ∨ψ),((φ→ψ)∧(ψ→φ)),¬Ki¬φ. We omit parentheses
from formulas unless confusion results.
ELKvris interpreted on epistemic models with assignments for the ele-
ments in D:M=hS, O, {∼i|i∈I}, V, VDiwhere Sis a non-empty set of
possible worlds, Ois a non-empty set of objects, ∼iis an equivalence relation
over S, and Vis a valuation function assigning a set of worlds V(p)⊆Sto each
p∈P, and VD:D×S→Ois a function assigning each d∈Dat each world
an object. In terms of first-order intensional logic [4], Mis an S5 intensional
model with a constant domain Oand assignments for non-rigid variables in D.
Note that for each d∈D,VD(d, ·) is a function from Sto Owhich can be
viewed as an intension as in [4]. The semantics is defined as follows:
M, s >always holds
M, s p⇔s∈V(p)
M, s ¬φ⇔ M, s 2φ
M, s φ∧ψ⇔ M, s φand M, s ψ
M, s Kiψ⇔for all tsuch that s∼it:M, t ψ
M, s Kvi(φ, d)⇔for any t1, t2∈Ssuch that s∼it1and s∼it2:
if M, t1φand M, t2φ, then VD(d, t1) = VD(d, t2)
Intuitively, Kvi(φ, d) is true at siff all the i-accessible φ-worlds agree on the
value of d. In other words, iknows what dis given φiff he is sure about d’s
value on φ-worlds. Based on this semantics, we can see clearly that Kvi(φ, d) is
indeed different from φ→Kvidand Ki(φ→Kvid). The condition φrestricts
the accessible worlds to be considered, and we then check whether dhas the
same value on these “relative alternatives”.
In [16], we give a complete axiomatization of the single agent ELKvrand
the following is the multi-agent version of that system which is an extension

Wang and Fan 5
of the multi-modality S5.
System ELKVr
Axiom Schemas
TAUT all the instances of tautologies
DISTK Ki(φ→ψ)→(Kiφ→Kiψ)
TKiφ→φ
4Kiφ→KiKiφ
5¬Kiφ→Ki¬Kiφ
DISTKvrKi(φ→ψ)→(Kvi(ψ, d)→Kvi(φ, d))
Kvr4Kvi(φ, d)→KiKvi(φ, d)
Kvr⊥Kvi(⊥, d)
Kvr∨ˆ
Ki(φ∧ψ)∧Kvi(φ, d)∧Kvi(ψ, d)→Kvi(φ∨ψ, d)
Rules
MP φ, φ →ψ
ψ
NECK φ
Kiφ
RE ψ↔χ
φ↔φ[ψ/χ]
where RE is the rule of replacement of equivalents, which plays an important
role in the later proofs. In the rest of the paper, we use `to denote the
derivation relation within ELKVr.
Note that Kvioperators do not behave like modalities in a normal modal
logic and the (obvious adaptations of) necessitation rule and the Kaxiom are
not valid for Kvi. Instead, we have the distribution axiom schema DISTKvr(note
the swap of ψand φin the consequent). Kvr4 is the counter part of the positive
introspection axiom 4, and Kvr⊥stipulates the effect of the absurd precondi-
tion. The most important axiom is Kvr∨which handles the composition of the
conditions: if all the possible φ-worlds agree on what dis and all the possible
ψ-worlds also agree on d, then the overlap between φpossibilities and ψpossi-
bilities implies that all the φ∨ψpossibilities also agree on what dis. We can
show that the above system is sound (cf. [16, Theorem 11]).
To facilitate the later proofs, we need the following propositions.
Proposition 2.1 (i) ` ¬Kiφ↔Ki¬Kiφ
(ii) The rule (RM): φ→ψ
ˆ
Kiφ→ˆ
Kiψis admissible in ELKVr.
(iii) ` ¬Kvi(φ, d)↔Ki¬Kvi(φ, d)
Proof (i) and (ii) are standard exercises in S5. The ←direction in (iii) is
trivial due to T. We show the other way around:
KiKvi(φ, d)↔Kvi(φ, d)T,Kvr4
¬KiKvi(φ, d)→Ki¬KiKvi(φ, d)5
¬Kvi(φ, d)→Ki¬Kvi(φ, d)RE
2
Note that the →half of (iii) can be viewed as the Kvicounterpart of 5thus
we denote it by Kvr5.
Another useful observation is that Kvr∨can be generalized to arbitrary
finite disjunctions as the following proposition shows.

