1 Introduction
Common knowledge, as the strongest concept among group epistemic notions, has been studied extensively in artificial intelligence
[parikh1986levels], epistemic logic [meyer2004epistemic], epistemology [Lewis1969Convention] and philosophy of language [Clark1981Definite]. Over Kripke semantics, common knowledge among group that , formulated as , is interpreted as: on every node that can be reached from the current node via the reflexivetransitive closure of relation ^{1}^{1}1 is the union of accessible relations of the agents in ., is satisfied. As the definition suggests, common knowledge is defined on the basis of classical knowledge, ‘knowing that’, to represent propositional knowledge in a group.There also exists a large amount of ‘wh’style knowledge^{2}^{2}2‘Knowing How’ is also considered as a ‘wh’style knowledge since it also describes a kind of nonpropositional knowledge. in natural language. To describe them formally, several works on epistemic logic have been undertaken, such as the logic of ‘knowing whether’[Fan2015CONTINGENCY], the logic of ‘knowing how’[Wang2016A][DBLP:conf/ijcai/FervariHLW17], and the logic of ‘knowing why’ [DBLP:journals/corr/XuW16]. Especially the notion of ‘knowing whether’, which means that an agent knows that is true or knows that is false, has been studied from distinct perspectives [friedell1969structure][Mccarthy1996First][Hart1994].
Considering that classical common knowledge is based on ‘knowing that’, the natural question arises of what the common knowledge based on ‘knowing whether’ is. There is no agreement on the definition of ‘commonly knowing whether’ yet and some possible definitions can currently only be expressed with an infinite language. In this paper, we suggest one of the definitions is the most plausible and deserves specific study.
In this paper, we give alternative definitions of ‘commonly knowing whether’ based on different intuitions and prove that they are not equivalent, following with a further study on one of them on its logical properties. In detail, Section 2 gives five different definitions of ‘commonly knowing whether’. Section 3 studies how these definitions are logically related over different frames. Section 4 discusses the expressivity of the language . Some properties of over two special classes of frames are investigated in Section 5. Due to the limitation on pages, we only show full proofs for these vital conclusions and proof sketches for some important results. As for other facts and lemmas, proof details are omitted.
2 Definitions of ‘Commonly Knowing Whether’ ()
One way of approaching standard common knowledge is an infinite conjunction of all finite iterations of ‘everyone knowing’. Before defining ‘commonly knowing whether’, two different definitions of ‘everyone knowing whether’ will be introduced. The following definitions of ‘everyone knowing whether’ and ‘commonly knowing whether’ are due to Yanjing Wang [Yanjing2017Commonlyknowingwhether]. The group concerned in this paper is finite. Since only one group is considered, we omit in notations for group knowledge in the remainder of the paper. Infinite conjunctions are temporarily used to define some of the following notions since they have not been expressed in any finite language yet. And , , refer to the classes of all Kripke frames, reflexive frames and equivalence relation frames, respectively. refers to the class of serial, transitive and Euclidean frames. ^{3}^{3}3For any set , contains all finite sequences only consisting of elements in . refers to the set of finite sequences only consisting of agents from , excluding the empty sequence. Let be a fixed set of propositional variables. An infinite language is introduced to define ‘commonly knowing whether’.
Definition 1
Given and , the infinite language is given as follows:
where , , and is a countably infinite set of formulas.
Definition 2
Intuitively, this definition imitates the definition of ‘knowing whether’, describing ‘everyone knows whether ’ as ‘everyone knows or everyone knows not ’.
Definition 3
, where the semantics of is defined over Kripke pointed models as: iff for any and with and , .
This definition is inspired by classical ‘everyone knows’, defining ‘everyone knows whether’ as ‘everyone in the group knows whether the proposition’.
On the basis of these definitions of ‘everyone knows whether’, five definitions of ‘commonly knows whether’ are given based on distinct interpretations.
Definition 4
A group commonly knows whether is possibly interpreted as they have common knowledge that or they have common knowledge that .
Definition 5
A group commonly knows whether plausibly means that everyone knows whether and it is a common knowledge of the group that everyone knows whether . Since there are two different definitions of ‘everyone knows whether’, we should separate into and .
Definition 6
‘Commonly knowing whether ’ can also be defined as everyone knows whether and everyone knows whether everyone knows whether , etc. We should again separate into and . As mentioned in Section 1, this is an infinitary form corresponding to intuition.
Definition 7
, that is
A group commonly knows whether possibly means for every member in the group, it is common knowledge of the group that she knows whether or it is common knowledge of the group that she does not know whether .
Definition 8
. For instance: if , then .
Listing all the possible inter‘knowing whether’ states among every subset of the group is an alternative way to define ‘commonly knowing whether’.
3 Implication Relations among the definitions of