6 Conditionally knowing what
Proposition 2.2 For any non-empty finite set Uof ELKvrformulas:
`ˆ
Ki(^U)∧^
φ∈U
Kvi(φ, d)→Kvi(_U, d).
Proof If |U|= 1 then the statement holds trivially. Suppose |U| ≥ 2, we prove
the statement by an inductive proof on |U|. The case of |U|= 2 is immediate
due to Kvr∨. Suppose the claim holds when |U|=k. Now consider the case
|U|=k+ 1, and let U=U0∪ {ψ}such that |U0|=k:
(i)ˆ
Ki(VU0)∧Vφ∈U0Kvi(φ, d)→Kvi(WU0, d) IH
(ii)Kvi(WU0, d)∧Kvi(ψ, d)∧ˆ
Ki(WU0∧ψ)→Kvi(WU, d)Kvr∨
(iii)ˆ
Ki(VU)∧Vφ∈UKvi(φ, d)TAUT
→ˆ
Ki(VU0)∧Vφ∈U0Kvi(φ, d)∧Kvi(ψ, d)∧ˆ
Ki(WU0∧ψ)DISTK,NECK
(iv)ˆ
Ki(VU)∧Vφ∈UKvi(φ, d)→Kvi(WU, d) (i)(ii)(iii)
2
The following proposition essentially says if dhas the same value over ac-
cessible φ- and ψ-worlds respectively, and there are a φ-world and a ψ-world
sharing the same dvalue, then dhas the same value over accessible φ∨ψ-worlds.
This proposition allows us to relax the antecedent of Kvr∨a little bit to make
it more useful.
Proposition 2.3 For any φ, ψ, χ ∈ELKvr, any d∈D:
`ˆ
Ki(φ∧χ)∧ˆ
Ki(ψ∧χ)∧Kvi(φ, d)∧Kvi(ψ, d)∧Kvi(χ, d)→Kvi(φ∨ψ, d)
Proof
(i)ˆ
Ki(φ∧χ)∧Kvi(φ, d)∧Kvi(χ, d)→Kvi(φ∨χ, d)Kvr∨
(ii)ˆ
Ki(ψ∧χ)∧Kvi(ψ, d)∧Kvi(χ, d)→Kvi(ψ∨χ, d)Kvr∨
(iii)φ∧χ→(φ∨χ)∧(ψ∨χ)TAUT
(iv)ˆ
Ki(φ∧χ)→ˆ
Ki((φ∨χ)∧(ψ∨χ)) RM,(iii)
(v)ˆ
Ki((φ∨χ)∧(ψ∨χ)) ∧Kvi(φ∨χ, d)∧Kvi(ψ∨χ, d)
→Kvi(φ∨ψ∨χ, d)Kvr∨
(vi)ˆ
Ki(φ∧χ)∧ˆ
Ki(ψ∧χ)∧Kvi(φ, d)∧Kvi(ψ, d)∧Kvi(χ, d)
→Kvi(φ∨ψ∨χ, d) (i)(ii)(iv)(v)
(vii)φ∨ψ→φ∨ψ∨χTAUT
(viii)Ki(φ∨ψ→φ∨ψ∨χ)NECK,(vii)
(ix)Ki(φ∨ψ→φ∨ψ∨χ)→Kvi(φ∨ψ∨χ, d)→Kvi(φ∨ψ, d)DISTKvr
(x)Kvi(φ∨ψ∨χ, d)→Kvi(φ∨ψ, d)MP,(viii)(ix)
(xi)ˆ
Ki(φ∧χ)∧ˆ
Ki(ψ∧χ)∧Kvi(φ, d)∧Kvi(ψ, d)∧Kvi(χ, d)
→Kvi(φ∨ψ, d) (vi)(x)
2
3 Completeness of multi-agent ELKVr
To prove the completeness, we need to build a canonical model such that each
maximal consistent set of ELKVris satisfied in it. The general difficulty is as

Wang and Fan 7
in the single-agent case: just using the maximal consistent sets as the states
in the canonical model is not enough, more information should be provided in
the states of the canonical model. As we mentioned, the Kvi(φ, d) formulas
can be viewed as ∃xKi(φ→d=x) where xis a rigid variable and dis a
non-rigid one. To build a canonical model for such a first-order intensional
logic, we need to include atomic formulas such as d=xand modal formulas
such as Ki(φ→d=x) and control their interactions by axioms. However,
those formulas are not expressible in ELKvrsince we simply cannot say what
dexactly is. Therefore, in the canonical model, we need to equip the ELKVr-
maximal consistent sets with information which can function as those atomic
formulas. Moreover, we need to specify how such extra information is related
to the ELKVr-maximal consistent sets. Note that since ELKVris very limited
we cannot enforce the extra information behave exactly like the intended first-
order intensional formulas. The real difficulty is to find the requirements which
are “just enough” to make sure the truth lemma holds and this is the most
fundamental idea behind our definition of the canonical model. It will also
become more clear why the single-agent case is much simpler (cf. Remark 3.6).
3.1 Canonical model
In the sequel we define our canonical model with the set of natural numbers N
as the constant domain of objects. 4
Definition 3.1 Let MCS be the set of maximal consistent sets w.r.t. ELKVr,
and let Nbe the set of natural numbers. The canonical model Mcof ELKVr
is a tuple hSc,N,{∼c
i|i∈I}, V c, V c
Diwhere:
•Scconsists of all the triples hΓ, f, gi ∈ MCS ×ND×(N∪ {?})I×ELKvr×D
that satisfy the following three conditions for any i∈I, any ψ, φ ∈ELKvr,
and any d∈D:
(i) g(i, ψ, d) = ?iff Kvi(ψ, d)∧ˆ
Kiψ /∈Γ,
(ii) If g(i, φ, d)6=?and g(i, ψ, d)6=?then: g(i, φ, d) = g(i, ψ, d)iff
there exists a χsuch that Kvi(χ, d)and ˆ
Ki(φ∧χ)and ˆ
Ki(ψ∧χ)are in Γ.
(iii) ψ∧Kvi(ψ, d)∈Γimplies f(d) = g(i, ψ, d).
For any s∈Sc, we write φ∈sif φis in the maximal consistent set of sand
write φ∈s∩tif φ∈sand φ∈t.fsand gsare used as the corresponding
functions in s, and gs(i)is the function from ELKvr×Dto N∪ {?}induced
by gsfixing a particular i∈I.
•s∼c
itiff {φ|Kiφ∈s} ⊆ tand gs(i) = gt(i)
•Vc
D(d, s) = fs(d)
Remark 3.2 Intuitively, fis roughly functioning as the collection of d=x
formulas, and gis roughly functioning as the collection of Ki(ψ→d=x)
formulas. Now for the intuitive ideas behind the three conditions:
4Note that this countable set is indeed big enough, since the (countable) language of ELKvr
can be translated into first-order intensional logic, which can be again translated into 3-sorted
first-order logic, which still enjoys L¨owenheim-Skolem property (cf. [1]).