The above five definitions of ‘commonly knowing whether’ do not boil down to the same thing, especially over and . In this section, we show how these definitions are related. The semantics presented in this paper uses Kripke models. A Kripke model is generally defined as , where is a nonempty set of nodes, , , , are accessibility relations for the agents in group and is the valuation which is a function . If , we write ; if , we write ; is the reflexivetransitive closure of .
3.1 Over
In order to clarify the relations among definitions of ‘commonly knowing whether’, we first give the logical relations between and .
Fact 1
The following statements hold:
;
;
;
.
Theorem 1
Over , the logical relationships among , , , and are shown in Figure 1.^{4}^{4}4In the following figures, the directed arrows between two definitions refer to the strict implications. The logical relations are surely transitive.
Every pair of these definitions are not equivalent over . Conversely, the ‘knowing that’versions of , and exactly correspond to three approaches to common knowledge [barwise2016three] which are logically equivalent.
We will show the proof sketches on an interesting case where implies but the converse does not hold.
Proof Sketch: prove with induction.
Consider in two cases:
The first case is . Let be an arbitrary pointed model, such that and . Then, for any node with , we have . For arbitrary , let , where . We show a stronger result: and for any such that , we have that .
By induction on the length of :
When , , where is an arbitrary agent. By , we have . And since for any with , there is . Thus there is .
Induction hypothesis: when , , there is and for any with , there is .
When , . By induction hypothesis, for any such that , for any , there is . Thus, for any , there is , saying, for any , we have . Now considering any with , let . By the definition of , . By induction hypothesis, we get . And since is an arbitrary node such that , we have for any , . That means, for any , there is .
Therefore, we proved that, for arbitrary and arbitrary with , there are and , which means . Thus, we have . Above all, we obtain .
For the second case where , since , we have . Therefore, we proved .
Proof Sketch: find a countermodel.
Consider the following model :
In this model, there is only one successor of . By semantics, we have . And since and , it follows that . Thus, .
3.2 Over
is considered to be the usual class of frames for doxastic (belief) logic [kraus1988knowledge, van2007dynamic]. The implication relations over among these five definitions are the same as those over .
Theorem 2
Over , the implication relations are shown as Figure 1.
3.3 Over and
The model of knowledge is generally defined over [van2007dynamic, fagin2004reasoning], which commits knowledge must be true via axiom . The implication relations among five definitions over or are given in Theorem 3.
Theorem 3
Over and , the implication relations are shown in Figure 2.
, , and boil down to the same thing once the frame is reflexive, which should be attributed to agents’ agreements on the values of . For example, if there is where is reflexive, the values of on all successors must agree with those on all successors since they share a common successor s. Comparatively, if is not reflexive, the case of is possible.
3.4 Singleagent Case
The results above concern the multiagent case. Considering the singleagent case, observe that which implies that is equivalent to and that is equivalent to . Moreover, since only one agent is involved, it is trivial to find that is equivalent to .
In terms of logical relationships, the five definitions share the same relations with the multiagent case over , , and , respectively.
The following sections concentrate on in the singleagent case.
4 The Case of
We focus on since the idea of it is inspired by the hierarchy of interknowledge of a group given in [parikh1986levels]. The language of ‘commonly knowing whether’ is defined in the same way as ^{5}^{5}5The language for the logic of ‘knowing whether’ defined in [Fan2015CONTINGENCY]., except introducing a new operator . We will investigate the expressivity between and in Section 4.3, so the language of classical common knowledge is also given here.
Definition 9
Given and , the language is given as follows:
where , .
Definition 10
Given and , the language is given as follows:
where , .
4.1 Some Valid or Invalid formulas
In the singleagent case, we consider some validities and invalidities over :
Fact 2
This valid formula illustrates that when is true at the current node, all successors of it agree the values of . However, the converse formula is invalid, saying .
Fact 3
This valid formula is established in the light of Almost Definability (AD) from [fan2014almost] , which is . AD shows that under the precondition for some , the classical knowledge operator can be defined in terms of . Since , we can replace with AD.
Fact 4
Although is the basic axiom of according to [Fan2015CONTINGENCY], its version is not valid.
Fact 5
Similarly, for another basic axiom in , the version is invalid.
Fact 6
This formula is invalid, which leads to the failure in defining with accessibility relations.
4.2 is Bisimulation Invariant
Definition 11
Let and be two Kripke models. A nonempty binary relation is called bisimulation between and , written as , if the following conditions are satisfied:
(i) If , then and satisfy the same proposition letters.
(ii) if and , then there is a such that and .
(iii) If and , then there exists such that and .
When is a bisimulation linking two states in and in , we say that two pointed models are bisimilar and write . If a language cannot distinguish any pair of bisimilar models, is bisimulation invariant.
Theorem 4
is bisimulation invariant.
Proof 1
By induction on formulas of .
When is a Boolean formula, the proof is classical.
When , we prove it in three cases. For arbitrary two bisimilar models and , we have:

if and for all with , . Since , for any with , there is an such that and . By induction hypothesis, is bisimulation invariant. Thus . So .

if and for all with , , similar to above case.

if , that means there are with and with , such that and . Since , there are with and with , such that and . By induction hypothesis, is bisimulation invariant. Thus and . Thus .
Thus, is bisimulation invariant.
When , assume two bisimular models and , such that and . That means there exists a sequence of agents , such that and . Let . So and , where . However, we have proved that for any formula of the form , they are bisimulation invariant. Thus, if , there must be . Contradiction.
Therefore, we proved that is bisimulation invariant.
4.3 and
Although is formed merely with , which can be defined by classical operator , surprisingly, cannot be defined with and .
Theorem 5
Over , is not weaker than in expressivity.
We prove Theorem 5 by defining two classes of models, which no formula can distinguish while there exists one formula that can distinguish them.
Definition 12
For every , define two sets of possible worlds and with induction:

;

If , then and ; if , then and

For every , ; for every ,

Besides defined above, has no more possible worlds; besides defined above, has no more possible worlds.
Then define the class of models as follows: for every ,



, where , .
Define the class of models with : for every :



, where ,
The first model and in and are shown as Figure 3.
The model is constructed by adding a new subtree rooted with to the root and just make unsatisfied on the leaf node whose index only consists of .
We will prove that no formula can distinguish and with the game^{6}^{6}6The definition of the game is given in Appendix.. If there is a winning strategy for duplicator in round games, and agree on all formulas whose modal depth is .
Theorem 6
For arbitrary , duplicator has a winning strategy in the game on and in rounds.
Proof 2
We describe Duplicator’s winning strategy case by case. Starting with the initial state , we mainly concerns the case where spoiler does a move. Otherwise, duplicator can move to a isomorphic submodel such that there must be a winning strategy in following rounds. Thus, the cases below exhaust all possibilities.

The initial state is :

If spoiler does a move or a move on reaching , then duplicator does a move or a move on to reach the corresponding . Since , there is a winning strategy after this move.

If spoiler does a move or a move on reaching , then duplicator does a move or a move on to reach the corresponding . Since , there is a winning strategy after this move.

If spoiler does a move on reaching , duplicator moves to .

If spoiler does a move on reaching an arbitrary node in , then duplicator does a move to reach . Since , there is a winning strategy after this move.


The current state is :

If spoiler does a move reaching or , then duplicator moves on to reach .

If spoiler does a move reaching on , then duplicator moves to on .

If spoiler does a move reaching , then duplicator moves to . Since , there is a winning strategy after this move.

If spoiler does a move reaching on , then duplicator moves to on .

If spoiler does a move reaching ^{7}^{7}7The notation is correct since the index for every node in begins with ., then duplicator moves to on . Since , there is a winning strategy after this move.


The current state is and : this means before the game gets to this state, both players have only done moves. In the current state, there have been at most rounds. Thus, and players can do next round as follows:

If spoiler does a move reaching or , then duplicator does a move to reach where since and there are and .

If spoiler does a move or a move reaching or , then duplicator does a move or a move to reach where since and there are and .

If spoiler does a move reaching , then duplicator does a move to reach . Since , there is a winning strategy after this move.

If spoiler does a move reaching , duplicator moves to on . Since , there is a winning strategy after this move.

Therefore, for arbitrary , there is a winning strategy for duplicator in the round game over and .
The above proof shows the nonexistence of some formula which can distinguish and since for any formula, its modal depth is a natural number which results in that it cannot distinguish and . Comparatively, there is a formula, such that and for every . Moreover, consider the following two pointed models, and :
We can find a formula to distinguish them, where and . But there is no any formula can distinguish and . It follows that is not weakly expressive than .
Therefore, together with Theorem 5, the following theorem is obtained:
Theorem 7
Language and are incomparable with respect to expressivity.
5 over Special Frames
Because of the invalidity of the formula , the operator is not normal, in the sense that it cannot be defined with accessibility relations. However, an interesting observation over binarytree models can be proved.
Definition 13
is a binarytree model with root if is a tree model with root and for any node in , has precisely two successors.
5.1 over Binary Trees
Theorem 8
Consider the singleagent case. If where is a binary tree with root , then on every layer of , the number of the nodes where is satisfied is even.
In order to prove Theorem 8, we need to prove a stronger theorem:
Theorem 9
For an arbitrary formula , if is a binarytree model, then iff the number of the satisfied nodes on the layer that can reach via relation is even.
Proof sketch
By induction on :

When , it is trivial.

(Induction Hypothesis) When , it holds.