8 Conditionally knowing what
•(i): We use ?to mark that the value of g(i, ψ, d) is irrelevant. If Kvi(ψ, d)6∈ Γ
then of course the value of g(i, ψ, d) is irrelevant. If Kvi(ψ, d)∈Γ but Ki¬ψ∈
Γ, then the condition ψis never possible for ithus the value of g(i, ψ, d) is
also irrelevant. Condition (i) is mainly for the technical convenience.
•(ii): This condition handles how the gvalues are inter-related. Intuitively, it
roughly says that x=yiff (Ki(ψ→d=x) and Ki(φ→d=y), and there
are some accessible ψ-world and φ-world which share the same value of d).
•(iii): Intuitively, this condition says that if Ki(ψ→d=x) and ψis indeed
true then d=x.
The definition of ∼c
iis in spirit the same as in the canonical model for
the standard epistemic logic. The extra condition gs(i) = gt(i) says the i-
indistinguishable worlds should satisfy the same Ki(ψ→d=x) formulas.
Condition (i), (ii), (iii) and the definition of ∼c
ispecify the minimal require-
ments of the extra information attached to maximal consistent sets. We need
to control them without using FOIL−formulas.
To show the above model is indeed an epistemic model, we need the follow-
ing proposition:
Proposition 3.3 For any i∈I,∼c
iis an equivalence relation.
Proof As a standard exercise in modal logic, by using T,4, and 5, we can
prove the following claim:
(for all φ:Kiφ∈simplies φ∈t) iff (for all φ:Kiφ∈siff Kiφ∈t) (∗).
Thus s∼c
itiff {φ|Kiφ∈s}={φ|Kiφ∈t}and gs(i) = gt(i). Then it is easy
to see ∼c
iis an equivalence relation.
2
Based on the above claim (∗), using Tand Kvr4, the following is immediate:
Proposition 3.4 For any two maximal consistent sets ∆and Γ, if {φ|Kiφ∈
∆} ⊆ Γ, then the following hold for all φ:
•Kiφ∈∆iff Kiφ∈Γ
•ˆ
Kiφ∈∆iff ˆ
Kiφ∈Γ
•Kvi(φ, d)∈∆iff Kvi(φ, d)∈Γ.
3.2 Completeness
In order to establish the truth lemma, the most difficult things are the existence
lemmas for Kiand Kvioperators. Since the states in the canonical models are
not merely maximal consistent sets, more efforts are required.
We first propose a general method to construct proper successors. This can
be viewed as some kind of Lindenbaum’s Lemma, though highly non-trivial in
this case, if we view gand fas collections of “hidden formulas”.

Wang and Fan 9
Proposition 3.5 Given a state s∈Sc, an agent i∈I, and a maximal consis-
tent set Γsuch that {φ|Kiφ∈s} ⊆ Γ, and any natural number x, we have a
deterministic method to construct t=hΓ, f, gibased on s,Γ, and xsuch that
t∈Scand s∼c
it.
Proof The construction and the proof are quite involved: we first build f
and build gby using finite approximations and show that the constructions
are well-defined; then we show that the hΓ, f, g isatisfy the three conditions of
states in Scand that s∼c
it. In the following, we fix a natural number x.
Let T0={hj, φ, di | j=ior φ∧Kvj(φ, d)∈Γ}. Let g0:T0→N∪ {?}be
defined as follows:
g0(j, φ, d) = 


gs(i, φ, d) if j=i
gs(i, ψ, d) if j6=iand ψ∧Kvi(ψ, d)∈Γ for some ψ
xif otherwise
We need to show that the second case is well-defined: the choice of ψdoes
not affect the value of gs(i, ψ, d), namely ψ∧Kvi(ψ, d)∈Γ and ψ0∧Kvi(ψ0, d)∈
Γ implies gs(i, ψ, d) = gs(i, ψ0, d).Suppose that ψ∧Kvi(ψ, d)∈Γ and ψ0∧
Kvi(ψ0, d)∈Γ, then the following formulas are also in Γ: ˆ
Kiψ0,ˆ
Kiψ, ˆ
Ki(ψ0∧
ψ) due to the contrapositive of Axiom T. By Proposition 3.4, the following
formulas are all in s:ˆ
Ki(ψ0∧ψ),ˆ
Kiψ0,ˆ
Kiψ,Kvi(ψ0, d), and Kvi(ψ, d). Now it
is clear that gs(i, ψ, d)6=?and gs(i, ψ0, d)6=?due to condition (i) of s. Let
χ=ψ0now we have Kvi(χ, d) and ˆ
Ki(ψ0∧χ) and ˆ
Ki(ψ∧χ) are all in s. By
condition (ii) of s,gs(i, ψ0, d) = gs(i, ψ, d).
Now we define fas follows:
f(d) = g0(j, φ, d) if Kvj(φ, d)∧φ∈Γ for some φand j
xif otherwise
We need to show that the first case is well-defined: the choices of φand jdo not
affect the value of g0(j, φ, d), namely φ∧Kvj(φ, d)∈Γ and ψ∧Kvk(ψ, d)∈Γ
implies g0(j, φ, d) = g0(k, ψ, d). To see this, consider four cases:
•j=iand k=i. Then due to the above proof and the first clause of the
definition of g0, we have g0(j, φ, d) = gs(i, φ, d) = gs(i, ψ, d) = g0(k, ψ, d).
•j6=iand k6=i. If there exists χsuch that χ∧Kvi(χ, d)∈Γ, then by the
second clause of g0, we have g0(j, φ, d) = gs(i, χ, d) = g0(k, ψ, d); otherwise,
by the third clause of g0, we have g0(j, φ, d) = x=g0(k, ψ, d).
•j=iand k6=i. From the first clause of g0, it follows that g0(j, φ, d) =
gs(i, φ, d). Due to the fact that φ∧Kvi(φ, d)∈Γ and the second clause of
g0, we have g0(k, ψ, d) = gs(i, φ, d), thus g0(j, φ, d) = g0(k, ψ, d).
•j6=iand k=i. Similar to the third case.
Now let ∆ be the set of the remaining non-i-triples:
∆ = {hj, φ, di | j6=i, φ ∧Kvj(φ, d)6∈ Γ, j ∈I, φ ∈ELKvr, d ∈D}.