When , suppose , which is equivalent to . Then we can use induction hypothesis. Suppose , which means on one successor, is satisfied, and on other one successor, is not satisfied. Then we can use induction hypothesis.
Proof details are shown in Appendix.
Remark 1
Theorem 8 can be extended into a more general conclusion considering the multiagent case: on any binarytree model^{8}^{8}8A binarytree model is a tree model where every node exactly has two successors for every . where is the root, iff for any sequence of agents in , on every layer of the subtree (of ) generated with ^{9}^{9}9A subtree (of some tree model ) generated with a sequence of agents is a subtree rooted with which only consists of all paths starting with in ., the number of the satisfied nodes is even.
5.2 over
By conclusions drawn in Section 3.2, is not equivalent to over , where still preserves its nontriviality.
Singleagent Case
When the group concerned is a singleton, there is nestings reducing for operator over . The proof idea of Theorem 10 is similar to the classical case proved in [meyer2004epistemic] where the nestings of operator can also be reduced over .
Theorem 10
Over , every formula^{10}^{10}10 is the language of with only one agent. is equivalent to some formula without nestings of the modal operator .
Corollary 1
In the singleagent case,
Multiagent Case
However, when more than one agent are involved, we cannot reduce nestings. Fortunately, there are still some interesting findings.
Theorem 11
For any sequence of agents, if there exists such that and , then for any formula .
The proof of Theorem 11 is given in Appendix.
Corollary 2
Over , where is a sequence of agents and there is no such that .
6 Conclusions
In this paper, we provide some initial results on definitions of ‘commonly knowing whether’, their relationships and properties, with special focus on . We also show that over , the languages and are incomparable with respect to expressivity.
There is much more expected work to be done in terms of ‘commonly knowing whether’. For instance, giving a Kripke semantics of , axiomatizing , proving the completeness of that axiomatization with respect to models of , studying other definitions of ‘commonly knowing whether’ ( is also a good choice), etc.
Acknowledgement
The authors are greatly indebted to Yanjing Wang for many insightful discussions on the topics of this work and helpful comments on earlier versions of the paper. Jie Fan acknowledges the support of the project 17CZX053 of National Social Science Fundation of China. Xingchi Su was financially supported by Chinese Scholarship Council (CSC) and we wish to thank CSC for its fundings.
References
Appendix A Proof of Theorem 9
Proof 3
Given a binary tree , where is the root, we firstly define the index of as follows: if there are nodes , , in and , , then define the index of as and the index of as .
Let be an arbitrary node in . Do induction on :

When ,

Assume . Since is a binary tree, there must be two nodes, and such that and . Since , we have ( and ) or ( or ). Thus, on the th layer, the number of nodes where is satisfied is 2 or 0, both of which are even.

Assume the number of the nodes on the th layer that can reach is even. That means there are only two possible cases: ( and ) or ( or ). Thus, we have .


Induction hypothesis: when , the number of the satisfied nodes on the layer that can reach via relation is even.

When ,

Assume , which is equivalent to .
Let be the set of nodes exactly consisting of all satisfied nodes on the th layer that can reach via relation . By induction hypothesis, is even. let . Thus, among all the successors of , the number of satisfied nodes is , where . is surely an even number. Let be a set of nodes only consisting of satisfied nodes on the th layer that can reach via relation . Since is a binary tree, let . For every node in has only one satisfied successor, among all the successors of , the number of satisfied nodes is . Thus, the number of satisfied nodes on the th layer is which must be even.

Assume , which means and . By induction hypothesis, the number of the satisfied nodes on the layer that can reach via relation is even. And the number of the satisfied nodes on the layer that can reach via relation
is odd. That means that the
satisfied nodes on the th layer that can reach via relation is an even number plus an odd number, which equals to an odd number.

Appendix B The definition of game
The Definition of Game
A game is a game with two players, duplicator and spoiler, playing on a Kripkemodel. Given two Kripke models and , from an arbitrary node in and an arbitrary node in , play games in rounds between duplicator and spoiler as following rules:

When , if the sets of satisfied formulas on node and are the same, then duplicator wins; otherwise, spoiler wins.

When ,

move: If spoiler starting from node does move to node which can be reached by , then duplicator starting from does move to a node in with the same set of satisfied propositional variables as . If spoiler starts from , then duplicator starts from with similar way to move.

move: If spoiler starting from node does move to node which can be reached by , then duplicator starting from does move to a node in with same set of satisfied propositional variables to . If spoiler starts from , then duplicator starts from with similar way to move.
In the game, for arbitrary and , we say or is a state of game.

Appendix C Proof of Theorem 11
A Corollary
For arbitrary formula , is valid over any , saying .
This is a corollary of the results in [Fan2015CONTINGENCY].
The Proof of Theorem 11
Let where there is an such that and . By the corollary above, , which means for an arbitrary formula . Since , we have
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