10 Conditionally knowing what
Due to the fact that D,Iand ELKvrare countable, we can enumerate ∆
as δ1, δ2, . . . and approximate gstep by step by extending the domain of gk
with δk+1. Let ∆kbe {δl|1≤l≤k}, and in particular ∆0=∅.Let Λkbe
{gk(δ)|δ∈∆k}and let max be the function which assigns to each non-empty
finite set of natural numbers its maximum. Let Dom(gk) be the domain of gk.
For k≥0, our construction of gk+1 :T0∪∆k+1 →N∪ {?}is as follows: let
gk+1(j, φ, d) = gk(j, φ, d) if hj, φ, di ∈ Dom(gk) = T0∪∆k, and for the only
new hj, φ, di 6∈ Dom(gk) (thus j6=i) we have:
gk+1(j, φ, d) =















?if ˆ
Kjφ∧Kvj(φ, d)6∈ Γ
gk(j, ψ, d) if ˆ
Kjφ∧Kvj(φ, d)∈Γ and there exist
χ, ψ such that hj, ψ, di ∈ Dom(gk) and
the following formulas are all in Γ :
ˆ
Kj(χ∧φ),ˆ
Kj(χ∧ψ),Kvj(ψ, d),Kvj(χ, d)
max(Λk∪ {f(d)}) + 1 if otherwise
Note that, we still need to show that the second case in the above definition
of gk+1 is well-defined. More precisely, we need to show for any k≥0 any
j6=i, the following (1) implies (2):
(1) there exist ψ, ψ0, χ, and χ0such that hj, ψ, diand hj, ψ0, diare in Dom(gk)
and the following formulas are all in Γ: ˆ
Kj(χ∧ψ),ˆ
Kj(χ∧φ),ˆ
Kj(χ0∧
ψ0),ˆ
Kj(χ0∧φ),Kvj(ψ, d),Kvj(ψ0, d),Kvj(χ, d),Kvj(χ0, d).
(2) gk(j, ψ, d) = gk(j, ψ0, d).
Induction on k:
•k= 0: hj, ψ , diand hj, ψ0, diare both in Dom(g0) = T0then according to
the definition of g0and the fact that j6=i,g0(j, ψ, d) = g0(j, ψ0, d).
•Induction Hypothesis: (1) implies (2) holds for all k≤n.
•k=n+ 1 : w.l.o.g, we assume that at least one of (j, ψ, d) and (j, ψ 0, d)
is not in Dom(g0), for otherwise the case is like the above one. Then we
can assume that there exists an m≤nsuch that hj, ψ, di ∈ Dom(gm),
hj, ψ0, di 6∈ Dom(gm), and hj, ψ0, diis added into Dom(gm+1 ) by our con-
struction. Assuming (1), let θbe χ∨χ0, we can show ˆ
Kj(θ∧ψ) and
ˆ
Kj(θ∧ψ0)∈Γ since ˆ
Kj(χ∧ψ) and ˆ
Kj(χ0∧ψ0) are in Γ. Moreover, since
ˆ
Kj(χ∧φ),ˆ
Kj(χ0∧φ),Kvj(χ, d),Kvj(χ0, d),Kvj(φ, d)∈Γ, we have Kvj(θ, d)∈
Γ by Proposition 2.3. Now we have ˆ
Kj(θ∧ψ), ˆ
Kj(θ∧ψ0), and Kvj(θ, d) are
all in Γ. According to our construction of gm+1 ,gm+1(j, ψ 0, d) = gm(j, ψ, d)
and the induction hypothesis guarantees the uniqueness of gm(j, ψ, d) since
m≤n. Therefore gk(j, ψ 0, d) = gm+1(j, ψ0, d) = gm(j, ψ, d) = gk(j, ψ , d).
Viewing each gkas a set of pairs hhj, φ, di, gk(j, φ, d)i, we let gbe Sk<ω gk.
Now we need to verify conditions (i), (ii) and (iii). Condition (iii) is trivial
by the definition of f. We verify condition (i) and (ii) below.
For condition (i): we first show that for the fixed iand any φ∈ELKvr,

Wang and Fan 11
any d∈D:g0(i, φ, d)6=?iff ˆ
Kiφ∧Kvi(φ, d)∈Γ. From right to left: suppose
that g0(i, φ, d) = ?then we have gs(i, φ, d) = ?thus ˆ
Kjφ∧Kvi(φ, d)6∈ s,
i.e., Ki¬φ∈sor ¬Kvi(φ, d)∈s. By Proposition 3.4, we have Ki¬φ∈Γ
or ¬Kvi(φ, d)∈Γ, i.e., ˆ
Kiφ∧Kvi(φ, d)6∈ Γ. From left to right: suppose
that g0(i, φ, d)6=?then ˆ
Kiφ∧Kvi(φ, d)∈sthus by Proposition 3.4 again,
ˆ
Kiφ∧Kvi(φ, d)∈Γ.
Now consider hj, φ, di ∈ Dom(g0) where j6=i. By definition, φ∧Kvj(φ, d)∈
Γ, thus ˆ
Kjφ∧Kvj(φ, d)∈Γ. By the construction of g0it is clear that
g0(j, φ, d)6=?, since x6=?and the fact that ˆ
Kiψ∧Kvi(ψ, d)∈Γ implies
g0(i, ψ, d)6=?which we have just proved. This concludes the proof for the
base case: for any hj, φ, di ∈ Dom(g0): g0(j, φ, d)6=?iff ˆ
Kjφ∧Kvj(φ, d)∈Γ.
The inductive case is obvious by the three cases of our construction of gk+1.
Condition (ii) is more complicated to verify and it requires an inductive
proof. We first claim the following:
Claim (◦): For each k≥0, and any hj, ψ, diand hj, φ, diin Dom(gk) such
that gk(j, ψ, d)6=?and gk(j, φ, d)6=?, the following two are equivalent:
(1) gk(j, φ, d) = gk(j, ψ, d)
(2) there exists a χsuch that Kvj(χ, d), ˆ
Kj(φ∧χ) and ˆ
Kj(ψ∧χ) are in Γ.
If claim (◦) holds then Condition (ii) holds too, since any hj, ψ, diand hj, φ, di
must both exist in Dom(gk) for some k. Now we prove the claim (◦).
•If k= 0 then both hj, ψ, diand hj, φ, diare in Dom(g0). There are two
subcases:
·If j=ithen we have g(j, φ, d) = g(j, ψ, d) iff g0(i, φ, d) = g0(i, ψ , d) iff
gs(i, φ, d) = gs(i, ψ, d) iff there exists a χsuch that Kvi(χ, d) and ˆ
Ki(φ∧χ)
and ˆ
Ki(ψ∧χ) are all in s(by condition (ii) of s). According to Proposi-
tion 3.4, the last statement is equivalent to that there exists a χsuch that
{Kvi(χ, d),ˆ
Ki(φ∧χ),ˆ
Ki(ψ∧χ)} ⊆ Γ.
·If j6=ithen clearly g(j, ψ, d) = g0(j, ψ, d) = g0(j, φ, d) = g(j, φ, d) by the
definition of g0. Now since g(j, ψ, d)6=?,Kvj(ψ, d)∈Γ due to condition
(i) of Γ which we have just verified. Since hj, ψ, diand hj, φ, diare both in
Dom(g0) = T0, we have φ, ψ ∈Γ, thus ˆ
Kj(φ∧ψ)∈Γ by axiom T. Finally
we have {Kvj(χ, d),ˆ
Kj(φ∧χ),ˆ
Kj(ψ∧χ)} ⊆ Γ given χ=ψ.
•Induction Hypothesis: the claim (◦) holds for k≤m.
•Suppose k=m+1 and at least one of hj, φ, diand hj, ψ , diis not in Dom(gm),
for otherwise it can be handled by IH. Then clearly j6=ifor otherwise
both triples are in Dom(g0) thus in Dom(gm). W.l.o.g, we can assume that
hj, ψ, di ∈ Dom(gm) and hj, φ, di 6∈ Dom(gm) but hj, φ, di ∈ Dom(gm+1 ),
i.e., hj, φ, diis added at step m+ 1. By assumption gm+1 (j, φ, d)6=?and
gm+1(j, ψ, d)6=?, then by condition (i) ˆ
Kjφ∧Kvj(φ, d)∈Γ and ˆ
Kjψ∧
Kvj(ψ, d)∈Γ. Now if there exists a χsuch that ˆ
Kj(χ∧ψ)∧ˆ
Kj(χ∧φ)∧
Kvj(χ, d)∈Γ then by the second clause of the definition of gm+1 we have

12 Conditionally knowing what
gm+1(j, φ, d) = gm(j, ψ , d) = gm+1(j, ψ, d). This proves that (2) implies (1).
For the other direction, suppose that gm+1(j, φ, d) = gm+1 (j, ψ, d) =
gm(j, ψ, d)6=?, due to the definition of gm+1 ,gm+1(j, φ, d) must be con-
structed according to the second clause, for the third clause can make sure
gm+1(j, φ, d)6=gm(j, ψ , d). To see this, note that if hj, ψ, di ∈ Dom(g0) then
gm(j, ψ, d) = g0(j, ψ, d) = f(d) by the definition of g0(note that j6=i).
The third clause guarantees that gm+1(j, φ, d)> f (d) = gm(j, ψ, d). If
hj, ψ, di/∈Dom(g0) then the third clause guarantees that gm+1 (j, φ, d)>
max(Λm)≥gm(j, ψ, d).
Now, if gm+1 (j, φ, d) is constructed by the second clause based on
gm(j, ψ, d) then (2) is immediate. Suppose otherwise that gm+1(j, φ, d) is
constructed based on gm(j, θ, d) for some θ6=ψ, such that gm(j, θ, d) =
gm(j, ψ, d)6=?, then there exists a ξsuch that ˆ
Kj(ξ∧θ)∧ˆ
Kj(ξ∧φ)∧
Kvj(ξ, d)∈Γ. Since gm(j, θ, d)6=?, by condition (i) we also have
Kvj(θ, d)∈Γ. Since gm(j, θ, d) = gm(j, ψ, d)6=?, by IH, there ex-
ists ξ0such that ˆ
Kj(ξ0∧ψ)∧ˆ
Kj(ξ0∧θ)∧Kvj(ξ0, d)∈Γ. Now we have
ˆ
Kj(ξ∧θ),ˆ
Kj(ξ0∧θ),Kvj(ξ, d),Kvj(ξ0, d) and Kvj(θ, d) all in Γ. By Proposi-
tion 2.3, Kvj(ξ∨ξ0, d)∈Γ. Let χ=ξ∨ξ0. Since ˆ
Kj(ξ∧φ) and ˆ
Kj(ξ0∧ψ)
are in Γ, we have ˆ
Kj(χ∧φ),ˆ
Kj(χ∧ψ) and Kvj(χ, d) are all in Γ, and this
completes the proof of claim (◦). Thus hΓ, f, gisatisfies the condition (ii).
In sum, hΓ, f, g i ∈ Sc, and s∼c
ihΓ, f, gidue to the facts that g(i) = gs(i)
(by the construction of g0) and the assumption that {φ|Kiφ∈s} ⊆ Γ. 2
Remark 3.6 To build an i-successor of s, we need to construct a proper gsuch
that it takes care of the information not only about ibut also about j6=i.
Note that if I={i}, then g(i) = gs(i) implies g=gs. In this case we do not
need the above construction, thus the single-agent case is much simper.
Important notation In the sequel, we refer to the above construction of f
as F(s, i, Γ, x) where xis a natural number as a parameter.
Now we are ready to prove two important existence lemmas:
Lemma 3.7 For any s∈Sc, any i∈I: Kiψ6∈ simplies there is a world t
such that s∼c
itand ¬ψ∈t.
Proof It is a standard exercise in modal logic to show that X={¬ψ} ∪ {φ|
Kiφ∈s}is consistent. Then by Lindenbaum Lemma for ELKvr, there exists
an MCS Γ including X. Now from Proposition 3.5 we can equip Γ with some
proper fand g, such that hΓ, f, gi ∈ Scand s∼c
ihΓ, f, gi.2
Lemma 3.8 For any s∈Sc, any i∈I:¬Kvi(φ, d)∈simplies there are two
states w, v in Scsuch that s∼c
iw,s∼c
iv,φ∈w∩v, and fw(d)6=fv(d).
The proof of the above lemma is again quite involved, we break it into
Proposition 3.9 and Proposition 3.10 below.
Proposition 3.9 Given any s∈Scand any i∈I, suppose there exist two
(possibly identical) maximal consistent sets Γ1and Γ2such that:

Wang and Fan 13
(a) {ψ|Kiψ∈s} ⊆ Γ1∩Γ2
(b) for any Kvi(θ, d)∈s,θ6∈ Γ1∩Γ2.
then Γ1and Γ2can be extended into two states w, v in Scsuch that s∼c
iw,
s∼c
ivand fw(d)6=fv(d).
Proof By condition (a) and Proposition 3.5, Γ1and Γ2can be extended into
two i-accessible states by using F(s, i, Γ1, x) and F(s, i, Γ2, y). We argue that
condition (b) and condition (ii) of sallow us to construct two states in Scthat
differ in the value of d. Consider the following cases:
•Suppose that there is no Kvi(χ, d)∧χ∈Γ1for any χ. Note that in this
case if fw=F(s, i, Γ1, x) then fw(d) = x. Now let fv=F(s, i, Γ2,0) and
let fw=F(s, i, Γ1, fv(d) + 1). Clearly fw(d) = fv(d)+1 6=fv(d). The
symmetric case when there is no Kvi(χ, d)∧χ∈Γ2for any χis similar.
•Suppose there exists Kvi(χ, d)∧χ∈Γ1for some χand there exists Kvi(χ0, d)∧
χ0∈Γ2for some χ0. Now let fw=F(s, i, Γ1,0) and fv=F(s, i, Γ2,0) we
have fw(d) = gs(i, χ, d) and fv(d) = gs(i, χ0, d). We need to show gs(i, χ, d)6=
gs(i, χ0, d). Towards contradiction suppose gs(i, χ, d) = gs(i, χ0, d) then by
condition (ii) of s, there exists θsuch that Kvi(θ, d) and ˆ
Ki(θ∧χ) and
ˆ
Ki(θ∧χ0) are in s. Note that due to Proposition 3.4, Kvi(χ, d) and Kvi(χ0, d)
are in s. Now by Proposition 2.3, Kvi(χ∨χ0, d)∈s. However, since χ∈Γ1
and χ0∈Γ2,χ∨χ0∈Γ1∩Γ2, which contradicts to the assumption (b).
2
In [16], we proved the following proposition in the single agent case. The
proof for the multi-agent version is almost the same.
Proposition 3.10 Given any s∈Scand any i∈I, suppose ¬Kvi(φ, d)∈s
then there are two (possibly identical) maximal consistent sets Γ1and Γ2such
that:
(a’) {φ}∪{ψ|Kiψ∈s} ⊆ Γ1∩Γ2
(b) for any Kvi(θ, d)∈s,θ6∈ Γ1∩Γ2.
Proof Let Z={ψ|Kiψ∈s} ∪ {φ}and let X={¬θ|Kvi(θ, d)∈s}. Note
that due to Kvr⊥,Xis non-empty. 5We want to build two consistent sets B
and Csuch that Z⊆B∩Cand X⊆B∪C. Then by a Lindenbaum-like
argument over countable language, we can extend Band Cinto the desired Γ1
and Γ2: (a0) is guaranteed by Z⊆B∩C, and (b) is guaranteed by X⊆B∪C
which says that for any Kvi(θ, d)∈s,¬θ∈Bor ¬θ∈Cthus θ6∈ Γ1or θ6∈ Γ2.
In the following we build Band C.
The idea is straightforward: simply adding the formulas in Xone by one
5Kvr⊥is indispensable in the proof system ELKVr. We can show that it is not provable
in ELKVr−Kvr⊥by designing an alternative semantics which coincides with the standard
semantics for Kvi-free formulas but falsifies all the Kvi(φ, d) formulas for any i,φand d. It
is not hard to see that ELKVr−Kvr⊥is sound w.r.t. this new semantics but Kvi(⊥, d) is not
valid, thus Kvr⊥is not provable in ELKVr.

14 Conditionally knowing what
into two copies of Zwhile keeping the consistency. Formally, we enumerate
formulas in Xas ¬θ0,¬θ1, . . . and let B0=Z∪ {¬θ0}and let C0=Zas the
starting points. Then we build Bn+1 and Cn+1 based on the already defined
Bnand Cnby adding ¬θn+1 into one of them:
(i) if ¬θn+1 is consistent with Bnthen Bn+1 =Bn∪{¬θn+1 }and Cn+1 =Cn;
(ii) if ¬θn+1 is not consistent with Bnthen Bn+1 =Bnand Cn+1 =Cn∪
{¬θn+1}.
Let B=Sn<ω Bn,C=Sn<ω Cnand we need to show that Band Care
consistent. Note that B(C) is consistent iff Bn(Cn) is consistent for each n,
since if B(C) is not consistent then there must be an nsuch that Bn(Cn) is
not consistent, due to the finitary nature of logical consistency. In the following
we show Bnand Cnare consistent by induction on n.
•n= 0: Suppose towards contradiction that B0is not consistent, then there
exist ψ1, . . . , ψm∈ {ψ|Kiψ∈s}such that `ψ1∧ · · · ∧ ψm∧φ→θ0, i.e.,
`ψ1∧ · · · ∧ ψm→(φ→θ0). Therefore `Kiψ1∧ · · · ∧ Kiψm→Ki(φ→θ0)
by DISTK,NECK and RE. Since Kiψ1,...,Kiψm∈s,Ki(φ→θ0)∈s. Now by
DISTKvrand the fact that Kvi(θ0, d)∈s(since Xis non-empty), it follows
that Kvi(φ, d)∈s, contradiction. Since C0⊆B0,C0is also consistent.
•n=k+ 1: by the induction hypothesis Bkand Ckare consistent. Accord-
ing to our construction of Bk+1 we just need to show that if ¬θk+1 is not
consistent with Bkthen it is consistent with Ck. Suppose not, then both
Bk∪ {¬θk+1}and Ck∪ {¬θk+1 }are inconsistent. In the sequel, to derive
a contradiction, we adopt the proof in [16, Lemma 19] for the multi-agent
setting .
Let U=Bk\Z,V=Ck\Z,U={θ| ¬θ∈U}, and V={θ| ¬θ∈V}.
Note that U, V, U , V are all finite and each formula in Vis not consistent
with Bkdue to the construction of Bk.
We claim: there exist ψ1, . . . , ψl, ψ0
1, . . . , ψ0
m,ψ00
1, . . . , ψ00
r∈ {ψ|Kiψ∈s}
such that
(i)`ψ1∧ · · · ∧ ψl∧φ∧VU→θk+1 ,
(ii)`ψ0
1∧ · · · ∧ ψ0
m∧φ∧VV→θk+1,
(iii)`ψ00
1∧ · · · ∧ ψ00
r∧φ∧VU→VV.
(i) and (ii) are immediate from the inconsistency of Bk∪ {¬θk+1}and
Ck∪ {¬θk+1}. For (iii), first recall that for any θ∈V,{¬θ} ∪ Bkis incon-
sistent due to the construction of Bk. Therefore for each θ∈Vthere exist
χ1, . . . , χh∈ {ψ|Kiψ∈s}such that:
`(χ1∧ · · · ∧ χh∧φ∧^U)→θ
Since Vis a finite set, we can collect all such χfor each θ∈Vto obtain (iii).
From (i)−(iii), NECK,DISTK,RE and the fact that
Kiψ1,...,Kiψl,Kiψ0
1,...,Kiψ0
m,Kiψ00
1,...,Kiψ00
n∈s,

Wang and Fan 15
we can show the following:
(iv)Ki((φ∧VU)→θk+1 )∈s,
(v)Ki((φ∧VV)→θk+1)∈s,
(vi)Ki((φ∧VU)→VV)∈s.
In the following, we will show that ˆ
Ki(θk+1 ∧VV)∈s. First we claim
ˆ
Ki(φ∧VU)∈s. Suppose not, then Ki¬(φ∧VU)∈s, thus ¬(φ∧VU)∈Bk.
Due to the construction of Bkwe know φand Uare in Bk, thus Bkis
inconsistent, contradicting the assumption. Therefore ˆ
Ki(φ∧VU)∈sthus
by (iv),(vi) we have ˆ
Ki(θk+1 ∧VV)∈s.
By our assumption, for any θ∈V∪ {θk+1}we have Kvi(θ, d)∈s. Now
based on this fact and ˆ
Ki(θk+1 ∧VV)∈s, we can use Proposition 2.2, and
obtain the following:
(vii)Kvi(θk+1 ∨WV , d)∈s.
Now using ` ¬ VV↔WV, let us change the from of (v) to the following:
(v0)Ki(φ→(WV∨θk+1)) ∈s,
Based on (v0),(vii) and DISTKvr, we have Kvi(φ, d)∈s, contradiction.
Therefore, Bk+1 and Ck+1 are consistent and this concludes the the inductive
proof.
In sum, Band Care consistent thus can be extended into Γ1and Γ2satisfying
(a0) and (b). 2
Clearly, (a0) in Proposition 3.10 implies (a) in Proposition 3.9, then
Lemma 3.8 is immediate.
Now we are ready to prove the truth lemma:
Lemma 3.11 (Truth Lemma) For any φ∈ELKvrand s∈Sc,φ∈siff
Mc, s φ.
Proof We only show the non-trivial cases of Kiψand Kvi(ψ, d).
•φ=Kiψ: If Kiψ∈s, then for any tsuch that s∼c
itwe have ψ∈tby
the definition of ∼c
i. Now by induction hypothesis (IH), Mc, s Kiψ. Now
suppose Kiψ /∈s, then by Lemma 3.7 and the IH, we have Mc, s ¬Kiψ.
•φ=Kvi(ψ, d): Suppose that Kvi(ψ, d)∈s,s∼c
it,s∼c
it0,ψ∈tand ψ∈t0.
It is easy to see that Kvi(ψ, d)∈t∩t0and gt(i) = gt0(i) = gs(i). Since ψ∈t
and ψ∈t0, according to condition (iii) and the fact that gt(i) = gt0(i):
VD(d, t) = ft(d) = gt(i, ψ, d) = gs(i, ψ, d) = gt0(i, ψ , d) = ft0(d) = VD(d, t0).
Now suppose Kvi(ψ, d)6∈ sthen by Lemma 3.8 and IH, Mc, s ¬Kvi(ψ, d).
2
From the above truth lemma, the completeness theorem almost follows. The
only missing piece is to show for each maximal consistent set there is indeed
at least one corresponding state in Mc.

16 Conditionally knowing what
Lemma 3.12 For every maximal consistent set Γ, there exist fand gsuch
that hΓ, f, gi ∈ Sc.
Proof The construction is very similar to the one in the proof of Proposi-
tion 3.5, though simpler. The only essential difference is the definition of g0,
thus the proofs related to g0need to be adapted.
Let xbe a natural number, and let T={hj, φ, di | Kvj(φ, d)∧φ∈Γ, j ∈
I, φ ∈ELKvr, d ∈D}.6Let g0:T→N∪ {?}be the constant function such
that g0(j, φ, d) = xfor all the triples in T. We define fas the constant function
such that f(d) = xfor all d∈D.Now redefine ∆ as the set of the remaining
triples:
∆ = {hj, φ, di | φ∧Kvj(φ, d)6∈ Γ, j ∈I, φ ∈ELKvr, d ∈D}.
As before, we can enumerate ∆ and build gk+1 by adding the new hj, φ, di/∈
Dom(gk) into the domain (where Λkis defined as before w.r.t. the new ∆):
gk+1(j, φ, d) =















?if ˆ
Kjφ∧Kvj(φ, d)6∈ Γ
gk(j, ψ, d) if ˆ
Kjφ∧Kvj(φ, d)∈Γ and there exist
χ, ψ such that hj, ψ, di ∈ Dom(gk) and
the following formulas are all in Γ :
ˆ
Kj(χ∧φ),ˆ
Kj(χ∧ψ),Kvj(ψ, d),Kvj(χ, d)
max(Λk∪ {f(d)}) + 1 if otherwise
Similar to the corresponding proof of Proposition 3.5, we can show that the
second clause in the above definition of gk+1 is well-defined (k= 0 case is now
obvious due to the definition of g0).
Now let g=Sk∈Ngk, we need to verify conditions (i), (ii) and (iii). Condi-
tion (iii) is immediate from the definition of f.
For condition (i), if hj, φ, di ∈ Dom(g0) = T, then g0(j, φ, d) = x6=?, and
we can see that Kvj(φ, d)∧ˆ
Kjφ∈Γ since Kvj(φ, d)∧φ∈Γ. Thus we have for
any hj, φ, di ∈ Dom(g0), g0(j, φ, d)6=?iff Kvj(φ, d)∧ˆ
Kjφ∈Γ. The inductive
case is obvious by the three clauses of gk+1.
For condition (ii), suppose that g(j, φ, d)6=?and g(j, ψ , d)6=?, we need
to prove claim (◦) inductively as before. For that, we only need to revise the
proof for the base case as follows:
Suppose k= 0 and thus both hj, φ, diand hj, ψ, diare in Dom(g0). By
definition of g0, it is easy to see that g(j, φ, d) = g0(j, φ, d) = x=g0(j, ψ, d) =
g(j, ψ, d). Moreover, we have φ∧Kvj(φ, d)∈Γ and ψ∧Kvj(ψ, d)∈Γ. Then
setting χ=φgives us Kvj(χ, d)∧ˆ
Kj(φ∧χ)∧ˆ
Kj(ψ∧χ)∈Γ. Then for any
hj, φ, diand hj, ψ, diin Dom(g0), g0(j, φ, d) = g0(j, ψ , d) iff there exists a χ
such that {Kvj(χ, d),ˆ
Kj(φ∧χ),ˆ
Kj(ψ∧χ)} ⊆ Γ.
In sum, hΓ, f, gi ∈ Sc.2
Based on Lemma 3.12 and Lemma 3.11 we can show the completeness.
6Note that Tmay be empty. In that case we start from the empty function g0.

Wang and Fan 17
Theorem 3.13 ELKVris sound and strongly complete for multi-agent
ELKvr.
Proof The soundness part can be found in [16, Theorem 11]. For the com-
pleteness part, we show that each consistent set of ELKvrformulas is sat-
isfiable. Given a consistent set ∆ of ELKvrformulas, by the Lindenbaum
Lemma for the countable language ELKvr, there exists a maximal consistent
set Γ such that ∆ ⊆Γ. Now Lemma 3.12 tells us that there exist f, g such
that hΓ, f, gi ∈ Sc. From Lemma 3.11, it follows that Mc,hΓ, f, giΓ thus
Mc,hΓ, f, gi∆. 2
In [16], we also discussed the logic of ELKvrextended with public an-
nouncement operators (call it PALKvr):
φ::= > | p| ¬φ|φ∧φ|Kiφ| hφiφ
As an immediate corollary of the above completeness theorem and Theorem
10 in [16], we can axiomatize multi-agent PALKvrby adding the following
reduction axioms to ELKVr(call the resulting system PALKVr):
!ATOM hψip↔(ψ∧p)
!NEG hψi¬φ↔(ψ∧ ¬hψiφ)
!CON hψi(φ∧χ)↔(hψiφ∧ hψiχ)
!K hψiKiφ↔(ψ∧Ki(ψ→ hψiφ))
!KvrhφiKvi(ψ, d)↔(φ∧Kvi(hφiψ, d))
Corollary 3.14 PALKVris sound and complete for multi-agent PALKvr.
4 Future work
In this paper, we showed that ELKVris sound and complete for multi-agent
ELKvr(over S5 frames). This is just a starting point of an unfolding story
about interesting modal fragments of first-order intensional logic.
For future work, the decidability of ELKvrdeserves a careful investigation.
The single-agent case is particularly promising, since we do have a neat canon-
ical model construction which only uses two copies of each maximal consistent
set, which may facilitate a finite filtration leading to to the small model prop-
erty. On the other hand, there are also hints for the undecidablity, for example,
in [3], it is shown that the quantifier-free fragment of S5-FOIL is undecidable,
where arbitrary relation symbols and arbitrary predicate abstractions are al-
lowed. Of course we may study the logic of ELKvron other weaker frame
classes, where decidability is more plausible according to [3].
Another natural question to ask is how to axiomatize the logic where Kvi
are the only primitive operators (call it PLKvr). In ELKVr, most of the
axioms involve interactions between “knowing that” and “knowing what”. We
are unsure if the system without these axioms can axiomatize PLKvr, though
it is unlikely.
As motivated in the introduction, ELKvrcan be used in a security setting
where the interaction between “knowing what” and “knowing that” is impor-
tant. To really handle epistemic reasoning in such scenarios, we need to express

18 Conditionally knowing what
statements like “I know that the message I just received is indeed the private
message that I sent before for authentication”, where equality is inevitable.
Due to our completeness proof method, we suspect that adding the equality
symbol (between d∈D) freely in ELKvrmay in turn ease the axiomatization,
since we have a better grip on the information we need in the canonical model.
Last but not least, on the philosophical side, the conditional versions of
other types of knowledge should be studied, probably in the context of rele-
vant alternative theory [2], since they may capture the common sense use of
knowledge better.
Acknowledgement
This work is partially supported by European Research Council (ERC) grant
EPS 313360. Yanjing Wang thanks National Social Science Foundation of
China (SSFC) for the grant 11CZX054 and the support from SSFC major
project 11&ZD088. Jie Fan acknowledges the support from China Scholarship
Council (CSC). We are also grateful to the anonymous referees of AiML2014
for their insightful comments on an early version of this paper.
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